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Here is a problem from the 2012 Moscow Olympiad:
There were n people at a meeting. It appears that any two people at the meeting shared exactly two common acquaintances.
- Prove that all the people have exactly the same number of common acquaintances at this meeting.
- Show that n can be greater than 4.
My question is: Why 4? I can answer that myself. If in a group of four people any two people share exactly two common acquaintances, then all four people know each other. So in this Olympiad problem, the author wanted students to invent a more intricate example.
Let's take this up a notch and work on a more difficult problem.
There were n people at a meeting. It appears that any k people at the meeting shared exactly k common acquaintances.
- Prove that all the people have exactly the same number of common acquaintances at this meeting.
- Is it possible that n can be greater than 2k?
I stumbled upon an article, Winners Live Longer, that says:
"When 524 nominees for the Nobel Prize were examined and compared to the actual winners from 1901 to 1950, the winners lived longer by 1.4 years. Why? It seems just having won and knowing you are on top gives you a boost of 1.8% to your life expectancy."
This goes on top of the pile of Bad Conclusions From Statistics. With any kind of awards where people can be nominated several times, winners on average would live longer. The reason is that nominees who die early lose their chance to be nominated again and to win.
I wonder what would happen if we were to compare Fields medal nominees and winners. There is a cut off age of 40 for receiving a Fields medal. If we compare the life span of Fields medal winners and nominees who survived past 40, we might get a better picture of how winning affects life expectancy.
Living a long life increases your chances of getting a Nobel Prize, but doesn't help you get a Fields medal.
I wish I could write four papers in three weeks. The title just means that I submitted four papers to the arXiv in the last three weeks—somehow, after the stress of doing my taxes ended, four of my papers converged to their final state very fast. Here are the papers with their abstracts:
This is the promised solution to the puzzle Integers and Sequences that I posted earlier. The puzzle is attached below.
Today I do not want to discuss the underlying math; I just want to discuss the puzzle structure. I'll assume that you solved all the individual clues and got the following lists of numbers.
Since the title mentions sequences, it is a good idea to plug the numbers into the Online Encyclopedia of Integer Sequences. Here is what you will get:
Your first "aha moment" happens when you notice that the sequences are in alphabetical order and each has exactly one number missing. The alphabetical order is a good sign that you are on the right track; it can also narrow down the possible names of the sequences that you haven't yet identified. Alphabetical order means that you have to figure out the correct order for producing the answer.
Did you notice that some groups above are as long as nine integers and some are as short as four? In puzzles, there is nothing random, so the lengths of the groups should mean something. Your second "aha moment" will come when you realize that, together with the missing number, the number of the integers in each group is the same as the number of letters in the name of the sequence. This means you can get a letter by indexing the index of the missing number into the name of the sequence.
So each group of numbers provides a letter. Now we need to identify the remaining sequences and figure out in which order the groups will produce the word that is the answer.
Let's go back and try to identify the remaining sequences. We already know the number of letters in the name of each sequence, as well as the range within the alphabet. The third sequence might represent a challenge as its numbers are small and there might be many sequences that fit the pattern, but let's try. The results are below with the capitalized letter being the one that is needed for the answer.What is going on? There are two sequences that fit the pattern of the third group and the sequence for the sixth group has many names, two of which fit the profile but produce different letters. Now we get to your third "aha moment": you have already seen some of the sequence names before, because they are in the puzzle. This will allow you to disambiguate the names.
Now that we have all the letters, we need the order. Sequences are mentioned inside the puzzle. You were forced to notice that because you needed the names for disambiguation. Maybe there is something else there. On closer examination, all but one of the sequence names are mentioned. Moreover, with one exception the clues for one sequence mention exactly one other sequence. Once you connect the dots, you'll have your last "aha moment:" the way the sequences are mentioned can provide the order. The first letter G will be from the pentagonal sequence, which was not mentioned. The clues for the pentagonal sequence mention the primeval sequence, which will give the second letter R, and so on.
The answer is GRANTER.
Many old-timers criticized the 2013 MIT Mystery Hunt. They are convinced that a puzzle shouldn't have more than one "aha moment." I like my "aha moments."
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Should I eat this piece of cake or not? I will certainly enjoy it very much. What harm will it do? Will this piece increase my weight? Maybe not. The next piece might, but this particular one looks harmless. Even if my weight increases by half a pound, it could be muscle weight. Yes, it probably would be due to muscle weight: I just went out of my house to throw away my garbage and this has to count as exercise.
Do you see the problem? Eating the cake provides an immediate reward, but the punishment is vague and in the far distant future. That is why I got excited when my son Alexey sent me the link to Beeminder, a company that creates an artificial non-vague and not far-in-the-future punishment for eating that piece of cake.
Here is how it works. You give them your target number — in my case my desired weight, but it could be any measurable goal — and the date by which you want to hit it. They draw a yellow path on a weight chart. You must weigh yourself every day. Whenever your weight is above your path, you have to pay real money to the company. Five dollars!
This is a great idea. Suddenly that piece of cake looks threatening. The only problem with using their system is that I have no clue how to lose weight. The company doesn't provide tools to lose weight: it just provides a commitment device. So it is difficult to stick with the weight-loss commitment without having a proven weight-loss plan.
The truth is that my son sent me the link, I laughed, and forgot about it. Besides, if I ever want to pay money for failing in my commitments, I would rather choose the beneficiary myself. Then I realized that I can use the yellow-road idea to try to lose weight while figuring out what works for me. I call my new plan the Adaptive Diet.
Starting from my actual weight on Day One, I drew a line that represents my target weight, assuming a daily decrease of 0.1 pounds. A deviation of one pound from my target weight on my daily weigh-in is what I call my Yellow Zone. When I am in the Yellow, I continue doing what I was doing before: trying to build new, healthier habits.
If I am more than one pound below my target weight, then I have entered what I call the Green Zone. When I am in the Green, I can allow myself to indulge my cravings. However, when I am one pound above my target weight, I call that the dreaded Red Zone. This Zone has different shades of red. If I am between 1 and 2 pounds above my target weight, I have to eat only apples after 8:00pm. If I am 2 to 3 pounds above my target weight, only-apples time starts at 6:00pm. And so on. Every extra pound above my target weight moves the cut-off time by two hours. That means that if I am 7 pounds above my target weight, I would have to eat apples all day long.
The system has to work: I do not like apples.
In skyscraper puzzles you have to put an integer from 1 to n in each cell of a square grid. Integers represent heights of buildings. Every row and column needs to be filled with buildings of different heights and the numbers outside the grid indicate how many buildings you can see from this direction. For example, in the sequence 213645 you can see 3 buildings from the left (2,3,6) and 2 from the right (5,6).
In mathematical terminology we are asked to build a Latin square such that each row is a permutation of length n with a given number of left-to-right and right-to-left-maxima. The following 7 by 7 puzzle is from the Eighth World Puzzle Championship.
Latin squares are notoriously complicated and difficult to understand, so instead of asking about the entire puzzle we discuss the mathematics of a single row. What can you say about a row if you ignore all other info? First of all, let us tell you that the numbers outside have to be between 1 and n. The sum of the left and the right numbers needs to be between 3 and n+1. We leave the proof as an exercise.
Let's continue with the simplest case. Suppose the two numbers are n and 1. In this case, the row is completely defined. There is only one possibility: the buildings should be arranged in the increasing order from the side where we see all of them.
Now we come to the question we are interested in. Given the two outside numbers, how many permutations of the buildings are possible? Suppose the grid size is n and the outside numbers are a and b. Let's denote the total number of permutations by fn(a, b). We will assume that a is on the left and b is on the right.
In a previous example, we showed that fn(n, 1) = 1. And of course we have fn(a, b) = fn(b, a).
Let's discuss a couple of other examples.
First we want to discuss the case when the sum of the border numbers is the smallest — 3. In this case, fn(1, 2) is (n−2)!. Indeed, we need to put the tallest building on the left and the second tallest on the right. After that we can permute the leftover buildings anyway we want.
Secondly we want to discuss the case when the sum of the border numbers is the largest — n+1. In this case fn(a,n+1-a) is (n-1) choose (a-1). Indeed, the position of the tallest building is uniquely defined — it has to take the a-th spot from the left. After that we can pick a set of a-1 buildings that go to the left from the tallest building and the position is uniquely defined by this set.
Before going further let us see what happens if only one of the maxima is given. Let us denote by gn(a) the number of permutations of n buildings so that you can see a buildings from the left. If we put the shortest building on the left then the leftover buildings need to be arrange so that you can see a-1 of them. If the shortest building is not on the left, then it can be in any of the n-1 places and we still need to rearrange the leftover buildings so that we can see a of them. We just proved that the function gn(a) satisfies the recurrence:
Actually gn(a) is a well-known function. The numbers gn(a) are called unsigned Stirling numbers of the first kind (see http://oeis.org/A132393); not only do they count permutations with a given number of left-to-right (or right-to-left) maxima, but they also count permutations with a given number of cycles, and they appear as the coefficients in the product (x + 1)(x + 2)(x + 3)...(x + n), among other places. (Another pair of exercises.)
We are now equipped to calculate fn(1, b). The tallest building must be on the left, and the rest could be arranged so that, in addition to the tallest building, b-1 more buildings are seen from the right. That is fn(1, b) = gn-1(b-1).
Here is the table of non-zero values of fn(1, b).
| b=2 | b=3 | b=4 | b=5 | b=6 | b=7 | |
|---|---|---|---|---|---|---|
| n=2 | 1 | |||||
| n=3 | 1 | 1 | ||||
| n=4 | 2 | 3 | 1 | |||
| n=5 | 6 | 11 | 6 | 1 | ||
| n=6 | 24 | 50 | 35 | 10 | 1 | |
| n=7 | 120 | 274 | 225 | 85 | 15 | 1 |
Now we have everything we need to consider the general case. In any permutation of length n, the left-to-right maxima consist of n and all left-to-right maxima that lie to its left; similarly, the right-to-left maxima consist of n and all the right-to-left maxima to its right. We can take any permutation counted by fn(a, b) and split it into two parts: if the value n is in position k + 1 for some 0 ≤ k ≤ n-1, the first k values form a permutation with a - 1 left-to-right maxima and the last n - k - 1 values form a permutation with b - 1 right-to-left maxima, and there are no other restrictions. Thus:
Let's have a table for f7(a,b), of which we already calculated the first row:
| b=1 | b=2 | b=3 | b=4 | b=5 | b=6 | b=7 | |
|---|---|---|---|---|---|---|---|
| a=1 | 0 | 120 | 274 | 225 | 85 | 15 | 1 |
| a=2 | 120 | 548 | 675 | 340 | 75 | 6 | 0 |
| a=3 | 274 | 675 | 510 | 150 | 15 | 0 | 0 |
| a=4 | 225 | 340 | 150 | 20 | 0 | 0 | 0 |
| a=5 | 85 | 75 | 15 | 0 | 0 | 0 | 0 |
| a=6 | 15 | 6 | 0 | 0 | 0 | 0 | 0 |
| a=7 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
We see that the first two rows of the puzzle above correspond to the worst case. If we ignore all other constrains there are 675 ways to fill in each of the first two rows. By the way, the sequence of the number of ways to fill in the most difficult row for n from 1 to 10 is: 1, 1, 2, 6, 22, 105, 675, 4872, 40614, 403704. The maximizing pairs (a,b) are (1, 1), (1, 2), (2, 2), (2, 2), (2, 2), (2, 3), (2, 3), (2, 3), (3, 3).
The actual skyscraper puzzles are designed so that they have a unique solution. It is the interplay between rows and columns that allows to reduce the number of overall solutions to one.
I just compared two searches on Google Trends:
The most personal puzzle that I wrote for the 2013 MIT Mystery Hunt was Integers and Sequences based on my Number Gossip database. I named it after the first lecture that I prepared after I decided to return to mathematics. It is still my most popular lecture.
Many of the clues in this puzzle are standard math problems that are very good for math competition training. Other clues are related to sequences and integer properties.
You might wonder why I often ask for the second largest integer with some property. Isn't the largest one more interesting than the second largest? I do think that the largest number is more interesting, but exactly for this reason the largest number is available on my Number Gossip website and therefore is googleable. For example, my Number Gossip properties for 3000 contain the fact that 3000 is the largest palindrome in Roman numerals. This is why in the puzzle I used a slightly different clue, i.e. "the second largest three-letter palindrome in Roman numerals."
It took me many hours to find non-googleable variations of interesting properties for this puzzle. Unfortunately, its non-googleability evaporated as soon as my solution was posted, right after the hunt. In any case some clues in this puzzle are useful for math competition training, and I plan to use them myself in my classes. The puzzle is attached below. I will post the solution in a couple of weeks.
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The ultimate goal of each MIT Mystery Hunt is to find a hidden coin. So it was highly appropriate that our 2013 team created a coin-weighing puzzle (written by Ben Buchwald, Darby Kimball, and Glenn Willen) as a final obstacle to finding the winning coin:
There are nine coins, one real and eight fake. Four of the fake coins weigh the same and are lighter than the real coin. The other four fake coins weigh the same and are heavier than the real coin. Find the real coin in seven weighings on the balance scale.
Actually, it is possible to find the real coin in six weighings. Can you do that?
My weight used to be my most guarded secret. In general, I am a very open person: I'll tell anyone anything about me, unless it involves other people. However, there were two exceptions, both of them numbers, interestingly enough: my age and my weight. The closest I came to revealing my weight was with my sister, because we often discuss our similar health issues. Unfortunately, she knows my age, so the only missing number is my weight. I am so tired of my struggle to lose weight, that I've stopped caring about keeping the number secret. I am ready to tell it to the whole world.
Let me start from the beginning. I grew up in a country and at a time when men liked plump women. I was never thin, and didn't have to worry about my weight like my thin girlfriends did. I'll never forget my high school boyfriend telling me, "Ninety percent of men like fat women, and the other ten percent like very fat women." When in college I weighed 70 kilograms (154 pounds) and I felt fine. I had my first child when I was 23. I gained 20 kilos while I was breastfeeding, reaching 90 kilos (200 pounds). My husband Andrey kept telling me that he liked Rubenesque women. I wasn't even slightly concerned about my weight. When we divorced in 1988, I felt that my world was crushed and I didn't want to go on living. As a result, I lost about 20 pounds.
By 1990 I recovered from my depression, married my next husband, and moved to the US to live with him. The US made me aware of my weight immediately. It didn't help that Andrey remarried a woman who was the opposite of Rubenesque. From this point on, I wanted to lose weight. After my second child was born, I gained 20 kilograms while breastfeeding, just as I had done with the first child. The result was that I weighed about 220 pounds, much more than I wanted.
I started to look around at what capitalist society had to offer. The pharmacy had many products. I tried Slim Fast, which started to kill my appetite immediately. However, I began to get depressed. The depression felt foreign. As a new mother, I had been very happy before using Slim Fast, and there had been no changes in my life other than consuming Slim Fast. I stopped using it and the depression disappeared. To make sure, I did an experiment. I started using Slim Fast again and the depression reappeared within three days. I stopped it and my depression disappeared. I was so desperate to lose weight that I repeated the experiment. But the result was the same. I stopped using it, and never used any slimming supplement since then. But within that whole process, I lost some weight.
I stayed slightly over 200 pounds for several years. The third time (after the divorce and the Slim Fast) that I lost a lot of weight was when I had my heart broken about 15 years ago. Since then I've been slowly gaining weight.
As you can see from my story, I was never able to lose weight when I wanted to. I lost it three times, but I can't and don't want to reproduce those circumstances. I actually do not know how to lose weight. For the past ten years I've been making changes in my eating habits that I hope, cumulatively, would help me lose weight. I do not buy soda or pizza. I significantly cut my consumption of sweets and starches. I eat more fruits and vegetables. I eat half of what I used to eat in a restaurant. I am still gaining weight.
he only thing I haven't tried is to be hungry. I am afraid of being hungry. Also I am scared that if I decide on a plan which might result in my being hungry, I will not be able to stick to it. I don't want to discover that I don't have enough will power. I am scared to be a failure. I hope that by writing and publishing this I'll gain the courage to replace my half-measures with a more drastic plan.
Oh! I forgot to tell you: I weigh 245 pounds.
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Grigori Perelman's theorem: There is no offer you can't refuse.
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A conversation between two Russians:
— Run to the store and fetch a couple bottles of vodka.
— How much is a couple?
— Seven.
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— Is it true that the Windows operating system was copied from a UFO computer that crashed in Roswell?
— All we know for sure is that the UFO that didn't crash had a different operating system.
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I saw our system administrator's shopping list. The first line was tomatoes.zip for ketchup.
I posted the puzzle In the Details two weeks ago. This is the most talked-about puzzle of the 2013 MIT Mystery Hunt. The author Derek Kisman invented this new type of puzzle and it is now called a Fractal Word Search. I anticipate that people will start inventing more puzzles of this type.
Let's discuss the solution. The words in the given list are very non-random: they are related to fractals. How do fractals work in this puzzle? The grid shows many repeating two-by-two blocks. There are exactly 26 different blocks. This suggests that we can replace them by letters and get a grid that is smaller, for it contains one-fourth of the number of letters. How do we choose which letters represent which blocks? We expect to see LEVEL ONE in the first row as well as many other words from the list. This consideration should guide us into the matching between letters and the two-by-two blocks.
The level one grid contains 18 more words from the list. But where are the remaining words? So we have level one, and the initial grid is level two. The substitution rule allows us to replace letters by blocks and move from level one to level two. When we do this again, replacing letters in level two by blocks, we get the level three grid. From there we can continue on to further levels. There are three words from the list on level three and one word on level four. But this is quickly getting out of hand as the size of the grid grows.
Let's step back and think about the next step in the puzzle. Usually in word search puzzles, after you cross out the letters in all the words you find, the remaining letters spell out a message. What would be the analogous procedure in the new setting of the fractal word search? In which of our grids should we cross out letters? I vote for grid number one. First, it is number one, and, second, we can assume that the author is not cruel and put the message into the simplest grid. We can cross-out the letters from words that we find in level one grid. But we also find words in other levels. Which letters in the level one grid should we cross out for the words that we find in other levels? There is a natural way to do this: each letter in a grid came from a letter in the previous level. So we can trace any letter on any level to its parent letter in the level one grid.
We didn't find all the words on the list, but the missing words are buried deep in the fractal and each can have at most three parent letters. I leave to the reader to explain why this is so. Because there are so few extra letters, it is possible to figure out the secret message. This is what my son Sergei and his team Death from Above did. They uncovered the message before finding all the words. The message says: "SUM EACH WORD'S LEVEL. X MARKS SPOT." Oh no! We do need to know each word's level. Or do we? At this point, the extra letters provide locations of the missing words. In addition, if a word on a deep level has three parents, then it has to be a diagonal word passing through a corner of one of the child's squares. So our knowledge of extra letters can help us locate the missing words faster.
Also, the message says that the answer to this puzzle will be on some level in the part of the grid that is a child of X. Luckily, but not surprisingly, there is only one letter X on level one. The child of X might be huge. But we could start looking in the center. Plus, we know from the number of blanks at the end of the puzzle, that the answer is a word of length 8. So Sergei and his team started looking for missing words and the answer in parallel. Then Sergei realized that the answer might be in the shape of X, so they started looking at different levels and found the answer before finding the last word on the list. The answer was hiding in the X shape in the center of the child of X on level 167: HUMPHREY.
H..Y .UE. .RM. H..P
Cambridge Waldo puzzle from the 2013 MIT Mystery Hunt was supposed to be easy. Its goal was to get people out of the building for some fresh air. I made this puzzle jointly with Ben Buchwald, Adam (Pesto) Hesterberg, Yuri Lin, Eric Mannes and Casey McNamara. The puzzle consists of 50 pictures of different locations in Cambridge; one of the above individuals was hiding in each picture. Let me use this opportunity to thank my friends for starring in my puzzle and being inventive while doing it.
The puzzle starts with a group picture of my stars. The caption to the picture gives their names. The fact that they are standing in alphabetical order is a clue.
Out of the 50 pictures, each person appears in exactly ten pictures. If you mark the locations of one person on a map, they look like a letter. For example, below are Ben's locations that form a letter "S". When you put the letters in the alphabetical order by people names you get the answer to the puzzle: SCAMP.
As you can see, you do not need all ten locations to recognize the letter. You might be able to recognize the letter with five locations, or at least significantly reduce possibilities for the letter. Besides, you do not need all the letters to recognize the answer. We thought that this was an easy puzzle.
And, to make it even easier, the order in which we posted the pictures was not completely random. The pictures of one person were in the order one might walk from one location to another. This played two important roles. If you recognize the person but do not recognize the location on the picture, you can make an approximate estimation of the location because it must be on the path between the previous and next locations. If you recognize the location but not the person, you can guess the person by checking whose path it fits better.
It was difficult to hide people, especially when there were no other people around. So we sometimes used props. We only used one prop per person. Here you can see Pesto with his sarongs in plain view. In the other picture (below) he is hiding under a white sarong. Yuri had a bicycle helmet. In one of the pictures, she hid so well that you couldn't see her — but you could see her helmet. Ben had a bear hat. In one of the pictures you can only see a shadow of a person, but this person was clearly wearing a hat with bear ears. Eric didn't have a prop, but my car was eager to make a cameo appearance at the Mystery Hunt, so I hid him in my car in one of the pictures.
As you might guess I made the pictures of different people at different times. So Ben Buchwald was the one who realized that solvers might differentiate people by looking at the data of the picture files. He carefully removed the original time stamps.
Despite our best intentions, our test-solvers decided not to leave their comfy chairs, but rather to use Google-StreetView. We strategically made some of the pictures not Google-StreetViewable, but our test-solvers still didn't leave the comfort of their chairs. They just became more inventive. I do not know all the things they did to solve the puzzle, but I heard about the following methods:
I have yet to understand why this puzzle was difficult for the Mystery Hunt teams.
Here is the solution to the Open Secrets puzzle I published recently. Through my discussion of this solution, you'll also get some insight into how MIT Mystery Hunt puzzles are constructed in general.
I've included the puzzle (below) so that you can follow the solution. The puzzle looks like a bunch of different cipher texts. Even before we started constructing this puzzle, I could easily recognize the second, the seventh, and the last ciphers. The second is the cipher used by Edgar Allan Poe in his story, The Gold-Bug. The seventh cipher is a famous pigpen (Masonic) cipher, and the last is the dancing men cipher from a Sherlock Holmes story. Luckily you do not need to know all the ciphers to solve the puzzle. You can proceed with the ciphers you do know. With some googling and substitution you will translate these three pieces of text into: COLFERR, OAOSIS OF LIFEWATER, and RBOYAL ARCH.
These look like misspelled phrases, each of which has an extra letter. However, there are no typos in good puzzles. Or, more precisely, "typos" are important and often lead to the answer. So now I will retype the deciphered texts with the extra letter in bold: COLFERR, OAOSIS OF LIFEWATER and RBOYAL ARCH. In the first word it is not clear which R should be bold, but we will come back to that later.
At this point you should google the results. You may notice that the "royal arch" leads you to the Masons, who invented the pigpen cipher. From this, you can infer the structure of the ciphers and the connections among them. Indeed, a translation of one cipher refers to another. So you should proceed in trying to figure out what the texts you have already deciphered refer to. Eoin Colfer is the author of the Artemis Fowl series that contains a Gnomish cipher, and Oasis of Lifewater will lead you to Commander Keen video games with their own cipher. When you finish translating all of the ciphers, you get the following list:
The bold letters do not give you any meaningful words. So there is more to this puzzle and you need to keep looking. You will notice that the translations are in alphabetical order. This is a sign of a good puzzle where nothing is random. The alphabetical order means that you need to figure out the meaningful order.
To start, the phrases reference each other, so there is a cyclic order of reference. More importantly, the authors of the puzzle added an extra letter to each phrase. They could have put this letter anywhere in the phrase. As there is nothing random, and the placements are not the same, the index of the bold letter should provide information. If you look closely, you'll see that the bold letters are almost all in different places. If we choose the second R in COLFERR as an extra R, then the bold letter in each text is in a different place. Try to order the phrases so that the bold letters are on the diagonal. You'll see that this order coincides with the reference order, which gives you an extra confirmation that you are on the right track. So, let's order:
Now the extra letters read BOKLORYFH. This is not yet meaningful, but what is this puzzle about? It is about famous substitution ciphers. The first and most famous substitution cipher is the Caesar cipher. So it is a good idea to use the Caesar cipher on the phrase BOKLORYFH. There is another hint in the puzzle that suggests using the Caesar cipher. Namely, there are many ways to clue the dancing men code. It could be Conan Doyle, Sherlock Holmes, and so on. For some reason the authors chose to use the word ELEMENTARY as a hint. Although this is a valid hint, you can't help but wonder why the authors of the puzzle are not consistent with the hints. Again, there is nothing random, and the fact that the clues are under-constrained means there might be a message here. Indeed, the first letters read ROMAN CODE, hinting again at the Caesar shift. So you have to do the shift to arrive at the answer to this puzzle, which is ERNO RUBIK.
The most difficult puzzle I wrote for the MIT Mystery Hunt 2013 was Turnary Reasoning. I can't take credit for the difficulty: I designed the checkers positions; they were expectedly the easiest. Timothy Chow created the chess positions, and Alan Deckelbaum created the MTG positions. As the name of the puzzle suggests, you need to find whose turn it is in each position or, as the flavor text suggests, decide that the position is impossible.
I tried to solve the chess positions myself and was charmed by their beauty. The most difficult one was the first chess puzzle presented below. Find whose turn it is or prove that the position is impossible.
The puzzle titled "50/50" was the most difficult puzzle in MIT Mystery Hunt 2013. It is a puzzle in which information is hidden in the probability distribution of coin flips. I consider it the most difficult puzzle of the hunt because it took the longest time to test-solve and we were not able to solve all four layers of the original puzzle. As a result, one of the layers was removed. I think this puzzle is very important and should be included in statistics books and taught in statistics classes. If I were ever to teach statistics, I would teach this puzzle. By the way, this elaborate monstrosity (meant as a compliment) was designed by Derek Kisman.
I am not sure that the puzzle is working on the MIT server. The puzzle is just a coin flip generator and gives you a bunch of Hs (heads) and Ts (tails). Here is the solution.
When you flip a coin, the first thing to check is the probability of heads. In this puzzle it is fifty percent as expected. Then you might check probabilities of different sequences of length 2 and so on. If you are not lazy, you will reach length 7 and discover something interesting: some strings of length seven are not as probable as expected. The two least probable strings are TTHHTTH and HTHHTTT, with almost the same probability. The two most probable strings are TTHHTTT and HTHHTTH, with the matching probability that is higher than expected. All oddly behaving strings of length 7 can be grouped in chunks of four with the same five flips in the middle. In my example above, the five middle flips are THHTT. Five flips is enough to encode a letter. The probabilities provide the ordering, so you can read a message. In the version I tested it was "TAXINUMBLOCKS." In the current version it is "HARDYNUMBLOCKS." Keep in mind that the message encoded this way has to have all different letters. So some awkwardness is expected. The message hints at number 1729, a famous taxicab number, which is a clue on what to do next in the second step.
What do you do with number 1729? You divide the data in blocks of 1729 and see how the k-th flip in one block correlates with the k-th flip in the next block. As expected, for most of the indices there is no correlation. But some of the indices do have correlation. These indices are close together: not more than 26 flips apart. Which means the differences will spell letters. Also, there is a natural way to find a starting point: the group of indices spans only a third of the block. In the original version the message was: "PLEASEHELPTRAPPEDINCOINFLIPPINGFACTORYJKHEREHAVEAPIECEOFPIE."
Now I want to discuss the original version, because its solution is not available online. Here is Derek's explanation of what happens in the third step:
So, this punny message is another hint. In fact the sequence of coin flips conceals pieces of the binary representation of Pi*e. These pieces are of length 14 (long enough to stand out if you know where to look, but not long enough to show up as significant similarities if you compare different sessions of flips), always followed by a mismatch. They occur every 1729 flips, immediately after the final position of the 1729-block message. The HERE in the message is intended to suggest looking there, but you can probably also find them (with more effort) if you search for matches with Pi*e's digits.
The 14-flip sequences start near the beginning of the binary representation of Pi*e and continue to occur in order. (ie, every 1729 flips, 14 of them will be taken straight from Pi*e.) However, between sequences, either 1, 3, or 5 digits will be skipped. These lengths are a sequence of Morse code (1=dot, 3=dash, 5=letter break) that repeats endlessly, with two letter breaks in a row to indicate the start:
- .... ..- ..- ..- - ..- -..- ..- ..- ..- .... ..- .... - ..- ..- .... ..- ....
Translated, this gives the message "THUUUTUXUUUHUHTUUHUH".
(Aside: I didn't use Pi or e individually, because one of the first things I expect some teams will try is to compare the sequence of flips with those constants!)
As I said before, we didn't solve the third step. So Derek simplified it. He replaced "PIECEOFPIE" by "BINARYPI", and made it the digits of Pi, rather than of Pi*e. We still couldn't solve it. So he changed the message from the second step to hint directly at the fourth step: "PLEASEHELPTRAPPEDINCOINFLIPPINGFACTORYJUSTKIDDINGTHUUUTUXUUUHUHTUUHUH." But the binary Pi was still trapped in the coin flipping factory.
Here is Derek’s explanation of the fourth step:
Almost there! This message looks like some sort of flip sequence, because it has several Ts and Hs in there, but what of the Us and Xs? Well, U just stands for "unknown", ie, we don't care what goes there. And there's only one X, so it seems significant!
The final step is to look for every occurrence of this pattern in the sequence. The flips that go where the X is are the final channel of information. You'll find that they repeat in an unvarying pattern (no noise!) with period 323=17*19. There's only one way to arrange this pattern into a rectangular image with a blank border, and it gives the following image:
................. ...X..XXX.XXX.... ...X..X...X.X.... ...X..XXX.XXX.... ...X..X...XX..... ...X..X...X.X.... ...X.....X....... ...X.....X....... ...X....XXX...... ...X.....X....... ...X............. ...X.....XXX...X. ...X....XXXXX.XX. ..X....XX.XXXXX.. .X......XXXXXX... .X...X.XXXXXXXX.. ..XXX...XXXXX.XX. .........XXX...X. .................
The final answer is the French word for fish, POISSON, a word heavily related to statistics!
The answer POISSON didn't fit in the structure of the Hunt. So Derek was assigned a different answer: MOUNTAIN. He changed the picture and it is now available in the official solution to the puzzle. He adjusted his code for coin flips so that the picture of a mountain is hidden there. But the digits of Pi are still trapped in the flips. They are not needed for the solution, but they are still there.
Derek kindly sent to me his C++ program for the latest version of the puzzle. So if the MIT website can't generate the flips, you can do it yourself. And play with them and study this amazing example of the use of statistics in a one-of-a-kind puzzle.
When the MIT Mystery Hunt was about to end, I asked my son Sergei, who was competing with the team "Death from Above," what his favorite puzzle was. I asked the same question to a random guy from team "Palindrome" whom I ran into in the corridor. Surprisingly, out of 150 puzzles they chose the same one as their favorite. They even used similar words to describe it. Calling it a very difficult and awesome puzzle, they both wondered how it was possible to construct such a puzzle.
The puzzle they were referring to is "In the Details" by Derek Kisman, which you can see below.
TWELEVELTWONSHELMUMUOERAIYRANL QAPIUNPIQAYDPEPIRPRPKVOYESOYOR ELRATFDTELDTTFDTBWNLMUTFONYDWJ PIOYJMHAPIHAJMHAAOORRPJMYDANFC MUOZCGTFBWIRYDHIRAIRTFNCUENCUE RPVQUHJMAOHKANJUOYHKJMZKBNZKBN IRONSHOZGOTFUEELTFOEELUEYDOETF HKYDPEVQDNJMBNPIJMKVPIBNANKVJM BWIYNLTFSHHIELTWGOYDONDTYDHIOE AOESORJMPEJUPIQADNANYDHAANJUKV SHDTYDRPBWUEBWIYTWTWTFYDMUELMU PEHAANAJAOBNAOESQAQAJMANRPPIRP ONTWELBWLMSHELTFUEBWBWLMOZEVHI YDQAPIAOGIPEPIJMBNAOAOGIVQUNJU DTCGUEYDRPEVNCIREVIRTWUEUETWON HAUHBNANAJUNZKHKUNHKQABNBNQAYD IRUERAMUTFELTWONTFOEOEEYDTNLYD HKBNOYRPJMPIQAYDJMKVKVHWHAORAN ELGORPNCTFDTYDSHYDELPKTFOZRACG PIDNAJZKJMHAANPEANPIDFJMVQOYUH DTMUWJOETFYDELMUMUGORAONIRDTCG HARPFCKVJMANPIRPRPDNOYYDHKHAUH
| BOUNDARY | HENON | LEVY DRAGON | SCALING |
| BROWNIAN | HILBERT | LYAPUNOV | SPACE |
| CAUCHY | HURRICANE | MANDELBROT | STRANGE |
| CURLICUE | ITERATE | NEURON | TAKAGI |
| DE RHAM | JULIA | NURNIE | TECTONICS |
| DIMENSION | LEIBNIZ | POWER LAW | T-SQUARE |
| ESCAPE | LEVEL ONE | RAUZY | WIENER |
| HAUSDORFF | LEVEL TWO | RIVER | YO DAWG |
_ _ _ _ _ _ _ _
The puzzle looks like a word search, but I can tell you up-front: you can't find all the words in the grid. You can only find six words there. So there is something else to this puzzle. I will discuss the solution later. Meanwhile I will ask you very pointed questions:
The easiest of the puzzles I made for the MIT Mystery Hunt 2013 was "Something in Common." I collaborated on this puzzle with Daniel Gulotta. Ironically, it was the most time-consuming of my puzzles to design — well over a hundred hours. I can't tell you why it took me so long without revealing hints about the solution, so I will wait until someone solves it.
I received a lot of critique from my editors for suggesting puzzles that were too easy. When, during the test-solve, I realized that this puzzle was one of the easiest in the hunt, I requested permission to make it harder. It would not actually have been difficult to make it harder: I could have just replaced some specific words with more general ones. Unfortunately, we didn't have time for a new test-solve, so the puzzle stayed as it was. That turned out to be lucky.
This puzzle was in the last round. By the time the last round opened, we knew that the hunt was much more difficult than we had anticipated. I was afraid that people were getting angry with difficult puzzles and so I was very happy that I hadn't changed this puzzle. By the time the teams started submitting answers to it, people were exhausted. Manic Sages increased the speed with which options to buy answers to puzzles were released. I was ecstatic that this puzzle was one of the few puzzles in the last round that was solved, not bought with options. Here is the puzzle:
While everyone was discussing which color to paint the room, the counselor was dreaming about a real chocolate sundae. A walk-in walked into the reception in honor of a regular phenomenon visible every four hours or so. The client just wanted to find his dog in order to provide a permanent shortcut to another part of space; instead, he found something else.
The Federation didn't want to buy a lemon, so their lawyer used a spider to type the password. They discovered that the creatures glow in the dark and the perceptive negotiator won the bid. While they were waiting for the results of the shuttle trip, they turned the lights off to hunt them. The bird-like sounds were coming from under the floor. They used alcohol to save the woman's life, which was a big mistake. She refused to be his conscience, instead she used her new untapped power of light to save the world; she already had a job as a counselor.
Their daughter disappeared two months ago, whereas they needed to keep the wounded survivor steady. The African activist was kidnapped by rebels when she was a teenager. While some people were building a boat on the beach, she escaped from her kidnappers. He said that the reward had been withdrawn to eliminate the financial incentive to lie. They needed to perform a blood transfusion, but they didn't have the proper instruments. By his false statement, he removed over-claimers: people lying to get attention. He needed to write vows. They read surprise at the unexpected labor in the woods in the boy's face. It was clear that the boy had never looked into the backpack.
She described horrors she had seen: rape, torture, massacres carried out by child soldiers, the doctor losing his own blood to help the wounded survivor. There was no deception leakage, but she seemed anxious, so the doctor couldn't go and deliver the baby. They needed to find the other girl before noon, and the crushed leg made blood transfusion impractical. His vows were self-referential. He talked about how he couldn't write the vows. While the doctor played 20 questions one man died and another was born. The girl wasn't answering, but the doctor read her micro-expressions. The writer touched her ear and exposed her lies. At the end the doctor suspected murder, and the book was withdrawn.
She took pictures in an abandoned house and jumped off the roof. Two girls came back to investigate. He was surprised by a basket of cookies and a statue that was in the wrong place. A new drug hit the streets making people see things, like old pictures of a woman who looked exactly like her friend. He started seeing his dead father: when people die in the past, they produce energy in the present.
They were the two most powerful people on the planet. For wedding preparations she got a list of 17 DVDs. His partner started hallucinating about her ex. People do not understand time. Naomi's ability caused all this mess: the box left them behind. The only place where she didn't use gloves was her garden. The dinner guests tried her chicken, and, as a consequence, they were stuck looking at each other forever. The drugged cookies were a gift to help resolve their issues. Because of that, she gave him the shorthand transcripts to complete the cycle.
Did she fall asleep while they were discussing the dead boy's name? The price tag was exorbitant for a witness to identify the defendant. The woman would be everything he wanted and needed. She was a security guard chewing gum. The witness changed her testimony: they went to the woods and were water-rafting, rock-climbing, and arrow-shooting. As she was not the first girl he brought to the woods, the jury gave a verdict of "Not Guilty."
The detective was in court, though it wasn't his case. She needed to stop talking and start running. Shit always rolls downhill. Poisoned water made her see things she was not supposed to remember like a shitty orange couch outside. In spite of the background check the client was a psychopath who was moved from uptown to the fucking low-rises. Shoulder to the wheel. Dead witness, mother-fucker.
A little girl was humming about the dark side. The Wicca group was boring, and the new invention was making whooshing sounds. While a laryngitis epidemic took over the town, the noise-eater got turned on by accident.
The princess's scream would kill the monsters, while the monster machine killed its own remote control. She and her future boyfriend discovered each other's secrets. The dark side is waiting.
The experiment in the gay club was very important to him. He was a semi-cute boy-next-door type and his shuttle trip was supposed to be easy. There is more to a guy than cock size, and he was being pulled. He was injected with something and started to understand everyone. They were a bald priestess and a warrior with a very nice butt. They offered you a discount if you bought the butt and the bulge together.
They were approaching a commerce planet in Pennsylvania. They escaped and were standing on the roof of a hospital. She was sending a transmission, while the older man was running his tongue along the young man's spine. The girl tried to protect him, believing his story, but was arrested for contamination. Fuck. He had a baby, two babies. He distracted the guards with a puzzle ring. And while they were painting his black car with a pink derogatory word, they escaped.
He was drinking before operating on his fiancée. What kind of a father kills people? Tonight was the night he would finally sleep, and drinking was the only way for him to stop his hands from shaking. The limo driver kidnapped him while she watched herself die in the mirror. While being sleep-deprived he conducted a test. Nothing was wrong, except he brought the wrong documents to court. The criminal walked, but, unexpectedly, he couldn't feel his leg and couldn't walk anymore. There was a lot of blood and the body was not filtering iodine. He took a picture of the actor falling into a coma with his blood-spatter camera.
His girlfriend wanted to buy a firm mattress to have sex. He preferred to sleep on a water-bed and told the secret of his bloody hobby to his baby. It was not an infection, it was an allergy, so he changed his plan of where to grab the victim. The case was solved because of the bubbles in the glass. He was killing for his son now. The patient was indeed allergic and crashed his car.
The woman died two weeks ago, but her car accident was yesterday. There was so much blood and, for some reason, ice. Jane Doe wasn't the girl's sister. Melting speed and surface tension could help the calculation. They came to pay their respects but the celebration was postponed. They brought personal effects to drop off. If the couple's daughter was like their mother, then they themselves were like their grandparents. The idea to ask the grandmother resulted in a kiss.
Their planet had been on the verge of a golden age, then everything fell apart. She brought some clothes, a toothbrush, shampoo and some other things to the detention center. "Don't let the history repeat itself," the worshiper of the second in command said. She didn't know what the red string was for. The teen's sister told him stories about another galaxy, and the war and the algorithm that might find the dead man's favorite coffee place and, consequently, solve the case. The kidney operation was in a hotel. They didn't expect to see the body, but there it was. She touched the body to reveal emptiness inside. At the end they gathered in his house for dinner as usual. He had to save Courtney's future body before everything blew up and put the red sticker on his license.
After retirement, he came back to an unsolved murder case; he remembered a kiss in flashbacks. He drove to the lake where the accidental drowning happened. He lured everyone involved to the lake, and, after having sex, he told her that he wanted a real marriage. They wanted their child to be born on the new planet. The doctor was one of the suspects. The couple was already in the cabin, and he doubted that he was the one who she really wanted.
The young man was murdered, and everyone thought that the detective was the real target. The man went to look for sneakers during the boxing match. The nosy lady-writer dropped names to persuade the sheriff to look into the cabin. The admiral challenged the chief to join him in the ring. They didn't find the sneakers, and they shouted this to the skies. The lady solved the murder, as usual, while the boxing match became too personal and people started leaving.
The second "instructioned" puzzle is Portals by Palmer Mebane. It is an insanely beautiful and difficult logic puzzle that consists of known puzzle types interconnected to each other through portals. Here Palmer Mebane explains how portals work:
"Each of the ten puzzles corresponds to a color, seen above the grid where the name of the puzzle is written. The grid contains nine square areas, one each of the other nine colors. These are portals that connect the puzzle to one of the other nine, as denoted by the portal color. Each puzzle's rules define which squares of their solution are "black". On the portal squares, the two puzzles must agree on which squares are black and which are not. For instance, if in the red grid the top left square of the blue portal is black, then in the blue grid the top left square of the red portal must also be black, and vice versa."
On the Portals puzzle page you can find the rules for how each individual game is played and how to shade areas. The puzzle requires a lot of attention. It took us a long time to test-solve it. If you make a mistake in one grid it will propagate and will lead to a contradiction in another grid, so it is difficult to correct mistakes. If you do make a mistake, you are not alone: we kept making mistakes during our test-solve. Because of the difficulty of tracing back to the source of the error, we just started anew, but this time making sure that every step was confirmed by two people. Working together in this way, we were able to finish it.
If you do not care about the extraction and the answer, ignore the letter grid in the middle and enjoy the logic of it.
There were a couple of puzzles during the MIT Mystery Hunt that were not so mysterious. Unlike in traditional hunt puzzles, these puzzles were accompanied by instructions. As a result you can dive in and just enjoy solving the logic part of the puzzle without bothering about the final phase, called the extraction, where you need to produce the answer.
The first puzzle with instructions is Random Walk by Jeremy Sawicki. I greatly enjoyed solving it. In each maze, the goal is to find a path from start to finish, moving horizontally and vertically from one square to the next. The numbers indicate how many squares in each row and column the path passes through. There are nine mazes in the puzzle of increasing difficulty. I am copying here two such mazes: the easiest and the toughest. The colored polyomino shapes are needed for the extraction, so you can ignore them here.
Today I have a special treat for you. Here is the first of several puzzles that I plan to present from the 150 that we used in the MIT Mystery Hunt 2013. Keep in mind that although the puzzles have authors, they were the result of a collaboration of all the team members. In many instances editors, test-solvers and fact-checkers suggested good ways to improve the puzzles.
I wrote the puzzle Open Secrets jointly with Rob Speer. The puzzle was in the opening round, which means it is not too difficult. By agreement the answers to the puzzles are words or phrases. I invite my readers to try this puzzle. I will post the explanation in about two weeks.
I dropped my blog for two months. Some of my readers got worried and wrote to ask if I was okay. Thanks for your concern.
I am okay. I was consumed by the MIT Mystery Hunt. My team, Manic Sages, won the hunt a year ago, and as a punishment — oops, I meant as a reward — we got to write the 2013 hunt, instead of competing in it. I myself ended up writing about ten problems for the hunt. This was in addition to test-solving about 150 problems my whole team prepared for the hunt.
I could only think about the hunt. My mind was full of ideas for the hunt so I was afraid to write in my blog about something that I might later want to use for my problems. Or even worse, I was nervous that my blog posts might be unconsciously revealing hunt secrets. Moreover, I didn't want to advertise the fact that I was working on the hunt, thereby drawing people to my blog to scrutinize my interests as they prepared for the hunt.
So I just disappeared.
I apologize; please forgive me.
The father of my son has four children. My son is my only child. How many children do we have in total together?
"I will win the next International Chopin Piano Competition."
No matter how good I am at positive affirmations, that won't work: I do not play piano.
I tried to read books on positive thinking, but they made me mistrust the genre. The idea that you can achieve anything by positive thinking makes no sense. For example:
Positive thinking might actually be harmful. I can invest tons of time into trying to change my natural eye color by using my thoughts, when instead I could just use my money and buy some colored contact lenses. Or, if I think myself rich, I might start spending more money than I have and end up bankrupt.
However, perhaps I should not have totally dismissed the idea of positive thinking. While it does have logical inconsistencies, such as those in my examples above, maybe there are ways in which positive thinking is helpful.
First, we should treat these beliefs not as a guarantee, but probabilistically. For example, if you think that you can win the piano competition, the judges will feel your confidence, and may give you slightly better marks.
Second, positive thinking can work, if we choose our affirmations correctly. I recently discovered that I am deceiving myself into believing that I am hungry when I'm not. I should be able to reverse that. I should be able to persuade myself that I am not hungry when I am.
I decided to start small. I tried to persuade myself that tiramisu doesn't really taste good. Once that seemed to be working, I got more serious. I bought a couple of CDs with affirmations for weight loss.
Unfortunately, they want me to lie down and relax. I do not have time to lie down. I could listen when I am driving or when I am cleaning my kitchen. Hey, does anyone know some good weight-loss affirmations CDs that do not require relaxation?
One of the 2012 PRIMES projects, suggested by Professor Jacob Fox, was about bounds on the number of halving lines. I worked on this project with Dai Yang.
Suppose there are n points in a general position on a plane, where n is an even number. A line through two given points is called a halving line if it divides the rest of n−2 points in half. The big question is to estimate the maximum number of halving lines.
Let us first resolve the small question: estimating the minimum number of halving lines. Let's take one point from the set and start rotating a line through it. By a continuity argument you can immediately see that there should be a halving line through any point. Hence, the number of halving lines is at least n/2. If the point is on the convex hull of the set of points, then it is easy to see that it has exactly one halving line through it. Consequently, if the points are the vertices of a convex n-gon, then there are exactly n/2 halving lines. Thus, the minimum number of halving lines is n/2.
Finding the maximum number of halving lines is much more difficult. Previous works estimated the upper bound by O(n4/3) and the lower bound by O(ne√log n). I think that Professor Fox was attracted to this project because the bounds are very far from each other, and some recent progress was made by elementary methods.
Improving a long-standing bound is not a good starting point for a high school project. So after looking at the project we decided to change it in order to produce a simpler task. We decided to study the underlying graph of the configuration of points.
Suppose there is a configuration of n points on a plane, and we are interested in its halving lines. We associate a graph to this set of points. A vertex in the configuration corresponds to a vertex in the graph. The graph vertices are connected, if the corresponding vertices in the set have a halving line passing through them. So we decided to answer as many questions about the underlying graph as possible.
For example, how long can the longest path in the underlying graph be? As I mentioned, the points on the convex hull have exactly one halving line through them. Hence, we have at least three points of degree 1, making it impossible for a path to have length n. The picture below shows a configuration of eight points with a path of length seven. We generalized this construction to prove that there exists a configuration with a path of length n−1 for any n.
We also proved that:
After we proved all these theorems, we came back to the upper bound and improved it by a constant factor. Our paper is available at arXiv:1210.4959.
Samuel Hansen has an unusual profession: he is a mathematics podcaster. He interviewed me for his Relatively Prime podcast titled 0,1,2,3,…, where we discussed my Number Gossip project. The podcast also includes interviews with Neil Sloane, Michael Shamos, and Alex Bellos.
My previous interview with Samuel is at acmescience.com. There I discuss both math education and gender in math issues.
When I listened to myself, I found it strange that I seemed to have a British accent on top of my Russian accent. Did you notice that too?
I discovered the following chess puzzle on a Russian blog for puzzle lovers. It is a helpmate-type puzzle. Black cooperates with White in checkmating himself. In this particular puzzle Black starts and helps White to win in one move.
Oops. Something is not quite right. There are not enough pieces on the board. To recover the missing pieces in order to solve the puzzle, you need to retrace your steps. If Black and White go back one move each, they will be able to cooperatively checkmate Black in one move. Find the position one move back and the cooperative checkmate.
I am Russian; I know how to play chess. My father taught me when I was three or four. We played a lot and he would always win. I got frustrated with that and one day, when I was five, I didn't announce my check. On the next move, I grabbed his king and claimed my victory.
He was so angry that he turned red and almost hit me. This frightened me so much that I lost my drive for chess that very moment.
I still understand its beauty and solve a chess problem about once a decade. Look for a cute chess puzzle in my next post.
For my every class I try to prepare a challenge problem to stretch the minds of my students. Here is a problem I took from Adam A. Castello's website:
There is a ceiling a hundred feet above you that extends for- ever, and hanging from it side-by-side are two golden ropes, each a hundred feet long. You have a knife, and would like to steal as much of the golden ropes as you can. You are able to climb ropes, but not survive falls. How much golden rope can you get away with, and how? Assume you have as many hands as you like.
The next problem I heard from my son Sergei:
You are sitting at the equator and you have three planes. You would like to fly around the equator. Each plane is full of gas and each has enough gas to take you half way around. Planes can transfer gas between themselves mid-air. You have friends, so that you can fly more than one plane at once. How do you fly around the equator?
* * *
Right clicking a file with a mouse will allow you to change it, check it for viruses, or revert to the previous version. I wonder where I can buy a mouse that can do the same thing with my husband.
* * *
— Let's have sex.
— Sure, but today I want to be the numerator.
* * *
Attention! We want to check that you are not a robot. Please, undress and turn on a web-camera.
* * *
In a drug store:
— I would like input and output cleaners.
— ???
— Toothpaste and toilet paper.
* * *
I used to recount the multiplication tables to delay my ejaculation. Now, each time I see the multiplication tables I get a hard on.
* * *
— Tonight my parents are away. Let's finally try a forbidden thing.
— Dividing by zero?
* * *
My friend put his mistress in his phone's contact list as 'low battery'.
In what base does 2012! have more trailing zeros: base 15 or 16?
Explain why the result shouldn't be too surprising.
I would like to report on my weight loss progress. Last time I added two new habits, walking my toy dog every day, and drinking more water from the enticing cute bottles I bought.
I named my stuffed dog Liza and I walk with her every day. I didn't expect immediate weight loss due to this new regime, because my first goal was to get out of the house every day, even if only for two seconds. The next step will be to increase walking time to ten minutes.
Drinking a lot of water doesn't work well. I spend too much time looking for bathrooms and panicking that I will not make it. I like the idea of drinking a lot of water, but I am not sure I can hold to it, if you understand what I mean.
Since taking on this challenge, I've gained two habits, but I haven't lost a pound.
Now I'm upping my game. Below is my analysis of why I eat. When I eat, I believe that I am hungry. But looking at this more objectively I think this is not always the case: sometimes there are other reasons. I am listing these other reasons so I can fight them face-to-face. Here we go:
Hmm. That was painful to write. My psychoanalyst taught me that pain means I am on the right track.
Davidson Institute for Talent Development announced their 2012 Winners. Out of 22 students, three were recognized for their math research. All three of them are ours: that is, they participated in our PRIMES and RSI programs:
I already wrote about Xiaoyu's project. Today I want to write about Sitan's project and what I do as the math coordinator for RSI.
RSI students meet with their mentors every day and I meet with students once a week. On the surface I just listen as they describe their projects. In reality, I do many different things. I cheer the students up when they are overwhelmed by the difficulty of their projects. I help them decide whether they need to switch projects. I correct their mistakes. Most projects involve computer help, so I teach them Mathematica. I teach them the intricacies of Latex and Beamer. I explain general mathematical ideas and how their projects are connected to other fields of mathematics. I never do their calculations for them, but sometimes I suggest general ideas. In short, I do whatever needs to be done to help them.
I had a lot of fun working with Sitan. His project was about the rank number of grid graphs. A vertex k-ranking is a labeling of the vertices of a graph with integers from 1 to k so that any path connecting two vertices with the same label passes through a vertex with a greater label. The rank number of a graph is the minimum possible k for which a k-ranking exists for that graph. When Sitan got the project, the ranking numbers were known for grid graphs of sizes 1 by n, 2 by n, and 3 by n. So Sitan started working on the ranking number of the 4 by n graph.
His project was moving unusually fast and my job was to push him to see the big picture. I taught him that the next step, once he finishes 4 by n graphs is not to do 5 by n graphs, as one might think. After the first step, the second step should be bigger. He should use his insight and understanding of 4 by n graphs to try to see what he can do for any grid graphs.
This is exactly what he did. After he finished the calculation of the rank number of the 4 by n graphs, he found a way to improve the known bounds for the ranking number of any grid graph. His paper is available at the arXiv.
I just looked at my notes for my work with Sitan. The last sentence: "Publishable results, a potential winner."
PRIMES-USA: A new MIT program for talented math students from across the country.
I've been working as a math coordinator for RSI, the most competitive summer program for high school juniors. RSI arranges for these select students to do scientific research. I only work with kids who do math, and usually we have a dozen of them. Every student has an individual mentor, usually a graduate student, with whom they meet daily. I supervise all the projects and meet with each high school student about once a week. My job was described as "going for the biggest impact": when the project is in trouble, I jump in to sort it out; when the project is doing well, I push it to further limits.
RSI is a great program: kids enjoy it and we produce interesting research. My biggest concern is that the program is too short. The kids do math for five weeks and they usually approach a good result, but at the end of RSI we generally see just a hint of what they could truly achieve. Kids who continue to work on their own after the program ends are more successful. Unfortunately most of the students stop working at the end of the program just as they are approaching a big theorem.
I discussed this dissatisfying trend with Pavel Etingof and Slava Gerovitch and we decided to do something about it. Pavel and Slava conceived and found funding for a new program called PRIMES that is similar to RSI, but runs for a year. From February through May, PRIMES students meet with their mentors weekly. In fact, we require on the application that the students commit to coming to MIT once a week, thereby limiting us to local students. Theoretically, someone from Detroit with a private jet who can fly to MIT weekly would be welcomed.
Before the first year, we wondered whether the smaller pool of local students would be weaker than national and international RSI students. To our delight, that wasn't the case. In the first year we got fantastic students. One explanation is that PRIMES is much more flexible. We do not mind when our students go to IMO in the summer or to math camps or when they go away on vacation with their parents. As a result, we get students who would never apply to RSI because of their summer schedules. Our PRIMES students have won so many prizes that I do not remember them all. We post our successes on the website.
Our success in PRIMES suggests that there are likely many talented kids in other states who never even apply to RSI because of a scheduling conflict. This led us to try to adapt PRIMES to national needs. So we created a new program called PRIMES-USA that will accept students from across the country. We will work with them through Skype. These students must commit to travel to MIT for a PRIMES conference in May. Because this is our pilot program, we will only accept five students.
Thank you to everyone who helped me to find a host for my Number Gossip website. Some readers and friends even offered me free hosting on their servers. I decided to pay for hosting because I have many specific requirements and that might be a burden on my friends.
On the basis of my readers' recommendations, I chose Dreamhost as my new webhosting provider. I apologize for the interruption in the flow of the gossip. I know that many people use Number Gossip for birthday gift ideas. I can tell you that on my previous birthday, you could have congratulated me on becoming prime and evil.
I visited Raymond Smullyan on my way home from Penn State. We went for lunch at Selena's Diner. What do two mathematicians do during lunch? Exchange magic tricks and jokes, of course. Here is a story Raymond told me:
Raymond: What is the date?
Stranger: I do not know.
Raymond: But you have a newspaper in your pocket!
Stranger: It's no use. It's yesterday's.
Consider central symmetry: squares and circles are centrally symmetric, while trapezoids and triangles are not. But if you have two trapezoids, which of them is more centrally symmetric? Can we assign a number to describe how symmetric a shape is?
Here is what I suggest. Given a shape A, find a centrally symmetric shape B of the largest area that fits inside. Then the measure of central symmetry is the ratio of volumes: B/A. For centrally symmetric figures the ratio is 1, and otherwise it is a positive number less than 1.
The measure of symmetry is positive. But how close to 0 can it be? The picture on the left is a shape that consists of five small disks located at the vertices of a regular pentagon. If the disks are small enough than the largest symmetric subshape consists of two disks. Thus the measure of symmetry for this shape is 2/5. If we replace a pentagon with a regular polygon with a large odd number of sides, we can get very close to 0.
What about convex figures? Kovner's theorem states that every convex shape of area 1 contains a centrally symmetric shape of area at least 2/3. It is equal to 2/3 only if the original shape is a triangle. That means every convex shape is at least 2/3 centrally symmetric. It also means that the triangle is the least centrally symmetric convex figure. By the way, a convex shape can have only one center of symmetry.
After I started writing this I discovered that there are many ways in which people define measures of symmetry. The one I have defined here is called Kovner-Besicovitch measure. The good news is that the triangle is the least symmetric planar convex shape with respect to all of these different measures.
I'm trying to lose weight. Many books explain that dieting doesn't work, that people need to make permanent changes in their lives. This is what I have been doing for several years: changing my habits towards a healthier lifestyle.
This isn't easy. I am a bad cook; I hate shopping; and I never have time. Those are strong limitations on developing new habits. But I've been a good girl and have made some real changes. Unfortunately, my aging metabolism is changing faster than I can adopt new habits. Despite my new and improved lifestyle, I am still gaining weight.
But I believe in my system. I believe that one day I will be over the tipping point and will start losing weight, and it will be permanent. Meanwhile I would like to share with you the great ideas that will work someday.
I have many other ideas that for different reasons haven't yet become habits. So I am thinking about tricks to turn them into habits.
I have many more ideas, but I gotta run now. I need to walk my dog.
I received the book Taking Sudoku Seriously by by Jason Rosenhouse and Laura Taalman for review and put it aside to collect some dust. You see, I have solved too many Sudokus in my life. The idea of solving another one made me barf. Besides, I thought I knew all there is to know about the mathematics of Sudoku.
One day out of politeness or guilt I opened the book — and couldn't stop reading.
The book is written for people who like Sudoku, but hate math. This is so strange. Sudoku is math. People who are good at Sudoku are good at math, or at least they are supposed to be. It seems that math education in the United States is so bad that people who were born to be good at math and to like math, hate it instead. So the goal of the book is to establish a bridge from Sudoku to math. And the book does a superb job of it.
This well-written book moves from puzzles to discussions in such a natural way that math becomes a continuation of puzzles.
Taking Sudoku Seriously covers a lot of fun material: methods to solve Sudoku, how to count the number of different Sudoku puzzles, and how to find the smallest number of clues that are needed for a unique puzzle. The book travels into the neighboring area of Latin and Greco-Latin squares. While discussing all those fun things it covers groups, symmetries, number theory, graph theory (including book thickness) and more.
I am not the target audience for this book, because I do not need convincing that math is fun. The best part for me was the hundred puzzles. Only a portion of them were standard Sudoku puzzles — and I skipped those. The others were either Sudoku with a twist or plain math puzzles.
The puzzles are all very different and I was so excited by them, that I went ahead and solved them, and caught up with reading the text later. And I enjoyed both: reading and solving.
Here is puzzle 91 from the book. Fill in the grid so that every row, column, and block contains 1-9 exactly once. In addition, each worm must contain entries that increase from tail to head. For blue worms you must figure out yourself which end is the head.
It's easy to judge who is the fastest runner or swimmer. Judges do not need to be runners and swimmers themselves. They simply need a stopwatch and a camera.
Other competitions are more difficult to judge. Take for example the Fields medal. The judges need to be mathematicians. Since they can't be experts in all the different areas of mathematics, they have to rely on recommendation letters. The mathematicians who write recommendation letters are biased, because they are interested in promoting their own field. The committee's job is not simple, not the least because it involves a lot of politics. It is easy to award the medal to Grigory Perelman. He solved a high-profile long-standing conjecture. But other cases are not that straightforward.
Imagine a genius mathematician with a new vision. He or she might be so far ahead of everyone else, that the Fields committee would fail to appreciate the new concept. I wish the math community would create a list of mathematicians who deserved the Fields medal, but were passed over. As time goes by, perhaps a new Einstein will emerge on this list.
The reason the Fields committee more or less works is that the judges do not need to be as talented mathematicians as the awardees. They do not need to create mathematics, they need to understand it. And the latter is easier than the former.
A completely different story happens with IQ tests. Someone has to write those tests. There is no reason to think that writers of the IQ tests are anywhere close to the end tail of the IQ distribution. Hence, the IQ tests are not qualified to find the IQ geniuses.
Now might be a good time to complain about the IQ test I took myself. Many years ago I tried an IQ test online through tickle.com. I was so disappointed with my non-perfect score that I never looked at my answers. Recently, while cleaning my apartment, I discovered the printout of the test. I made one mistake in the following question.
Which one of the designs is least like the other four?
The checkmark is the expected answer. They think that the circle is the odd one out because all the other shapes are polygons. The arrow points to my answer. I chose the right triangle because it is the only shape without symmetries. Who says that polygonality is more important than symmetry?
I recently received an invoice from Jumpline, Inc. requesting a payment for hosting www.tanya-khovanova-temp.com. I had never heard of Jumpline before and I didn't have a webpage with that address. So I thought that it was spam.
Because the invoice had my name and address, I decided to call them and check what was going on. It appeared that Jumpline had swallowed Hosting Rails, the company that was hosting my Number Gossip page. Still, I didn't have a clue what the invoice was about.
I asked the representative whether the web address was related to Number Gossip, and he said no. So I canceled the hosting. My work schedule is the busiest in July, so I forgot about the invoice and didn't check my website.
Then I received a letter from Christian, a Number Gossip fan, who told me that the website was down. I called Jumpline again.
It appears that the representative didn't know what he was doing and misled me. The web address www.tanya-khovanova-temp.com was an internal name for my Number Gossip site. They had deleted all the files and were unable to restore my website.
Now I have to decide what to do. I do not want to go back to Jumpline as they are very unprofessional in these ways:
Can anyone suggest a company that can host a website that is written in Ruby on Rails?
The Fibonacci sequence is all about addition, right? Indeed, every element Fn of the Fibonacci sequence is the sum of the two previous elements: Fn = Fn-1 + Fn-2. Looking closer we see that the Fibonacci sequence grows like a geometric progression φn, where φ is the golden ratio. In addition, the Fibonacci sequence is a divisibility sequence. Namely, if m divides n, then Fm divides Fn.
My point: we define the sequence through addition, and then multiplication magically appears by itself. What would happen if we tweak the rule and combine addition and multiplication there?
John Conway did just that: namely, he invented a new sequence, or more precisely a series of sequences depending on the pair of the starting numbers. The sequences are called Conway's subprime Fibonacci sequences. The rule is: the next term is the sum of the two previous terms, and, if the sum is composite, it is divided by its least prime factor.
Let me illustrate what is going on. First we start with two integers. Let's take 1 and 1 as in the Fibonacci sequence. Then the next term is 2, and because it is prime and we do not divide by anything. The next two terms are 3 and 5. After that the sum of two terms is 8, which is now composite and it is divided by 2. So the sequence goes: 1, 1, 2, 3, 5, 4, 3, 7, 5, 6, 11 and so on.
The subprime Fibonacci sequences excite me very much. Not only does adding some multiplication to the rule make sense to me, but also, the sequences are fun to play with. I got so excited that I even coauthored a paper about these sequences titled, not surprisingly, Conway's Subprime Fibonacci Sequences. The paper is written jointly with Richard K. Guy and Julian Salazar, and is available at the arXiv:1207.5099.
We can start a subprime Fibonacci sequence with any two positive numbers. You can see that such a sequence doesn't grow fast, because we divide the terms too often. We present a heuristic argument in the paper that allows us to conjecture that no subprime Fibonacci sequence grows indefinitely, but they all start cycling. The conjecture is not proven and I dare you to try.
Meanwhile, the sequences are a lot of fun and I suggest a couple of exercises for you:
By the way, a trivial cycle is the boring thing that happens if we start a sequence with two identical numbers n bigger than one: n, n, n, n, ….
Have fun.
Kvant is a very popular Russian math and physics journal for high school-children. My favorite page is the one with puzzles directed to younger readers. Here are two puzzles from the latest online issue: 2012 number 3.
The first one, by N. Netrusova, is optimistic about the next year.
An astrologist believes that a year is happy if its digit representation contains four consecutive digits. For example, the next year, 2013, will be happy. When was the previous happy year?
The second problem is by L. Mednikov and A. Shapovalov. It confused me at first. For a moment I thought that the best answer is 241 rubles:
A big candle lasts one hour and costs 60 rubles. A small candle lasts 11 minutes and costs 11 rubles. Can you measure a minute by spending not more than a) 200 rubles, b) 150 rubles?
My American friends often ask me for insights into why Grigory Perelman refused the one million dollar Clay prize for his proof of the Poincaré conjecture. They are right to ask me: my life experience was very similar to Perelman's.
I went to a high school for children gifted in math. I was extremely successful in competitions. I got my gold medal at IMO and went to college without entrance exams. I received my undergraduate and graduate degrees in one of the best math academic centers in Soviet Russia. Perelman traveled a similar path.
Without ever having met Perelman, I can suggest two explanations of why he might reject the money.
First explanation. To have it publicly known that you have suddenly come into money is very dangerous in Russia. Perelman's life expectancy would have dropped immediately after accepting the million dollars. Russians that have tons of money either hide their wealth or build steel doors way before they make their first million. In addition to being a life hazard, money attracts a lot of bother. He would have been chased by all types of acquaintances asking for help or suggesting marriage proposals.
Second explanation. We grew up in a communist culture where money was scorned and math was idolized. The goal of research was research. Proving the conjecture was the prize itself. In his mind, receiving the award money might diminish the value of what he did. I understand this way of thinking, but I am personally too practical to follow such feelings and would accept the prize.
My first explanation has a flaw. Though valid, it doesn't explain why he rejected the Fields medal. So I reached for the book abour Perelman, Perfect Rigor: A Genius and the Mathematical Breakthrough of the Century by Masha Gessen. I like Gessen's explanation of why he rejected the Fields medal:
His objection to the Fields Medal, though never stated as clearly, seemed to have been twofold: first: he no longer considered himself a mathematician and hence could not accept a price intended for the encouragement of midcareer researchers; and second, he wanted no part of ICM, with all the attendant publicity, speeches, ceremony, and king of Spain.
The reasons are specifically related to the medal, so the Clay prize rejection might not be connected to the medal rejection. This argument slightly rehabilitates my first explanation.
I liked the book. It is a tremendous undertaking — writing about a person who doesn't want to talk to anyone. After reading it, I have one more possible explanation of his refusal of the prize.
Perelman is a loner. One of the closest people to him was his math Olympiad coach. The coaches tend to understand the solutions on the spot, mostly because they already know them. If in his mind Perelman expected all mathematicians to be like his coach, then he might have expected a parade in his honor the day after he solved the conjecture. Instead, he got silence and attempts to steal the prize from him.
Can you imagine doing the century's best math work without receiving congratulations for many years? The majority of mathematicians waited for the judgment of the experts, as did Perelman. The experts were busy and much slower than Perelman expected. The conjecture was extremely difficult, and it was a high-profile situation — after all, $1 million was attached to its solution. So the experts were very cautious in their pronouncements.
Finally, instead of congratulating Grigory, they said that the proof seemed to be correct and that they had not yet found any mistakes. If like Perelman, I was certain of my proof, I would have found this a painfully under-whelming conclusion.
Perelman expected to feel proud, but instead he probably felt unappreciated and attacked. Instead of the parade he may have hoped for, he had to wait for a long time, only to face disappointment and frustration. This reminds me of an old joke:
A genie is trapped in a lantern at the bottom of the sea. He vows, "I will give one million dollars to the person who frees me." One thousand years pass. He changes his vow, "I will give any amount of money to the one who frees me." Another thousand years pass. He ups the ante, "I will give any amount of money and two more wishes to the person who frees me." Another thousand years pass. He promises, "I will kill the one who frees me."
Third explanation. Perelman was profoundly disappointed in the math community. Unlike the genie, Perelman didn't want to kill anyone, but he did want to express his disillusionment. Perhaps that is why he rejected a million dollars.
Do you like challenging puzzles? Are you tired of sudoku? Here's your chance to try your hand at puzzles that are designed for world puzzle championships.
I've already done the homework for you — and it turned out to be more complicated than I anticipated. The world puzzle federation has a website, but unfortunately they are lazy or secretive. It is difficult to find puzzles there. A few puzzles are available in the World Puzzle Federation Newsletters.
Since I am stubborn, I spent a lot of time looking for championship puzzles. I found them in books. Here is the list I compiled so far. If you too are interested in high-level puzzles, this ought to make your search a lot easier. The book titles are confusing, so I added a description of what's in them.
One of my favorite puzzle types is Easy as ABC. You have to fill one of A, B, C, and D in each row and column. The letters outside the grid indicate which letter you see first from that direction. Here is one from the 2011 newsletter:
* * *
Engraved on a mathematician's tombstone: "Q.E.D."
* * *
—You act very brave on the Internet. But could you repeat this looking into my eyes?
—Sure. Send me your picture.
* * *
—Your birthday?
—December 26th.
—What year?
—Every year.
* * *
Teacher: "How much do we get if we cut eight into two halves?"
Student: "Two threes, if we cut vertically; and two zeros, if we cut horizontally."
Two girls. One is older and more experienced. The other is younger and more naive. Which of these two girls will the unnamed male narrator choose? What a great plot for a math book.
I am talking about Hiroshi Yuki's book Math Girls
The book starts discussing sequences and patterns. Can you guess the pattern behind the sequence: 1, 2, 3, 4, 6, 9, 8, 12, 18, 27, …? Can you explain how the beginning of this sequence might be very deceptive?
For the answer, you can read the book, which also discusses tons of fun topics: prime numbers, sum of divisors, absolute values, rotations and oscillations, De Moivre's formula, generating functions, arithmetic and geometric means, differential and difference operators, Catalan numbers, infinite series, harmonic numbers, zeta function, Taylor series, partitions, and more.
I usually do not like math fiction, but this is more math than fiction. It's quite superior to most other math books I've read, for it shows the unity of mathematics. It allows the readers to discover connections among different parts of mathematics, and it accomplishes this in a very thrilling way. Frankly, more thrilling than the romantic sections.
The fictional element brings an additional value to the book. The author uses dialogue to discuss points that are usually skipped in regular text books. The two girls give the narrator an opportunity to explore math on different levels: to talk about heavy stuff with Miruka and to provide explanations with Tetra.
I expected to be more interested in the sections dealing with advanced math. But the book is so well-written that the simpler things were a lot of fun, too. For example, I never before noticed that the column notation for n choose k is exactly the same as for a 2d vector with coordinates n and k. And I will never ever shout "zero" because the exclamation makes it "one".
I have a problem with my binocular vision. The muscles that are responsible for moving my eyes outwards are very weak, much weaker than the muscles that move my eyes inwards. When I am very tired, I can't focus on people or things that are far away. I start seeing doubled monsters with extra eyes and noses.
Luckily, instead of looking scary, the monsters look familiar. In fact, they look exactly like Picasso's portraits. I bet Picasso had problems with his eye muscles.
More than ten years ago I went through a process of psychotherapy which, although very painful, was extremely successful. When I tell my friends about this, they are interested in knowing what can be gained through psychotherapy, so here's my story.
I was living in Princeton, NJ, and I was very tired all the time. My primary care doctor told me that I was depressed and needed to do psychotherapy. A friend of mine recommended Dr. Ella Friedman. During my first visit Ella told me that I block my negative emotions. I protested. All my life I truly tried to be honest with myself. She insisted. I had nothing to lose because I had to solve the problem of my constant exhaustion and I had no other potential solutions. Besides, I liked her very much. So I decided to play along and started my search looking for negative emotions.
For some time I tried to convince Ella that if my best friend broke my favorite mug I wouldn't get angry with her. Ella tried to convince me otherwise. She pushed me back in time to the source of my beliefs and feelings. After several months of therapy, I discovered that I had a strong underlying belief that for my mother to love me, I must be a good girl who is always fair. Since my friend who broke the mug didn't do it on purpose, I wasn't allowed to be angry with her. I repressed all my angry feelings.
It took a lot of time for Dr. Friedman to rewire me and persuade me that my negative emotions do not mean that I am a bad girl. My actions define my goodness, not my emotions. I resisted. She had already convinced me that I might have negative emotions, but I didn't want to look at them. The power forcing me to block my emotions was the threat that my mother would withdraw her love if I wasn't a good girl. Dr. Friedman converted me. I started to believe her and continued more vigorously searching for my hidden emotions. Finally one day I collapsed in the shower. I actually felt my blocked emotions flooding me.
Negative emotions protect us. If someone treats you badly you need to be able to recognize it and get away from the danger. Because I didn't see my emotions I stayed in situations, like toxic relationships, that caused me great pain, without realizing it.
My psychotherapy didn't stop then. We started working on how to understand my emotions and how to process them. Now when someone is talking to me, I listen not only with my ears, but also with my gut. Suppose someone tells me, "I am so glad to see you," but I feel a strange tightness in my stomach. I start wondering what the tightness is about, and usually can figure it out. For the first time I was able to hear my gut and it was more illuminating than what I was hearing with my ears. All my life I processed information as text. Now the sentence "I am so glad to see you" has many different meanings.
The therapy changed my life. It feels as if I added a new sense to my palette of senses. I feel as if I was color blind for many years and at last I can see every color. Now that I've learned to recognize my pain, I can do something about it. I am so much happier today than I ever was before. While my friends may not have consciously recognized the big change in me, they have stopped calling me clueless and now often come to me for advice.
Did this solve my problem of tiredness? When Ella Friedman told me that I was no longer depressed, I still felt tired. I started investigating it further. It turns out that the depression was a result of the tiredness, not the other way around. It seems that I have a sleeping disorder and an iron problem.
SEAHOP created a practice puzzle, called "Making Connections," that includes me. It seems I am making connections.
I found a strange piece of paper in an old pile. I believe that it is a visual proof of the following statement:
If ∞ = 1/0, then 0 = 1/∞.
Proof. Assume ∞ = 1/0. Rotate each side of the equation counterclockwise 90 degrees. We get 8 = −10. Subtract 8, getting 0 = −18. Then rotate both parts back: 0 = 1/∞. QED.
I recently posted a short article on plagiarism. Did you notice that not a word of it was mine?
Baron Münchhausen is famous for his tall tales. My co-author Konstantin Knop wants to rehabilitate him and so invents problems where the Baron is proven to be truthful from the start. We already wrote a paper about one such problem. Here is a new problem by Konstantin:
Kostya has a black box, such that if you put in exactly 3 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 5 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than 3 times?
Actually, Konstantin designed a more complicated problem that was given at the Euler Olympiad, 2012 in Russia.
Let n be a fixed integer. Kostya has a black box, such that if you put in exactly 2n+1 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 4n+1 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than n+2 times?
Note that Kostya can't just put 4n+1 coins in the box. The box accepts exactly 2n+1 coins. The problem that I started with is for n = 1. Even such a simple variation was a lot of fun for me to solve. So, have fun.
Once upon a time there was a land where the only antidote to a poison was a stronger poison, which needed to be the next drink after the first poison. In this land, a malevolent dragon challenges the country's wise king to a duel. The king has no choice but to accept.
By bribing the judges, the dragon succeeds in establishing the following rules of the duel: Each dueler brings a full cup. First they must drink half of their opponent's cup and then they must drink half of their own cup.
The dragon wanted these rules because he is able to fly to a volcano, where the strongest poison in the country is located. The king doesn't have the dragon's abilities, so there is no way he can get the strongest poison. The dragon is confident of winning because he will bring the stronger poison.
The only advantage the king has is that the dragon is dumb and straightforward. The king correctly predicts what the dragon will do. How can the king kill the dragon and survive?
There is an array containing all the integers from 1 to n in some order, except that one integer is missing. Suggest an efficient algorithm for finding the missing number.
A friend gave me the problem above as I was driving him from the airport. He had just been at a job interview where they gave him two problems. This one can be solved in linear time and constant space.
But my friend was really excited by the next one:
There is an array containing all the integers from 1 to n in some order, except that one integer is missing and another is duplicated. Suggest an efficient algorithm for finding both numbers.
My friend found an algorithm that also works in linear time and constant space. However, the interviewer didn't know that solution. The interviewer expected an algorithm that works in n log n time.
The company claims that they are looking for the smartest people in the world, and my friend had presented them with an impressive solution to the problem. Despite his excitement, I predicted that they would not hire him. Guess who was right?
I reacted like this because of my own story. Many years ago I was interviewing for a company that also wanted the smartest people in the world. At the interview, the guy gave me a list of problems, but said that he didn't expect me to solve all of them — just a few. The problems were so difficult that he wanted to sit with me and read them together to make sure that I understood them.
The problems were Olympiad style, which is my forte. While we were reading them, I solved half of them. During the next hour I solved the rest. The interviewer was stunned. He told me of an additional problem that he and his colleagues had been trying to solve for a long time and couldn't. He asked me to try. I solved that one as well. Guess what? I wasn't hired. Hence, my reaction to my friend's interview.
The good news: I still remember the problem they couldn't solve:
A car is on a circular road that has several gas stations. The gas stations are running low on gas and the total amount of gas available at the stations and in the car is exactly enough for the car to drive around the road once. Is it true that there is a place on the road where the car can start driving, stopping to refuel at each station, so that the car completes a full circle without running out of gas? Assume that the car's tank is large enough not to present a limitation.
Sid Dhawan was one of our RSI 2011 math students. He was studying interlocking polyominoes under the mentorship of Zachary Abel.
A set of polyominoes is interlocked if no subset can be moved far away from the rest. It was known that polyominoes that are built from four or fewer squares do not interlock. The project of Dhawan and his mentor was to investigate the interlockedness of larger polyominoes. And they totally delivered.
They quickly proved that you can interlock polyominoes with eight or more squares. Then they proved that pentominoes can't interlock. This left them with a gray area: what happens with polyominoes with six or seven squares? After drawing many beautiful pictures, they finally found the structure presented in our accompanying image. The system consists of 12 hexominoes and 5 pentominoes, and it is rigid. You cannot move a thing. That means that hexominoes can be interlocked and thus the gray area was resolved.
You can find the proofs and the details in their paper "Complexity of Interlocking Polyominoes". As you can guess by the title, the paper also discusses complexity. The authors proved that determining interlockedness of a a system that includes hexominoes or larger polyominoes is PSPACE hard.
The Fomenko drawing on the left is from the original Russian edition of Homotopic Topology by Fuks, Fomenko and Gutenmacher. Dmitry Fuchs signed this book for me after my success in the USSR Math Olympiad when I was in the 9th grade. For many years I didn't know what the picture meant and was mystified by it. Now the book has been republished with explanations and is available in English at a non-affordable price. You can find this picture and many other Fomenko drawings in his book called Mathematical Impressions, which is affordable, although the comments accompanying the illustrations are confusing. So I have my own explanation for the meaning of this illustration.
The bracelet is made out of shells. Each shell is a hollow cone whose vertex is glued to a point on the rim of the cone's opening, thus giving each hollow cone its own handle. In a part of another drawing (at left), Fomenko shows how the bracelet is built by an army of tiny slaves. First they build the shells and then they connect them together.
How do they connect the shells to each other? The rim of the next shell is glued to the handle of the previous shell. Let me remind you that a straight line connecting a point on the rim to the vertex of a cone is called a generatrix. Imagine a generatrix that connects a vertex of a cone to the point on the rim to which this vertex is glued. This generatrix becomes a circle in a shell, which I call the handle circle. So the rim of the next shell is glued to the handle circle of the previous shell.
Now consider the fundamental group of a shell. The rim can be contracted to the handle circle. Moreover, the cone itself can be contracted to the handle circle. If we glue several shells together, the result is contractible to the handle circle of the last shell.
Now let's go back to the bracelet. The shells become smaller in both directions and end in two points. The front end point is more interesting topologically than the one in back. Every point other than the front end has a contractible neighborhood, while the front end point does not. Or in scientific terms: The bracelet gives an example of a space with a point at which the space is "1-lc" but with no open neighborhoods on which every (Cech) 1-cycle bounds.
My son Sergei invented the following game a couple of years ago. Two people, Alice and Bob, agree on a number, say, four. Alice takes a clean Rubik's cube and secretly makes four moves. Bob gets the resulting cube and has to rotate it to the initial state in not more than four moves. Bob doesn't need to retrace Alice's moves. He just needs to find a short path back, preferably the shortest one. If he is successful, he gets a point and then it is Alice's turn.
If they are experienced at solving Rubik's cube, they can increase the difficulty and play this game with five or six moves.
By the way, how many moves do you need to solve any position on a Rubik's cube if you know the optimal way? The cube is so complicated that people can't always know the optimal way. They think that God can, so they called the diameter of the set of all possible Rubik's cube positions, God's Number. It was recently proven that God's Number is 20. If Alice and Bob can increase the difficulty level to 20, that would mean that they can find the shortest path back to the initial state from any position of the cube, or, in short, that they would master God's algorithm.
One hundred people play the following game. Their names are written on pieces of paper and put into 100 labeled boxes at random. Each box is labeled with a number from 1 to 100 and one name has been placed inside each box. The boxes are placed on a table in a separate room. The players go into the room one by one and each has to open 99 boxes one after another. After each player finishes and leaves the room, the boxes are closed again. The players are not allowed to communicate with each other in any way, although they have been given one day before the event to discuss their strategies. They only win if every one of the one hundred players avoids opening the box with his or her own name. What is the optimal strategy?
Let me first discuss a simpler version of the game. Each player has to open exactly one box and they win if each one of them finds their name. After each player finishes and leaves the room, the boxes are closed again and the room is re-set.
If all of them decide to open box number 42, they are guaranteed to lose. They can try to open random boxes, then they win with probability (1/100)100. Can they use a joint strategy that is better than random?
Yes, they can. Clearly, two people shouldn't open the same box. So on the day before, if each agrees to open a box with a different assigned number, their probability to win is one over 100!. I leave it to my readers to prove that this is the best strategy.
What is the difference between this problem and the original problem? Isn't choosing the last box the same as choosing the first? Aha! When they open 99 boxes they see the names, so they can use this information as part of their strategy.
I hope that this new version is so intriguing that you will start solving this puzzle right away.
* * *
Decimals have a point.
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During the show "Are You Smarter Than a 5th Grader?" the following question was asked:
What is superfluous in the following list: a carrot, an onion, a potato, a Lexus?
A smart 5th grader answered: a carrot, an onion, and a potato.
* * *
If you buy 3 DVDs for the price of 4, you will get one more as a bonus.
* * *
Only yesterday, today was tomorrow.
* * *
By definition, one divided by zero is undefined.
* * *
Finally artificial intelligence has caught up with humans: when filling out electronic forms, many humans need several tries to prove they are not robots.
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Be back in 5 minutes. If I am late, reread this sms.
* * *
— We'll split the money 50-50.
— I want 70.
— Okay, 70-70!
I recently published my new favorite math problem:
A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic's goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic's assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.
If the psychic is only allowed to look at the backs of the cards, then the amount of transmitted information is 236, which is the same amount of information as suits for 18 cards. This number of guesses is achievable: the backs of every two cards can clue in the suit of the second card in the pair. This way the psychic can guess the suits of all even-numbered cards in the deck. So the problem is to improve on that. Using the info from the cards that the psychic is permitted to turn over can help too.
The problem is from the book Moscow Mathematical Olympiads, 2000-2005. The book and Russian blog discussions provide many different ideas on how to guess more than half of the deck.
Here is the list of ideas.
Idea 1. Counting cards. If you count cards you will know the suits of the last cards.
Idea 2. Trading. As we discussed before, the psychic can correctly guess the suits of even-numbered cards. By randomly guessing the odd-numbered cards she can correctly guess on average the suits of 4.5 additional cards. Unfortunately, this is not guaranteed. But wait. What if we trade the knowledge of the second card's suit for the majority suit among odd-numbered cards?
Idea 3. Three cards. Suppose we have three cards. Three bits can provide the following knowledge: the majority color, plus the suit of the first and of the second cards in the majority color. Thus, three bits of information will allow the psychic to guess the suits of two cards out of three.
Idea 4. Which card. Suppose the assistant signals the suits of even-numbered cards. With no loss, the psychic can guess the even-numbered card and repeat the same suit for the next card. If this is the plan, the assistant can choose which of the two cards to describe. Which card of the two matches the psychic's guess provides an additional bit of information.
Idea 5. Surprise. Suppose we have a strategy to inform the psychic about some cards. Suppose the assistant deliberately fails on one of the cards. Then the index of this card provides info to the psychic.
I leave it to my readers to use these ideas to find the solution for 19, 23, 24 and maybe even for 26 cards.
Imagine a slice of buttered white bread with a heap of sugar on top. That was my favorite lunch when I was a kid. My mom was working very hard, I was the oldest sister, and this was what I would make for myself almost every day.
Later someone told me that sugar is brain food. I believed that sugar and chocolate helped me do mathematics, so my love for sugar got theoretical support. I finally figured out the source of this love when my first son was born. To teach my son to stop requesting milk at night, my mother pushed me to give him sugar-water instead. At that moment, I realized that I developed my love of sugar with my mother's milk. Or, more precisely, instead of my mother's milk.
Now there is more and more evidence that the love of my life is a mistake. See for example Is Sugar Toxic?. Will I ever be able to break my oldest bad habit, the one I developed before I can remember myself doing it?
This is how my ex-husband Joseph Bernstein used to start his courses in representation theory.
Suppose there is a four-armed dragon on every face of a cube. Each dragon has a bowl of kasha in front of him. Dragons are very greedy, so instead of eating their own kasha they try to steal kasha from their neighbors. Every minute every dragon extends four arms to the neighboring cube's faces and tries to get the kasha from the bowls there. As four arms are fighting for every bowl of kasha, each arm manages to steal one-fourth of what is in the bowl. Thus each dragon steals one-fourth of of the kasha of each of his neighbors, while all of his own kasha is stolen too. Given the initial amounts of kasha in every bowl, what is the asymptotic behavior of the amounts of kasha?
You might ask how this relates to representation theory. First, it relates to linear algebra. We can consider the amounts of kasha as a six-dimensional vector space and the stealing process as a linear operator. As mathematicians, we can easily assume that a negative amount of kasha is allowed.
Now to representation theory. The group of rotations of the cube naturally acts on the 6-dimensional vector space of kashas. And the stealing operator is an intertwining operator of this representation. Now for a spoiler alert: I'm about to finish the solution, so stop here if you want to try it on your own.
The intertwining operator acts as a scalar on irreducible representations of the group. Thus we should decompose our representation into irreducible ones. The group has five irreducible representations with dimensions 1, 1, 2, 3, and 3.
We can decompose the kasha into the following three representations:
We see that asymptotically every dragon will have the same amount of kasha.
Now it is your turn to use this method to solve a similar problem, where there are n dragons sitting on the sides of an n-gon. Each dragon has two arms, and steals half of the kasha from his neighbors. Hey, wait a minute! Why dragons? There are people around the table stealing each other's kasha. But the question is still the same: What is the asymptotic behavior of the amounts of kasha?
Students should use a different strategy for the AIME than for the AMC. So students who are approaching the AIME for the first time need to question the habits they have developed after years of doing multiple choice tests. Here are some suggestions.
Checking. I've noticed that the accuracy level of students who take the AIME for the first time drops significantly. It seems that they are so used to multiple choice questions that they rely on multiple choices as a confirmation that they are right. So when someone solves a problem, they compare their answer to the given choices and if the answer is on the list they assume that the answer must be correct. Their pattern is broken when there are no choices. So they arrive at an answer and since there is no way to check it against choices, they just submit it. Because of this lack of confirmation, checking their answer in other ways becomes more important.
Timing.
At the AMC we have 3 minutes per problem. At the AIME — 12. That means the timing strategies need to be different. Indeed, the AMC is so fast-paced that it is reasonable to save time by not reading a problem twice. If you read it, you either solve it or skip it and go on. The student who is not trying to achieve a perfect score can decide in advance not to read those final, highly-difficult problems.
For the AIME it is not expensive, in relative terms of time, to read all the problems. The student can read the problems and choose the most promising ones to start with, knowing that if there is time they can always come back to other problems.
Guessing. Guessing at the AMC is very profitable if you can exclude three choices out of the given five. Guessing for the AIME is a waste of time because the answers are integers between 000 and 999. So the probability of a random guess is one in a thousand. Actually, this is not quite right, because the problem writers are human and it is much easier to write a problem with an answer of 10 than one with an answer of 731. But the AIME designers are trying very hard to make answers that are randomly distributed. So the probability of a random guess is not one in a thousand, but it is very close. You can improve your chances by an intelligent guess. For example, you might notice that the answer must be divisible by 10. But guessing is still a waste of time. Thinking about a problem for two minutes in order to increase the probability of a correct guess to one in a 100 means that your expected gain is 1/200 points per minute. Which is usually much less than the gain for checking your answers. You can play the guessing game if you have exhausted your other options.
What saddens me is that the students who are not trained in checking use their first guess to make their life choices. But this is a subject for a separate discussion.
I have already written about how American math competition are illogically structured, for the early rounds do not prepare students for the later rounds. The first time mathletes encounter proofs is in the third level, USAMO. How can they prepare for problems with proofs? My suggestion is to look East. All rounds of Russian math Olympiads — from the local to the regional to the national — are structured in the same way: they have a few problems that require proofs. This is similar to the USAMO. At the national All-Russian Olympiad, the difficulty level is the same as USAMO, while the regionals are easier. That makes the problems from the regionals an excellent way to practice for the USAMO. The best regional Olympiad in Russia is the Moscow Olympiad. Here is the problem from the 1995 Moscow Olympiad:
We start with four identical right triangles. In one move we can cut one of the triangles along the altitude perpendicular to the hypotenuse into two triangles. Prove that, after any number of moves, there are two identical triangles among the whole lot.
This style of problems is very different from those you find in the AMC and the AIME. The answer is not a number; rather, the problem requires proofs and inventiveness, and guessing cannot help.
Here is another problem from the 2002 Olympiad. In this particular case, the problem cannot be adapted for multiple choice:
The tangents of a triangle's angles are positive integers. What are possible values for these tangents?
The problems are taken from two books: Moscow Mathematical Olympiads, 1993-1999, and Moscow Mathematical Olympiads, 2000-2005. I love these books and the problems they present from past Moscow Olympiads. The solutions are nicely written and the books often contain alternative solutions, extended discussion, and interesting remarks. In addition, some problems are indexed by topics, which is very useful for teachers like me. But the best thing about these books are the problems themselves. Look at the following gem from 2004, which can be used as a magic trick or an idea for a research paper:
A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic's goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic's assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.
Do you remember how to divide three apples among four people? Make apple sauce, of course. In the following two puzzles you are not allowed to cut apples. Here is an old riddle:
There are four apples in a basket. How can you divide them among four people, so that one apple remains in the basket?
Here is a variation from Konstantin Knop's blog:
There are four apples in a basket. How can you divide them among three people, so that no one has more than the others and one apple remains in the basket?
Imagine a magician who comes on stage and performs the following magic trick:
He asks someone in the audience to think of a two-digit number, then subtract the sum of its digits. He waves his wand and guesses that the result is divisible by nine. Ta-Da!
This is not magic. This is a theorem. To make it magical we need to disguise the theorem.
First, there are many ways to hide the fact that we subtract the sum of the digits. For example, we can ask to subtract the digits one by one, while chatting in between. It is better to start with subtracting the first digit. Indeed, if we start with subtracting the second digit, the audience might notice that the result is divisible by 10 and start suspecting that some math is involved here. You can be more elaborate in how you achieve the subtraction of the sum of digits. For example, subtract twice the first digit, then the second, then add back the original number divided by 10.
Second, we need to disguise that the result is divisible by 9. A nice way to do this is implemented in the online version of this trick. The website matches the resulting number to a gift that is described on the page in pale letters. Paleness of letters is important as it is difficult to see that the same gift reappears in a pattern. In my work with students I use the picture on the left. At the end I tell them, "Ta-Da! the resulting number is blue." Here is the full sized version of the same picture that you can download.
My students are too smart. They see through me and guess what is going on. Then I ask them the real question, "Why do I have some cells with question marks and other symbols?" To give you a hint, I can tell you that the symbols are there for the same reason some blue numbers are not divisible by 9.
We got this problem from Rados Radoicic:
A 7 by 7 board is covered with 38 dominoes such that each covers exactly 2 squares of the board. Prove that it is possible to remove one domino so that the remaining 37 still cover the board.
Let us call a domino covering of an n by n board saturated if the removal of any domino leaves an uncovered square. Let d(n) be the number of dominoes in the largest saturated covering of an n by n board. Rados' problem asks us to prove that d(7) < 38.
Let's begin with smaller boards. First we prove that d(2) = 2. Suppose that 3 dominoes are placed on a 2 × 2 board. Let us rotate the board so that at least two of the dominoes are horizontal. If they coincide, then we can remove one of them. If not, they completely cover the board and we can remove the third one. Similarly, you can check all the cases and show that d(3) = 6.
Now consider a saturated domino covering of an n × n board. We can view the dominoes as vertices of a graph, joining two if they share a cell of the board. No domino can share both cells with other dominoes, or we could remove it. Hence, each domino contains at most one shared cell. This means that all the dominoes in a connected component of the graph must overlap on a single shared cell. Hence, the only possible connected components must have the following shapes:
The largest shape in the picture is the X-pentomino. We can describe the other shapes as fragments of an X-pentomino, where the center and at least one more cell is intact. We call these shapes fragments.
A saturated covering by D dominoes corresponds to a decomposition of the n × n board into F fragments. Note that a fragment with k cells is made from k − 1 dominoes. Summing over the dominoes gives: D = n2 − F. Thus, in order to make D as large as possible, we should make F as small as possible. Let f(n) be the minimal number of fragments that are required to cover an n by n board without overlap. Then d(n) = n2 − f(n).
Consider the line graph of the n by n board. The vertices of the line graph correspond to cells in the original board and the edges connect vertices corresponding to neighboring cells. Notice that in the line graph our fragments become all star graphs formed by spokes coming out from a single central node. Thus a decomposition of a rectangular board into fragments corresponds to a covering of its line graph by star graphs. Consider an independent set in the line graph. The smallest independent set has the same number of elements as the smallest number of stars that can cover the graph. This number is called a domination number.
Now let's present a theorem connecting domino coverings with X-pentomino coverings.
Theorem. f(n) equals the smallest number of X-pentominoes that can cover an n by n board allowing overlaps and tiles that poke outside, which is the same as the domination number of the corresponding line graph.
The proof of this theorem and the solution to the original puzzle is available in our paper: "Saturated Domino Coverings." The paper also contains other theorems and discussions of other boards, not to mention a lot of pictures.
The practical applications of star graph coverings are well-known and widely discussed. We predict a similar future for saturated domino coverings and its practical applications, two examples of which follow:
First, imagine a party host arranging a plate of cookies. The cookies must cover the whole plate, but to prevent the kids sneaking a bite before the party, the cookies need to be placed so that removal of just one cookie is bound to expose a chink of plate. This means the cookies must form a saturated covering of the plate. Of course the generous host will want to use a maximal saturated covering.
For the second application, beam yourself to an art museum to consider the guards. Each guard sits on a chair in a doorway, from where it is possible to watch a pair of adjacent rooms. All rooms have to be observed. It would be a mistake to have a redundant guard, that is, one who can be removed without compromising any room. Such a guard might feel demotivated and then who knows what might happen. This means that a placement of guards must be a saturated domino covering of the museum. To keep the guards' Union happy, we need to use a maximal saturated covering.
We would welcome your own ideas for applications of saturated coverings.
Several years ago my son Sergei started a new movement: Sergeism. Followers of this philosophy seek to maximize Sergei's happiness. Since Sergei's happiness involves everyone being happy, becoming happy is a consequential goal of his followers.
Let me explain why this might be a perfect religion for many people, not the least myself. My parents didn't teach me to love myself. They taught me to sacrifice myself and put other peoples' interests ahead of my own. After reading tons of books and spending hours in therapy, I've learned to love myself — well, somewhat. But the truth is, I still feel guilty when I pamper myself. Sergeism eliminates this guilt. I can freely invest in my happiness as a committed member of this movement.
I became a Sergeist when I lost all hope of losing weight. I realized that my own health wasn't a strong enough motivation. But I'm always glad to skip a cookie in tribute to Sergeism. If, like me, you put others ahead of yourself and never find the time to exercise or the will to refuse deserts, join me. Become a Sergeist and lose weight for Sergei.
I recently posted an essay Binary Bulls without Cows with the following puzzle:
The test Victor is taking consists of n "true" or "false" questions. In the beginning, Victor doesn't know any answers, but he is allowed to take the same test several times. After completing the test each time, Victor gets his score — that is, the number of his correct answers. Victor uses the opportunity to re-try the test to figure out all the correct answers. We denote by a(n) the smallest numbers of times Victor needs to take the test to guarantee that he can figure out all the answers. Prove that a(30) ≤ 24, and a(8) ≤ 6.
There are two different types of strategies Victor can use to succeed. First, after each attempt he can use each score as feedback to prepare his answers for the next test. Such strategies are called adaptive. The other type of strategy is one that is called non-adaptive, and it is one in which he prepares answers for all the tests in advance, not knowing the intermediate scores.
Without loss of generality we can assume that in the first test, Victor answers "true" for all the questions. I will call this the base test.
I would like to describe my proof that a(30) ≤ 24. The inequality implies that on average five questions are resolved in four tries. Suppose we have already proven that a(5) = 4. From this, let us map out the 24 tests that guarantee that Victor will figure out the 30 correct answers.
As I mentioned earlier, the first test is the base test and Victor answers every question "true." For the second test, he changes the first five answers to "false," thus figuring out how many "true" answers are among the first five questions. This is equivalent to having a base test for the first five questions. We can resolve the first five questions in three more tests and proceed to the next group of five questions. We do not need the base test for the last five questions, because we can figure out the number of "true" answers among the last five from knowing the total score and knowing the answers for the previous groups of five. Thus we showed that a(mn) ≤ m a(n). In particular, a(5) = 4 implies a(30) ≤ 24.
Now I need to prove that a(5) = 4. I started with a leap of faith. I assumed that there is a non-adaptive strategy, that is, that Victor can arrange all four tests in advance. The first test is TTTTT, where I use T for "true" and F for "false." Suppose for the next test I change one of the answers, say the first one. If after that I can figure out the remaining four answers in two tries, then that would mean that a(4) = 3. This would imply that a(28) ≤ 21 and, therefore, a(30) ≤ 23. If this were the case, the problem wouldn't have asked me to prove that a(30) ≤ 24. By this meta reasoning I can conclude that a(4) ≠ 3, which is easy to check anyway. From this I deduced that all the other tests should differ from the base test in more than one answer. Changing one of the answers is equivalent to changing four answers, and changing two answers is equivalent to changing three answers. Hence, we can assume that all the other tests contain exactly two "false" answers. Without loss of generality, the second test is FFTTT.
Suppose for the third test, I choose both of my "false" answers from among the last three questions, for example, TTFFT. This third test gives us the exactly the same information as the test TTTTF, but I already explained that having only one "false" answer is a bad idea. Therefore, my next tests should overlap with my previous non-base tests by exactly one "false" answer. The third test, we can conclude, will be FTFTT. Also, there shouldn't be any group of questions that Victor answers the same for every test. Indeed, if one of the answers in the group is "false" and another is "true," Victor will not figure out which one is which. This uniquely identifies the last test as FTTFT.
So, if the four tests work they should be like this: TTTTT, FFTTT, FTFTT, FTTFT. Let me prove that these four tests indeed allow Victor to figure out all the answers. Summing up the results of the last three tests modulo 2, Victor will get the parity of the number of correct answers for the first four questions. As he knows the total number of correct answers, he can deduce the correct answer for the last question. After that he will know the number of correct answers for the first four questions and for every pair of them. I will leave it to my readers to finish the proof.
Knop and Mednikov in their paper proved the following lemma:
If there is a non-adaptive way to figure out a test with n questions by k tries, then there is a non-adaptive way to figure out a test with 2n + k − 1 questions by 2k tries.
Their proof goes like this. Let's divide all questions into three non-overlapping groups A, B, and C that contain n, n, and k − 1 questions correspondingly. By our assumptions there is a non-adaptive way to figure out the answers for A or B using k tries. Let us denote subsets from A that we change to "false" for k − 1 non-base tests as A1, …, Ak-1. Similarly, we denote subsets from B as B1, …, Bk-1.
Our first test is the base test that consists of all "true" answers. For the second test we change the answers to A establishing how many "true" answers are in A. In addition we have k − 1 questions of type Sum: we switch answers to questions in Ai ∪ Bi ∪ Ci; and type Diff: we switch answers to (A ∖ Ai) ∪ Bi. The parity of the sum of "false" answers in A − Ai + Bi and Ai + Bi + Ci is the same as in A plus Ci. But we know A's score from the second test. Hence we can derive Ci. After that we have two equations with two unknowns and can derive the scores of Ai and Bi. From knowing the number of "true" answers in A and C, we can derive the same for B. Knowing A and Ai gives all the answers in A. Similarly for B. QED.
This lemma is powerful enough to answer the original puzzle. Indeed, a(2) = 2 implies a(5) ≤ 4, and a(3) = 3 implies a(8) ≤ 6.
The following variation of a Bulls and Cows problem was given at the Fall 2008 Tournament of the Towns:
A test consists of 30 true or false questions. After the test (answering all 30 questions), Victor gets his score: the number of correct answers. Victor doesn't know any answer, but is allowed to take the same test several times. Can Victor work out a strategy that guarantees that he can figure out all the answers after the 29th attempt? after the 24th attempt?
Let's assume that we have a more general problem. There are n questions, and a(n) is the smallest number of times we need to take the test to guarantee that we can figure out the answers. First we can try all combinations of answers. This way we are guaranteed to know all the answers after 2n attempts. The next idea is to start with a baseline test, for example, to say that all the answers are true. Then we change answers one by one to see if the score goes up or down. After changing n − 1 answers we will know the answers to the first n − 1 questions. Plus we know the total number of true answers, so we know the answers to all the questions. We just showed that a(n) ≤ n.
This is not enough to answer the warm-up question in the problem. We need something more subtle.
Let's talk about the second part of the problem. As we know, 24 = 4 ⋅ 6. So to solve the second part, on average, we need to find five correct answers per four tests. Is it true that a(5) ≤ 4? If so, can we use it to show that a(30) ≤ 24?
The following three cases are the most fun to prove: a(5) = 4, a(8) ≤ 6, and a(30) ≤ 24. Try it!
By the way, K. Knop and L. Mednikov wrote a paper (available in Russian) where they proved that a(n) is not more than the smallest number k such that the total number of ones in the binary expansion of numbers from 1 to k is at least n − 1. Which means they proved that a(30) ≤ 16.
On the left is a puzzle from the 2000 Qualifying Test for USA and Canada teams to compete in the world puzzle championship. A set of all 21 dominoes has been placed in a 7 by 6 rectangular tray. The layout is shown with the pips replaced by numbers and domino edges removed. Draw the edges of the dominoes into the diagram to show how they are positioned.
We would like to discuss the mathematical theory behind this puzzle using a toy example below. Only three dominoes: 1-1, 1-2, 2-2 are positioned on the board and the goal is to reconstruct the positioning:
Let's connect adjacent numbers with segments to show potential dominoes and color the segments according to which domino they represent. The 1-1 edge is colored green, the 1-2 — blue, and the 2-2 — red. Now our puzzle has become a graph, where the numbers are vertices, the segments are edges, and the edges are colored. In this new setting, the goal of the puzzle is to find edges of three different colors so that they do not share vertices.
The next picture represents the line graph of the previous graph. Now the colors of the vertices correspond to different potential dominoes. Vertices are connected if the corresponding dominoes share a cell. In the new setting finding dominoes that do not share a cell is equivalent to finding an independent set. The fact that we need to use all possible dominoes means that we want the most colorful independent set.
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— If a black cat crosses in front of you and then crosses back, what does it mean? Is your bad luck doubled or canceled?
— Is this a scalar or a vector cat?
— Huh?
— A scalar cat doubles and a vector cat cancels.
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Unbuttered bread, unable to cause the usual harm, tries to fall on the dirtiest spot.
* * *
Chance is a design carefully planned by someone else.
* * *
Wikipedia: I know everything.
Google: I can find anything.
Facebook: I know everyone.
Internet: You are nothing without me.
Electricity: Shut up, jerks.
* * *
Yesterday I bought pills to increase my IQ. Couldn't open the jar.
* * *
Today I opened my desktop's case and finally understood whither my trash is emptied.
I just discovered a Russian Internet Linguistics Olympiad. Even though most linguistics problems are not translatable, this time we are lucky. My favorite problem from this Olympiad is related to chemical elements — their names in Russian have the same logical structure as in English. Keep in mind, the problem doesn't assume any knowledge of chemistry. Here is the problem:
The formulae for chemical elements and their names are given below in mixed order:
C3H8, C4H6, C3H4, C4H8, C7H14, C2H2;
Heptene, Butine, Propane, Butene, Ethine, Propine.
- Match the formulae with their names. Explain your solution.
- Write the names of the elements with the following formulae: C2H4, C2H6, C7H12.
- Write the formulae for the following elements: Propene, Butane.
I have two admirers, Alex and Mike. Alex lives next to my home and Mike lives next to my MIT office. I have a lousy memory, so I invented the following system to guarantee that I see both of my friends and also manage to come to my office from time to time. I have a sign hanging on the inside of my home door that says Office on one side and Alex on the other. When I approach the door, I can see right away where I went last time. So I flip the sign and that tells me where next to go. I have a similar sign inside my office door that tells me to go either to home or to Mike. Every evening I spend with one of my admirers discussing puzzles or having coffee. Late at night I come home to sleep in my own bed. Now let's see what happens if today my home sign shows Office and the office sign shows Mike:
After three days the signs return to their original positions, meaning that the situation is periodic and I will repeat this three-day pattern forever.
Let's get back to reality. I am neither memory-challenged nor addicted to coffee. I invented Alex and Mike to illustrate a rotor-router network. In general my home is called a source: the place where I wake up and start the day. There can only be one source in the network. My admirers are called targets and I can have an infinite number of them. The network needs to be constructed in such a way that I always end up with a friend by the end of the day. There could be many other places that I can visit, other than my office: for example, the library, the gym, opera and so on. These places are other vertices of a network that could be very elaborate. Any place where I go, there is a sign that describes a pattern of where I go from there. The sign is called a rotor.
The patterns at every rotor might be more complicated than a simple sign. Those patterns are called rotor types. My sign is called 12 rotor type as it switches between the first and the second directions at every non-friend place I visit.
The sequence of admirers that I visit is called a hitting sequence and it can be proved that the sequence is eventually periodic. Surprisingly, the stronger result is also true: the hitting sequence is purely periodic.
The simple 12 rotor is universal. That means that given a set of friends and a fancy periodic schedule that designates the order I want to visit them in, I can create a network of my activities where every place has a sign of this type 12 and where I will end up visiting my friends according to my pre-determined periodic schedule.
It is possible to see that not every rotor type is universal. For example, palindromic rotor types generate only palindromic hitting sequences, thus they are not universal. The smallest such example, is rotor type 121. Also, block-repetitive rotor types, like 1122, generate block-repetitive hitting sequences.
It is a difficult and an interesting question to describe universal rotor types. My PRIMES student Xiaoyu He was given a project, suggested by James Propp, to prove or disprove the universality of the 11122 rotor type. This was the smallest rotor type the universality of which was not known. Xiaoyu He proved that 11122 is universal and discovered many other universal rotor types. His calculations support the conjecture that only palindromic or block-repetitive types are not universal. You can find these results and many more in his paper: On the Classification of Universal Rotor-Routers.
My co-author Konstantin Knop wrote a charming book, Weighings and Algorithms: from Puzzles to Problems. The book contains more than one hundred problems. Here are a couple of my favorites that I translated for you:
There is one gold medal, three silver medals and five bronze medals. It is known that one of the medals is fake and weighs less than the corresponding genuine one. Real medals made of the same metal weigh the same and from different metals do not. How can you use a balance scale to find the fake medal in two weighings?
There are 15 coins, out of which not more than seven are fake. All genuine coins weigh the same. Fake coins might not weigh the same, but they differ in weight from genuine coins. Can you find one genuine coin using a balance scale 14 times? Can you do it using fewer weighings?
You might get the impression that the latter problem depends on two parameters. Think about it: It is necessary that the majority of the coins are genuine in order to be able to solve the problem. In fact, the number of weighings depends on just one parameter: the total number of coins. Denote a(n) the optimal number of weighings needed to find a genuine coin out of n coins, where more than half of the coins are genuine. Can you calculate this sequence?
Hint. I can prove that a(n) ≤ A011371(n-1); that is, the optimal number of weighings doesn't exceed n − 1 − (number of ones in the binary expansion of n−1).
We all heard this paradoxical statement:
This statement is false.
Or a variation:
True or False: The correct answer to this question is 'False'.
Recently we received a link to the following puzzle, which is similar to the statement above, but has a cute probabilistic twist:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 60%
- 25%
There are four answers, so you can choose a given answer with probability 25%. But oops, this answer appears twice. Is the correct answer 50%? No, it is not, because there is only one answer 50%. You can see that none of the answers are correct, hence, the answer to the question—the chance to be correct—is 0. Now is the time to introduce our new puzzle:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 0%
- 25%
This fractal was designed by Ross Hilbert and is named "Weathered Steel Weave." You can find many other beautiful pictures in his fractal gallery.
The fractal is based on iterations of the following fractal formula znew = cos(c zold), where the Julia Constant c is equal to −0.364444444444444+0.995555555555556i. To produce the image, you need to start with a complex value of z and iterate it many times using the formula above. The color is chosen based on how close the iteration results are to the border of the unit circle.
I found a new Russian Olympiad for high schools related to universities. I translated my favorite problems from last year's final round. These are the math problems:
8th grade. In a certain family everyone likes their coffee with milk. At breakfast everyone had a full cup of coffee. Given that Alex consumed a quarter of all consumed milk and one sixth of all coffee, how many people are there in the family?
8th grade. How many negative roots does the equation x4 − 5x3 − 4x2 − 7x + 4 = 0 have?
10th grade. Find a real-valued function f(x) that satisfies the following inequalities for any real x and y: f(x) ≤ x and f(x+y) ≤ f(x) + f(y).
I liked the physics problems even more:
8th grade. Winnie-the-Pooh weighs 1 kg. He hangs in the air with density 1.2kg/m3 next to a bee hive. He is holding a rope connected to a balloon. Estimate the smallest possible diameter of the balloon, assuming that this happens on Earth.
10th grade. Two containers shaped like vertical cylinders are connected by a pipe underneath them. Their heights are the same and they are on the same level. The cross-sectional area of the right container is twice bigger than the left's. The containers are partially filled with water of room temperature. Someone put ice into both containers: three times more ice into the right one than into the left one. After that, the containers are closed hermetically. How will the water level will change after the ice melts completely:
- The levels will not change.
- The level on the left will be higher than on the right.
- The level on the left will be lower than on the right.
- The answer depends on the initial volume of water in the containers.
Once I talked to my friend Michael Plotkin about IQ tests, which we both do not like. Michael suggested that I run an experiment and send a standard IQ question for children to my highly-educated friends. So I sent a mass email asking:
What's common between an apple and an orange?
I believe that the expected answer is that both are fruits.
Less than half of my friends would have passed the IQ test. They gave four types of answer. The largest group chose the expected answer.
The second group related the answer to language. For example, apples and oranges both start with a vowel and they both have the letters A and E in common.
The third group connected the answer to what was on their minds at the time:
And the last group were people who just tried to impress me:
If my friends with high IQs have given so many different answers, I would expect children to do the same. The variety of answers is so big that no particular one should define IQ. By the way, my own well-educated kids' answers are quoted above — and they didn't go with the standard answer. I'm glad they never had IQ tests as children: I'm sure they would never have passed.
I have an idea for a start-up medical insurance company for Massachusetts. My insurance will have an infinite deductible. That means you pay your own bills. The cost of insurance can be very low, say $100 a year, as I do not need to do anything other than to send you a letter confirming that you have medical insurance. People who otherwise will be fined up to $900 for being uninsured will run in droves to buy my insurance.
I have an even better idea. For an extra fee, I will negotiate with doctors so that you will pay the same amount as medical insurance companies pay to them, which is often three times less than you would pay on your own.
Who am I kidding? I am not a business person, I can't build a company. But I am looking to buy the insurance I just described.
I am just wondering:
What is the largest integer consisting of distinct digits such that, in its English pronunciation, all the words start with the same letter?
I continue to wonder:
What is the largest integer consisting of the same digit such that, in its English pronunciation, all the words start with distinct letters?
When you name your child there are many considerations to take into account. For example, you should always check that your kids' initials don't embarrass them. For example, if the Goldsteins want to name their son Paz, because it means golden in Biblical Hebrew, the middle name shouldn't be Isaak, or anything starting with I.
Contemporary culture adds another consideration: how easy would it be to find your child on the Internet? I personally find it extremely convenient to have a rare name, because my fans can find my webpage and blog just by googling me. Parents need to decide whether they want their children to be on the first page of the search engine or hidden very far away when someone googles them.
When I named my son Sergei, I knew that there was another mathematician named Sergei Bernstein. But I didn't think about the Internet. As a result, I confused the world: is my son more than a hundred years old or did Sergei Natanovich Bernstein compete at Putnam?
I decided to see the film The Oxford Murders
At the core of the movie are sequences of numbers and symbols. When the characters started a discussion about how to continue a sequence, I immediately tensed up. Why? Because when people ask what the next element in the sequence is, I get ready to confront them, by explaining that there are many ways to continue a sequence. For example, the sequence — 1, 2, 4 — could be powers of two, or could be Tribonacci numbers, or any of 10,000 sequences that the Online Encyclopedia of Integer Sequences spills out if you plug in 1, 2, 4. That is, if we do not count the infinity of sequences that are not in the Encyclopedia.
To my surprise and relief, the logic Professor, one of the main characters in the movie, explained that there is no unique way to continue a sequence. From that moment on, I relaxed and fell in love with the movie.
The movie is a detective story with a lot of twists and turns. The crimes are related to symbols. The first two symbols are in the picture below. Can you guess the next symbol?
I cannot. There is an irony in the film at this point, because the Professor and the student need to guess the sequence in order to solve the crimes. But the Professor has already explained that there is no unique way to continue. So illogical for a movie about logic.
And what's worse, the sequence of symbols they finally discover doesn't make sense. I guess I fell in love with this movie too quickly.
* * *
— I've noticed that fools are always sure of themselves, while clever people are doubtful.
— No doubt.
* * *
— What happened to your girlfriend, that really cute math student?
— She's no longer my girlfriend. I caught her cheating on me.
— I don't believe that she cheated on you!
— Well, a couple of nights ago I called her, and she told me that she was in bed wrestling with three unknowns.
* * *
A programmer calls the library:
— Can I talk to Kate please?
— She's in the archive.
— Can you unzip her?
* * *
To protect the population from airplane disasters, Congress has ratified an addendum to the law of gravity.
* * * (invented by David Bernstein)
Energy conservation: it's not just a good idea; it's the law.
* * *
— Your computer is such a mess.
— It got a nasty virus.
— And it poured coffee on your keyboard?
* * *
After little Tom learned to count, his father had to start dividing dumplings evenly.
* * *
In spite of the crisis, inflation, and erratic fluctuations of the market, Russian mathematicians promised the president to keep number Pi between 3 and 4 until at least the end of the year.
* * *
A logician rides an elevator. The door opens and someone asks:
— Are you going up or down?
— Yes.
My webpage and my blog generate a lot of emails. I love receiving most of the emails, but if I reply to them, I won't have time to work on my blog. My favorite type of message is one that is full of compliments, with a note that the writer doesn't expect a reply.
I am grateful to people who send me things I requested, like pictures of Russian plates, or some interesting number properties. I apologize that it takes me so long to reply.
The emails that I don't enjoy reading contain amazing elementary proofs of Fermat's last theorem, or any other theorem on the Millennium list, for that matter. I also do not like when my readers ask me for help with their homework.
Like most people, I'm already dealing with spammers who want to enlarge the body parts I do not have or to slim the ones I do have. However, if you do need to send me millions of dollars that I won in your lottery, there is no reason to waste time on email exchanges: you can process them through my "donate" button.
You are welcome to contact me, but ….
I already gave an example of the kinds of problems that were given to Jewish people at the oral entrance exam to the math department of Moscow State University. In fact, I have a whole page with a collection of such problems, called Jewish problems or Coffins. That page was one of the first pages I created when I started my website more than ten years ago.
When my son Alexey was in high school, I asked him to help me type these problems into a file and to recover their solutions from my more than laconic notes, and solve the problems that I didn't have notes for. He did the job, but the file was lying dormant on my computer. Recently I resurrected the file and we prepared some of the solutions for a publication.
The problems that were given during these exams were very different in flavor: some were intentionally ambiguous questions, some were just plain hard, some had impossible premises. In our joint paper "Jewish Problems" we presented problems with a special flavor. These are problems that have a short and "simple" solution, that is nonetheless very difficult to find. This way the math department of MSU was better protected from appeals and complaints.
Try the following problem from our paper:
Find all real functions of real variable F(x) such that for any x and y the following inequality holds: F(x) − F(y) ≤ (x − y)2.
I will give a talk on the subject for UMA at MIT on October 18, at 5pm.
What's "plagiarism"? It's when you take someone else's work and claim it's your own. It's basically STEALING.
Ideas improve. The meaning of words participates in the improvement. Plagiarism is necessary. Progress implies it. It embraces an author's phrase, makes use of his expressions, erases a false idea, and replaces it with the right idea.
Perhaps the Russians have done the right thing, after all, in abolishing copyright. It is well known that conscious and unconscious appropriation, borrowing, adapting, plagiarizing, and plain stealing are variously, and always have been, part and parcel of the process of artistic creation. The attempt to make sense out of copyright reaches its limit in folk song. For here is the illustration par excellence of the law of Plagiarism. The folk song is, by definition and, as far as we can tell, by reality, entirely a product of plagiarism.
If you copy from one author, it's plagiarism. If you copy from two, it's research.
In my essays The Oral Exam and A Math Exam's Hidden Agenda, I gave some examples of math problems that were used during the entrance exams to Moscow State University. The problems were designed to prevent Jewish and other "undesirable" students from studying at the University. My readers might have supposed that an occasional bright student could, by solving all the problems, get in. Here is the story of my dear friend Mikhail (Misha) Lyubich; it shows that being extremely bright was not enough.
Misha passed the first three exams and was facing his last exam: oral physics. He answered all the questions. None of his answers were accepted: all of them were declared wrong. Misha insisted that he was right and requested that the examiners explain themselves. Every time their reply was the same:
This is not a consultation, it's an exam.
Misha failed the exam. The solution to the last problem was a simple picture: a document that seemed to be impossible to deny, so Misha decided that he had grounds for an appeal. The person in charge denied the appeal. When Misha requested an explanation, can you guess the answer?
This is not a consultation, it's an appeal.
Misha ended up studying at Kharkov State University. Now he is a professor at Stony Brook and the director of the Institute for Mathematical Sciences at Stony Brook.
You know that the negation of a true statement is a false statement, and the negation of a false statement is a true statement. You also know that you can negate a sentence by preceding it with "It is not true that …."
Now look at the following statement and its negation, invented by David Bernstein. Which one is true?
How about this pair?
My son Alexey taught me to always plug unused power strips into themselves, so that we can call them "The Rings of Power." These are my Borromean Rings of Power:
Let's call a projection of a body L onto a hyperplane a shadow. Here is a mathematical way to hide behind. An object K can hide behind an object L if in any direction the shadow of K can be moved by a translation to be inside the corresponding shadow of L. If K can hide inside L, then obviously K can hide behind L. Dan Klain drew my interest to the following questions. Is the converse true? If K can hide behind L can it hide inside L? If not, then if K can hide behind L, does it follow that the volume of K is smaller than that of L?
We can answer both questions for 2D bodies by using objects with constant width. Objects with constant width are ones that have the same segment as their shadow in every direction. The two most famous examples are a circle and a Reuleaux triangle:
Let's consider a circle and a Reuleaux triangle of the same width. They can hide behind each other. Barbier's Theorem states that all objects of the same constant width have the same perimeter. We all know that given a fixed perimeter, the circle has the largest area. Thus, the circle can hide behind the Reuleaux triangle which has smaller area and, consequently, the circle can't hide inside the Reuleaux triangle. By the way, the Reuleaux triangle has the smallest area of all the objects with the given constant width.
To digress. You might have heard the most famous Microsoft interview question: Why are manhole covers round? Presumably because round manhole covers can't fall into slightly smaller round holes. The same property is true for manhole covers of any shape of constant width. On the picture below (Flickr original) you can see Reuleaux-triangle-shaped covers.
Let's move the dimensions up. Dan's questions become both more difficult and more interesting, because the shadows are not as simple as segments any more.
Before continuing, I need to introduce the concept of "Minkowski sums." Suppose we have two convex bodies in space. Let's designate the origin. Then a body can be represented as a set of vectors from the origin to the points in the body. The Minkowski sum of two bodies are all possible sums of two vectors corresponding to the first body and the second body.
Another way to picture the Minkowski sum is like this: Choose a point in the second body. Then move the second body around by translations so that the chosen point covers the first body. Then the area swept by the second body is the Minkowski sum of both of them.
Suppose we have two convex bodies K and L. Their Minkowski interpolation is the body tK + (1-t)L, where 0 ≤ t ≤ 1 is a scaling coefficient. The picture below made by Christina Chen illustrates the Minkowski interpolation of a triangle and an inverted triangle.
If two bodies can hide behind L, then their Minkowski interpolation can hide behind L for any value of parameter t. In particular if K can hide behind L, then the Minkowski interpolation tK + (1-t)L can hide behind L, for any t.
In my paper co-authored with Christina Chen and Daniel Klain "Volume bounds for shadow covering", we found the following connection between hiding inside and volumes. If L is a simplex, and K can hide behind it, but can't hide inside L, then there exists t such that the Minkowski interpolation tK + (1-t)L has a larger volume than the volume of L.
In the paper we conjecture that the largest volume ratio V(K)/V(L) for a body K that can hide behind another body L is achieved if L is a simplex and K is a Minkowski interpolation of L and an inverted simplex. The 3D object that can hide behind a tetrahedron and has 16% more volume than the tetrahedron was found by Christina Chen. See her picture below.
The main result of the paper is a universal constant bound: if K can hide behind L, then V(K) ≤ 2.942 V(L), independent of the dimension of the ambient space.
Question 1. Holodeck. After a long and difficult assignment on an uninhabited planet, Commander Riker went to Holodeck III to unwind. While there he ate three cheeseburgers generated by the holodeck program. Is Commander Riker hungry after he ends the program?
Question 2. Relativity. We know that speed in space is relative, there is no absolute speed. What does Captain Picard mean when he orders a "full stop"?
Question 3. The Replicator. Captain Picard approached a replicator and requested: "Tea, Earl Grey. Hot." The replicator immediately created a glass with hot Earl Grey tea. How much energy would the Enterprise have saved in seven years if they used a dish-washing machine, rather than creating glasses from atoms each time and dissolving them afterwards?
Question 4. Contractions. Commander Data hasn't mastered contractions in English speech. In what year do you think the first program was written to convert formal English into English with contractions?
Question 5. Data. Commander Data is fully functional and absolutely superior to a vibrator. Given that there are more than a thousand people on board the Enterprise, estimate how many times a year on average Data will receive sexual requests.
The next two questions are related to particular episodes.
Question 6. "Up The Long Ladder". Mariposans reproduce by cloning. Why do all the identical sets of clones appear to be the same age? Does it mean that upon the reproduction the clone is the age of the host? If so, they all should be 300 years old.
Mariposans steal sample DNA from Commander Riker and Dr. Pulaski. If Riker and Pulaski didn't destroy their maturing clones what age would those clones be? Would they know how much two plus two is when they awaken? If clones awaken as adults, what is their life span?
Question 7. "Force of Nature". Serova sacrifices herself to save her world from the effects of warp drive, but in doing so, she herself creates the rift that will destroy her world. Explain the logic.
* * *
Logic: if an empty yogurt container is in the sink, a spoon is in the garbage can.
* * *
Logically, a wireless mouse should be called a hamster.
* * *
— I started a new life today.
— You quit smoking and drinking?
— No, I changed my email and Facebook accounts.
* * *
— The reviewer has rejected your paper submitted to our math journal because it doesn't contain any theorems or fomulae or even numbers.
— Wait a minute. Your reviewer is mistaken. There are page numbers on every page.
* * *
A kyboard for sal: only on ky dosn't work.
* * *
My computer always beats me in chess. In revenge, I always beat it in a boxing match.
* * *
— Were your parents married when you were born?
— 50%.
— 50%?
— Yes, my father was married and my mother was not.
* * *
Two programmers are talking:
— I can't turn on my oven.
— What's the error message?
Have you ever been punished for being too good at spider solitaire? I mean, have you ever been stuck because you collected too many suits? Many versions of the game don't allow you to deal from the deck if you have empty columns, nor do they allow you to get back a completed suit. If the number of cards left on the table in the middle of the game is less than ten — the number of columns — you are stuck. I always wondered what the probability is of being stuck. This probability is difficult to calculate because it depends on your strategy. So I invented a boring version of spider solitaire for the sake of creating a math problem. Here it goes:
You start with two full decks of 104 cards. Initially you take 54 cards. At each turn you take all full suits out of your hand. If you have less than ten cards left in your hand, you are stuck. If not, take ten more cards from the leftover deck and continue. What is the probability that you can be stuck during this game?
Let us simplify the game even more by playing the easy level of the boring spider solitaire in which you have only spades. So you have a total of eight full suits of spades. I leave it to my readers to calculate the total probability of being stuck. Here I would like to estimate the easiest case: the probability of being stuck before the last deal.
There are ten cards left in the deck. For you to be stuck, they all should have a different value. The total number of ways to choose ten cards is 104 choose 10. To calculate the number of ways in which these ten cards have different values we need to choose these ten values in 13 choose 10 ways, then multiply by the number of ways each card of a given value can be taken from the deck: 810. The probability is about 0.0117655.
I will leave it to my readers to calculate the probability of being stuck before the last deal at the medium level: when you play two suits, hearts and spades.
No, I will not tell you how many times I played spider solitaire.
Is there a way to put a sequence of numbers to music? The system that comes immediately to mind is to match a number to a particular pitch. The difference between any two neighboring integers is the same, so it is logical to assume that the same tone interval should correspond to the same difference in integers. After we decide which tone interval corresponds to the difference of 1, we need to find our starting point. That is, we need to choose the pitch that corresponds to the number 1. After that, all numbers can be automatically matched to pitches.
After we know the pitches for our numbers, to make it into music we need to decide on the time interval between the notes. The music should be uniquely defined by the sequence, hence the only logical way would be to have a fixed time interval between two consecutive notes.
We see that there are several parameters here: the starting point, the pitch difference corresponding to 1, and the time interval between notes. The Online Encyclopedia of Integer Sequences offers the conversion to music for any sequence. It gives you freedom to set the parameters yourself. The sequences do not sound melodic because mathematical sequences will not necessarily follow rules that comply with a nice melody. Moreover, there are no interesting rhythms because the time interval between the notes is always the same.
One day I received an email from a stranger named Michael Blake. He sent me a link to his video on YouTube called "What Pi Sounds Like." He converted the digits of Pi to music. My stomach hurt while I was listening to his music. My stomach hurts now while I am writing this. He just numbered white keys on the piano from 1 to 9 starting from C. Then he played the digits of Pi. Clearly, Michael is not a mathematician, as he does not seem to know what to do with 0. Luckily for him the first 32 digits of Pi do not contain zero, so Michael played the first several digits over and over. My stomach hurts because he lost the basic math property of digits: the difference between the neighboring digits is the same. In his interpretation the digits that differ by one can have a tone interval of minor or major second in a random order corresponding to his random starting point.
I am not writing this to trash Michael. He is a free man in a free country and can do whatever he wants with the digits of Pi. Oops, I am sorry, he can't do whatever he wants. Michael's video was removed from YouTube due to an odd copyright infringement claim by Lars Erickson, who wrote a symphony using the digits of Pi.
Luckily for my readers Michael's video appears in some other places, for example at the New Scientist channel. As Michael didn't follow the symmetry of numbers and instead replaced the math rules with some music rules, his interpretation of Pi is one of the most melodic I've heard. The more randomly and non-mathematically you interpret digits, the more freedom you have to make a nice piece of music. I will say, however, that Michael's video is nicely done, and I am glad that musicians are promoting Pi.
Other musicians do other strange things. For example, Steven Rochen composed a violin solo based on the digits of Pi. Unlike Michael, he used the same tone interval for progressing from one number to the next, like a mathematician would do. He started with A representing 1 and each subsequent number corresponded to an increase of half a tone. That is, A# is 2 and so on. Like Michael Blake he didn't know what to do with 0 and used it for rest. In addition, when he encountered 10, 11, and 12 as part of the decimal expansion he didn't use them as two digits, but combined them, and used them for F#, G, G# respectively. To him this was the way to cover all possible notes within one octave, but for me, it unfortunately caused another twinge in my stomach.
In August I visited my son Alexey Radul, who currently works at the Hamilton Institute in Maynooth, Ireland. One of the greatest Irish attractions, Broom Bridge, is located there. It's a bridge over the railroad that connects Maynooth and Dublin. One day in 1843, while walking over the bridge, Sir William Rowan Hamilton had a revelation. He understood how the formulae for quaternions should be written. He scratched them into a stone of the bridge. Now the bridge has a plaque commemorating this event. The plaque contains his formulae. I don't remember ever seeing a plaque with math, so naturally I rushed off to make my pilgrimage to Broom Bridge.
Quaternions have very pronounced sentimental value for me, since my first research was related to them. Let's consider a simple graph. We can construct an algebra associated with this graph in the following way. For each vertex we have a generator of the algebra. In addition we have some relations. Each generator squared is equal to −1. If two vertices are connected the corresponding generators anti-commute, and they commute otherwise. The simplest non-commutative algebra associated with a graph corresponds to a graph with two vertices and one edge. If we call the generators i and j, then the we get the relations: i2 = j2 = −1, and ij = −ji. I we denote ij as k, the algebra as a vector space has dimension 4 and a basis: 1, i, j, k. These are exactly the quaternions. In my undergraduate research I studied such algebras related to Dynkin diagrams. Thirty years later I came back to them in my paper Clifford Algebras and Graphs. But I digress.
I was walking on the bridge hoping that like Hamilton I would come up with a new formula. Instead, I was looking around wondering why the Broombridge Station didn't have a ticket office. I already had my ticket, but I was curious how other people would get theirs. I asked a girl standing on the platform where to buy tickets. She said that there is no way to buy tickets there, so she sometimes rides without a ticket. The fine for not having tickets is very high in Ireland, so I expressed my surprised. She told me that she just says that she is from the town of Broombridge if she is asked to present her ticket.
Being a Russian I started scheming: obviously people can save money by buying tickets to Broombridge and continuing without a ticket wherever they need to go. If the tickets are checked, they can claim that they are traveling from Broombridge. Clearly Ireland hasn't been blessed with very many Russians visitors.
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and is offered a bet. She may pay $550 in which case she will get $1000 if the coin was tails. If the coin was tails, she is put back to sleep with her memory erased, and awakened on Tuesday and given the same bet again. She knows the protocol. Should she take the bet?
As we discussed in our first essay about Sleeping Beauty, she should take the bet. Indeed, if the coin was heads her loss is $550. But if the coin was tails her gain is $900.
To tell you the truth, when Beauty is offered the bet, she dreams: "It would be nice to know the day of the week. If it were Tuesday, then the coin must have been tails and I would gladly take the winning bet."
In our next variation of the riddle her dream comes true.
Every time she is awakened she is offered to buy the knowledge of the day of the week. How much should she be willing to pay to know the day of the week?
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning and instead of the expected questions she is offered a bet. She may pay $600 in which case she will get $1000 if the coin was tails. Should she take the bet?
We know that tripling the triangular number 1 yields the triangular number 3. The figure shows how we can use this fact to conclude that tripling the triangular number 15 yields the triangular number 45.
Using this new fact, can you modify the figure to find even larger examples of tripling triangles?
This post is inspired by the following problem:
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning: What is her probability that the coin was heads?
Some people argue: asleep or awake, the probability of a fair coin being heads is one half, so her probability should be one half.
Other people, including us, argue that those people didn't study conditional probability. On the information of the setup to the problem and the information of having awakened, the three situations "Coin was heads and it is Monday", "Coin was tails and it is Monday", and "Coin was tails and it is Tuesday" are symmetric and therefore equiprobable; thus the probability that the coin was tails is, on this information, two thirds.
So who is right? We are, of course. A good way to visualize probability judgements is to turn them into bets. Suppose each time Beauty wakes up she is offered the following bet: She pays $600 and gets $1000 if the coin was tails. Should she take it? If her probability of the coin being tails were one half, then obviously not; if her probability of the coin being tails were two thirds, obviously yes. So which is it? Consider the situation from her perspective as of Sunday. She can either always take this bet or always refuse it. If she always refuses, she gets nothing. If she always accepts: If the coin turns up heads, she will be asked the question once and will lose $600. If the coin turns up tails, she will be asked the question twice and will gain $800. So on average she will win, so she should take the bet. By this thought experiment, her probability of tails is clearly not one half.
To make matters more interesting, let's try another bet. Suppose she is given the above bet just once, in advance, on Sunday. She pays $600, and she gets paid $1000 on Wednesday if the coin was tails. This has nothing to do with sleeping and awakening. If she takes the bet she loses $600 with probability one half and gains $400 otherwise. So she shouldn't take the bet. Her probability on Sunday that the coin will come up heads is, of course, one half. The point is that just as these two bets are different bets, the sets of information Beauty has on Sunday vs at awakening are different, and lead to different conclusions. On Sunday she knows that the next time she wakes up it will be Monday, but when she then wakes up, she doesn't know that it's Monday.
Parting thought: The phenomenon of predictably losing information leads to the phenomenon of predictably changing one's assessments. Suppose for some reason she decided to take that unprofitable bet on Sunday. When she wakes up during the experiment, should she feel happy or sad? From her perspective during the experiment, the odds of gaining $400 vs losing $600 are two to one, so she should be happy. Given that she knows on Sunday how she will (with complete certainty!) feel about this bet on Monday, should she take it, even given her Sunday self's assessment that it's a bad bet?
If you buy one Mega Millions ticket, your probability of hitting the jackpot is one in 175,000,000. For all practical purposes it is zero. When I give my talk on lotteries, there is always someone in the audience who would argue that "but someone is winning and so can I." The fact that someone is winning depends on the number of people buying tickets. It is difficult to visualize the large number of people buying tickets and the miniscule odds of winning. For example, the probability of you dying from an impact with a meteorite is larger than the odds of winning the jackpot.
I receive a lot of emails from strangers asking me to advertise their websites on my blog. I always check out their websites and I often find them either unrelated to math or boring. That is why I was pleasantly surprised when I was asked to write about a useful website: Understanding Big Numbers. In each post Liam Gray takes a big number and puts it into some perspective. For example, he estimates Mark Zuckerberg's Hourly Wage by dividing Mark's estimated wealth in 2011 by the number of hours Mark might have worked on Facebook. Facebook has existed for 7 years and, assuming 10 hours of work a day every day, we get 25,000 work hours. That is more than half a million dollars an hour.
Imagine someone calls Mark Zuckerberg and asks to talk to him for a minute. Mark wouldn't be out of line to request nine thousand dollars for that. Lucky am I, that I do not need to talk to Mark Zuckerberg.
Here is a game that John Conway popularizes. It is called "Finchley Central," which is a station of the London Underground. The game goes as follows. Alice and Bob take turns naming London Underground stations, in any order. The first person to say "Finchley Central" wins.
Alice, who starts, can just name the station. But then Bob will give her a look. It is not fun to win a game on the first turn. To avoid appearing rude, Alice will not start with "Finchley Central." It would be impolite of Bob to take advantage of Alice's generosity, so he also won't say "Finchley Central." The game might continue like this for a while.
The game has a hidden agenda: winning it after 10 turns will supply many more bragging rights than winning it right away would. We can make this hidden agenda explicit by assigning a value to the honor of continuing the game. For example, suppose every time Alice (or Bob) says a station, she puts one dollar into the pile. The person who says "Finchley Central" first takes all the money from the pile. The implicit goal of the game becomes explicit: you want to say "Finchley Central" right before your opponent says it.
By the way, Finchley Central is not actually a particularly central station — it is the station between Finchley East and Finchley West, serving the relatively small place called Finchley; and is not even under ground. It has the distinction of being one of the oldest still-standing pieces of London Underground physical plant, because plans to rebuild it were interrupted on account of World War II and never resumed. It also has the distinction of having served the home of the guy (an employee of the Underground system) who had the brilliant idea that since the Underground was, indeed, mostly under ground, the right way to map it was topologically, rather than geographically.
Here is another way to model the game. Alice writes an odd number on a piece of paper, and Bob writes an even number. When they compare, the person who wrote a smaller number wins that number of dollars. This version loses the psychological aspect. When you take turns, it is to your advantage to read the non-verbal signs of your opponent to see when s/he is getting ready to drop the bomb.
People play this game in real life. Here are Alice and Bob looking at the last piece of a mouth-watering Tiramisu:
At this point Alice wins with some extra brownie points for being polite.
We can model the honor points differently. We can say you will be the most proud of the game if you name the station write before you opponent is about to do so. Then the model is: everyone writes down their next move; if your move is Finchley Central when your opponent's next move was going to be Finchley Central, then you win.
Here we suggest another game that we call "Reverse Finchley Central." Alice and Bob name London Underground stations in turns and the person who names "Finchley Central" first loses. This game can continue until all the stations are exhausted, if the players are forbidden to repeat them, or it can continue indefinitely otherwise. But this is quite tiresome. The hidden agenda would be to not waste too much time. Clearly the person who values time less will win.
But let us model this game. We want to fix the value of winning. Let us set aside ten dollars for the winner. On their turn, each player puts one dollar into the pile, and as soon as one of the players says "Finchley Central," the other one wins and takes the ten dollars. The pile goes to charity. Alternatively, Alice and Bob can each write a number. The person with the larger number wins the prize, while both have to pay the smaller number to charity.
We play this game with our parents. They nag us to do the dishes. We resist. Then they give up and do the dishes themselves. They lose, but we all pay with our nerves for nagging or being nagged at. Later our parents get their revenge when we have children of our own.
At the 2011 IMO, Lisa Sauermann received yet another gold medal. Now she tops the Hall of Fame of the IMO with four gold medals and one silver medal.
In addition, in 2011 she achieved the absolute best individual result and was the only person with a perfect score. In previous years, there were several girls who tied for first place, but she is the first girl ever to have an absolute rank of 1.
I told you so. In my 2009 essay Is There Hope for a Female Fields Medalist?, I predicted that a girl will soon become an absolute champion of the IMO.
In that essay I draw a parallel between the absolute champion of IMO and a Fields medalist. Indeed, we get one of each per year. Lisa Sauermann is the best math problem solver in her year. Will she grow up to receive a Fields medal? I am not so sure: the medal is still unfriendly to women. Lisa Sauermann is the best math problem solver ever. Will she grow up to be the best mathematician of our century? I wonder.
My e-friend and coauthor, Konstantin Knop, designed the following problem for the 2011 All-Russia Olympiad:
Some cells of a 100 by 100 board have one chip placed on them. We call a cell pretty if it has an even number of neighboring cells with chips. Neighbors are the cells that share a side. Is it possible for exactly one cell to be pretty?
The problem is not easy. Only one person at the Olympiad received full credit for it.
From time to time my female colleagues share stories with me of great unfairness or horrible sexual harassment in the world of mathematics. I can't reciprocate — certainly not on that level.
I do not have any horror stories to tell. Generally I am treated with great respect, at least to my face. In fact, some men have told me that I am the smartest person they ever met.
The stories I want to share are not about harassment. No single incident is a big deal. But when these things happened time after time after time, I realized: this is gender bias.
First story. A guy told me, "Your proof is unbelievably amazing."
What can I say? It is just a compliment. Though I am not sure why the word "unbelievable" was included. Is it difficult to believe that I can produce an amazing proof? I encounter surprise too often to my taste.
Second story. Another guy tells me after I explain a solution to a math problem, "I didn't realize it was so simple."
Actually it wasn't simple. When I explained the solution, it may have seemed simple, but that was because I was able to explain it to him with such clarity. People tend to downgrade their opinion of the problem, rather than upgrade their opinion of my ability. It actually affects my reputation as a mathematician.
Third story. Another guy said to me (and I quote!), "I am so dumb. I tried for a week to write the program that computes these numbers and you did it in one hour. I feel so dumb. I didn't expect myself to be so dumb. Why am I so dumb?"
After the fourth "dumb", I started wondering what it was all about. Many guys try to compete with me. And they hate losing to a woman. It creates a strong motivation for them to discard my brilliance and to explain away my speed, even if they have to claim temporary dumbness.
Fourth story. Someone asked me, "What is the source of the solutions and math ideas in your blog? Can you refer me to the literature?"
I do invest extra effort in citing the sources of the math puzzles I discuss. Everything else — the solutions, the ideas, new definitions, new sequences — I invent myself. I have even started inventing math puzzles. This is my blog. I thought of it myself, I wrote it myself. Has anyone ever asked Terence Tao where he takes the solutions for his blog from?
Unfortunately, this attitude damages my career. When people think that my ideas come from someone else, they do not cite me.
But all these stories however minor happen all the time, not only to me but to all my female colleagues. Gender bias is real. Next time someone tells me how unbelievably amazing my proof is, I will explode.
Recently I stumbled on a cute xkcd comic with the hidden message:
Wikipedia trivia: if you take any article, click on the first link in the article text not in parentheses or italics, and then repeat, you will eventually end up at "Philosophy".
Naturally, I started to experiment. The first thing I tried was mathematics. Here is the path: Mathematics — Quantity — Property — Modern philosophy — Philosophy.
Then I tried physics, which led me to mathematics: Physics — Natural science — Science — Knowledge — Fact — Information — Sequence — Mathematics.
Then I tried Pierre de Fermat, who for some strange reason led to physics first: Pierre de Fermat — French — France — Unitary state — Sovereign state — State — Social sciences — List of academic disciplines — Academia — Community — Living — Life — Objects — Physics.
The natural question is: what about philosophy? Yes, philosophy goes in a cycle: Philosophy — Reason — Rationality — philosophy.
The original comic talks about spark plugs. So I tried that and arrived at physics: Spark plug — Cylinder head — Internal combustion engine — Engine — Machine — Machine (mechanical) — Mechanical system — Power — Physics.
Then I tried to get far away from philosophy and attempted sex, unsuccessfully: Sex — Biology — Natural science. Then I tried dance: Dance — Art — Sense — Physiology — Science.
It is interesting to see how many steps it takes to get to philosophy. Here is the table for the words I tried:
| Word | # Steps |
|---|---|
| Mathematics | 4 |
| Physics | 11 |
| Pierre de Fermat | 24 |
| Spark plug | 19 |
| Sex | 12 |
| Dance | 13 |
Mathematics wins. It thoroughly beats all the other words I tried. For now. Fans of sex might be disappointed by these results and tomorrow they might change the wiki essay about sex to start as:
Modern philosophy considers sex …
Not personally. Someone hacked into my website.
I would like to thank my readers Qiaochu Yuan, Mark Rudkin, "ano" and Paul who alerted me to the problem. Viewers who were using the Google Chrome browser and who tried to visit my website got this message: "This site contains content from howmanyoffers.com, a site known to distribute malware."
It took me some time to figure out what was going on. It appears that on June 19 someone from 89-76-135-50.dynamic.chello.pl hacked into my hosting account and added a script to all my html files and to my blog header. It seems that the script was dormant and wasn't yet doing bad things.
As soon as I grasped what was going on, I replaced all the affected files.
I have had my website for many years without changing my hosting password. Unfortunately, passwords, not dissimilar to humans, have this annoying tendency to become weaker with age. I wasn't paying attention to the declining strength of my password and so I was punished.
Now I have fixed the website and my new password is: qwP35q2054uWiedfj052!@#$%.
Just kidding.
I recently had the following chat with a particular calculator:
It seems odd to me that putting a few more e's down the bottom should result in it thinking there were the same number of extra 10s at the bottom. In fact, I've never seen a calculator answer in this form at all. I'm especially intrigued that the final power of ten seems to be the same in all three cases, so it can't even just be estimating. Do you have any thoughts on what screwy counting could be behind these particular answers?
May the Mass times the Acceleration be with you!
John Conway taught me how to tell time at night. But first I need to explain the notions of the "time in the sky" and the "time in the year."
The clock in the sky. Look at Polaris and treat it as the center of a clock. The up direction corresponds to 12:00. Now we need to find a hand. If you find Polaris the way I do, first you locate the Big Dipper. Then you draw a line through the two stars that are furthest away from the Big Dipper's handle. The line passes through Polaris and is your "hour" hand. Now you can read the time in the sky.
The hand of the clock in the sky makes a full rotation in approximately 24 hours. So if you stare at the sky for a long time, you can calculate the time you spent staring. Keep in mind that the hand in the sky clock is twice as slow as the hour hand, and it turns counter-clockwise. So to figure out how long you're looking into the sky, take the sky-time when you start staring, subtract the sky-time when you stop staring and multiply the result by 2.
To calculate the absolute time, we need to adjust for the day in the year.
The clock in the year. A year has twelve months and a clock has twelve hours. How convenient. You can treat each month as one hour. In addition as a month has about 30 days and an hour has exactly 60 minutes, we should count a day as two minutes. Thus, January 25 is 1:50.
Fact: on March 7th at midnight the clock in the sky shows 12:00. March 7th corresponds to 3:15. So to calculate the solar time you need to add up the time in the sky and the time in the year and multiply it by 2. Then subtracting the result from 6:30, which is twice 3:15, you get the solar time.
You are almost ready. You might need to adjust for daylight savings time or for peculiarities of your time zone.
This time formula is not very precise. But if you are looking into the sky and you do not have your watch or cell phone with you, you probably do not need to know the time precisely.
In my life as a female mathematician I have quite often encountered a mathematician's wife who, despite not knowing me, already hated me. It was clear that it had nothing to do with me personally, so being clueless and naive, I assumed that most men were cheaters and that their wives were extremely insecure and jealous.
Then one day one of the wives decided to be frank about her feelings. It wasn't about cheating, she told me. It was that she felt distant from her husband. He lived in a world of mathematics from which she was excluded. I on the other hand shared this world with him.
It was very sad. It meant that I incurred their jealousy, not because of my sins, but because I am a female mathematician.
Let me tell you another story that helped me realize how all-encompassing this world of mathematics can be for some people. Once I had a very close friend who we will call Jack. I do not want to name him as he is a famous mathematician. Jack told me that the strongest emotions he feels are related to mathematics. He can only feel close to someone if he can share a mathematical discussion with them.
Now I understand the wives better. Husbands like Jack invest so much more in their math world and their colleagues than they do in their home life, that it is not surprising the wives are jealous. Because women mathematicians are scarce, when I appear in their husbands' world, it adds another layer of worry.
Another thing that Jack told me is that he gets such a euphoric feeling when he discovers a new math idea that it is better than any orgasm. Of course, this statement made me question the quality of Jack's orgasms, but in any case, for some mathematicians math is an aphrodisiac.
If math is an aphrodisiac, then tattooing a formula on the lover's body may well enhance the orgasm. I just remembered the movie by Ed Frenkel. But I digress.
If math is an aphrodisiac, then I understand jealous wives even better. Without sex I can give their husbands pleasure they can't.
* * *
I am taking my dog to tweet. He'll check other dog's posts at every pole and will leave his comments.
* * *
Not many people know that 1000 chameleons is a chabillion.
* * *
The Internet paradox: it connects people who are far apart, and disconnects those who are close.
* * *
We bought a cell phone for our TV set. We attached it to the remote control, so that we can call our TV when the remote is lost.
* * *
Mary's mom failed arithmetic. Actually, that is why Mary was born.
* * *
Your call is very important to us. Please, hold. And in the meantime, to protect your health, our customer care team encourages you to drink a glass of water at least every two hours.
* * *
Who is your favorite computer game character?
The stick from Tetris.
* * *
Our new boss invited everyone to bring their keyboards to his office. He kept the employees who had worn letters and laid off the ones with worn arrows.
* * *
My son will be a hacker. He started his career before he was born: he found a flaw in the condom.
I recently stumbled upon some notes (in Russian) of a public lecture given by Vladimir Arnold in 2006. In this lecture Arnold defines a notion of complexity for finite binary strings.
Consider a set of binary strings of length n. Let us first define the Ducci map acting on this set. The result of this operator acting on a string a1a2…an is a string of length n such that its i-th character is |ai − a(i+1)| for i < n, and the n-th character is |an − a1|. We can view this as a difference operator in the field F2, and we consider strings wrapped around. Or we can say that strings are periodic and infinite in both directions.
Let's consider as an example the action of the Ducci map on strings of length 6. Since the Ducci map respects cyclic permutation as well as reflection, I will only check strings up to cyclic permutation and reflection. If I denote the Ducci map as D, then the Ducci operator is determined by its action on the following 13 strings, which represent all 64 strings up to cyclic permutation and reflection: D(000000) = 000000, D(000001) = 000011, D(000011) = 000101, D(000101) = 001111, D(000111) = 001001, D(001001) = 011011, D(001011) = 011101, D(001111) = 010001, D(010101) = 111111, D(010111) = 111101, D(011011) = 101101, D(011111) = 100001, D(111111) = 000000.
Now suppose we take a string and apply the Ducci map several times. Because of the pigeonhole principle, this procedure is eventually periodic. On strings of length 6, there are 4 cycles. One cycle of length 1 consists of the string 000000. One cycle of length 3 consists of the strings 011011, 101101 and 110110. Finally, there are two cycles of length 6: the first one is 000101, 001111, 010001, 110011, 010100, 111100, and the second one is shifted by one character.
We can represent the strings as vertices and the Ducci map as a collection of directed edges between vertices. All 64 vertices corresponding to strings of length 6 generate a graph with 4 connected components, each of which contains a unique cycle.
The Ducci map is similar to a differential operator. Hence, sequences that end up at the point 000000 are similar to polynomials. Arnold decided that polynomials should have lower complexity than other functions. I do not completely agree with that decision; I don't have a good explanation for it. In any case, he proposes the following notion of complexity for such strings.
Strings that end up at cycles of longer length should be considered more complex than strings that end up at cycles with shorter length. Within the connected component, the strings that are further away from the cycle should have greater complexity. Thus the string 000000 has the lowest complexity, followed by the string 111111, as D(111111) = 000000. Next in increasing complexity are the strings 010101 and 101010. At this point the strings that represent polynomials are exhausted and the next more complex strings would be the three strings that form a cycle of length three: 011011, 101101 and 110110. If we assign 000000 a complexity of 1, then we can assign a number representing complexity to any other string. For example, the string 111111 would have complexity 2, and strings 010101 and 101010 would have complexity 3.
I am not completely satisfied with Arnold's notion of complexity. First, as I mentioned before, I think that some high-degree polynomials are so much uglier than other functions that there is no reason to consider them having lower complexity. Second, I want to give a definition of complexity for periodic strings. There is a slight difference between periodic strings and finite strings that are wrapped around. Indeed, the string 110 of length 3 and the string 110110 of length 6 correspond to the same periodic string, but as finite strings it might make sense to think of string 110110 as more complex than string 110. As I want to define complexity for periodic strings, I want the complexity of the periodic strings corresponding to 110 and 110110 to be the same. So this is my definition of complexity for periodic strings: let's call the complexity of the string the number of edges we need to traverse in the Ducci graph until we get to a string we saw before. For example, let us start with string 011010. Arrows represent the Ducci map: 011010 → 101110 → 110011 → 010100 → 111100 → 000101 → 001111 → 010001 → 110011. We saw 110011 before, so the number of edges, and thus the complexity, is 8.
The table below describes the complexity of the binary strings of length 6. The first column shows one string in a class up to a rotation or reflections. The second column shows the number of strings in a class. The next column provides the Ducci map of the given string, followed by the length of the cycle. The last two columns show Arnold's complexity and my complexity.
| String s | # of Strings | D(s) | Length of the end cycle | Arnold's complexity | My complexity |
|---|---|---|---|---|---|
| 000000 | 1 | 000000 | 1 | 1 | 1 |
| 000001 | 6 | 000011 | 6 | 9 | 8 |
| 000011 | 6 | 000101 | 6 | 8 | 7 |
| 000101 | 6 | 001111 | 6 | 7 | 6 |
| 000111 | 6 | 001001 | 3 | 6 | 5 |
| 001001 | 3 | 011011 | 3 | 5 | 4 |
| 001011 | 12 | 011101 | 6 | 9 | 8 |
| 001111 | 6 | 010001 | 6 | 7 | 6 |
| 010101 | 2 | 111111 | 1 | 3 | 3 |
| 010111 | 6 | 111001 | 6 | 8 | 7 |
| 011011 | 3 | 101101 | 3 | 4 | 3 |
| 011111 | 6 | 100001 | 6 | 9 | 8 |
| 111111 | 1 | 000000 | 1 | 2 | 2 |
As you can see, for examples of length six my complexity doesn't differ much from Arnold's complexity, but for longer strings the difference will be more significant. Also, I am pleased to see that the sequence 011010, the one that I called The Random Sequence in one of my previous essays, has the highest complexity.
I know that my definition of complexity is only for periodic sequences. For example, the binary expansion of pi will have a very high complexity, though it can be represented by one Greek letter. But for periodic strings it always gives a number that can be used as a measure of complexity.
I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here's his version:
The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?
When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:
A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.
* * * A Generic Limerick (submitted by Michael Chepovetsky)
There once was an X from place B,
Who satisfied predicate P,
The X did thing A,
In a specified way,
Resulting in circumstance C.
* * *
I just learned that 4,416,237 people got married in the US in 2010. Not to nitpick, but shouldn't it be an even number?
* * *
We are happy to announce that 100% of Russian citizens are computer-savvy and use the Internet on a regular basis (according to a recent Internet survey).
* * *
Two math teachers had a fight. It seems they couldn't divide something.
* * *
Do you know that if you start counting seconds, once you reach 31,556,926 you discover that you have wasted a whole year?
* * *
What I need after a visit to the hairdresser is a "Save" button.
* * *
— Hello! Is this a fax machine?
— Yes.
* * *
— I am not fat at all! My girlfriend tells me that I have a perfect figure.
— Your girlfriend is a mathematician. For her a perfect figure is a sphere.
* * *
A: Hi, how are you?
B: +
A: Will you come to classes today?
B: -
A: You will be kicked out!
B: =
A: Are you using your calculator to chat?
I wonder what the largest number is that can be represented with one character. Probably 9. How about two characters? Is it 99? What about three or four?
I guess I should define a character. Let's have two separate cases. In the first one you can only use keyboard characters. In the second one you can use any Unicode characters.
I'm awaiting your answers to this.
The last time I talked to John H. Conway, he taught me to walk up the stairs. It's not that I didn't know how to do that, but he reminded me that a nerd's goal in climbing the steps is to establish the number of steps at the end of the flight. Since it is boring to just count the stairs, we're lucky to have John's fun system.
His invention is simple. Your steps should be in a cycle: short, long, long. Long in this case means a double step. Thus, you will cover five stairs in one short-long-long cycle. In addition, you should always start the first cycle on the same foot. Suppose you start on the left foot, then after two cycles you are back on the left foot, having covered ten stairs. While you are walking the stairs in this way, it is clear where you are in the cycle. By the end of the staircase, you will know the number of stairs modulo ten. Usually there are not a lot of stairs in a staircase, so you can easily estimate the total if you know the last digit of that number.
I guess I am not a true nerd. I have lived in my apartment for eight years and have never bothered to count the number of steps. That is, until now. Having climbed my staircase using John's method, I now know that the ominous total is 13. Oh dear.
The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year's Olympiad:
Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?
Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, "My age is equal to the sum of the digits of my birth year." "Mine too," replied Paul. By how many years is Paul younger than Peter?
Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?
Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?
Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.
Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.
A mathematician is someone who pauses when asked "How much is two and two?"
Indeed, the answer might be:
I already wrote about the research of my friend Olga Amosova who studied the sickle-cell anemia mutation. She and her colleagues needed to store short fragments of hemoglobin genes for their experiments. All the fragments were identical. They noticed that with time the fragments always broke down in the same place. It was a mystery. When good scientists stumble on a mystery, they start digging.
They found that one of the nucleotides rips off the DNA fragment at the site of the Sickle-cell mutation. That place on the DNA becomes fragile and later breaks down. These sites need to be repaired. The repair is very error-prone and often leads to a mutation.
When DNA strands are left unattended, they want to pair up. There are four types of nucleotides: A, C, G and T. So mathematically the fragment of DNA is a string in the alphabet A, C, G, T. These nucleotides are matched to each other. When two DNA strands pair up, A on one strand always matches T and C matches G. So it is logical that if there are two complementary DNA pieces on the same fragment, they will find each other and pair up. They form a hydrogen bond. For example, a piece AACGT matches perfectly another piece TTGCA. Suppose a substring of DNA consists of a piece AACGT and somewhere later the reverse of the match: ACGTT. Such a string is called an inverted repeat. The DNA fragment I mentioned contains a string AACGT****ACGTT. Two pieces AACGT and ACGTT are complementary and not too far from each other in space. So it is easy for them to find each other and to bond to form a so-called stem-loop or a hairpin structure. The site of Sickle-cell mutation falls into the loop.
Olga and her colleagues discovered that for some particular loops the orientation in space becomes awkward and one of the nucleotides rips off. Such a rip off is called depurination. In further investigation, Olga found examples of when depurination happens. The first sequence of the pair that will bond later has to have at least five nucleotides and has to end in T. Correspondingly the second part in the pair has to begin with A. In the middle there needs to be four nucleotides GTGG. The first G flies away. Enzymes rush like a first aid squad to repair it and introduce mistakes that lead to mutation and diseases like cancer.
DNA was thought to be simply a passive information storage system, not capable of any action. Now we see that DNA is capable of action. DNA can damage itself. Damage provokes a mutation. For all practical purposes it is self-mutilation. Olga and her colleagues scanned the human genome for other sequences that are capable of self-mutilation. They found that such sequences are overwhelmingly present. They are present in much higher numbers than would be expected statistically. The pieces that are capable of damaging themselves occur 40 times more often than would occur if the nucleotides were distributed randomly. They are especially overrepresented in genes linked to cancer.
Self-damaging shouldn't happen in normal situations. It can be provoked by the environment, for example, the chemistry of the cell. That means, that our cancers are not only in our genes but also in our life-style. There was, for example, a suggestion in a recent NY Times article, Is Sugar Toxic?, that too much sugar in a diet might provoke cancer. If the rate of mutation depends on the environment, we can influence it and prolong our lives.
It is not clear why the ability to self-mutilate survives in the evolutionary process. It is quite possible that if something very bad happens to our planet, we need our genes to be able to mutate very fast in order to adjust to the environment so that humans can survive.
Though I never tried to donate my sperm to a sperm bank, because of my inability to produce it, I know that sperm banks look for people who have ancestors who lived for a very long time. Such sperm is in bigger demand as everyone wants their children to live longer. I wonder if this tendency is a mistake. Global warming is upon us. People with longevity genes might not be flexible enough for their children to survive the changing of the Earth.
As you might have guessed from the title, this essay is about domino tilings.
Suppose a subset of a square grid has area N, and the number of possible domino tilings is T. Let's imagine that each cell is contributing a factor of x tilings to the total independently of the others. Then we get that xN = T. This mental exercise suggests a definition: we call the nth root of T the degree of freedom per square for a given region.
Let's consider a 1 by 2k rectangle. There is exactly one way to tile it with dominoes. So the degree of freedom per square of such a rectangle is 1. Now consider a 2 by k rectangle. It has the same area as before, and we know that there should be more than one tiling. Hence, we expect the degree of freedom to be larger than the one in the previous example. The number of tilings of a 2 by k rectangle is Fk-2, where Fk is kth Fibonacci number. So the degree of freedom for large k will be approximately the square root of the golden ratio, which is about 1.272.
You might expect that squares should give larger degrees of freedom than rectangles of the same area. The degree of freedom for a large square is about 1.3385. You can find more information in the beautiful paper Tilings by Federico Ardila and Richard P. Stanley.
Let's move from rectangles to Aztec diamonds. They are almost like squares but the side of the diamond is aligned with diagonals of the dominoes rather than with their sides. See the sample diamonds in the picture above, which Richard Stanley kindly sent to me for this essay.
It is easier to calculate the degree of freedom for Aztec diamonds than for regular squares. The degree is the fourth root of 2, or 1.1892…. In the picture below created by James Propp's tiling group you can see a random tiling of a large Aztec diamond.
Look at its colors: horizontal dominoes are yellow and blue; vertical ones are red and aquamarine. You might wonder what rule decides which of the horizontal dominoes are yellow and which are blue. I will not tell you the rule; I will just hint that it is simple.
Back to freedom. As you can see from the picture, freedom is highly non-uniform and depends on where you live. Freedom is concentrated inside a circle called the arctic circle, perhaps because the areas outside it are frozen for lack of freedom.
Now I would like to expand the notion of freedom to give each cell its own freedom. For a large Aztec diamond, I will approximate freedom with a function that is one outside the arctic circle and is uniform inside. The Aztec diamond AZ(n) consists of 2n(n+1) squares, shaped like a square with side-length n√2. So the area of the circle is πn2/2. Hence we can calculate the freedom inside the circle as the πth root of 2, which is about 1.247. This number is still much less than the degree of freedom of a cell in a large square.
Jorge Tierno sent me a link to the following puzzle:
There is a certain country where everybody wants to have a son. Therefore, each couple keeps having children until they have a boy, then they stop. What fraction of the children are female?
If we assume that a boy is born with probability 1/2 and children do not die, then every birth will produce a boy with the same probability as a girl, so girls will comprise half of all children.
Now, I wonder why everyone would want a boy? Y-chromosomes are much shorter than X-chromosomes. If a man wants to pass his genes to the next generation, a daughter should be preferable as she keeps more genes from the father. I am a mother of two boys, so my granddaughters will have my X-chromosome while my grandsons will have my ex-husband's Y-chromosome, so to keep my genes in the pool I should be more interested in granddaughters.
But I digress. I started writing this essay because in the original puzzle link the answer was different from mine. Here is how the other argument goes:
Half of all families have zero girls, a quarter have 1/2 girls, 1/8 have 2/3 girls, and so on. If we sum this up the expected ratio of girls to boys is (1/2)0 + (1/4)(1/2) + (1/8)(2/3) + (1/16)(3/4) + ... which adds to 1 − ln 2, which is about 30%.
What's wrong with this solution?
I found this cute problem in the Russian book Sharygin Geometry Olympiad by Zaslavsky, Protasov and Sharygin.
Find numbers p and q that satisfy the equation: x2 + px + q = 0.
The book asks you to find a mistake in the following solution:
By Viète's formulae we get a system of equations p + q = − p, pq = q. Solving the system we get two solutions: p = q = 0 and p = 1, q = −2.
What is wrong with this solution?
I wrote how the written entrance exam was used to keep Jewish students from studying at Moscow State University, but the real brutality happened at the oral exam. Undesirable students were given very difficult problems. Here is a sample "Jewish" problem:
Solve the following equation for real y:
Here is how my compatriots who studied algebra in Soviet high schools would have approached this problem. First, cube it and get a 9th degree equation. Then, try to use the Rational Root Theorem and find that y = 1 is a root. Factoring out y − 1 gives an 8th degree equation too messy to deal with.
The most advanced students would have checked if the polynomial in question had multiple roots by GCDing it with its derivative, but in vain.
We didn't study any other methods. So the students given that problem would have failed it and the exam.
Unfortunately, this problem is impossible to appeal, because it has an elementary solution that any applicant could have understood. It goes like this:
Let us introduce a new variable: x = (y3 + 1)/2. Now we need to solve a system of equations:
This system has a symmetry which we can exploit. The graphs of the functions x = (y3 + 1)/2 and y = (x3 + 1)/2 are reflections of each other across the line x = y. As both functions are increasing, the solution to the system of equations should lie on the line x = y. Hence, we need to solve the cubic y = (y3 + 1)/2, one of whose roots we already know.
Now I offer you another problem without telling you the solution:
Four points on a plane used to belong to four different sides of a square. Reconstruct the square by compass and straightedge.
Recently I asked my readers to look at the 1976 written math exam that was given to applicants wishing to study at the math department of Moscow State University. Now it's time to reveal the hidden agenda. My readers noticed that problems 1, 2, and 3 were relatively simple, problem 4 was very hard, and problem 5 was extremely hard. It seems unfair and strange that problems of such different difficulty were worth the same. It is also suspicious that the difficult problems had no opportunity for partial credit. As a result of these characteristics of the exam, almost every applicant would get 3 points, the lowest passing score. The same situation persisted for many years in a row. Why would the best place to study math in Soviet Russia not differentiate the math abilities of its applicants?
In those years the math department of Moscow State University was infamous for its antisemitism and its efforts to exclude all Jewish students from the University. The strange structure of the exam accomplished three objectives toward that goal.
1. Protect the fast track. There was a fast track for students with a gold medal from their high school who got 5 points on the written exam. The structure of the exam guaranteed that very few students could solve all 5 problems. If by chance a Jewish student solved all 5 problems, it was not much work to find some minor stylistic mistake and not count the solution.
2. Avoid raising suspicion at the next exams. The second math entrance exam was oral. At such an exam different students would talk one-on-one with professors and would have to answer different questions. It was much easier to arrange difficult questions for undesirable students and fail all the Jewish students during the oral exam than during the written exam. But if many students with perfect scores on the written exam had failed the oral exam, it might have raised a lot of questions.
3. Protect appeals. Despite these gigantic efforts, there were cases when Jewish students with a failing score of 2 points were able to appeal and earn the minimum passing score of 3. If undesirable students managed to appeal all the exams, they would only get a half-passing grade at the end and would not be accepted because the department was allowed to choose from the many students that the exams guaranteed would have half-passing scores.
I have only heard about one faculty member who tried to publicly fight the written exam system. It was Vladimir Arnold, and I will tell the story some other time.
The following problem appeared at the Gillis Math Olympiad organized by the Weizmann Institute:
A foreign government consists of 12 ministers. Each minister has 5 friends and 6 enemies amongst the ministers. Each committee needs 3 ministers. A committee is considered legitimate if all of its members are friends or all of its members are enemies. How many legitimate committees can be formed?
Surprisingly, this problem implies that the answer doesn't depend on how exactly enemies and friends are distributed. This meta thought lets us calculate the answer by choosing an example. Imagine that the government is divided into two factions of six people. Within a faction people are friends, but members of two different factions dislike each other. Legitimate committees can only be formed by choosing all three members from the same faction. The answer is 40.
We would like to show that actually the answer to the problem doesn't depend on the particular configuration of friendships and enmities. For this, we will count illegitimate committees. Every illegitimate committee has exactly two people that have one enemy and one friend in the committee. Let's count all the committees from the point of view of these "mixed" people. Each person participates in exactly 5*6 committees as a mixed person. Multiply by 12 (the number of people), divide by 2 (each committee is counted twice) and you get the total 180. This gives an answer of 40 for the number of legitimate committees without using a particular example.
What interests us is the fact that the number of illegitimate, as well as legitimate, committees is completely defined by the degree distribution of friends. For any set of people and who are either friends or enemies with each other, the number of illegitimate committees can be calculated from the degree distribution of friends in the same way as we did above.
Any graph can be thought of as representing friendships of people, where edges connect friends. This cute puzzle tells us that the sum of the number of 3-cliques and 3-anti-cliques depends only on the degree distribution of the graph.
As a non mathematical comment, the above rule for legitimate committees is not a bad idea. In such a committee there is no reason for two people to gang up on the third one. Besides, if at some point in time all pairs of friends switch to enemies and vice versa, the committees will still be legitimate.
In 1976 I was about to become a student in the math department at Moscow State University. As an IMO team member I was accepted without entrance exams, but all of my other classmates had to take the exams. There were four exams: written math, oral math, physics, and an essay.
The written math exam was the first, and here are the problems. I want my non-Russian readers to see if they notice anything peculiar about this exam. Can you explain what is peculiar, and what might be the hidden agenda?
Problem 1. Solve the equation
Problem 2. Solve the inequality
Problem 3. Consider a right triangle ABC with right angle C. Angle B is 30° and leg CA is equal to 1. Let D be the midpoint of the hypotenuse AB, so that CD is a median. Choose F on the segment BC so that the angle between the hypotenuse and the line DF is 15°. Find the area of CDF. Calculate its numeric value with 0.001 precision.
Problem 4. Three balls, two of which are the same size, are tangent to the plane P, as well as to each other. In addition, the base of a circular cone lies on the plane P, and its axis is perpendicular to the plane. All three balls touch the cone from the outside. Find the angle between a generatrix of the cone and the plane P, given that the triangle formed by the points of tangency of the balls and the plane has one angle equal to 150°.
Problem 5. Let r < s < t be real numbers. If you set y equal to any of the numbers r, s or t in the equation x2 − (9 − y)x + y2 − 9y + 15 = 0, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that −1 < r < 1.
Let me describe some background to this exam. Applicants who solve fewer than two problems fail the exam and are immediately rejected. People who solve two or three problems are given 3 points. Four problems earn 4 points, and five problems earn 5 points.
If you still do not see the hidden agenda, here is another clue. People who get 5 points on the first exam and, in addition, have a gold medal from their high school (that means all As) are admitted right after the first exam. For the others, if they do not fail any of the exams, points are summed up with their GPAs to compute their scores. The so-called half-passing score is then calculated. Scores strictly higher than the half-passing score qualify applicants for admission. However, there are too many applicants for the available openings with at least the half-passing score. As a result only some people with exactly the half-passing score are accepted, at the discretion of the department.
Now my readers have enough information to figure out the hidden agenda behind that particular exam.
We worked for several years with RSI where we supervised summer math research projects by high school students. Now, we've started an additional program at MIT's math department called PRIMES, where local high school students do math research during the academic year. In this essay we would like to discuss what makes a good math research project for a high school student.
A doable project. The project should not be believed to be extremely difficult to yield at least results. It is very discouraging for an aspiring mathematician not to produce anything during their first project.
An accessible beginning. The student should be able to start doing something original soon after the start of the project. After all, they don't come to us for coursework, but for research.
Flexibility. It is extremely important to offer them a project that is adjustable; it should go in many directions with many different potential kinds of results. Since we do not know the strength of incoming students in advance, it is useful to have in mind both easier and harder versions of the project.
Motivation. It is important for the project to be well motivated, which means related to other things that have been studied and known to be interesting, to research of other people, etc. Students get more excited when they see that other people are excited about their results.
A computer component. This is not a must for a good project. But modern mathematics involves a lot of computation and young students are better at it than many older professors. Such a project gives young students the opportunity to tackle something more senior people are interested in but might not have enough computer skills to solve. In addition, through computer experiments students get exposed to abstract notions (groups, rings, Lie algebras, representations, etc.) in a more "hands-on" way than when taking standard courses, and as a result are less scared of them.
A learning component. It is always good when a project exposes students to more advanced notions.
The student should like their project. This is very difficult to accomplish when projects are chosen in advance before we meet the students. However, we try to match them to great projects by using the descriptions they give of their interests on their applications. It goes without saying that mentors should like their project too.
Having stated the desired properties of a good project, let us move on to giving examples: bad projects and good projects. We start with a bad one:
Prove that the largest power of 2 that doesn't contain 0 is 286.
The project satisfies only one requirement: it contains a computer component. Otherwise, it doesn't have an accessible beginning. It is not very flexible: if the student succeeds, the long-standing conjecture will be proven; if s/he doesn't, there is not much value in intermediate results. The question is not very interesting. The only motivation is that it has been open for a long time. Also, there is not much to learn. Though, almost any theoretical question can be made flexible. We can start with the question above and change its direction to make it more promising and enticing.
Another bad example is a project where the research happens after the programs are written. This is bad because it is difficult to estimate the programming abilities of incoming students. It doesn't have an accessible beginning and there is no flexibility until the programming part is finished. If the student can't finish the programming quickly, s/he will not have time to look at the results and produce conjectures. For example, almost any project in studying social networks may fall into this category:
Study an acquaintance graph for some epic movies or fiction, for example Star Wars or The Lord of the Rings. In this graph people are vertices and two people are connected by an edge if they know each other. The project is to compare properties of such graphs to known properties of other social networks.
Though the networks in movies are much smaller than other networks that people study, the amount of programming might be substantial. This project can be a good project for a person with a flexible time frame or a person who is sure in advance that there will be enough time for him/her to look at the data.
Now on to an example of a good project. Lynnelle Ye and her mentor, Tirasan Khandhawit, chose to analyze the game of Chomp on graphs during RSI 2009.
Given a graph, on each turn a player can remove an edge or a vertex together with all adjacent edges. The player who doesn't have a move loses. This game was previously solved for complete graphs and forest graphs, so the project was to analyze the game for other types of graphs.
It is clear how to analyze the game for any particular new graph. So that could be a starting point providing an accessible beginning. After that the next step could be to analyze other interesting sets of graphs. The flexibility is guaranteed by the fact that there are many sets of graphs that can be used. In addition, the project entails learning some graph theory and game theory. And the project has a computational component.
Lynnelle Ye successfully implemented this project and provided a complete analysis of complete n-partite graphs for arbitrary n and all bipartite graphs. She also gave partial results for odd-cycle pseudotrees. The paper is available at the arxiv. Not surprisingly, Lynelle got fourth place in the Intel Science Talent Search and second place in the Siemens Competition.
Suppose that we choose all families with two children, such that one of them is a son named Luigi. Given that the probability of a boy to be named Luigi is p, what is the probability that the other child is a son?
Here is a potential "solution." Luigi is a younger brother's name in one of the most popular video games: Super Mario Bros. Probably the parents loved the game and decided to name their first son Mario and the second Luigi. Hence, if one of the children is named Luigi, then he must be a younger son. The second child is certainly an older son named Mario. So, the answer is 1.
The solution above is not mathematical, but it reflects the fact that children's names are highly correlated with each other.
Let's try some mathematical models that describe how the parents might name their children and see what happens. It is common to assume that the names of siblings are chosen independently. In this case the first son (as well as the second son) will be named Luigi with probability p. Therefore, the answer to the puzzle above is (2-p)/(4-p).
The problem with this model is that there is a noticeable probability that the family has two sons, both named Luigi.
As parents usually want to give different names to their children, many researchers suggest the following naming model to avoid naming two children in the same family with the same name. A potential family picks a child's name at random from a distribution list. Children are named independently of each other. Families in which two children are named the same are crossed out from the list of families.
There is a problem with this approach. When we cross out families we may disturb the balance in the family gender distributions. If we assume that boys' and girls' names are different then we will only cross out families with children of the same gender. Thus, the ratio of different-gender families to same-gender families will stop being 1/1. Moreover, it could happen that the number of boy-boy families will differ from the number of girl-girl families.
There are several ways to adjust the model. Suppose there is a probability distribution of names that is used for the first son. If another son is born, the name of the first son is crossed out from the distribution and following that we proportionately adjust the probabilities of all other names for this family. In this model the probability of naming the first son by some name and the second son by the same name changes. For example, the most popular name's probability decreases with consecutive sons, while the least popular name's probability increases.
I like this model, because I think that it reflects real life.
Here is another model, suggested by my son Alexey. Parents give names to their children independently of each other from a given distribution list. If they give the same name to both children the family is crossed-out and replaced with another family with children of the same genders. The advantage of this model is that the first child and the second child are named independently from each other with the same probability distribution. The disadvantage is that the probability distribution of names in the resulting set of families will be different from the probability distribution of names in the original preference list.
I would like my readers to comment on the models and how they change the answer to the original problem.
I am reading the book Eat to Live by Joel Fuhrman. It contains a formula that as a math formula doesn't make any sense. But as an idea, it felt like a revelation. Here it is:
HEALTH = NUTRIENTS/CALORIES
The idea is to choose foods that contain more nutrients per calorie. The formula doesn't make sense for many reasons. Taken to its logical conclusion, the best foods would be vitamins and tea. The formula doesn't provide bounds: it just emphasizes that your calories should be nutritious. However, too few calories — nutritious or not — and you will die. And too many calories — even super nutritious — are still too many calories. In addition the formula doesn't explain how to balance different types of nutrients.
Let's see why it was a revelation. I often crave bananas. I assumed that I need bananas for some reason and my body tells me that. Suppose I really need potassium. As a result I eat a banana, which contains 800 milligrams of potassium and adds 200 calories as a bonus. If I ate spinach instead, I would get the same amount of potassium at a price of only 35 calories.
The book suggests that if I start eating foods that are high in nutrients, I will satisfy my need for particular nutrients, and my cravings will subside. As a result I will not want to eat that much. If I start my day eating spinach, that might eliminate my banana desire.
I've been following an intuitive eating diet. I am trying to listen to my body hoping that my body will tell me what is better for it. It seems that my body sends me signals that are not precise enough. It's not that my body isn't communicating with me, but it is telling me "potassium" and all I hear is "bananas." What I need to do is use my brain to help me decipher what my body really, really wants to tell me.
As Dr. Fuhrman puts it, we are a nation of overfed and malnourished people. But Fuhrman's weight loss plan is too complicated and time-consuming for me, so I designed my own plan based on his ideas:
I will start every meal with vegetables, as they are the most nutritious. I hope that vegetables will provide the nutrients I need. That in turn will make me less hungry by the next meal, at which time I'll take in fewer calories. I will report to my readers whether or not my plan works. I'm off to shop for spinach. Will I ever love it as much as bananas?
The Cookie Monster is a peculiar creature that appeared in The Inquisitive Problem Solver (Vaderlind, Guy & Larson, MAA, P34). Presented with a set of cookie jars, the Cookie Monster will try to empty all the jars with the least number of moves, where a move is to select any subset of the jars and eat the same number of cookies from each jar in the subset.
Even an untalented Cookie Monster would be able to empty n jars in n moves: to fulfill this strategy the Monster can devour all the cookies of one jar at a time. If the Monster is lucky and some jars have the same number of cookies, the Monster can apply the same eating process to all these identical jars. For example, if all the jars have the same number of cookies, the Monster can gulp down all of them in one swoop.
Now, let us limit our discussion to only cases of n non-empty jars that contain distinct numbers of cookies. If indeed all the numbers are distinct, can the Monster finish eating faster than in n moves?
The answer depends on the actual number of cookies in each jar. For example, if the number of cookies in jars are different powers of 2, then even the most talented Monsters can't finish faster than in n steps. Indeed, suppose the largest jar contains 2N cookies. That would be more than the total number of cookies in all the other jars together. Therefore, any strategy has to include a step in which the Monster only takes cookies from the largest jar. The Monster will not jeopardize the strategy if it takes all the cookies from the largest jar in the first move. Applying the induction process, we see that we need at least n steps.
On the other hand, sometimes the Monster can finish the jars faster. If 2k−1 jars contain respectively 1, 2, 3, …, 2k−1 cookies, the Cookie Monster can empty them all in k steps. Here is how. For its first move, the Monster eats 2k-1 cookies from each of the jars containing 2k-1 cookies or more. What remains are 2k-1−1 pairs of identical non-empty jars containing respectively 1, 2, 3, …, 2k-1−1 cookies. The Monster can then continue eating cookies in a similar fashion, finishing in k steps. For instance, for k=3 the sequences of non-empty jars are: 1,2,3,4,5,6,7 → 1,1,2,2,3,3 → 1,1,1,1 → all empty.
Now we would like to prove a theorem that shows that the example above is the lowest limit of moves even for the most gifted Cookie Monsters.
Theorem. If n non-empty jars contain distinct numbers of cookies, the Cookie Monster will need at least ⌈log2(n+1)⌉ steps to empty them all.
Proof. Suppose that n jars contain distinct numbers of cookies and let f(n) be the number of distinct non-empty jars after the first move of the Cookie Monster. We claim that n ≤ 2f(n)+1. Indeed, after the first move, there will be at least n − 1 non-empty jars, but there cannot be three identical non-empty jars. That means, the number of jars plus 1 can't decrease faster than twice each time.
Now here is something our readers can play with. Suppose a sequence of numbers represents the number of cookies in the jars. Which sequences are interesting, that is, which can provide interesting solutions for the Cookie Monster problem?
I read an interesting article on the paradoxes involved in allocating seats for the Congress. The problem arises because of two rules: one congressperson has one vote, and the number of congresspeople per state should be proportional to the population of said state.
These two rules contradict each other, because it is unrealistic to expect to be able to equally divide the populations of different states. Therefore, two different congresspeople from two different states may represent different sizes of population.
Let me explain how seats are divided by using as an example a country with three states: New Nevada (NN), Massecticut (MC) and Califivenia (C5). Suppose the total number of congresspeople is ten. Also suppose the population distribution is such that the states should have the following number of congresspeople: NN — 3.33, MC — 3.34 and C5 — 3.33. As you know states generally do not send a third of a congressperson, so the situation is resolved using the Hamilton method. First, each state gets an integer portion of the seats. In my example, each state gets three seats. Next, if there are seats left they are allocated to states with the largest remainders. In my example, the remainders are 0.33, 0.34 and 0.33. As Massecticut has the largest reminder it gets the last seat.
This is not fair, because now each NN seat represents a larger population portion than each MC seat. Not only is this not fair, but it can also create some strange situations. Suppose there have been population changes for the next redistricting: NN — 3.0, MC — 3.4 and C5 — 3.6. In this case, NN and MC each get 3 seats, while C5 gets the extra seat for a total of 4. Even though MC tried very hard and succeeded in raising their portion of the population, they still lost a seat.
Is there any fair way to allocate seats? George Szpiro in his article suggests adding fractional congresspersons to the House of Representatives. So one state might have three representatives, but one of those has only a quarter of a vote. Thus, the state's voting power becomes 2 1/4.
We can take this idea further. We can use the Hamilton method to decide the number of representatives per state, but give each congressperson a fractional voting power, so the voting power of each state exactly matches the population. This way we lose one of the rules that each congressperson has the same vote. But representation will be exact. In my first example, NN got three seats, when they really needed 3.33. So each congressperson from New Nevada will have 1.11 votes. On the other hand MC got four seats, when they needed 3.34. So each MC representative gets 0.835 votes.
Continuing with this idea, we do not need congresspeople from the same state to have the same power. We can give proportional voting power to a congressperson depending on the population in his/her district.
Or we can go all the way with this idea and lose the districts altogether, so that every congressperson's voting power will be exactly proportionate to the number of citizens who voted for him/her. This way the voting power will reflect the popularity — rather than the size of the district — of each congressperson.
Lionel Levine invented a new hat puzzle.
The sultan decides to torture his hundred wise men again. He has an unlimited supply of red and blue hats. Tomorrow he will pile an infinite, randomly-colored sequence of hats on each wise man's head. Each wise man will be able to see the colors of everyone else's hats, but will not be able to see the colors of his own hats. The wise men are not allowed to pass any information to each other.
At the sultan's signal each has to write a natural number. The sultan will then check the color of the hat that corresponds to that number in the pile of hats. For example, if the wise man writes down "four," the sultan will check the color of the fourth hat in that man's pile. If any of the numbers correspond to a red hat, all the wise men will have their heads chopped off along with their hats. The numbers must correspond to blue hats. What should be their strategy to maximize their chance of survival?
Suppose each wise man writes "one." The first hat in each pile is blue with a probability of one-half. Hence, they will survive as a group with a probability of 1 over 2100. Wise men are so wise that they can do much better than that. Can you figure it out?
Inspired by Lionel, I decided to suggest the following variation:
This time the sultan puts two hats randomly on each wise man's head. Each wise man will see the colors of other people's hats, but not the colors of his own. The men are not allowed to pass any info to each other. At the sultan's signal each has to write the number of blue hats on his head. If they are all correct, all of them survive. If at least one of them is wrong, all of them die. What should be their strategy to maximize their chance of survival?
Suppose there is only one wise man. It is clear that he should write that he has exactly one blue hat. He survives with the probability of one-half. Suppose now that there are two wise men. Each of them can write "one." With this strategy, they will survive with a probability of 1/4. Can they do better than that? What can you suggest if, instead of two, there is any number of wise men?
* * *
Today I saw an ad — "A printer for sale" — handwritten. Hmm.
* * *
What do you call a motherboard on your spouse's computer?
The motherboard-in-law.
In a puzzle book by Mari Berrondo (in Russian), I found the following logic problem:
Alfred, Bertran and Charles are asked about their profession. One of them always lies; another one always tells the truth; and the third one [who I will refer to as a "half-liar"] sometimes lies and sometime tells the truth. Here are their answers:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The solution in the book is the following. As Alfred is right about what Bertran would say, Alfred can't be a liar. If Alfred is a half-liar then the other two people would give the opposite statements, since one will be a truth-teller and the other a liar. But they both say that Alfred is a piano-tuner, therefore Alfred must be a truth-teller. Hence, Alfred's statement about everyone's profession must be the truth. Now we know that Charles is an insurance agent. As Charles confirms that, thus telling the truth in this instance, we recognize that he must be a half-liar. The answer to the problem is that the half-liar is an insurance agent.
But I have a problem with this problem. You see, a liar can say many things. He can say that he is a conductor, a mathematician, a beekeeper or whatever. So there is no way of knowing what a person who decides to lie can say. Let's just analyze the statement by Alfred: "Concerning Bertran, if you ask him, he will tell you that he is a painter."
If Alfred tells the truth about what Bertran would say, he needs to know for sure that Bertran will say that he is a painter. Hence, Bertran must be a truth-teller and a painter. If Alfred lies, he needs to be sure that Bertran won't say that he is a painter. So Bertran must be either a truth-teller and not a painter, or a liar and a painter. Bertran can't be a half-liar, because a half-liar can say that he is a painter as well as he can say something else, no matter what his real profession.
There is one interesting aspect of this that many people overlook. There are different types of people who are half-liars. In some books half-liars are introduced as people who, before making a statement, flip a coin to decide whether to lie or to tell the truth. Such a person needs to know in advance exactly what other people are saying, in order to construct a statement about what those people might say that corresponds to the coin flip. On the other hand, other types of half-liars exist. One half-liar can say something and then see later whether it is true. If Alfred is a half-liar who doesn't care in advance about the truth of his statement, he can say that Bertran will claim that he is a painter.
I leave it to my readers to finish my analysis and see that the problem doesn't have a solution. To end my essay on a positive note, I decided to slightly change the problem, so that there is no contradiction. In the same setting:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is not a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The data from annual surveys carried out by the American Mathematical Society shows the same picture year after year: the percentage of females in different categories decreases as the category level rises. For example, here is the data for 2006:
| Category | Percentage of Women |
|---|---|
| Graduating Math Majors | 41 |
| PhDs Granted | 32 |
| Fresh PhD hires in academic jobs | 27 |
| Full-time Faculty | 27 |
| Full-time tenured or tenure-track faculty | 12 |
The high percentage of female math majors means that a lot of women do like mathematics. Why aren't women becoming professors of mathematics? In the picture to the left, little Sanya fearlessly took her first integral. I hope, even as an adult, she will never be afraid of integrals.
I am one of the organizers of the Women and Mathematics Program at the Institute for Advanced Study at Princeton In 2009 we had a special seminar devoted to discussing this issue. Here is the report of our discussion based on the notes that Rajaa Al Talli took during the meeting.
Many of us felt, for the following three reasons, that the data doesn't represent the full picture.
First, the different stages correspond to women of different ages; thus, the number of tenured faculty should be compared, not to the number of current math majors, but rather to women who majored in math many years ago. The percentage of female PhDs in mathematics has been increasing steadily for the past several years. As a result, we expect an eventual increase in the number of full-time female faculty.
Second, international women mathematicians might be having a great impact on the numbers. Let's examine a hypothetical situation. If many female professors come to the US after completing their studies in other countries, it would be logical to assume that they would raise the numbers. But since the numbers are falling, we might be losing more females than we think. Or, it could be the opposite: international graduate students complete a PhD in mathematics in the USA and then go back to their own countries. In this case we would be losing fewer females to professorships than the numbers seem to suggest. Unfortunately, we can't really say which case is true as we do not know the data on international students and professors.
Third, many women who major in mathematics also have second majors. For example, the women who have a second major in education probably plan to become teachers instead of pursuing an academic career. It would be interesting to find the data comparing women who never meant to have careers in science with those women who left because they were discouraged. If we are losing women from the sciences because they decide not to pursue scientific careers, then at least that is their choice.
It is also worth studying why so few women are interested in careers in mathematics in the first place. Changing our culture or applying peer pressure in a different direction might change the ambitions of a lot of people.
We discussed why the data in the table doesn't represent the full picture. On the other hand, there are many reasons why women who can do mathematics and want to do mathematics might be discouraged from pursuing an academic career:
Our group proposed many solutions to help retain women in mathematics:
At the end of our meeting, everyone accepted Ingrid Daubechies' proposal that we do the following:
Each woman in mathematics should take as her responsibility the improvement of the mathematical environment in which she works. If every woman helps change what's going on in her university or the school where she teaches, that will help solve the problem on the larger scale.
The book How to Drive Your Man Wild in Bed
The list of bad features also includes professions to avoid. Can you guess the first profession on the list? OK, I think you should be able to meta-guess given the fact that I am writing about it. Indeed, the list on page 64 starts:
Avoid, on the whole, mathematicians…
I am an expert on NOT avoiding mathematicians: in fact, I've married three of them and dated x number of them. That isn't necessarily because I like mathematicians so much; I just do not meet anyone else.
When I was a student I had a theory that mathematicians are different from physicists. My theory was based on two conferences on mathematical physics I attended in a row. The first one was targeted for mathematicians and the second for physicists. The first one was very quiet, and the second one was all boozing and partying. So I decided that mathematicians are introverts and physicists are extroverts. I was sure then that my second husband chose a wrong field, because he liked booze and parties.
By now, years later, I've met many more mathematicians, and I have to tell you that they are varied. It is impossible and unfair to describe mathematicians as a type. One mathematician even became the star of an erotic movie. I write this essay for girls who are interested in dating mathematicians. I am not talking about math majors here, I am talking about mathematicians who do serious research. Do I have a word of advice?
I do have several words of caution. While they don't apply to all mathematicians, it's worth keeping them in mind.
First, there are many mathematicians who, like my first husband, are very devoted to mathematics. I admire that devotion, but it means that they plan to do mathematics on Saturday nights and prefer to spend vacation at their desks. If they can only fit in one music concert per year, it is not enough for me. Of course, this applies to anyone who is obsessed by his work.
Second, there are mathematicians who believe that they are very smart. Smarter than many other people. They expand their credibility in math to other fields. They start going into biology, politics and relationships with the charisma of an expert, when in fact they do not have a clue what they are talking about.
Third, there are mathematicians who enjoy their math world so much that they do not see much else around them. The jokes are made about this type of mathematician:
What is the difference between an extroverted mathematician and an introverted one? The extroverted one looks at your shoes, rather than at his own shoes.
Yes, I have met a lot of mathematicians like that. Do you think that their wives complain that their husbands do not notice their new haircuts? No. Such triviality is not worth mentioning. Their wives complain that their husbands didn't notice that the furniture was repossessed or that their old cat died and was replaced by a dog. My third husband was like that. At some point in my marriage I discovered that he didn't know the color of my eyes. He didn't know the color of his eyes either. He wasn't color-blind: he was just indifferent. I asked him as a personal favor to learn the color of my eyes by heart and he did. My friend Irene even suggested creating a support group for the wives of such mathematicians.
While you need to watch out for those traits, there are also things I like about mathematicians. Many mathematicians are indeed very smart. That means it is interesting to talk to them. Also, I like when people are driven by something, for it shows a capacity for passion.
Mathematicians are often open and direct. Many mathematicians, like me, have trouble making false statements. I stopped playing —Mafia— because of that. I prefer people who say what they think and do not hold back.
There is a certain innocence among some mathematicians, and that reminds me of the words of the Mozart character in Pushkin's poetic drama, Mozart and Salieri: —And genius and villainy are two things incompatible, aren't they?— I feel this relates to mathematicians as well. Many mathematicians are so busy understanding mathematics, they are not interested in plotting and playing games.
Would I ever date a mathematician again? Yes, I would.
33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?
The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.
Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.
First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.
As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.
Let's look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.
Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.
We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.
Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.
If we continue this, we'll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.
We get two interesting new sequences out of this. The sequence of the fastest rider's speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.
The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.
I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.
One day we may all face the necessity of hiring a lawyer. If the case is tricky the lawyer must be smart and inventive. I am collecting puzzles to give to a potential lawyer during an interview. The following puzzle is one of them. It was given at the second Euler Olympiad in Russia:
At a local Toyota dealership, you are allowed to exchange brand new cars. You can exchange three Camrys for one Prius and one Avalon, and three Priuses for two Camrys and one Avalon. Assuming an unlimited supply of cars at the dealership, can collector Vasya exchange 700 Camrys for 400 Avalons?
The beauty of this puzzle is that the answer I may find acceptable from a mathematician is not the same as I want from my future lawyer.
Have I intrigued you? Get to work and send me the solutions.
I am sitting in front of my computer and scheming, or, more precisely, scamming. I am inventing scams as a way of raising awareness of how probability theory can be used for deception.
My first scam is my lottery project. Suppose I create and run a private lottery. I will award minor payments to some participants, while promising a grand prize of one hundred million dollars. However, there will be a very small probability that anyone will win the big payout. My plan is to live lavishly on my proceeds, hoping no one ever wins the big ticket.
The beauty of this scheme is that nobody will complain until someone scores the top prize. After all, everyone has been receiving what I promised, and no one realizes my fraud. If nobody wins the big award until I retire, I will have built my life style on deception without having been caught.
Suppose someone wins the hundred million dollars. Oops. I am in big trouble. On the other hand, maybe I can avoid jail time. I could tell the winner that the money is gone and if s/he complains to the police, I will declare bankruptcy and we will all lose. Alternatively, I can suggest a settlement in exchange for silence. For example, we could share future proceeds. Probability theory will help me run this lottery with only a small chance of being exposed.
But even a small chance of failure will cause me too much stress, so I have come up with an idea for another scam. I will write some complicated mathematical formulas with which to persuade everyone that global warming will necessarily produce earthquakes in Boston in the near future. Then I'll open an insurance company and insure everyone against earthquakes. As I really do not expect earthquakes in my lifetime, I can spend the money. I'll just need to keep everyone scared about earthquakes. This time I can be sure that I won't be caught as no one will have a reason to complain. The only danger is that someone will check my formulas and prove that I used mathematics to lie.
Perhaps I need a scam that covers up the lie better. Instead of inventing an impossible catastrophe, I need to insure against a real but rare event. Think Katrina. I collect the money and put aside money for payouts and pocket the rest. But I actually tweak my formulas and put aside less than I should, boosting my bank account. I will be wealthy for many years, until this event happens. I might die rich but if this catastrophe happens while I'm still alive, I'll declare bankruptcy.
Though I was lying to everyone, I might be able to avoid jail time. I might be able to prove that it was an honest mistake. Mathematical models include some subjective parameters; besides, everyone believes that nature is unpredictable. Who would ever know that I rigged my formulas in my favor? I can claim that the theory ended up being more optimistic than reality is. Who could punish me for optimism?
Maybe I can be accused of lying if someone proves that I knew that the optimistic model doesn't quite match the reality. But it is very difficult for the courts to punish a person for a math mistake.
When I started writing this essay, I wanted to write about the financial crisis of 2008. I ended up inventing scams. In a way, I did write about the financial crisis. My scams are simplified versions of what banks and hedge funds did to us. Will we ever see someone punished?
Most movies related to mathematics irritate me because of simplifications. I especially do not like when a movie pretends to be intelligent and then dumbs it down. I recently watched the Spanish movie Fermat's Room
The movie opens with people receiving invitations to attend a meeting for geniuses. To qualify for the meeting they need to solve a puzzle. Within ten days, they must guess the order underlying the following sequence: 5, 4, 2, 9, 8, 6, 7, 3, 1. Right away, at the start of the movie, I was already annoyed because of the simplicity of the question. You do not have to be a genius to figure out the order, not to mention how easy it would be to plug this sequence into the Online Encyclopedia of Integer Sequences to find the order in five minutes.
The participants were asked to hide their real names, which felt very strange to me. All famous puzzle solvers compete in puzzle championships and mystery hunts and consequently know each other.
The meeting presumably targets the brightest minds and promises to provide "the greatest enigma." During the meeting they are given seven puzzles to solve. All of them are from children's books and the so-called "greatest enigma" could easily be solved by kids. Though I have to admit that these were among the cutest puzzles I know. For example:
There are three boxes: one with mint sweets, the second with aniseed sweets, and the last with a mixture of the two. The boxes are labeled, but all the labels are wrong. What is the minimum number of sweets you need to taste to correctly re-label all the boxes?
Another of the film's puzzles includes a light bulb in a room and three switches outside, where you have to correctly find the switch that corresponds to the bulb, but you can only enter the room once. In another puzzle you need to get out of prison by deciding which of two doors leads to freedom. You are allowed to ask exactly one question to one of the two guards, one of whom is a truth-teller and the other is a liar.
The other four puzzles are similar to these three I have just described. To mathematicians they are not the greatest enigmas. They are nice material for a children's math club. For non-mathematicians, they may be fascinating. Certainly it's a good thing that such tasteful puzzles are being promoted to a large audience. But they just look ridiculous as "the greatest enigmas."
So what is it about this film that I so enjoyed?
The intensity of the movie comes from the fact that the people are trapped in a room that starts shrinking when they take more than one minute to solve a puzzle.
I well remember another shrinking room from Star Wars: A New Hope. When Princess Leia leads her rescuers to a room, it turns out to be a garbage compactor. The bad guys activate the compactor and two opposite walls start moving in. In contrast, Fermat's room is shrinking in a much more sophisticated way: all four walls are closing in. Each of the walls in the rectangular room is being pressured by an industrial-strength press. The walls in the corners do not crumble, but rather one wall glides along another. I was more puzzled by this shrinking room than I was by the math puzzles. I recommend that you try to figure out how this can be done before seeing the movie or its poster.
However, the best puzzle in the movie is the plot itself. Though I knew all the individual puzzles, what happened in between grabbed me and I couldn't wait to see what would happen next. I saw the movie twice. After the first time, I decided to write this review, so I needed to check it again. I enjoyed it the second time even better than the first time. The second time, I saw how nicely the plot twists were built.
Maybe I shouldn't complain about the simplicity and the familiarity of the puzzles. If they were serious new puzzles I would have started solving them instead of enjoying the movie. The film's weakness might be its strength.
I collect hats puzzles. A puzzle about hats that I hadn't heard before appeared on the Konstantin Knop's blog (in Russian):
The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat — for both of which he has an inexhaustible supply — on every wizard's head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan's signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide on a strategy to maximize the number of survivors. Suggest a strategy for them.
I'll start the discussion with a rather simple idea: Each wizard writes down a color randomly. In this case the expected number of survivors is 50. Actually, if each wizard writes "red", then the expected number of survivors is 50, too. Can you find a better strategy, with a greater expected number of survivors or prove that such a strategy doesn't exist?
As a bonus question, can you suggest a strategy that guarantees 50 survivors?
Now that you've solved that issue, here's my own variation of the problem.
The wizards are all very good friends with each other. They decide that executions are very sad events and they do not wish to witness their friends' deaths. They would rather die themselves. They realize that they will only be happy if all of them survive together. Suggest a strategy that maximizes the probability of them being happy, that is, the probability that all of them will survive.
The following two problems appeared together in Martin Gardner's Scientific American column in 1959.
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Many people, including me and Martin Gardner, wrote a lot about Mr. Smith. In his original column Martin Gardner argued that the answer to the first problem is 1/3. Later he wrote a column titled "Probability and Ambiguity," where he corrected himself about Mr. Smith.
… the answer depends on the procedure by which the information "at least one is a boy" is obtained.
This time I would like to ignore Mr. Smith, as I wrote a whole paper about him that is now under consideration for publication at the College Mathematics Journal. I would rather get back to Mr. Jones.
Mr. Jones failed to stir a controversy from the start and was forgotten. Olivier Leguay asked me about Mr. Jones in a private email, reminding me that the answer to the problem about his children also depends on the procedure.
One of the reasons Mr. Jones was forgotten is that for many natural procedures the answer is 1/2. For example, the following procedures will produce an answer of 1/2:
There are many other procedures that lead to the answer 1/2. However, there are many procedures that lead to other answers.
Suppose I know Mr. Jones, and also know that he has two children. I meet Mr. Jones at a mall, and he tells me that he is buying a gift for his older daughter. Most probably I would assume that the other child is a daughter, too. In my experience, people who have a son and a daughter would say that they are buying a gift for "my daughter." Only people with two daughters would bother to specify that they are buying a gift for "my older daughter."
In some sense I didn't forget about Mr. Jones. I wrote about him implicitly in my essay Two Coins Puzzle. His name was Carl and he had two coins instead of two children.
My blog is getting more famous. Now I don't need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov's Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:
Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?
Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:
Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?
Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.
86 is conjectured to be the largest power of 2 not containing a zero. This simply stated conjecture has proven itself to be proof-resistant. Let us see why.
What is the probability that the nth power of two will not have any zeroes? The first and the last digits are non-zeroes; suppose that other digits become zeroes randomly and independently of each other. This supposition allows us to estimate the probability of 2n not having zeroes as (9/10)k-2, where k is the number of digits of 2n. The number of digits can be estimated as n log102. Thus, the probability is about cxn, where c = (10/9)2 ≈ 1.2 and x = (9/10)log102 ≈ 0.97. The expected number of powers of 2 without zeroes starting from the power N is cxN/(1-x) ≈ 40 ⋅ 0.97N.
Let us look at A007377, the sequence of numbers such that their powers of 2 do not contain zeros: 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86. Our estimates predicts 32 members of this sequence starting from 6. In fact, the sequence has 30 conjectured members. Similarly, our estimate predicts 2.5 members starting from 86. It is easy to check that the sequence doesn't contain any more numbers below 200 and our estimate predicts 0.07 members after 200. As we continue checking larger numbers and see that they do not belong to the sequence, the probability that the sequence contains more elements vanishes. With time we check more numbers and become more convinced that the conjecture is true. Currently, it has been checked up to the power 4.6 ⋅ 107. The probability of finding something after that is about 1.764342396 ⋅10-633620.
Let us try to approach the conjecture from another angle. Let us check the last K digits of powers of two. As the number of possibilities is finite, these last digits eventually will start cycling. If we can show that all the elements inside the period contain zeroes, then we need to check the finite number of powers of two until this period starts. If we can find such K, we can prove the conjecture.
Let us look at the last two digits of powers of two. The sequence starts as: 01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04. As we would anticipate, it starts cycling. The cycle length is 20, and 90% of numbers in the cycle don't have zeroes.
Now let's continue to the last three digits. The period length is 100, and 19 of them either start with zero or contain zero. The percentage of elements in the cycle that do not contain zero is 81%.
The cycle length for the last n digits is known. It is 4 ⋅ 5n-1. In particular the cycle length grows by 5 every time. The number of zero-free elements in these cycles form a sequence A181610: 4, 18, 81, 364, 1638, 7371, 33170. If we continue with our supposition that the digits are random, and study the new digits that appear when we move from the cycle of the last n digits to the next cycle of the last n+1 digits, we can expect that 9/10 of those digits will be non-zero. Indeed, if we check the ratio of how many numbers do not contain zero in the next cycle compared to the previous cycle, we get: 4.5, 4.5, 4.49383, 4.5, 4.5, 4.50007. All of these numbers are quite close to our estimation of 4.5. If this trend continues the portion of the numbers in the cycle that don't have zeroes tends to zero; however, the total of such numbers grows exponentially. We can even estimate that the expected growth is 4 ⋅ 4.5n-1. From this estimation, we can derive the conjecture:
Conjecture. For any number N, there exists a power of two such that its last N digits are zero-free.
Indeed, the last N digits of powers of two cycle, and there are an increasing number of members inside that cycle that do not contain zeroes. The corresponding powers of two don't have zeroes among N rightmost digits.
So, how do we combine the two results? First, the expected probability of finding the power of two larger than 86 that doesn't contain zero is minuscule. And second, we most certainly can find a power of two that has as many zeroless digits at the end as we want.
To combine the two results, let us look at the sequences A031140 and A031141. We can deduce from them that for the power 103233492954 the first zero from the right occupies the 250th spot. The total number of digits of that power is 31076377936. So 250 is a tiny portion of the digits.
As time goes by we grow more and more convinced that 86 is the largest power of two without zeroes, but it is not at all clear how we can prove the conjecture or whether it can be proven at all.
My son, Sergei, suggested that I claim that I have a proof of this conjecture, but do not have enough space in the margin to fit my proof in. The probability that I will ever be shamed and disproven is lower than the probability of me winning a billion dollars in the lottery. Though then, if I do win the big bucks, I will still care about being shamed and disproven.
Janet Mertz encouraged me to find IMO girls and compare their careers to that of their teammates. I had always wanted to learn more about the legendary Lida Goncharova — who in 1962 was the first girl to win an IMO gold medal. So I located her, and after an interview, wrote about her. Only 14 years later, in 1976, did the next girl get a gold medal. That was me. I was ranked overall second and had 39 points out of 40.
As I did in the article about Lida, I would like to compare my math career to that of my teammates.
I got my PhD in 1988 and moved to the US in 1990. My postdoc at MIT in 1993 was followed by a postdoc at Bar-Ilan University. In 1996 I got a non-paying visiting position at Princeton University. In 1998 I gave up academia and moved to industry, accepting an offer from Bellcore. There were many reasons for that change: family, financial, geographical, medical and so on.
On the practical level, I had had two children and raising them was my first priority. But there was also a psychological element to this change: my low self-esteem. I believed that I wasn't good enough and wouldn't stand a chance of finding a job in academia. Looking back, I have no regrets about putting my kids first, but I do regret that I wasn't confident enough in my abilities to persist.
I continued working in industry until I resigned in January 2008, due to my feeling that I wasn't doing what I was meant to do: mathematics. Besides, my children were grown, giving me the freedom to leave a job I did not like and return to the work I love. Now I am a struggling freelance mathematician affiliated with MIT. Although my math blog is quite popular and I have been publishing research papers, I am not sure that I will ever be able to find an academic job because of my non-traditional curriculum vitae.
The year 1976 was very successful for the Soviet team. Out of nine gold medals our team took four. My result was the best for our team with 39 points followed by Sergey Finashin and Alexander Goncharov with 37 points and by Nikita Netsvetaev with 34 points.
Alexander Goncharov became a full professor at Brown University in 1999 and now is a full professor at Yale University. His research is in Arithmetic Algebraic Geometry, Teichmuller Theory and Integral Geometry. He has received multiple awards including the 1992 European Math Society prize. Sergey Finashin is very active in the fields of Low Dimensional Topology and Topology of Real Algebraic Varieties. He became a full professor at Middle East Technical University in Ankara, Turkey in 1998. Nikita Netsvetaev is an expert in Differential Topology. He is a professor at Saint Petersburg State University and the Head of the High Geometry Department.
Comparing my story to that of Lida, I already see a pattern emerging. Now I'm curious to hear the stories of other gold-winning women. I believe that the next gold girl, in 1984, was Karin Gröger from the German Democratic Republic. I haven't yet managed to find her, so can my readers help?
I wrote a series of essays about AMC competitions:
This essay is next in the series. Although it is not strictly about AMC, it should be useful during any test when you need to check your answers. There are several important rules which are helpful.
Rule 0. Checking is important. If wrong answers are punished, then correcting a mistake brings more points than solving a new problem. In addition, problems that were solved are often easier than problems yet to be solved, so finding a mistake might be faster than solving a new problem.
Rule 1. Your checking methods must be fast. The tests are generally timed. This means that in order to check your answers, you need to sacrifice your work on the next problem.
Rule 2. Customize how you check according to your strengths and weaknesses. For example, if you tend to jump to conclusions about what the question is going to be, and as a result answer your anticipated question instead of the one that is actually on the test, then when you are checking you should start reading the problem from the question. Or, if you usually make mistakes in geometry problems, you should allocate more time to geometry problems when you are checking. If you never make mistakes in arithmetic problems then you do not need to check those.
Rule 3. Mark problems that might need checking. If you do not have enough time to check all the problems, check only those you are not sure about.
Rule 4. Do not repeat your solution when you check. While solving the problem your brain often creates a pathway from start to finish. If on this pathway your brain decided to believe that two plus two is five, very often during checking, your brain will make the same mistake again. Because of that it is crucial to use other methods for checking than repeating your reasoning. In case you can't find a way to check your answers using a different method and have to repeat your reasoning, you should repeat it in a different order.
This rule is so important, that I am providing some methods to change your brain pathway when you are checking your answers.
Plug in. Plugging in the answer you found is much faster than finding it. Use this method whenever possible. It is perfect for problems like this one below from 2004 AMC10-A:
What is the value of x if |x – 1| = |x – 2|?
Plug in an intermediate result. Sometimes you can't plug in the answer, but you can plug in an intermediate result. In the following problem from 2004 AMC10-B you can plug in the number of nickels and dimes:
Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?
Calculate something else related to your answer. For example a negation. Here is a problem from 2004 AMC10-B:
How many two-digit positive integers have at least one 7 as a digit?
If you calculated the answer directly, to check it you may want to calculate the number of two-digit positive integers that do not contain 7.
Create an example. Sometimes you solve a problem by reasoning, but to check it you might create a particular example. Here is a problem from 2001 AMC10:
Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N?
If we denote the tens digit by a and the units digit by b, then N = 10a + b, P(N) = a*b, and S(N) = a + b. We get an equation a(b+1) = 10a, from which the answer is 9. To check the answer we do not need to repeat the reasoning. It is enough to check that 19 is the sum of the product of its digits plus the digits.
Here is another problem from 2001 AMC10:
Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?
The list of choices is: 6, 14, 21, 28, 42. Your solution might go like this: the product of three consecutive numbers is divisible by 6. Hence, n is divisible by 42. So, the answer must be 28. To check you might consider a product of three consecutive numbers: 5*6*7=210 and see that it is not divisible by 4, hence it is not divisible by 28.
Rule 5. Embrace the partial check. It is very important to check your answers fast. Sometimes you can gain speed if you do not check the problem completely, but check it partially. For example, you can check that your answer is one of the two correct answers. There are many methods for partial checking.
Try an example. Sometimes an example doesn't guarantee that your choice is correct, but it increases your confidence in your answer. Here is another problem from 2001 AMC10:
The sum of two numbers is S. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
The choices are: 2S + 3, 3S + 2, 3S + 6, 2S + 6, 2S + 12. You can reason that increasing each summand by 3, increases the sum by 6. After that doubling each summand increases the resulting sum twice, so the answer is 2S + 12. To check the answer you can use an example. Usually an example doesn't guarantee the confirmation of your answer, but it might help you eliminate some of the wrong answers. For example, if you choose zero and zero as your initial two numbers, then S = 0, and your transformation brings the result to 12, which confirms your answer 2S + 12. In this particular case, a very easy specific example excluded all the wrong answers.
Divisibility. Sometimes it is faster to calculate the remainder of the answer by some number.
For example, look at the following problem from 2003 AMC10:
What is the units digit of 132003?
The choices are 1, 3, 7, 8, 9. We can immediately say that the answer must be an odd number.
Approximation check. One important example of a partial check is an approximation check. By estimating an approximate answer you might exclude most of the wrong answers. Consider this problem from 2001 AMC12:
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
The divisibilities by 3, 4 or 5 shouldn't correlate with each other. Approximately one third of those number are multiples of 3 and one quarter are multiples of 4. Let's say that one twelfth are multiples of both 3 and 4. Hence, we estimate the portion of numbers that are multiples of 3 or 4 as 1/3 + 1/4 – 1/12 = 1/2. We have about 1,000 such numbers. The number of numbers that are, in addition, not divisible by 5, are less than that. So out of the given choice of (A) 768, (B) 801, (C) 934, (D) 1067, (E) 1167, we can immediately confirm that the answer is among the first three.
The methods above can be useful even if you do not have multiple choices. But if you do…
Rule 6. Use given choices as extra information. In the previous examples you saw how to use a partial check to exclude some of the choices. Here is a specific example from 2006 AMC10-A of how to exclude choices:
What non-zero real value for x satisfies (7x)14 = (14x)7?
The choices are: 1/7, 2/7, 1, 7, 14. If you solved the problem directly, to check it you can reason why other choices do not work. In this particular case it can be done very fast. 1/7 doesn't work because the left part of the equation becomes 1 when the right is clearly not. 1 and 7 do not work because the left part is odd and the right is even; 14 doesn't work because the left is clearly bigger than the right.
Rule 7. Use meta considerations. If you get into the mind of the designers you can better anticipate when you should check more thoroughly. Consider this problem from 2006 AMC10-A:
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
The most common mistake would be to assume that 12:59 supplies the largest sum, which is 17. But look at the choices: 17, 19, 21, 22, 23. When the designers are asking to find the largest number with some property, they assume that some students will make a mistake and chose a smaller number over a larger one. That means the designers would include this potential mistake among the choices. So the answer is extremely unlikely to be the smallest number on the list of choices. Thus, if you think the answer is 17, understanding how these problems are constructed should alert you to thoroughly check your answer. Indeed, the correct answer is 23 which corresponds to 9:59. Not surprisingly, it is the largest on the list of choices.
AMC 10/12 is coming on February 8 and HMMT on February 12. Happy checking.
I've been celebrating my 29th birthday for many years. Once, when I was actually 45 and wanted to have a big party, I invited everyone to the 5th anniversary of my 29th birthday.
Last week my son turned 29 and I realized that it is time to drop this beautiful, prime, evil, deficient, lazy-caterer number, that in addition is the largest power of two to have all different digits. No more celebrating 29.
For my next age, I picked 42. Not because it is the smallest abundant odious number, but rather because it is the answer to life, the universe and everything.
Thank you everyone who congratulated me on my birthday two days ago. For your information, from now on I am 42.
As I did for 2010 and for previous years, here are math-related puzzles from the MIT Mystery Hunt 2011.
Two more puzzles deserve a special mention for their nerdiness. My teammates loved them.
Tired of the same old sudoku? Here's an opportunity to try many variations of it. Thomas Snyder and Wei-Hwa Huang wrote a book called Mutant Sudoku
The book contains about 180 fun puzzles. Look at the variety:
Wei-Hwa Huang kindly sent me this sample Thermo Sudoku puzzle from the book to use on my blog. The grey areas represent thermometers. Every particular thermometer has to have numbers in increasing order (not necessarily consecutive) starting from the bulb.
The second book by the same authors Sudoku Masterpieces: Elegant Challenges for Sudoku Lovers
I bought both books and immediately started scribbling in the first one. My bad handwriting would seem so out of place in the beautiful second book that I have not even started working in it yet. Maybe I will give it as a gift to someone with better penmanship.
Two days ago I threw at my readers the following problem:
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
The purpose of throwing this problem was to discuss the nature of the implicit assumptions that we are asked to make when solving math problems, and the implicit assumptions we teach our children to make when we teach them to solve math problems. This is especially important for problems like this, that are phrased in terms of a situation in the real world. The real world is too complex to model all of; the great power of mathematics is that sufficiently idealized situations are predictable. But which idealizations are appropriate? How does one choose? How does one teach youngsters what to choose?
Before I get to the actual discussion, however, I want to re-throw this problem at my readers, in an effort to highlight what originally jumped out at me as being wrong with it.
Neglecting the effects of altitude, differential wind, acceleration, relativity, measurement error, finite size and non-superimposability of the planes, and the Earth's deviations from perfect sphericity,
- Find how much time it takes them to become 2000 miles apart, assuming that the planes are starting from Boston and the distance is measured as
- a straight line in 3-space.
- the shortest surface distance.
- How far from the closest pole may the starting point be located, so that the answer to the problem is "never"? Solve separately for
- the 3D distance.
- the shortest surface distance.
- What portion of the Earth's surface do the "never"-locations of the previous question occupy?
- under the 3D distance?
- under the shortest surface distance?
Hint: The easiest question is 2b.
My Jewish ex-father-in-law, Naum Bernstein, is 96 years old and is full of life. He has a joke for every situation. In the last decade he wrote several volumes of memoirs in Russian. One of the books was a collection of his favorite jokes and his explanations of them. I decided to retell some of the jokes from his selection.
An arithmetic teacher calls the student Rabinovich to the blackboard. "It is known that from 1 kilogram of sour cream you can make 200 grams of butter. Imagine, Rabinovich, that your father bought 2 kilograms of sour cream. How much butter can he make?"
"Five hundred grams," Rabinovich replies.
The teacher frowns, "Rabinovich, you do not know arithmetic!"
Rabinovich answers, "Sir, you do not know my father."
An astronomy teacher explains that in the future the Earth will lose its heat energy, continents will collide, and solar radiation will increase. In six billion years life will be extinct. A student looking really scared raises his hand and asks, "In how many years will life become extinct?"
"In about six billion years," the teacher repeats.
"Whew," says the student, "you got me so scared. I thought you said six million."
Two professors are chatting while watching a soccer game. The first one says, "They say that soccer players have their brains in their legs. So their heads are really empty."
"Not quite," the second professor replies. "The player on the right passed my exam yesterday."
The first professor expresses interest, so the second one elaborates. "As a rule, I ask two questions. If the student gives a correct answer to one of them, he passes."
"So, what did you ask that guy?"
"My first question was 'What color are red blood cells?' He answered 'Yellow.' That was an incorrect answer. The second question was 'How is sulfuric acid produced?' To this he replied, 'I do not know,' which was absolutely true, so he passed."
A Russian literature teacher asks a pupil, "Who wrote Eugene Onegin?" The pupil gets scared that he is being blamed for something and replies, "No, not me! I swear I didn't write it!" Everyone laughs. The teacher decides that the pupil disrupted the class on purpose and asks for his father to come by.
The father arrives and after the teacher explains what happened, the father says, "Maybe he is not guilty; maybe he really didn't write it. I doubt that he is capable of writing anything."
The teacher is stunned and later tells the whole story in the teachers' lounge to her colleagues. An astronomy teacher comes home and retells the story to her husband who works for the KGB. The husband comments, "Do not worry, we are on it. Three people already confessed to writing it."
A hardcore anti-Semite was dying. As he got weaker he made a last request. He wanted to convert to Judaism. Everyone was extremely surprised, but decided not to interfere. After the conversion, his wife summoned the courage to ask him what was going on. "Do you think you were mistaken, hating Jews all your life?"
"No," he replied happily, "But now with my death, the world will get rid of one more Jew."
An old Jew comes to a Rabbi and asks if he can shave his beard off, because his children think that he is old-fashioned. The Rabbi tells him that by Jewish law he is not allowed to shave. The old man turns to go home when he realizes that the Rabbi himself doesn't have a beard. He stops and asks, "Dear Rabbi, you just forbade me to shave my beard, but how come you are clean-shaven yourself?"
The Rabbi replies, "I didn't ask anyone's permission."
When Rabinovich came to a bureaucrat with a request, the bureaucrat replied, "Come back tomorrow." Rabinovich returned the next day and received the same reply. Rabinovich was very persistent and returned day after day.
Finally, the bureaucrat lost his patience and attacked Rabinovich, "This is outrageous! Don't you understand simple language? I keep telling you to come tomorrow and you keep coming today."
The communist committee of a supermarket in the USSR received a lot of complaints about the rudeness of their salespeople. The committee decided to improve the quality of service and provided special training in which salespeople were taught politeness. The training emphasized what to do in case a particular item was unavailable. The salespeople were supposed to politely explain that the item is temporarily unavailable and to offer a substitute.
The next thing one of their salespeople said to a customer was "I am very sorry, we are temporarily out of toilet paper. May I offer some sandpaper?"
There is panic in an apartment on the 13th floor. The wife recognizes the sound of her husband's approach, even though he was supposed to be on a business trip. The lover asks, "What should I do, honey?"
"What do people do in such cases? Jump out the window!"
"But we are on the 13th floor!"
"This is no time for superstition!"
A young man had smelly feet, plus he always forgot to change his socks. His girlfriend got tired of it and asked him to promise that he would always change his socks before coming to see her.
Next visit the young man smelled as bad as ever. Outraged, the girl said, "But you promised to change your socks!"
The young man answered, "I did as I promised."
"I don't believe you, you smell awful."
"I was sure you wouldn't believe me. Good thing I brought my dirty socks with me as proof."
At the 50th anniversary of a very happy couple, someone asked the husband for their secret. He said that right before the wedding they agreed that the husband would decide all the crucial and very important things, and the wife would be responsible for all minor decisions. "For example," he continued, "yesterday I decided that the US should withdraw their troops from Iraq, and my wife decided where to buy our vacation house."
Two long-time girlfriends meet after several years without being in touch. "How are your children?" asks one of them.
The other replies, "My daughter is fine, she married a nice young man, who is providing for her. He also helps her with chores and even brings her coffee in bed every morning."
"What about your son?"
"It's a disaster. I don't know what to do. He married a really lazy woman. Even though she's not working, she wants him to help her with the chores. Can you imagine that? She even dared to ask him to bring her coffee in bed every morning."
Two friends are walking along a street very late at night. Robbers attack them with guns, demanding their wallets. One of the friends asks the robbers, "Can you give me 30 seconds?" The robbers agree. He takes out $100 from his wallet and gives it to his friend, "Remember I owed you $100? I am paying back my debt in front of witnesses."
Life is a struggle. Before lunch with hunger, after lunch with sleepiness.
A mother says to her son, "Please, close the window — it's cold outside."
The son replies, "Do you think it will get warmer outside if I close the window?"
How are pessimists and optimists different from normal people?
A pessimist uses both a belt and suspenders, an optimist uses neither.
In a cemetery there is a beautiful monument with a picture of a bald, wrinkled old man. He is smiling, showing his perfect white teeth. His epitaph says:
Here lies Mr. X, who lived more than 100 years, lost his hair, became all wrinkled, but kept his perfect teeth. That is because he always used our company's toothpaste.
A nearby monument has a picture of an old toothless woman with beautiful, voluminous hair. The inscription explains which brand of shampoo she used.
Many other tombstones with ads are scattered throughout the cemetery. But in the middle there is a huge mausoleum with an inscription reading:
No one is buried here and no one ever will be, because his or her parents used condoms made by our company.
A Russian man marries an American woman. After a while he writes a letter home.
My wife must be very dirty. She showers every day.
Rabinovich was asked why he didn't attend the last committee meeting. He replied, "If I knew it was the last, I would certainly have come."
I stumbled upon the following problem in Mathematics Teacher v.73 (September 1980):
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
Ooh, boy!
Question for my readers: explain my reaction.
I have a leased Toyota Corolla, and I am happily enrolled in AutoCheck payments with Toyota's Financial Services. So I do not even look at my bills. Once I opened my bill and noticed that the requested payment was twice as high as I expected. I looked closer and the bill had a car tax included in it. I looked even closer and read that:
Your Current Payment Due will be automatically withdrawn from your checking or savings account on the above Payment Due Date or the next banking day.
I decided that everything was taken care of and continued my relaxed life. After several months I checked my bill again, and the car tax was still there. After more careful study of my bill I discovered that Toyota's "Current Payment Due" doesn't include my car tax. Obviously they assume that their definition of "Current Payment Due" is crystal clear to everyone.
I got worried about this delayed car tax payment and went online to pay it. I tried to make this payment, but Toyota's website rejected it. The website informed me that because I am enrolled in AutoCheck, I am not allowed to make separate online payments. I couldn't believe it: to do so, I would have to de-enroll first!
So I just wrote a check.
In one day my feelings for my Toyota Corolla were turned around. If their financial system is designed so stupidly, what can we say about their car designs? Suddenly the sound of my brakes and the squeak in my steering wheel worry me.
A year ago I posted a chessboard puzzle. Recently I stumbled on a September 2008 issue of "Math Horizons" where it was presented as a magic trick.
When the magician leaves the room, the trickees lay out eight coins in a row deciding which side is turned up according to their whim. They also think of a number between 1 and 8 inclusive. The magician's assistant then flips exactly one of the coins, before inviting the magician back in. The magician looks at the coins and guesses the number that the trickees thought of.
The magician's strategy can be derived from the solution to the chessboard puzzle. The assistant numbers the coins from zero to seven from left to right. Then s/he flips the coin so that the parity addition (XORing) of all the numbers corresponding to heads is the number that the magician needs to guess. For this trick to work, the number of coins needs to be a power of 2.
Andrey Zelevinsky posted (in Russian) a cool variation of this trick with two decks of cards.
The magician has two identical card decks and he is out of the room for now. A random person from the audience thinks of a card. Next, the audience chooses several cards from the first deck. Then the assistant adds one card from the second deck to the set of chosen cards, lays them on a table, and then invites the magician back. The magician looks at the cards on the table and guesses the card that was thought of.
Unlike in the coin trick above, the number of cards in the deck doesn't need to be a power of 2. This flexibility is due to the fact that the magician has two decks of cards, as opposed to one set of coins. Having the second deck is equivalent to the assistant in the coin trick being allowed to flip one or ZERO coins.
Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.
An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?
The other one is from 1995.
There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles' orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?
Perfidy is to parity as odious is to odd and evil is to even. As a reminder, odious numbers are numbers with an odd number of ones in their binary expansions. From here you can extrapolate what the evil numbers are and the fact that the perfidy of an integer is the parity of the number of ones in its binary expansion. We live in a terrible world: all numbers are perfidious.
So why are we writing about the perfidy of negative numbers? One would expect it to be a natural extension of the perfidy of positive numbers, but it turns out that the naive way of defining it doesn't work at all. Is there hope? Could negative numbers be innocent of evil and free of odiousness? Is zero an impenetrable bulwark against perfidy? Not quite, but something interesting does happen to evil as it tries to cross zero. Read on.
To define perfidy for negative numbers, let us study how perfidy behaves for positive numbers. It is easiest to think about the perfidies of power-of-two-sized chunks of non-negative integers at a time. Let us denote by Tn the string of perfidies of the integers from 0 to 2n−1, with evil being zero and odious being 1. So T0 = 0, T1 = 01, T2 = 0110, T3 = 01101001, …. The recurrence relation governing the Tn is Tn+1 = TnTn, where T is the bitwise negation of the string T, and juxtaposition is concatenation. The limit of this as n tends to infinity is the (infinite) sequence of perfidies of non-negative integers. This sequence is called the Thue-Morse sequence: 0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,….
So defining the perfidy of negative numbers is equivalent to extending the Thue-Morse sequence to the left. If we are to define "the" perfidy of negative numbers, that definition should preserve most of the properties of the Thue-Morse sequence after extension.
So, let's see. We asked around, and most people said that the binary expansion of a negative integer should be the binary expansion of its absolute value, but with a minus sign. Defining perfidy as parity of number of ones in this binary expansion corresponds to the following extended Thue-Morse sequence in which we mark values corresponding to negative indices with bold font: … 0, 1, 1, 0, 1, 1, 0, ….
One of the major properties of the Thue-Morse sequence is its fractal property: if you replace every zero of the Thue-Morse sequence by 0,1 and every one by 1,0, you will get the Thue-Morse sequence back. Clearly, our new extended sequence doesn't have this property.
Another set of properties for the Thue-Morse sequence, called avoidance properties, is a long list of patterns that the sequence avoids. For example, the Thue-Morse sequence doesn't contain any overlapping squares — patterns axaxa, where a is a character and x is a word. But you can see above, our first extension contains it. So this definition is wrong, not just once but twice (and two wrongs only make a right under very unusual circumstances). Perfidy is stymied by the cross-over from zero to minus one. Are negative numbers protected from the ravages of evil? (and odiousness?)
Unfortunately, there are many people, for example John Conway, who inadvertently extend the reach of perfidy by arguing that the binary expansion of a negative integer should be different. Indulge in a flight of fancy and imagine the binary expansion that consists of infinitely many ones to the left: …1111. What happens when you add 1 to it? The carry gets pushed infinitely far away, and you get …000000 — zero. So it is quite reasonable to let …1111 be the binary expansion of −1. Similarly, the string …1110 represents −2, …1101 represents −3, etc. Continuing this we see that the binary expansion of a negative integer −n is the bitwise negation of the binary expansion of n − 1 (including the leading zeros). This is called the Two's complement representation.
Why is two's complement a reasonable representation? Suppose you were trying to invent a binary notation for negative numbers, but you wanted to pursue uniformity by not using a minus sign. The problem is that the standard definition of the binary representation allows you to represent only positive numbers. But you can solve this problem with modular arithmetic: modulo any fixed N, every negative number is equivalent to some positive number (by adding enough multiples of N), so you can just represent it by representing that positive number. If you choose N to be a power of two, modding out by it is just truncation of the binary representation. If you let those powers of two tend to infinity, you get the two's complement representation described above.
Aside: When you are building a computer, uniformity is money, because special cases cost special transistors. The two's complement idea lets one build arithmetic units that just operate on positive numbers with some number of bits (effectively doing arithmetic modulo 2k), and leave the question of negativeness to the choice of representatives of those equivalence classes.
If we take two's complement as the binary expansion of negative numbers, how will we define the perfidy? Is the number of ones in the infinite string …1111 corresponding to −1 even or odd?
We can't answer that question, but we know for every binary expansion of negative numbers the parity of the number of zeroes. Thus we can divide all negative integers in two classes with different perfidy. We just do not know which one is which.
Let us consider two cases. In the first case we consider a negative number odious if the number of zeroes in its binary expansion is odd. The corresponding extended Thue-Morse sequence is: … 0, 1, 1, 0, 0, 1, 1, 0, …. The negative half is the reflection of the classical Thue-Morse sequence. In the second case we consider a negative number odious if the number of zeroes in its binary expansion is even. The corresponding extended Thue-Morse sequence is: … 1, 0, 0, 1, 0, 1, 1, 0, …. The negative half is the bitwise negation of the reflection of the classical Thue-Morse sequence.
Can we say that one of the sequences is better than the other? Both of them respect the fractal property of the classical Thue-Morse sequence. Let us look at the avoidance properties. The avoidance properties are symmetric with respect to switching zeroes with ones and with respect to changing the direction of the sequence. Hence, the negation, the reflection, and the reflection of the negation of the Thue-Morse sequence will continue to respect these properties.
Thus, we only need to check the avoidance properties of the finite subsequences that span both negative and non-negative indices. We claim that for both definitions of perfidy, any finite middle subsequence of the extended Thue-Morse sequence occurs as a subsequence in the classical Thue-Morse sequence. So any avoidance properties that are true for the Thue-Morse sequence will also be true for both extensions.
Indeed, it is easy to show that the strings T2n defined above are palindromes. So for the first definition of perfidy the string in the middle will be a substring of T2nT2n for some large n, and for the second definition a substring of T2nT2n. But the classical Thue-Morse sequence contains the subsequence T2nT2nT2nT2nT2nT2nT2nT2n. So whichever way we extend the Thue-Morse sequence to the left any finite middle part will always be a repetition of a piece in the classical Thue-Morse sequence. Thus, all the avoidance properties will hold.
We see that there are two logical ways to define perfidy for negative integers. There are two clear groups of numbers with the same perfidy, but which is called evil and which odious is interchangeable. So evil doesn't stop at zero after all, but at least it gets an identity crisis.
* * *
A note posted on the door of the tech-support department:
"Theory — you know everything, but nothing works. Practice — everything works, but nobody knows why. In our department we merge theory with practice: nothing works and nobody knows why.
* * *
A plus is two minuses at each others' throats.
I was teaching my students PIE, the Principle of Inclusion and Exclusion. This was the last lesson of 2010 and it seemed natural to have a party and bring some pie. It appears that the school has a new rule. If I want to bring any food to class, I need to submit a request that includes all food ingredients. The administrators send it to the parents asking them to sign a permission slip and then, if I receive all the slips back in time, I can bring pie to school. We had to study PIE without pie.
Our most important task as parents and teachers is to teach kids to make their own decisions. They are in high school; they know by now about their own allergies and diets; they should be able to avoid foods that might do them harm. I understand why schools create such rules, but we are treating the students like small children. We can't protect them forever; they need to learn to protect themselves.
Next semester, we will study the mathematics of fair division. I will have to teach them how to cut a cake without a cake.
I once wrote a story about a mistake that my medical insurance CIGNA made. They had a typo in the year of the end date of my insurance coverage in their system. As a result of this error, they mistakenly thought they had paid my doctors after my insurance had expired and tried to get their money back. While I was trying to correct all this mess, an interesting thing happened.
To help me explain, check out the following portion of my bill. (If it looks a bit funny, it's because I cut out some details including the doctor's name).
On the bill you can see that I had a mammography for which I was charged $493.00, but CIGNA paid only $295.80. The remaining $197.20 was removed from the bill as an adjustment, as frequently happens because of certain agreements between doctors and insurance companies. A year later when CIGNA made their mistake, they requested that the payment be returned. You can see on the bill that once the payment was reversed, my doctors reversed the adjustment too.
When CIGNA fixed the typo, they repaid the doctors, but the adjustment stayed on the bill, which the doctors then wanted me to pay. And that was only one of many such bills. It took me a year of phone calls to get the adjustments taken off, but this is not what I am writing about today.
If not for this mistake, I would have never seen these bills and the revealing information on the different amounts doctors charge to different parties, and how much they really expect to receive. As you can see my doctors wanted 67% more for my mammogram than they later agreed to.
The difference in numbers for my blood test was even more impressive. I was charged $173.00, and the insurance company paid $30.28 — almost six times less.
If I ever need a doctor and I don't have insurance, I will take these bills with me to support my request for a discount. I do not mind if you use this article for the same purpose.
I met Ed (Edik) Frenkel 20 years ago at Harvard when he was a brilliant math student of my now ex-husband, and a handsome young man. Now, at 42, he is a math professor at Berkeley and he is even hotter. He made a bizarre move for a mathematician: he produced and starred in an erotic short movie, Rites of Love and Math. If he wants to be known as the sexiest male mathematician alive, he just might get the title.
The movie created a controversy when Mathematical Sciences Research Institute (MSRI) withdrew its sponsorship for the first screening after a lot of objections based on the trailer. My interest was piqued by a painting that dominated the visual of the trailer's erotica scene. The black and white amateur painting is of the integral sign with Russian letters stylized as math symbols that spell the word "Truth". In addition, the name of the woman in the movie, Mariko, means "truth" in Japanese. Though it felt pretentious, I was hoping that the movie would be symbolic. When I heard that the actors do not talk in the movie, my expectations of symbolism grew. I love movies that are open to interpretation. So I bought the movie, watched it and wrote the following review. Before getting to the review itself I would like to thank Ed Frenkel for sending me the photos and giving me permission to use them in my frank assessment of his work.
Here is the plot:
A Mathematician, hoping to serve humanity, discovers a formula of Love. Bad guys find an evil way to use the formula to destroy humanity and are hunting for the Mathematician, who is hiding in his lover Mariko's home. The Mathematician fears for his own life. Although it would make sense to destroy all the papers with the formula, the Mathematician loves his formula even more than his lover and himself. He wants to preserve the formula and tattoos it on her body with her consent.
There is much about the film that I like, including the slow pace and the visuals, with their minimalistic background and palette of black, white and red. The camera work is superb.
I welcomed the idea of a Love formula, because mathematics is ready to broaden the scope of its models, including venturing into love. Of course, some mathematical models of relationships already exist.
As I mentioned, I was looking forward to the movie, hoping that it would encourage the imagination of viewers in their interpretations. To my disappointment, every scene in the movie is preceded by text that describes the plot, removing any flexibility of interpretation. Besides that, the emotions portrayed didn't quite match the written plot, in no small part because Ed Frenkel is not a good actor.
The idea of preserving a formula by tattooing it on someone is beyond strange. He could have used a safe-deposit box. Or put the formula in an envelope and given it to the lover to keep, or just encrypted it, etc. With narcissistic lack of consciousness, the Mathematician seems unaware of the implications of his action of imprinting this dangerous secret on Mariko. She can never go swimming, or go to the gym, or be intimate with anyone else. Moreover, if the bad guys discover that Mariko is the Mathematician's lover, her life will be in grave danger. Not to mention that tattooing is painful.
Something that could have been interesting and watchable in a historic movie, in this contemporary movie seems pointlessly cruel, dehumanizing and senseless.
I know for sure that Ed Frenkel is not stupid, so what are his reasons for constructing the plot in this way? Before investigating his reasons, I have a mathematical complaint about the movie. Every mathematician and teacher knows that when asserting a formula you need to indicate its interpretation: what its symbols refer to in the real world. For example, suppose I tell you my own great Formula of Love: Cn = (2n)!/(n+1)!n!. You may recognize Cn as the Catalan numbers, but what does this have to do with Love? To give the formula meaning I need to tell you that Cn is the number of ways you can seat n loving couples at a round table with 2n chairs, so that each couple can join hands (assuming the arms are long enough to reach across the table) without any two pairs of arms crossing. Assigning an interpretation makes the Catalan numbers part of the world's growing body of romantic research.
Writing a formula without mentioning what the variables mean fails to preserve it for the future. Ed Frenkel knows that. Wait a minute. The formula in the movie is actually not the Formula of Love, but a real formula from Ed's paper on instantons. It's right there, formula 5.7 on page 74. Every variable is explained in the paper. Ah-ha! So his movie isn't actually about art, but rather about Ed's formula. Indeed, there is no real Formula of Love. In such situations in other movies, they have simply shown fragments of a formula. However, in Rites of Love and Math, Frenkel's formula — which has nothing to do with Love — is shot in full view, zooming in slowly.
The movie is a commercial. Ed is using our fascination with sex to popularize his formula, and using his formula and his scientific standing to advertise his body.
I was so disappointed that the default interpretation of the movie was imposed on me by those pre-scene texts, that I decided to watch the movie for a second time, trying to ignore the text, hoping to find some new meaning.
If you decide to see the movie, you'll probably come up with your own interpretation of the plot. I actually came up with several. I had a funny one and an allegorical one, but the most interesting task for me was to try create an interpretation matching the emotions portrayed:
Mariko knows that something is wrong in her sex life with the Mathematician. But she still loves him and writes him a love letter. The Mathematician comes to Mariko's place. He is distant and cold. They cuddle. He explains to her that sex doesn't bring him pleasure anymore and that moreover, he can't even perform. He tells her that the only thing that brings him joy is mathematics and suggests that his sexual dysfunction and lack of pleasure will be fixed if they tattoo his favorite formula on her body. She agrees, but first they decide to give sex a last try. They try real hard. But he can't relax and he doesn't enjoy it, so she agrees to the tattoo. He does get excited during the tattooing process itself, but once he finishes his whole formula, he is no longer turned on. Mariko's suffering has been in vain.
Janet Mertz wrote several papers about the gender gap in mathematics. One of her research ideas was to find girls who went to the International Math Olympiad (IMO) and compare their fate to that of their teammates with a similar score. She asked me to find Soviet and Russian IMO girls. All my life I had heard about Lida Goncharova, the first girl on a Soviet team, and the first girl in the world who took a gold medal, but I had never dared to reach out to her. A little push by Janet Mertz was enough for me to find Lida's phone number in Moscow and call her.
Lida got interested in mathematics when she was five years old. Luckily, many of her relatives were mathematicians and she started bugging them for math puzzles.
Her involvement with math was interrupted by the death of her parents — her mother when she was seven, and her father when she was nine. She ended up living with her sister, but felt very lonely.
After several years of personal turmoil, she renewed her pursuit of mathematics. Lida started discussing math with her mother's first husband. She joined a math circle which was run at Moscow State University. When she was 13 she went to a summer camp and found a mentor there to study trigonometry. Eventually she ended up at School Number 425, one of the first schools in Moscow that opened for children gifted in math.
At the end of high school she went to the IMO as part of the Soviet team and won a gold medal there. After that she enrolled in the most prestigious Soviet institute for the study of math — Moscow State University (MSU).
Half of her high school classmates went to MSU, including her high school sweetheart Alexander Geronimus. Lida married Alexander when she was a sophomore and they had their first son in her fourth year of undergraduate school.
Meanwhile, she wasn't doing as well in her studies as she had hoped. Lida was very fast to pick up math ideas during conversations, but she had difficulty reading books. As ideas were becoming more complicated and involved, this became a problem. She started feeling that she was falling behind her friends. When her friends gathered together to discuss mathematics she couldn't understand everything. She wanted to ask questions, but was too shy. Plus, she didn't want to impose on them. She made a decision to be silent. As a result she started ignoring the conversations of others and became discouraged as she fell behind.
She had her second child at the beginning of graduate school, where she studied under the supervision of Dmitry Fuchs. Lida was already losing her self-esteem and so she chose a self-contained problem that didn't require a lot of outside knowledge. The solution involved some combinatorial methods, but Lida didn't quite understand the big picture and the problem's goal.
I contacted Dmitry Fuchs and asked him about Lida's thesis. He told me that Lida's main result is extremely important and widely cited. It is called Goncharova's theorem.
Meanwhile, her husband finished his PhD in math and secured a great job in an academic institution. They had started as peers, but her work was interrupted by having their children. Lida finished her PhD a couple of years after her husband and got a very boring job as an algorithm designer. She even wrote some papers at the job, but she was not much interested. She continued her attempts to do mathematics and continued asking everyone for problems, but it didn't go anywhere. Her friends were not very interested in her calculations and after the birth of her third child she began to lose hope in her research.
When Lida and her husband entered graduate school they became religious. Ten years later, Alexander decided to pursue the Russian Orthodox religion as a career and got a parish 600 km from Moscow. They didn't want to move their children away from Moscow, with its educational and cultural opportunities. So they started living in two places with long commutes. This didn't help her math either.
Eight years after the third son, the fourth son was born. Although Lida sporadically continued her calculations, she still didn't talk about them to anyone.
When the older children went to high school, Lida enjoyed solving their math problems tremendously. In 1990 perestroika started and Lida lost her job. She got an offer to create a private school and teach there. By this time she had had two more children, a son and a daughter. Lida continued working for the private school until her six children grew out of it. Lida enjoyed teaching and inventing methods to teach mathematics. The school ended in 2004. But she continues working with kids sharing with them her joy of mathematics.
Lida believes that she has had an extremely lucky life in many ways. The only exception was her unsuccessful math career. She can't live without math, and will continue working with kids, solving fun problems and doing her private research.
When I first called her and said I wanted to talk about her and math, she told me: —There is nothing to talk about. I stopped doing math after my PhD. Almost.— That —almost— kept me asking questions.
Janet Mertz was considering a serious research project comparing the fates of IMO medal girls with the fates of their teammates, to see whether gender plays a role later. However, due to the language and cultural differences and the fact that most of the girls changed their last name, it was difficult to locate them. So Mertz put this research project on hold.
She had asked me to find and contact the Russian women and I was so fascinated with Lida's story that I decided to write it up in this article. And because the research is on hold, I decided to include the fates of Lida's teammates.
Lida Goncharova got her gold medal in the 1962 IMO with 42 points and was ranked third. The teammate with the closest score was Joseph Bernstein with another gold medal and 46 points. I don't even have to check Wikipedia to tell you about Joseph, as I was once married to him. He used to be a professor at Harvard University and is now a professor at Tel-Aviv University. He is a member of the Israel Academy of Sciences and Humanities and the United States National Academy of Sciences. He achieved a lot and is greatly respected by his peers.
Joseph Bernstein might not be the best person to compare Lida to as he had a perfect score. Some might argue that a perfect score indicates that he might have done better if the problems had been more difficult.
The two Soviet teammates whose scores were the closest to that of Lida, but below her, were Alexey Potepun with 37 points and Grigory Margulis with 36 points.
Alexei Potepun got a PhD in mathematics and is now a professor at Saint-Petersburg University. He has published eleven papers.
Next to Alexey Potepun is Grigory Margulis, who is a professor at Yale and was awarded the Fields Medal and the Wolf Prize. He is a member of the U.S. National Academy of Sciences.
You might notice that the two people who moved to the US are much more famous than those who stayed in Russia. You might say that moving to the US is a better predictor of success than gender. Sure, living in a free country helps, but Margulis got his Fields medal while he was in the USSR. And Bernstein invented his famous D-Modules while in Russia also.
My conversation with Lida was personally inspiring. I loved the tone of her voice when she talked about mathematics. There were many elements that prevented her from having the mathematical career she might have had: the untimely death of her parents, her shyness, raising six children, many years of long commutes. When we look at the achievements of her closest teammates, we can't help but wonder what kind of mathematics we lost.
This conversation was very encouraging for me. I felt there were similarities between Lida and myself in more ways than I expected. What we share most of all is a love for mathematics. I could hear that in her voice.
I gave my students a problem from the 2002 AMC 10-A:
Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, …, 10}. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is: (A) 2/5, (B) 9/20, (C) 1/2, (D) 11/20, (E) 24/25.
Here is a solution that some of my students suggested:
On average Tina gets 6. The probability that Sergio gets more than 6 is 2/5.
This is a flawed solution with the right answer. Time and again I meet a problem at a competition where incorrect reasoning produces the right answer and is much faster, putting students who understand the problem at a disadvantage. This is a design flaw. The designers of multiple-choice problems should anticipate mistaken solutions such as the one above. A good designer would create a problem such that a mistaken solution leads to a wrong answer — one which has been included in the list of choices. Thus, a wrong solution would be punished rather than rewarded.
Readers: here are three challenges. First, to ponder what is the right solution. Second, to change parameters slightly so that the solution above doesn't work. And lastly, the most interesting challenge is to explain why the solution above yielded the correct result.
This is a version of the standard charades game that my son, Sergei Bernstein, invented.
Unlike in regular charades, the person who acts out the phrase doesn't know what the phrase is and has to guess it. The viewers on the other hand, know the phrase but they are not allowed to talk.
So the actor is blindfolded and the viewers are not just watching; they are actively moving the actor and his/her body parts around to communicate the phrase. For example, if the actor is on the right track, since the viewers can't say, "Yes, good!", they might communicate it by nodding the actor's head.
Sounds like fun, especially for people who enjoy touching and being touched.
Once again I am one of the organizers of the Women and Math Program at the Institute for Advanced Study in Princeton, May 16-27, 2011. It will be devoted to an exciting modern subject: Sparsity and Computation.
In case you are wondering about the meaning of the picture on the program's poster (which I reproduce below), let us explain.
The left image is the original picture of Fuld Hall, the main building on the IAS campus. The middle image is a corrupted version, in which you barely see anything. The right image is a striking example of how much of the image can be reconstructed from the corrupted image using clever algorithms.
Female undergraduates, graduates and postdocs are welcome to apply to the program. You will learn exactly how the corrupted image was recovered and much more. The application deadline is February 20, 2011.
Eugene Brevdo generated the pictures for our poster and agreed to write a piece for my blog explaining how it works. I am glad that he draws parallels to food, as the IAS cafeteria is one of the best around.
The three images you are looking at are composed of pixels. Each pixel is represented by three integers corresponding to red, green, and blue. The values of each integer range between 0 and 255.
The image of Fuld hall has been corrupted: some pixels have been replaced with all 0s, and are therefore black; this means the pixel was not "observed". In this corrupted version, 85% of the pixel values were not observed. Other pixels have been modified to various degrees by stationary Gaussian noise (i.e. independent random noise). For the 15% observed pixel values, the PSNR is 6.5 db. As you can see, this is a badly corrupted image!
The really interesting image is the one on the right. It is a "denoised" and "inpainted" version of the center image. That means the pixels that were missing were filled in and the observed pixel integer values were re-estimated. The algorithm that performed this task, with the longwinded name "Nonparametric Bayesian Dictionary Learning," had no prior knowledge about what "images should look like". In that sense, it's similar to popular wavelet-based denoising techniques: it does not need a prior database of images to correct a new one. It "learns" what parts of the image should look like from the original image, and fills them in.
Here's a rough sketch of how it works. The idea is to use a new technique in probability theory — the idea that a a patch, e.g. a contiguous subset of pixels, of an image is composed of a sparse set of basic texture atoms (from the "Dictionary"). Unfortunately for us, the number of atoms and the atoms themselves are unknowns and need to be estimated (the "Nonparametric Learning" part). In a way, the main idea here is very similar to Wavelet-based estimation, because while Wavelets form a fixed dictionary, a patch from most natural images is composed of only a few Wavelet atoms; and Wavelet denoising is based on this idea.
We make two assumptions that allow us to simplify and solve this problem, which is unwieldy-sounding and vague when the texture atoms have to be estimated. First, there may be many atoms, but a single patch is a combination of only a sparse subset of them. Second, because each atom appears in part in many patches, even if we observe some noisily, once we know which atoms appear in which patches, we can invert and average together all of the patches associated with an atom to estimate it.
To explain and programmatically implement the full algorithm that solves this problem, probability theorists like to explain things in terms of going to a buffet. Here's a very rough idea. There's a buffet with a (possibly infinite) number of dishes. Each dish represents a texture atom. An image patch will come up to the buffet and, starting from the first dish, begins to flip a biased coin. If the coin lands on heads, the patch takes a random amount of food from the dish in front of it (the atom-patch weight), and then walks to the next dish. If the coin lands on tails, the patch skips that dish and instead just walks to the next. There it flips its coin with a different bias and repeats the process. The coins are biased so the patch only eats a few dishes (there are so many!). When all is said and done, however, the patch has eaten a random amount from a few dishes. Rephrased: the image patch is made from a weighted linear combination of basic atoms.
At the end of the day, all the patches eat their own home-cooked dessert that didn't come from the buffet (noise), and some pass out from eating too much (missing pixels).
If we know how much of each dish (texture atom) each of the patches ate and the biases of the coins, we can estimate the dishes themselves — because we can see the noisy patches. Vice versa, if we know what the dishes (textures) are, and what the patches look like, we can estimate the biases of the coins and how much of a dish each patch ate.
At first we take completely random guesses about what the dishes look like and what the coins are, as well as how much each patch ate. But soon we start alternating guesses between what the dishes are, the coin biases, and the amounts that each patch ate. And each time we only update our estimate of one of these unknowns, on the assumption that our previous estimates for the others is the truth. This is called Gibbs sampling. By iterating our estimates, we can build up a pretty good estimate of all of the unknowns: the texture atoms, coin biases, and the atom-patch weights.
The image on the right is our best final guess, after iterating this game, as to what the patches look like after eating their dishes, but before eating dessert and/or passing out.
I've heard many fun problems in which blindfolded parachutists are dropped somewhere and they need to meet up once they're on the ground. They can't shout or purposefully leave traces behind. They will recognize each other as soon as they bump into each other. Their goal is to get to the same assembly point. They can design their strategy in advance.
Here is the first problem in a series that gets increasingly difficult:
Two parachutists are dropped at different locations on a straight line at the same time. Both have an excellent sense of direction and a good geographical memory, so both know where they are at any moment with respect to their starting point on the line. What's their strategy?
The strategy is that the first person stands still and the second one goes forward and back repeatedly, increasing the distance of each leg until they collide.
In the next variation, both are required to execute the same program, that is, if one stands still, then both stand still. To compensate for this increased difficulty, they are allowed to leave their parachutes anywhere. And both of them will recognize the other's parachute if they bump into it.
In the third variation, the set-up is similar to the previous problem, but they are not allowed to change the direction of their movement. To their advantage, they know which way East is.
I recently heard a 2-D version from my son Sergei in which the parachutists are ghosts. That means that when they bump into each other they go through each other without even recognizing the fact that they met:
Several blindfolded men are sleeping at different locations on a plane. Each wakes up, not necessarily at the same time. At the moment of waking up, each of them receives the locations of all the others in relation to himself at that moment. They are not allowed to interact, nor will they receive any further information as time passes. They need to get together in one place. How can they do that, if they are allowed to decide on their strategy in advance?
They do not know where North is. So they can't go to the person at the most Northern point. Also they do not know how locations correspond to people, so they can't all go to where, say, Peter is. Let us consider the case of two men. Suppose they decide to go to the middle of the segment of two locations they receive when they awake. But they get different locations because they wake up at different times. Suppose the first person wakes up and goes to the middle. The fact that he walks while the other is sleeping, means that he changes the middle. So when the second person wakes up, his calculated middle is different from the one calculated by the first person. Consequently, they will never manage to meet. Hence, the solution should be different.
Actually Sergei gave me a more difficult problem:
Not only do they need to meet, but they need to stay together for a predefined finite time period.
Here is as bonus problem.
If there are three parachutists, it is possible to end up in a meeting place and stay there indefinitely. For four people it is often possible too.
Darth Maul killed Qui-Gon Jinn. Obi-Wan Kenobi killed Darth Maul. Palpatine killed Mace Windu. Darth Vader killed Obi-Wan Kenobi and Palpatine. I am mentally drawing the kill graph of Star Wars, where people are vertices and kills are edges. The graph is not very interesting. In movies where no one gets resurrected, the kill graph is a forest.
I'm interested in studying social networks in the movies and how they differ from social networks in real life. As we saw, the kill graph is not very exciting mathematically.
Now let's try the acquaintance graph, where edges mark two people who know each other. Unfortunately, in the movies there are often many nameless people and we learn very little about their acquaintances. On the other hand, all the "nameful" people usually know each other, thus their acquaintance graph is a complete graph. The richest acquaintance graphs would be for epic movies like Star Wars, in which the events span two generations and many planets. As a result, there are characters who never meet each other. For example, Leia, Luke and Han from the original trilogy never meet people who died in the prequel, such as Anakin's mother and Count Dooku.
But I think that the most intriguing type of filmic social network is the fight graph, where edges represent characters who fight each other. Usually such graphs are bipartite, reflecting the division between bad guys and good guys. When an epic film is more complex and has traitors, the fight graph is no longer bipartite. Consider Darth Vader who fought and killed a lot of good guys including Obi-Wan Kenobi as well as many bad guys including Count Dooku and the Emperor.
I would like to immortalize Darth Vader in mathematics. He did restore the balance to the Force. If there is a graph which is not bipartite and can become bipartite by removing one highly connected node, I would like to name such a node Darth Vader.
Dear Nita Palmira,
I do not recall your name and I'm not sure where you got my email address from, but I really appreciate you contacting me. I am excited by your Two-Procedures-For-The-Price-Of-One offer. I am really looking forward to my enlarged penis and my DDD breasts.
Meanwhile, I can give you a unique group discount on IQ tests. I can test the IQ of all your company employees for the price of one test. Moreover, you do not need to waste even a minute. Actually, no one even needs to answer any questions. You can send me your $500 check to the address below and I will promptly send you the IQ report, the accuracy of which I can guarantee.
Sincerely yours.
I am coaching my AMSA students for math competitions. Recently, I gave them the following problem from the 1964 MAML:
The difference of the squares of two odd numbers is always divisible by:
A) 3, B) 5, C) 6, D) 7, E) 8?
The fastest way to solve this problem is to check an example. If we choose 1 and 3 as two odd numbers, we see that the difference of their squares is 8, so the answer must be E. Unfortunately this solution doesn't provide any useful insight; it is just a trivial calculation.
If we remove the choices, the problem immediately becomes more interesting. We can again plug in numbers 1 and 3 to see that the answer must be a factor of 8. But to really solve the problem, we need to do some reasoning. Suppose 2k + 1 is an odd integer. Its square can be written in the form 4n(n+1) + 1, from which you can see that every odd square has remainder 1 when divided by 8. A solution like this is a more profitable investment of your time. You understand what is going on. You master a method for solving many problems of this type. As a bonus, if students remember the conclusion, they can solve the competition problem above instantaneously.
This is why when I am teaching I often remove multiple choices from problems. To solve them, rough estimates and plugging numbers are not enough. To solve the problems the students really need to understand them. Frankly, some of the problems remain boring even if we remove the multiple choices, like this one from the 2009 AMC 10.
One can holds 12 ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?
It's a shame that many math competitions do not reward deep analysis and big-picture understanding. They emphasize speed and accuracy. In such cases, plugging in numbers and rough estimates are useful skills, as I pointed out in my essay Solving Problems with Choices.
In addition, smart guessing can boost the score, but I already wrote about that, too, in How to Boost Your Guessing Accuracy During Tests and To Guess or Not to Guess?, as well as Metasolving AMC 8.
As the AMC 10 fast approaches, I am bracing myself for the necessity to include multiple choices once again, thereby training my students in mindless arithmetic.
Many people ask me when is a good time to teach kids math. In my experience, it can never be too early. You just need to keep some order. Multiplication should be taught after addition, and negative numbers after subtraction. Kids should remember multiplication by heart at the age of seven. They can understand negative numbers as early as four.
In the picture I am explaining Platonic solids to four-month-old Eli, the son of my friends. His homework is to chew on a dodecahedron.
Suppose Romeo is encouraged by love and attention. If Juliet likes him, his feelings for Juliet grow and flourish. If she doesn't like him, he loses his interest in her.
Juliet, on the other hand, is the opposite. If Romeo doesn't like her, she needs to win him over and her attraction for him grows. If he likes her, she feels that her task is accomplished and she loses her interest in him. Juliet likes the challenge more than the relationship.
Steven Strogatz used differential equations to model the dynamics of the relationship between Romeo and Juliet. This is a new and fascinating area of applied mathematical research; you can read more about the roller-coaster relationship between Romeo and Juliet in Steven Strogatz's Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering
Mathematicians like symmetry: in math literature they switch the roles between Romeo and Juliet randomly. So in some papers they give Romeo the role of preferring a challenge over love and in some papers they give that role to Juliet.
When I teach this subject of love, Alexander Pushkin's famous quote always pops into my mind. The quote comes from the first lines of Chapter Four of Eugene Onegin, and in Russian it is:
Чем меньше женщину мы любим,
Тем легче нравимся мы ей…
I didn't like the English translations that I found, so I asked my son Alexey to provide a more literal translation:
The less we love a woman, the more she likes us in return…
I blame Pushkin for my tendency to always pick Juliet as the character who thrives on the challenge, even though men are often assumed to be the chasers. I'd like to ask my readers to comment on these roles: Do you think both genders play these roles equally? If not, then who is more prone to be into the chase?
Let's return to mathematical models. In the original model, the reactions of Romeo and Juliet are a linear function of feelings towards them. I would like to suggest two other roles, in which people react to the absolute value of feelings towards them. They do not care if it is love or hate: they care about intensity.
First, there is the person, like my friend Connie, who feeds on the emotions of other people. She's turned on by guys who love her as well as by guys who hate her. If they're indifferent, she's turned off.
Second, there is the opposite type, like my colleagues George, Joseph, David and many others. They hate emotion and prefer not to be involved. They lose all interest in people who feel strongly about them and they like people who are distant. I know the name for this role: it's a mathematician!
My friend Olga Amosova worked as a molecular biologist at Princeton University. Last time I visited her, we talked about her research.
She told me that she and her group designed a repair for a DNA mutation that is highly localized. "What's the point," I asked her, "of repairing DNA mutation in one cell?"
I was amazed to learn that not only is there a practical use to her research, but that there is something urgent that I myself must do.
There are many diseases that are caused by localized (so called "point") mutations. The most famous one is Sickle-cell disease. In Sickle-cell disease, defective hemoglobin causes erythrocytes to adopt a sickle shape that makes it difficult to pass through blood vessels. It is a very painful and debilitating disease. However, it turns out that the results of the research of Olga and her group could make the lives of people with such mutations much easier.
Stem cells have two amazing abilities. They grow fast and they can be turned into any type of cells in the human body. If the mutation is repaired in just one stem-cell, it can be selected and turned into a blood progenitor cell. These progenitor cells produce erythrocytes that actually transport oxygen. If these repaired cells are added to the patient's blood, they would produce good hemoglobin for half a year. This would improve the patient's quality of life tremendously.
So what do the rest of us learn from Olga's research? That we must save all left-over stem-cells that are produced in childbirth, like the umbilical cord and the placenta. It's not only Sickle-cell, but many other diseases that could benefit from using stem-cells. Research is moving so fast that these frozen stem-cells might become relevant in surprising ways — not only for the child, but also for relatives of the child — like you yourself!
So what's the urgent thing I must do? My son recently got married, so I must finish this post and send it to my son in case they get pregnant.
Mikhail Zhvanetsky is the most prolific and famous Russian humorist. Here are my own translations of some of his best lines.
I love the TV series of Angel
Where do vampires take their energy from? Usually oxygen is the fuel for the muscles of living organisms, but vampires do not breathe. Vampires are not living organisms, and yet they have to get their energy from somewhere.
When you kill a vampire, it turns to dust. If organisms are 60% water, then a 200-pound vampire should generate 80 pounds of dust. So why, in the series, do you get just a little puff of dust whenever someone plunges a stake into a vampire? Plus 120 pounds of water apparently evaporates instantly during staking. Can someone who is less lazy than me please calculate the energy needed to evaporate 120 pounds of water in one second? Because my first reaction is that you would need an explosion, not just one stab with Buffy's stake.
All these unscientific elements do not actually bother me that much. What does bother me are inconsistencies in logic. For example, at the end of Season One of Buffy, Angel refuses to give Buffy CPR, claiming that as a vampire he can't breathe. But then how can Spike and other vampires smoke? If they can smoke that means they are capable of inhaling and exhaling. Not to mention that these vampires talk: wouldn't they need an airflow through their throats to produce sounds?
It would make more sense for the show to state that vampires do not need to breathe, but are nonetheless capable of inhaling and exhaling. So Angel should have given Buffy CPR. It would have created a great plot twist: Angel saves Buffy at the end of Season One, only for her to send him to the hell dimension at the end of Season Two.
Back to breathing. I remember a scene in "Bring On the Night" in which Spike was tortured by Turok-Han holding his head in water. But if Spike can't breathe, why is this torture?
Another thing that bothers me in the series is not related to what happens but to what doesn't happen. For example, vampires do not have reflections. So I don't understand why every vampire-aware person didn't install a mirror on the front door of their house to check for reflections before inviting anyone in.
Also, it looks like producers do not care about backwards compatibility. Later in the series we get to know that vampires are cold. Watch the first season of Buffy with that knowledge. In the very first episode, Darla is holding hands with her victim, but he doesn't notice that she is cold. Later Buffy kisses Angel, before she knows that he is a vampire, and she doesn't notice that he's cold either. Unfortunately, the series also isn't forward compatible. In the second season of Angel in the episode "Disharmony", when we already know that vampires are cold, Harmony is trying to reconnect with Cordelia. They hug and touch each other. Such an experienced demon fighter as Cordelia should have noticed that Harmony is cold and, therefore, dead.
Finally, let's look at Spike in the last season of Angel. Spike is non-corporeal for a part of the season; we see him going through walls and standing in the middle of a desk. Yet, one time we see him sitting on a couch talking to Angel. In addition, he can take the stairs. He can go through the elevator wall to ride in an elevator instead of falling down through its floor. And what about floors? Why isn't he falling through floors? Some friends of mine said that we can assume that floors are made from stronger materials. But, if there is a material that can prevent Spike from penetrating it, they ought to use this material to make a weapon for him.
I've never been involved in making a show, but these producers clearly need help. Perhaps they should hire a mathematician like me with an eye for detail to prevent so many goofs.
The jokes are a rough translation from a Russian collection, except the last one I invented myself.
* * *
— Moishe, do you know how many cuckolds are there in Odessa not counting you?
— What? What do you mean by saying "not counting you"?
— Sorry. Okay then, how many counting you?
* * *
At a very prestigious Russian nursery school a teacher talks to a four-year-old applicant.
"Mike, can you count for me?"
Mike counts very fast and with a lot of enthusiasm, "Fifty-nine, fifty-eight, fifty-seven…"
"Super," says the teacher, "But how did you learn to count backwards?"
Mike replies proudly, "I can heat my own lunch — in the microwave."
* * *
The curl of the curl equals the gradient of the divergence minus the Laplacian. Why do I remember this shit that I never need, but can't remember where I put my keys yesterday?
* * *
In a bike store:
Customer: "Can you show me your finest helmet? I've already spent $200,000 on my head, so I don't want to take any risks."
Clerk, sympathetically: "You had a head trauma?"
Customer: "No, I went to college."
* * *
A topologist walks into a cafe:
— Can I have a doughnut of coffee, please.
In my old essay I presented the following coin problem.
We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weigh the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale. Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.
Here is my solution to this problem. Let us start with small values of N. For one coin you can't do anything. For two coins there isn't much you can do either. I will leave it to the readers to solve this for three coins, while I move on to four coins.
Let us compare two coins against the other two. The weighing has to unbalance. Then put aside the two coins from the right pan and compare one coin from the left pan with the other coin from the left pan. If they balance, then the right pan in the first weighing contained the fake coin. If they are unbalanced then the left pan in the first weighing contained the fake coin. Knowing where the fake coin was in the first weighing gives us the answer.
It is often very useful to go through the easy cases. For this problem we can scale the solution for three and four coins to get a solution for any number of coins that is divisible by three and four by just grouping coins accordingly. Thus we have solutions for 3k and 4k coins.
For any number of coins we can try to merge the solutions above. Divide all coins into three piles of size a, a and b, where a ≤ b ≤ 2a. In the first weighing compare the first two piles. If they balance, then the fake coin must be among the b remaining coins. Now pick any b coins from both pans in the first weighing and compare them to the remaining b coins. If the first weighing is unbalanced, then the remaining coins have to be real. For the second weighing we can pick a coins from the remaining pile and compare them to one of the pans in the first weighing.
The solution I just described doesn't cover the case of N = 5. I leave it to my readers to explain why and to solve the problem for N = 5.
Among ten given coins, some may be real and some may be fake. All real coins weigh the same. All fake coins weigh the same, but have a different weight than real coins. Can you prove or disprove that all ten coins weigh the same in three weighings on a balance scale?
When I first received this puzzle from Ken Fan I thought that he mistyped the number of coins. The solution for eight coins was so easy and natural that I thought that it should be eight — not ten. It appears that I was not the only one who thought so. I heard about a published paper with the conjecture that the best you can do is to prove uniformity for 2n coins in n weighings.
I will leave it to the readers to find a solution for eight coins, as well as for any number of coins less than eight. I'll use my time here to explain the solution for ten coins that my son Sergei Bernstein suggested.
First, in every weighing we need to put the same number of coins in both pans. If the pans are unbalanced, the coins are not uniform; that is, some of them are real and some of them are fake. For this discussion, I will assume that all the weighings are balanced. Let's number all coins from one to ten.
Consider two sets. The first set contains only the first coin and the second set contains the second and the third coins. Suppose the number of fake coins in the first set is a and a could be zero or one. The number of fake coins in the second set is b where b is zero, one or two. In the first weighing compare the first three coins against coins numbered 4, 5, and 6. As they balance the set of coins 4, 5, and 6 has to have exactly a + b fake coins.
In the second weighing compare the remaining four coins 7, 8, 9, and 10 against coins 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among the coins 7, 8, 9, and 10 is 2a + b.
For the last weighing we compare coins 1, 7, 8, 9, and 10 against 2, 3, 4, 5, and 6. The balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. This in turn means that either a = b = 0 and all the coins are real, or that a = 1, and b = 2 and all the coins are fake.
Now that you've solved the problem for eight and less coins and that I've just described a solution for ten coins, can we solve this problem for nine coins? Here is my solution for nine coins. This solution includes ideas of how to use a solution you already know to build a solution for a smaller number of coins.
Take the solution for ten coins and find two coins that are never on the same pan. For example coins 2 and 10. Now everywhere where we need 10, use 2. If we need both of them on different pans, then do not use them at all. The solution becomes:
The first weighing is the same as before with the same conclusion. The set containing the coin 1 has a fake coins, the set containing the coins 2 and 3 has b fake coins and the set containing coins 4, 5, and 6 has to have exactly a + b fake coins.
In the second weighing compare the four coins 7, 8, 9, and 2 against 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among 7, 8, 9, and 2 is 2a + b.
For the last weighing we compare coins 1, 7, 8, and 9 against 3, 4, 5, and 6. If we virtually add the coin number 2 to both pans, the balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. Which in turn means, similar to above, that either all the coins are real or all of them are fake.
It is known (see Kozlov and Vu, Coins and Cones) that you can solve the same problem for 30 coins in four weighings. I've never seen an elementary solution. Can you provide one?
I am starting yet another part-time job as the Head Mentor at PRIMES, a new MIT research program for high schoolers. I am very excited about this program, for it will be valuable not only to kids who want to become researchers, but also to kids who just want to see what research is like. Kids who want to learn to think in a new way will also find it highly useful.
PRIMES is in many ways similar to RSI, which it augments and complements. There are also a lot of differences. Keep in mind that I am only comparing PRIMES to the math part of RSI, with which I was working as a coordinator for two years. I do not know how RSI handles other sciences.
Different time scale. RSI lasts six weeks; PRIMES will take about a year. I already wrote about some peoples' skepticism towards RSI in my piece called "Fast Food Research?." PRIMES creates a more natural pace for research.
Choices. Because of the time schedule at RSI, students get their project as soon as they start. Students who realize by the end of the second week that they do not like their project are at a disadvantage: if they do not change their project, they're stuck with something that does not inspire them or is too difficult, and if they do change their project, they won't have enough time to do a great job. At PRIMES students will have time to talk to the mentors before starting their project, so that they can participate in choosing their project. Depending on how it goes later, they'll have time to try several different directions. I believe that the best research comes from the heart: students who have the time and opportunity to shape their choices will be more invested in their project.
Application process. At RSI, The Center for Excellence in Education reviews the applications. Even though they usually do a superb job at sending us great students, I believe it would be an advantage if mentors were able to influence the review process, for they might find even better matches to their projects. At PRIMES, the mentors will have this opportunity to review the applications.
Geography. RSI accepts students from all over the US and from some other countries. PRIMES can only accept local students — those who live close enough to visit MIT once a week for four months. Because of this restriction, PRIMES is recruiting from a smaller pool of students than RSI. But for local students it means that it will be easier to get accepted to PRIMES than to RSI.
Coaching. At RSI, students get a lot of coaching. I think that every student is in close contact with four adults. Two of them are from the math department — mentor and coordinator (that's me!) — and two tutors from CEE. PRIMES will have less coaching. A student will have a mentor and me, the head mentor. In addition, mentors might arrange for students to talk to the professors who originated their projects.
Immersion. RSI students are physically present. They are housed at MIT with the expectation that they completely devote their time to their research. Students at PRIMES will be integrating their research into the rest of their lives and their commitments. That will require good organizational skills and a lot of self-discipline. RSI students have discipline imposed on them by their situation — which may be an advantage to them.
Olympiads. While they are at RSI, students can't go to IMO or other summer activities. This is why many strong Olympiad students choose not to go to RSI, or they turn down an RSI acceptance if in the meantime they have gotten on to an Olympic team. At PRIMES you can do both. It is possible to go to an Olympiad, in addition to writing a paper.
Grade. RSI students have to be juniors. There are no grade limitations for PRIMES. Thus, it is possible to go to PRIMES in one's senior year. In this case, it may be too late to use PRIMES on college applications, but it is perfectly fine for the sake of research itself. Or it might be possible to go to PRIMES as a sophomore, and then apply for RSI the next year. This will strengthen the student's application for RSI.
RSI is well-established and has proven itself. PRIMES is new and hopefully will offer young mathematicians additional opportunities to try research.
I think that the American system of education creates a lot of pressure for teachers to drill their students for standardized tests and multiple choice questions. This blocks creative thinking. Every program like PRIMES is very good for unleashing students' creativity and contributing to the development of the future thinkers of American society.
Silvio Micali taught me cryptography. To explain one-way functions, he gave the following example of encryption. Alice and Bob procure the same edition of the white pages book for a particular town, say Cambridge. For each letter Alice wants to encrypt, she finds a person in the book whose last name starts with this letter and uses his/her phone number as the encryption of that letter.
To decrypt the message Bob has to read through the whole book to find all the numbers. The decryption will take a lot more time than the encryption. If the book increases in size the time it takes Alice to do the encryption almost doesn't increase, but the decryption process becomes more and more draining.
This example is very good for teaching one-way functions to non-mathematicians. Unfortunately, the technology changes and the example that Micali taught me fifteen years ago isn't so cute anymore. Indeed you can do a reverse look-up online of every phone number in the white pages.
I still use this example, with an assumption that there is no reverse look-up. I recently taught it to my AMSA students. And one of my 8th graders said, "If I were Bob, I would just call all the phone numbers and ask their last names."
In the fifteen years since I've been using this example, this idea never occurred to me. I am very shy so it would never enter my mind to call a stranger and ask for their last name. My student made me realize that my own personality affected my mathematical inventiveness.
Since modern technology is murdering my 15-year-old example, I would like to ask my readers to suggest other simple examples of one-way functions or ways to resurrect the white pages example.
I received a message at the beginning of October: "This month has 5 Fridays, 5 Saturdays and 5 Sundays; this only happens every 823 years."
Wait a minute. The Gregorian calendar cycles every 400 years. Where is the figure of 823 coming from?
Wait another minute. Within a century the calendar repeats itself every 28 years. So we are guaranteed that October 2038 will be the same as October 2010.
Wait one more minute. To have a month with five Fridays, Saturdays and Sundays, we need a month that has 31 days and starts on a Friday. There are seven months a year with 31 days, so on average we would expect to have such a month once a year.
If you study the calendar you can see that the seven long months start on six different days. This means that two of the months start on the same day and one of the days is skipped altogether. We see this in both leap years and non-leap years.
Ironically, 2010 is the year with two long months starting on Friday — October and January. Despite the claims of the email about this only happening every 823 years, in fact the same phenomenon occurred twice this year. The next time this will happen is in July 2011.
For those people who get all excited when a month has five Fridays, five Saturdays and five Sundays, I have good news for you. The month following each of these months has to start on Monday. And unless it is a February of a non-leap year, it will have five Mondays.
I asked my son Alexey Radul what exactly he is doing for his postdoc at the Hamilton Institute in Ireland. Here is his reply:
The short, jargon-loaded version: We are building an optimizing compiler for a programming language with first-class automatic differentiation, and exploring mathematical foundations, connections, applications, etc.
Interpretation of jargon:
Automatic differentiation is a technique for turning a program that computes a function into a program that computes that function together with its derivative; with a constant factor overhead. This is better than the usual symbolic differentiation that, say, Mathematica does because there is no intermediate-expression bulge. For example, if your function is a large product
Product f1(x) f2(x) ... fn(x),
the symbolic derivative has size n2
Sum (Product f1'(x) f2(x) ... fn(x)) (Product f1(x) f2'(x) ... fn(x)) ... (Product f1(x) f2(x) ... fn'(x))
automatic differentiation avoids that cost. Automatic (as opposed to symbolic) differentiation also extends to conditionals, data structures, higher-order functions, and all the other wonderful things that distinguish a computer program from a mathematical expression.
First-class means that the differentiation operations are normal citizens of the programming language. This is not the case with commonly used automatic differentiation systems, which are all preprocessors that rewrite C or Fortran source code. In particular, we want to be able to differentiate any function written in the language, even if it is a derivative of something, or contains a derivative of something, etc. The automatic differentiation technique works but becomes more complicated in the presence of higher order, multivariate, or nested derivatives.
We are building an optimizing compiler because the techniques necessary to get good performance and correct results with completely general automatic differentiation are exactly the techniques used to produce aggressive optimizing compilers for functional languages, so we might as well go all the way.
It appears that the AD trick (or at least half of it) is just an implementation of synthetic differential geometry in the computer. This leads one to hope that a good mathematical foundation can be found that dictates the behavior of the system in all the interesting special cases; there is lots of math to be thought about in the vicinity of this stuff.
Applications are also plentiful. Any time you want to optimize anything with respect to real parameters, gradients help. Any time you are dealing with curves, slopes help. Computer graphics, computer vision, physics simulations, economic and financial models, probabilities — there's so much stuff to apply a high quality such system to that we don't know where to begin.
I usually give a lot of lectures and I never used to announce them in my blog. This time I will give a very accessible lecture at the MIT "Women in Mathematics" series. It will be on Wednesday October 6th at 5:30-6:30 PM in room 2-135. If you are in Boston, feel free to join. Here is the abstract.
I will discuss several coin-weighing puzzles and related research. Here are two examples of such puzzles:
1. Among 10 given coins, some may be real and some may be fake. All real coins weigh the same. All fake coins weigh the same, but have a different weight than real coins. Can you prove or disprove that all ten coins weigh the same in three weighings on a balance scale?
2. Among 100 given coins, four are fake. All real coins weigh the same. All fake coins weigh the same, but they are lighter than real coins. Can you find at least one real coin in two weighings on a balance scale?
You are not expected to come to my talk with the solutions to the above puzzles, but you are expected to know how to find the only fake coin among many real coins in the minimum number of weighings.
In 2009 I was working at MIT coordinating math research for Research Science Institute for high school students. One of our students Jacob Hurwitz got a project on decycling graphs.
"Decycling" means removing vertices of a graph, so that the resulting graph doesn't have cycles. The decycling number of a graph is the smallest number of vertices you need to remove.
Decycling is equivalent to finding induced forests in a graph. The set of vertices of the largest induced forest is a complement to the smallest set of vertices you need to remove for decycling.
Among other things, Jacob found induced trees and forests of the highest densities on graphs of all semi-regular tessellations. On the pictures he provided for this essay, you can see an example of a tessellation, a corresponding densest forest, and a corresponding densest tree. The density of the forest and the tree is 2/3, meaning that 1/3 of the vertices are removed.
To motivate RSI students I tried to come up with practical uses for their projects. When I was talking to Jacob about decycling, the only thing I could think of was terrorists. When terrorists create their cells, they need to limit connections among themselves, in order to limit the damage to everyone else in the cell if one of them gets busted.
That means the graph of connections of a terrorist cell is a tree. Suppose there is a group of people that we suspect, and we know the graph of their contacts, then the decycling number of the graph is the number of people that are guaranteed to be innocent.
Have you noticed how Facebook and LinkedIn are reasonably good at suggesting people you might know? The algorithm they use to analyze the data is fairly effective in revealing potential connections. Recently, someone was able to download all of the Facebook data, which means that any government agency ought to be able to do the same thing. They could analyze such data to discover implicit connections. As a byproduct of looking for terrorists, they would also discover all of our grudges.
Oh dear. What are they going to think when they find out I'm not connected to my exes?
In the name of privacy, I have changed the names of the men I did not marry. But there is no point in changing the names of my ex-husbands, as my readers probably know their names anyway.
I received my first marriage proposal when I was 16. As a person who was unable to say "no" to anything, I accepted it. Luckily, we were not allowed to get married until I was 18, the legal marriage age in the USSR, and by that time we broke up.
To my next proposal, from Sasha, I still couldn't say "no", and ended up marrying him. The fact that I was hoping to divorce him before I got married at 19 shows that I should have devoted more effort in learning to say "no". I decided to divorce him within the first year.
My next proposal came from Andrey, I said yes, with every intention of living with Andrey forever. We married when I was 22 and he divorced me when I was 29.
After I recovered from my second divorce, I had a fling with an old friend, Sam, who was visiting Moscow on his way to immigrate to Israel.
Sam proposed to me in a letter that was sent from the train he took from the USSR to Israel. At that point I realized I had a problem with saying "no". The idea of marrying Sam seemed premature and very risky. I didn't want to say yes. I should have said no, but Sam didn't have a return address, so I didn't say anything.
That same year I received a phone call from Joseph. Joseph was an old friend who lived in the US, and I hadn't seen or heard from him for ten years. He invited me to visit him in the US and then proposed to me the day after my arrival. The idea of marrying Joseph seemed premature and very risky, but in my heart it felt absolutely right. I said yes, and I wanted to say yes.
I was very glad that I hadn't promised anything to Sam. But I felt uncomfortable. So even before I called my mother to notify her of my marriage plans, I located Sam in Israel and called him to tell him that I had accepted a marriage proposal from Joseph. I needed to consent to marry someone else as a way of saying "no" to Sam.
After I married Joseph, I came back to Russia to do all the paperwork and pick up my son, Alexey for our move to the US. There I met Victor. I wasn't flirting with Victor and was completely disinterested. So his proposal came as a total surprise. That was the time I realized that I had a monumental problem with saying "no". I had to say "no" to Victor, but I couldn't force myself to pronounce the word. Here is our dialogue as I remember it:
My sincere attempt at saying "no" didn't work. I moved to the US to live with Joseph and I soon got pregnant. Victor was the first person on my list to notify — another rather roundabout way to reject a proposal.
The marriage lasted eight years. Sometime after I divorced Joseph, I met Evan who invited me on a couple of dates. I wasn't sure I wanted to get involved with him. But he proposed and got my attention. I was single and available, though I had my doubts about him.
Evan mentioned that he had royal blood. So I decided to act like a princess. I gave him a puzzle:
I have two coins that together make 15 cents. One of them is not a nickel. What are my coins?
He didn't solve it. In and of itself, that wouldn't be a reason to reject a guy. But Evan didn't even understand my explanation, despite the fact that he was a systems administrator. A systems administrator who doesn't get logic is a definite turn-off.
So I said "no"! That was my first "no" and I have mathematics to thank.
When the Women and Math program at IAS was coming to an end, Ingrid Daubechies invited me to a picnic at her place for PACM (The Program in Applied and Computational Mathematics at Princeton University). I accepted with great enthusiasm for three reasons: I was awfully tired and needed a rest; PACM was my former workplace, so I was hoping to meet old acquaintances; and most of all, I loved the chance to hang out with Ingrid.
Ingrid is a great cook, so she prepared some amazing deserts for the picnic. While I was helping myself to a second serving of her superb lemon mousse, a man asked me if I was Ingrid's sister. Ingrid overheard this and laughingly told him that we are soul-sisters.
I admire Ingrid, so at first I took this as a compliment and felt all warm and fuzzy. But when my critical reasoning returned, I had to ask myself: Why would someone think I am Ingrid's sister with my Eastern European round face, my Russian name and my Russian accent?
I started talking to the man. He asked me what I do. I told him that I am a mathematician. He was stunned. What is so surprising in meeting a mathematician at a math department picnic?
Now I think I understand what happened. It never occurred to him that I was a mathematician. I was clearly unattached, so he couldn't place me as someone's wife. As the picnic was at Ingrid's house, he must have concluded that I had to be Ingrid's relative. Very logical, but very gender biased.
I heard this problem many years ago when I was working for Math Alive.
Three men are fighting in a truel. Andrew is the worst shot; he misses 2/3 of the time. Bob is better; he misses 1/3 of the time. Connor is the best shot; he always hits. Each of the three men have an infinite number of bullets. Each shot is either a kill or a miss. They have to shoot at each other in order until two of them are dead. To make it more fair they decide to start with Andrew, followed by Bob, and then Connor. We assume that they choose their strategies to maximize their probability of survival. At whom should Andrew aim for his first shot?
This is a beautiful probability puzzle, and I will not spoil it for you by writing a solution. Recently, I watched an episode of Numb3rs: The Fifth Season
Imagine a duel between three people. I'm the worst shot; I hit the target once every three trials. One of my opponents [Charlie] is better; hits it twice every three shots. The third guy [Colby] is a dead shot; he never misses. Each gets one shot. As the worst I go first, then Charlie, then Colby. Who do I aim for for my one shot?
This is a completely different problem; it's no longer about the last man standing. Colby doesn't need to shoot since the other two truelists have already expended their only shots. If Charlie believes that Colby prefers nonviolence, all else being equal, then Charlie doesn't need to shoot. Finally, the same is true of Marshall. There is no point in shooting at all.
To make things more mathematically interesting, let's assume that the truelists are bloodthirsty. That is, if shooting doesn't decrease their survival rate, they will shoot. How do we solve this problem?
If he survives, Colby will kill someone. If Charlie is alive during his turn, he has to shoot Colby because Colby might kill him. What should Marshall do? If Marshall kills Colby, then Charlie misses Marshall (hence Marshall survives) with probability 1/3. If Marshall kills Charlie, then Marshall is guaranteed to be killed by Colby, so Marshall survives with probability 0. If he doesn't kill anyone, things look much better: with probability 2/3, Colby is killed by Charlie and Marshall survives. Even if Colby is alive, he does not necessarily shoot Marshall, so Marshall survives with probability at least 2/3. Overall, Marshall's chances of staying alive are much better if he misses. He should shoot in the air!
The sad part of the story is that Charlie Eppes completely missed. That is, he completely missed the solution. In the episode he suggested that Marshall should shoot Charlie. If Marshall shoots Charlie, he will be guaranteed to die.
It is disappointing that a show about math can't get its math right.
From the 1966 Moscow Math Olympiad:
Prove that you can choose six weights from a set of weights weighing 1, 2, …, 26 grams such that any two subsets of the six have different total weights. Prove that you can't choose seven weights with this property.
Let us define the sequence a(n) to be the largest size of a subset of the set of weights weighing 1, 2, …, n grams such that any subset of it is uniquely determined by its total weight. I hope that you agree with me that a(1) = 1, a(2) = 2, a(3) = 2, a(4) = 3, and a(5) = 3. The next few terms are more difficult to calculate, but if I am not mistaken, a(6) = 3 and a(7) = 4. Can you compute more terms of this sequence?
Let's see what can be said about upper and lower bounds for a(n). If we take weights that are different powers of two, we are guaranteed that any subset is uniquely determined by the total weight. Thus a(n) ≥ log2n. On the other hand, the total weight of a subset has to be a number between 1 and the total weight of all the coins, n(n+1)/2. That means that our set can have no more than n(n+1)/2 subsets. Thus a(n) ≤ log2(n(n+1)/2).
Returning back to the original problem we see that 5 ≤ a(26) ≤ 8. So to solve the original problem you need to find a more interesting set than powers of two and a more interesting counting argument.
Fifteen years ago I attended Silvio Micali's cryptography course. During one of the lectures, he asked me to close my eyes. When I did, he wrote a random sequence of coin flips of length six on the board and invited me to guess it.
I am a teacher at heart, so I imagined a random sequence I would write for my students. Suppose I start with 0. I will not continue with zero, because 00 looks like a constant sequence, which is not random enough. So my next step would be sequence 01. For the next character I wouldn't say zero, because 010 seems to promise a repetitive pattern 010101. So my next step would be 011. After that I do not want to say one, because I will have too many ones. So I would follow up with 0110. I need only two more characters. I do not want to end this with 11, because the result would be periodic, I do not want to end this with 00, because I would have too many zeroes. I do not want to end this with 01, because the sequence 011001 has a symmetry: reversing and negating this sequence produces the same sequence.
During the lecture all these considerations happened in the blink of an eye in my mind. I just said: 011010. I opened my eyes and saw that Micali had written HTTHTH on the board. He was not amused and may even have thought that I was cheating.
Many teachers, when writing a random sequence, do not flip a coin. They choose a sequence that looks "random": it doesn't have too many repetitions and the number of ones and zeroes is balanced (that is, approximately the same). When they write it character by character on the board, they often choose a sequence so that any prefix looks "random" too.
As a result, the sequence they choose stops being random. Actually, they're making a choice from a small set of sequences that look "random". So the fact that I guessed Micali's sequence is not surprising at all.
If you have gone to many math classes, you've seen a lot of professors choosing very similar-looking "random" sequences. This discriminates against sequences that do not look "random". To restore fairness to those under-represented sequences, I have decided that the next time I need a random sequence, I will choose 000000.
Two month ago I made a minor rearrangement of my math humor page. The traffic to that page tripled. Would you like to know what I did? I collected all the suggestive jokes in one chapter and named it Dirty Math Jokes.
Mathematics is so far from sex that stumbling on a math sex joke is always a special treat.
Combinatorialists do it discretely.
When those jokes were randomly placed in my joke file, it was easy to miss flirtatious connotations.
She was only a mathematician's daughter, and she sure learned how to multiply using square roots.
So I decided to collect them together in one place.
Math Problem: A mother is 21 years older than her son. In 6 years she will be 5 times as old as her son. Where is the father?
Sergei Bernstein and Nathan Benjamin brought back a variation of the "Rock, Paper, Scissors" game from the Mathcamp. They call it "Rock, Paper, Scissor." In this variation one of the players is not allowed to play Scissors. The game ends as soon as someone wins a turn.
Can you suggest the best strategy for each player?
They also invented their own variation of the standard "Rock, Paper, Scissors." In their version, players are not allowed to play the same thing twice in a row.
If there is a draw, then it will remain a draw forever. So the game ends when there is a draw. The winner is the person who has more points.
They didn't invent a nice name for their game yet, so I am open to suggestions.
Let me describe a variation of Nim that is at the same time a variation of Chomp. Here's a reminder of what Nim and Chomp are.
In the game of Nim, there are several piles of matches and two players. Each of the players, in turn, can take any number of matches, but those matches must come from the same pile. The person who takes the last match wins. Some people play with a different variation in which the person who takes the last match loses.
Mathematicians do not differentiate between these two versions since the strategy is almost the same in both cases. The classic game of Nim starts with four piles that have 1, 3, 5 and 7 matches. I call this configuration "classic" because it is how Nim was played in one of my favorite movies, "Last Year at Marienbad". Recently that movie was rated Number One by Time Magazine in their list of the Top 10 Movies That Mess with Your Mind.
In the game of Chomp, also played by two people, there is a rectangular chocolate bar consisting of n by m squares, where the lowest left square is poisoned. Each player in turn chooses a particular square of the chocolate bar, and then eats this square as well as all the squares to the right and above. The player who eats the poisonous square loses.
Here is my Nim-Chomp game. It is the game of Nim with an extra condition: the piles are numbered. With every move a player is allowed to take any number of matches from any pile, with one constraint: after each turn the i-th pile can't have fewer matches than the j-th pile if i is bigger than j.
That was a definition of the Nim-Chomp game based on the game of Nim, so to be fair, here is a definition based on the game of Chomp. The game follows the rules of Chomp with one additional constraint: the squares a player eats in a single turn must all be from the same row. In other words, the chosen square shouldn't have a square above it.
The game of Nim is easy and its strategy has been known for many years. On the other hand, the game of Chomp is very difficult. The strategy is only known for 2 by m bars. So I invented the game of Nim-Chomp as a bridge between Nim and Chomp.
More and more I stumble upon the claim that the difference in individual personal choices between men and women is one of the main contributors to the gender gap in mathematical careers. Let me tell you some stories that I've heard that illustrate some choices that women made. The names have been changed.
Ann got her PhD in math at the same time as her husband. They both got job offers at places very far from each other. As Ann was pregnant, she decided not to accept her offer and to follow her husband. In two years she was ready to go back to research and she started to search for some kind of position at the University where her husband worked. At that time there was a nuptial rule that prevented two spouses from working at the same place. There was no other college nearby, and Ann's husband didn't yet have tenure and freaked out at the thought of changing his job. They decided to have a second child. After two more years of staying home, she felt completely disqualified and dropped the idea of research.
Olga was very passionate about her children's education. She felt that public education was insufficient and that parents needed to devote a lot of time to reading and playing with their kids. Her husband insisted that the children would be fine on their own and he refused outright to read to them or to participate in time-consuming activities. Olga took upon herself the burden she was hoping to share. At that same time, she started a tenure track position. In addition to all of this, she took her teaching very seriously. Her students loved her, but her paper-writing speed declined. She didn't get tenure and quit academia. Later, she realized that in fact her husband had shared her sense of the importance of their children's education, but he had played a power-game which left her doing all the work.
Maria told her husband Alex that she wanted a babysitter for their two children now that she was ready to get back to mathematics. Alex sat down with her to look at their finances. Before looking for a job Maria needed to finish her two papers. That meant they had to pay a babysitter even though Maria wasn't bringing in any money. Maria agreed to postpone her comeback until the kids went to kindergarten. She somehow finished her papers and found her first part-time job as an adjunct lecturer. Alex sat down with her again to discuss their finances. They calculated how much time she would need to prepare to teach after such a long break. The babysitter would cost more than her income. In addition, they would have to buy a second car and some professional clothes for Maria. They summed everything up: it appeared that they couldn't afford for her to take this job. Maria rejected the offer.
Two years later, Alex was offered an additional job as a part-time mathematical consultant. Instead of accepting it Maria's husband suggested that the company interview Maria, who was longing to return to work. Maria got an offer, but at half the fee her husband was offered. The manager explained this difference by pointing to her husband's superior work record. Maria and Alex sat down together again and calculated that it would more profitable for them as a couple if he took the job and dropped all his household responsibilities. Maria couldn't find a way to argue.
I recently wrote about my own decision to quit academia twelve years ago. Although what I really wanted to do was to work in academia, my family responsibilities took precedence.
Yes, personal choices are a great contributor to the gender gap in mathematical careers. I just do not like when people assume that women chose freely. Some choices we were cornered into making.
I was somewhat taken aback by the popularity of my earlier essay "Divisibility by 7 is a Walk on a Graph." Tanya tells me it got a good number of hits. The graph in that article is rather crude, and takes a bit of care to use, because the arrows go off in random directions from each node. So taking a hint from a commenter on the first graph, I redrew the graph, sacrificing planarity in favor of ease of use. Specifically, I arranged the black arrows in a counterclockwise circle, which makes them easy to follow.
The graph is used in the same way as the first graph. To find the remainder on dividing a number by 7, start at node 0, for each digit D of the number, move along D black arrows (for digit 0 do not move at all), and as you pass from one digit to the next, move along a single white arrow.
For example, let n = 325. Start at node 0, move along 3 black arrows (to node 3), then 1 white arrow (to node 2), then 2 black arrows (to node 4), then 1 white arrow (to node 5), and finally 5 black arrows (to node 3). Finishing at node 3 shows that the remainder on dividing 325 by 7 is 3.
I fancy it to be a little animal face.
I am a milk person. I can easily drink half a gallon of milk a day. The problem with half a gallon of milk a day is that it is about half of my target calorie intake. That is why I switched to the reduced-fat milk. It didn't taste good, but I was very proud of myself. That is, I was proud for about a year until I finally decided to read the labels. One serving of whole milk is 150 calories, while one serving of reduced-fat milk is 130 calories. All this year-long suffering saved me an insignificant 13%. Not to mention that the amount of sugar is the same.
I decided to look into this more closely. I went to my nearest super-market and checked the milk. The low-fat milk is 110 calories per serving, while non-fat milk is 90 calories.
If I had just reduced my milk intake by half, I would have consumed fewer calories than by replacing whole milk with non-fat milk, but I would have enjoyed it so much more. The lesson? Read the labels and do the math!
Forty years ago, it took about 18 months for us to find the rules that eventually became the Game of Life. We thought in terms of birth rules and death rules. Maybe one day's death rule would be a bit too strong compared to its birth rule. So the next day at coffee time we'd either try to weaken the death rule or strengthen the birth rule, but either way, only by a tiny bit. They had to be extremely well-balanced; if the death rule was even slightly too strong then almost every configuration would die off. And conversely, if the birth rule was even a little bit stronger than the death rule, almost every configuration would grow explosively.
What's wrong with that, you might ask. Well, if the "radius" grows by 1 unit per generation, then after 9 or 10 moves, it's off the (19 by 19) Go board. We can probably find more Go boards, of course, but after another 20 or so moves it will outflow the coffee table and then it is awfully hard to keep track. We wanted to be able to study configurations for much longer than that, which meant that we had to disallow rules that might lead to linear growth. Of course, we weren't interested in rules that usually led to collapse.
Who were "we"? Well, I was the chief culprit and had an aim in mind — to find a simple set of rules that would lead to a system able to simulate a universal computer. Von Neumann had already shown that this was possible, but his system had 29 states and a very complicated set of rules. The rest of "us" were mostly graduate students who had no higher aim than amusing themselves. Every now and then some rather older colleagues or visitors took an interest.
So my plan was, first, to find a set of rules that almost always prevented explosive growth and catastrophic collapse. Second, I wanted to study it long enough to learn how it could be "programmed". I hoped to find a system whose rules were much simpler than Von Neumann's, preferably with only two states (on and off) per cell, rather than his 29.
I'll just describe the last few rule fiddles. We had in fact given up on finding a two-state system, in favor of one with three states: 0, A, B. State 0 represented an empty cell, and it was natural to think of A and B as two sexes, but we only found their proper names when Martin Huxley walked by and said, "Actresses and bishops!"
Perhaps I should explain this. There is a British anecdote that starts like this:
"The actress sat on the left side of the bed, and removed her stockings. The bishop, on the right side of the bed, removed his gaiters. Then she unbuttoned her blouse and he took off his shirt…"
You are supposed to be getting excited, but it all ends quite tamely, because it turns out that the bishop was in his palace, while the actress was in her bedsit near the theater. There are lots of stories in England about the actress and the bishop, and if a person says something that has a salacious double meaning, it's standard to respond "as the actress said to the bishop," or "as the bishop said to the actress".
Okay, back to Life! To inhibit explosive growth, we decided to imitate biology by letting death be a consequence of either overcrowding or isolation. The population would only grow if the number of neighbors was neither too large nor too small. Rather surprisingly, this turned out to mean that children had to have three or more parents. Let's see why. If two parents could give birth, then in the figure below, the parents A and B, who are on the border of the population, would produce children A' and B' at the next time step, followed by grandchildren A" and B" and so on, thus giving us linear growth!
So we moved to threesomes. Children were born to three parents, made up of both sexes. Moreover, the sex of the child was determined by the sex of its parents — two bishops and one actress would give birth to a little actress, while two actresses and one bishop would produce a tiny bishop. This was "the weaker-sex birth rule," and it was accompanied by "the sexual frustration death rule," which made death the punishment for not touching somebody of the opposite sex!
However, the weaker-sex birth rule lived up to its name, by being weaker than the death rule. Remember we weren't interested in rules that led to disappointingly swift collapses, as the actress said to the bishop. Therefore, we strengthened the birth rule by allowing same-sex conception, but again by applying the weaker-sex rule — so that three actresses would produce a bishop or three bishops an actress. However this strengthened the birth rule too much, causing us to apply the death penalty more often.
We decided to apply the death penalty to those who weren't touching at least two other people, whatever their sex. At first sight it was not obvious that this was stronger than the sexual frustration rule, but in fact it was, because the weaker-sex rule ensured that the sexes were fairly evenly mixed, so if you were touching at least two other people, there was a good chance that one of them would be of the opposite sex.
According to our new set of rules, the sex of parents played no role except to determine the sex of the children, so we abolished sex. After all, according to the bishop, Life without sex is much cleaner.
This is now called the Game of Life and these rules, at last, turned out to be clean and well-balanced.
The article Daring to Discuss Women in Science by John Tierney in the New York Times on June 7, 2010 purports to present a dispassionate scientific defense of Larry Summers's claims, in particular by reviewing and expanding his argument that observed differences in the length of the extreme right tail of the bell curves of men's and women's test scores indicate real differences in their innate ability. But in fact any argument like this has to acknowledge a serious difficulty: it is problematic to assume without comment that the abilities of a group can be inferred from the tail of a bell curve. We are so used to invoking bell curves to talk about group abilities, we don't notice that such arguments usually use only the mean of the curve. Using the tail is a totally different story.
Think about it: it is reasonable to question whether a single data point — the test score of an individual person — is a true indication of his/her ability. It might not be. Maybe a single test score represents a dunce with hyper-overachieving parents who push him to study all the time. So does that single false reading destroy the validity of the curve? No of course not: because some other kid might have been a super-genius who was drunk last night and can barely keep his eyes open during the test. One is testing above his "true ability" and the other is testing below his "true ability," and the effect cancels out. Thus the means of curves are a good way to measure the ability of large groups, because all the random false readings average out.
But tails are not. On the tail this "canceling out" effect doesn't work. Look at the extreme right tail. The relatively slow but hyper-motivated kids are not canceled out by the hoard of far-above-the-mean super geniuses who had drunken revels the night before. There just aren't that many super-geniuses and they just don't party that much.
Or let's look at it another way: imagine that you had a large group which you divided in half totally at random. At this point their bell curve of test scores looks exactly the same. Lets call one of the group "boys" and the other group "girls". But they are two utterly randomly selected groups. Now lets inject the "boys" with a chemical that gives the ones who are very good already a burning desire to dominate any contest they enter into. And let us inject the "girls" with a chemical that makes the ones who are already good nonetheless unwilling to make anyone feel bad by making themselves look too good. What will happen to the two bell curves? Of course the upper tail of the "boys'" curve will stretch out, while the "girls'" tail will shrink in. It will look like the "boys" whipped the "girls" on the right tail of ability hands down, no contest. But the tail has nothing to do with ability. Remember they started out with the same distribution of abilities, before they got their injections. It is only the effect of the chemicals on motivation that makes it look like the "boys" beat the "girls" at the tail.
So, when you see different tails, you can't automatically conclude that this is caused by difference in underlying innate ability. It is possible that other factors are at play — especially since if we were looking to identify these hypothetical chemicals we might find obvious candidates like "testosterone" and "estrogen".
The possibility of alternative explanations for these findings calls into question Tierney and Summers' claims to superior dispassionate scientific objectivity. Moving from the mean to the tail of a bell curve makes systematic effects on averages irrelevant, true, but it is instead susceptible to systematic effects on deviations, which are irrelevant at the mean. An argument that uses this trick to dodge gender differences in averages cannot claim the mantle of scientific responsibility without accounting for gender differences in deviations. I am deeply disappointed that Tierney and Summers did not accompany their assertions with a suitable reminder of this fact.
I recently published a puzzle about wizards, hats of different colors and rooms. Unfortunately, I was too succinct in my description and didn't explicitly mention several assumptions. Although such assumptions are usual in this type of puzzle, I realize now, from your responses, that I should have listed them and I apologize.
The Sultan decided to test the wisdom of his wizards. He collected them together and gave them a task. Tomorrow at noon he will put hats of different colors on each of the wizard's heads.
The wizards have a list of the available colors. There are enough hats of each color for every wizard. The wizards also have a list of rooms. There are enough rooms to assign a different room for every color.
Tomorrow as the Sultan puts hats on the wizards, they will be able to see the colors of the hats of the other wizards, but not the color of their own. Without communicating with each other, each wizard has to choose a room. The challenge comes when two wizards have the same hat color, for they must choose the same room. On the other hand, if they have different hat colors, they must choose different rooms. Wizards have one day to decide on their strategy. If they do not all complete their task, then all of their heads will be chopped off. What strategy would you suggest for the wizards?
In his comment on the first blog about this problem, JBL beautifully described the intended solution for the finite number of wizards, and any potentially-infinite number of colors. I do not want to repeat his full solution here. I would rather describe his solution for two wizards and two colors.
Suppose the colors are red and blue. The wizards will designate one of the rooms as red and another as blue. As soon as each wizard sees the other wizard's hat color, he chooses the room of the color he sees. The beauty of this solution is that if the colors of hats are different, the color of the rooms will not match the color of the corresponding hats: the blue-hatted wizard will go to the red room and vice versa. But the Sultan's condition would still be fulfilled.
JBL's solution doesn't work if the number of wizards is infinite. I know the solution in that case, but I do not like it because it gives more power to the axiom of choice than I am comfortable with. If you are interested, you can extrapolate the solution from my essay on Countable Wise Men with Hats which offers a similar solution to a slightly different problem.
The International Math Olympiad started in Eastern Europe in 1959. Romania was the first host country. The Olympiad grew and only in 1976 did it move outside the Eastern bloc. The competition was held in Austria.
I was on the Soviet team in 1975 and 1976, so I was able to compare competitions held in Eastern vs. Western countries. Of course, the Austrian Olympiad was much better supported financially, but today I want to write about the differences in how our team was prepped.
Before our travel to Austria the Soviet team members were gathered in a room with strangers in suits for a chat. I assumed that we were talking to the KGB. They gave us a series of instructions. For example, they told us not to leave the campus during the competition, to always walk in groups, and to avoid talking to kids from countries that are enemies of the USSR. They warned us that they would be watching, and I was scared to death.
Now that I am older and wiser, I understand that their goal was to frighten us. Our team traveled with adult supervisors, who were trusted by the KGB. But for several days during the grading period of the competition, our supervisors were not allowed to see us. So the KGB wanted us to be too afraid to be very adventurous when we were left on our own.
In addition, the KGB had a Jewish problem. In general, Jews were not allowed to go abroad. I had many Jewish friends who qualified for the pre-IMO math camp where the team was chosen, but who were not able to get on the IMO because of delays with their travel documents. Some local bureaucrats were eager to impress the KGB and therefore held up visas for Jewish students, preventing them from being on the team. But the team selection process itself wasn't yet corrupt in 1976. So every year despite the efforts of the system, some young Jewish mathematicians would end up on the team.
Before 1976, the Olympiad was in the Eastern bloc, so the KGB wasn't quite so concerned about having Jewish members on the team. But Austria was not only a Western country, it was also the transition point for Jewish refugees from the Soviet Union. The speed with which the IMO moved their competition to a Western country was much faster than the Soviet bureaucratic machine could build a mechanism for completely preventing Jews from joining the team.
One very strong candidate, Yura Pass, didn't get his documents, but two other Jewish boys made it on to the team that was going to Austria. They were joking that they would be the only Soviet Jews who would go to Austria and actually come back. They did come back, only to go forward later: both are now math professors working in the US.
Because we had Jewish members on our team, it gave the KGB a special extra reason to scare us. But the biggest pressure was to win. We were told that 1976 was the most important year for the Soviet team to be the best. We were told that capitalist countries spread rumors that the judges in Eastern bloc countries favored the Soviet team and that the relative success of the Soviet team throughout the years had not been fully deserved. Now that the competition was in Austria, the suits told us, the enemies of the USSR were hoping for the downfall of the Soviet team. Our task was to prove once and for all that the Soviet students were the best at math, and that the rumors were unfounded. We had to win the team competition not only to prove ourselves, but also to clear the name of the Soviet team for all the previous years.
We did have a very strong team. The USSR came out first with 250 points, followed by the UK with 214 points and the USA with 188 points. Out of nine gold medals, we took four.
We could have gotten one more gold medal if Yura Pass had been allowed on the team. Yura was crushed by the machine's treatment of Jews and soon afterwards quit mathematics.
I just received a mass email on how to live longer and it made these points:
Tip 1. Delay your retirement. Studies show that people who retire at 65 live longer than people who retire at 60.
Tip 2. Sex makes you younger. Studies show that older people who have sex twice a week look ten years younger than their peers who do not have sex at all.
People who draw conclusions from such studies usually do not understand statistics. Correlation doesn't mean causality. Let me use the above-mentioned studies to reach different conclusions by reversing the causality assumption of the unknown writer of the mass email. You can compare results and make your own decisions.
Case 1. Studies show that people who retire at 65 live longer than people who retire at 60. Reversed causality: People who live longer are healthier, so they are able to keep working and to retire later in life.
Case 2. Studies show that older people who have sex twice a week look ten years younger than their peers who do not have sex at all. Reversed causality: Older people who look ten years younger than their peers can get laid easier, so they have sex more often.
Sergei just came back from MOP — Mathematical Olympiad Summer Program, where he was a grader. The first question I asked him was, "What was your favorite math problem there?" Here it is:
There are wisemen, hats and rooms. Hats are of different color and there are enough hats of each color for every wisemen. There are enough rooms, so that you can assign a different room for every color. At some moment in time the sultan puts hats on the wisemen's heads, so as usual they see all other hats, but do not see their own hat color. At the same time, each wiseman has to choose a room to go to. If two wisemen have the same hat color, they should go to the same room. If they have different hat colors, they should go to different rooms. What strategy should the wisemen decide upon before this event takes place?
Oh, I forgot to mention the most interesting part of this problem is that you shouldn't assume that the number of wisemen or hats or rooms is finite. You should just assume that they have the power of choice.
My son Alexey Radul and I were discussing the Tuesday's child puzzle:
You run into an old friend. He has two children, but you do not know their genders. He says, "I have a son born on a Tuesday." What is the probability that his second child is also a son?
Here is a letter he wrote me on the subject. I liked it because unlike many other discussions, Alexey not only asserts that different interpretations of the conditions in the puzzle form different mathematical problems, but also measures how different they are.
If you assume that boys and girls are symmetric, and that days of the week are symmetric (and if you have no information to the contrary, assuming anything else would be sheer presumption on your part), then you can be in one of at least two states.
1) You say that "at least one son born on a Tuesday" is all the information you have, in which case your distribution including this information is uniform over consistent cases, in which case your answer is 13/27 boy and your information entropy is
− ∑27 (1/27) log(1/27) = − log(1/27) = 3.2958.
2) You say that the information you have is "The guy might have said
any true thing of the form 'I have at least one {boy/girl} born on a
{day of the week}', and he said 'boy', 'Tuesday'." This is a
different mathematical problem with a different solution. The
solution: By a symmetry argument (see below [*]) we must assign uniform
probability of him making any true statement in any particular
situation. Then we proceed by Bayes' Rule: the statement we heard is
d, and for each possible collection of children
− ∑ p log p = − 1/14 log 1/14 − 26/28 log 1/28 = 3.2827
Therefore that assumed additional structure really is more information, which is only present at best implicitly in the original problem. How much more information? The difference in entropies is 3.2958 - 3.2827 = 0.0131 nats (a nat is to a bit what the natural log is to the binary log). How much information is that? Well, the best I can do is to reproduce an argument of E.T. Jaynes', which may or may not really apply to this situation. Suppose you have some repeatable experiment with some discrete set of possible outcomes, and suppose you assign probabilities to those outcomes. Then the number of ways those probabilities can be realized as frequencies counted over N trials is proportional to eNH, where H is the entropy of the distribution in nats. Which means that the ratio by which one distribution is easier to realize is approximately eN(H1-H2). In the case of N = 1000 and H1 - H2 = 0.0131, that's circa 5x105. For each way to get a 1000-trial experiment to agree with version 2, there are half a million ways to get a 1000-trial experiment to agree with version 1. So that's a nontrivial assumption.
[*] The symmetry argument: We are faced with the following probability assignment problem
Suppose our subject's first child is a boy born on a Tuesday, and his second child is a girl born on a Friday. What probability must we assign to him asserting that he has at least one boy born on a Tuesday?
Good question. Let's transform our coordinates: Let Tuesday' be Friday, Friday' be Tuesday, boy' be girl, girl' be boy, first' be second and second' be first. Then our problem becomes
Suppose our subject's second' child is a girl' born on a Friday', and his first' child is a boy' born on a Tuesday'. What probability must we assign to him asserting that he has at least one girl' born on a Friday'?
Our transformation necessitates p(boy Tuesday) = p(girl' Friday'), and likewise p(girl Friday) = p(boy' Tuesday'). But our state of complete ignorance about what's going on with respect to the man's attitudes about boys, girls, Tuesdays, Fridays, and first and second children has the symmetry that question and question' are the same question, and must, by the desideratum of consistency, have the same answer. Therefore p(boy Tuesday) = p(boy' Tuesday') = p(girl Friday) = 1/2.
My classmate Misha gave me a math problem. Although I liked the math part, I hated the setup. Here is the problem using the new setup:
You are hoping to get very rich one day and you base your hopes on your new invention: a diet pill. The pill works beautifully and doesn't have side effects. Patients simply take a pill when they get hungry. Within one hour they will fall asleep and will be unable to awake for exactly two hours, at which time they will awake by themselves feeling completely sated.
You have just arrived at your lab, when you realize that one of your interns has misplaced your bottle of wonder pills. After some investigation, you come to the conclusion that your bottle is on the shelf with 239 other bottles that contain a placebo. Unfortunately, those placebo bottles were custom-designed for your statistical tests to exactly match your medicine bottle.
You need to bring the medicine bottle to your boss in two hours. While you panic, all your five interns volunteer for experiments, hoping to be mentioned in your paper. Can you find your bottle in two hours?
The problem Misha gave me had 240 barrels of wine, one of which contained deadly poison and five slaves who could be spared.
I do not like killing people even when they are imaginary. But while I was slow in inventing my own setup, the original version of the puzzle started making the rounds on the Internet. So I decided to kill the problem by writing the solution.
The strategy is to give out some pills immediately, wait for one hour and see who falls asleep. The next step is to give some other pills at the beginning of the next hour to some of the interns who are awake.
Let's count the information you can get out of this. Each intern will experience one of three different situations: falling asleep in the first hour, in the second hour, or not falling asleep at all. Thus, you can have a total of 35 = 243 different outcomes.
If you had more bottles than 243, there would be no way to distinguish between them. The fact that you have 240 bottles might mean that 243 will work too, but apparently the designer of the puzzle didn't want to hint into powers of three and picked the largest round number below 243. These considerations should increase your willingness to look into this problem base 3.
Let us label the bottles with different 5-character strings containing three characters 0, 1, and 2. Now we can use the label as instructions. The first character will be associated with the first intern and so on. Suppose the fourth character on the bottle's label is 0, then the fourth intern doesn't need to struggle with digesting a pill from this particular bottle. If the fourth character is 1, then the fourth intern gets the pill in the first hour. If the fourth character is 2, the fourth intern gets the pill in the second hour.
Note this minor detail: Suppose the fourth character on a bottle is 2, but the fourth intern is asleep by the second hour. That means, the bottle doesn't contain the medicine, and we can put it aside.
At the end of two hours you know who fell asleep and when. This data will exactly match the label on the bottle with the medicine.
Once I read a book in Russian that mentioned a study of the children of Soviet military personnel who had to move often. The conclusion was that frequent relocation is very damaging for children's psyche. The children had to build new friendships, which they would lose the next time they had to move. After several moves they would stop making friends; later, as adults, they would be afraid of getting close to anyone.
In September 1996, my husband, my two children and I came to Princeton from Israel for my husband's month-long visit to the Institute for Advanced Study. After the visit we were supposed to go back to Israel, but that didn't happen. My husband returned alone and I stayed in Princeton with my children. That's a long and sentimental story for another time.
Meanwhile, my older son Alexey started going to Princeton High School. By this time he had attended seven schools in three different countries. In light of the evidence presented in that book about the impact on children of moving, I felt very guilty. Alexey was entering 10th grade. Moving him again not only would further damage his ability to make friends, but would also screw up his college chances. He needed a stable environment leading up to college. For example, recommendation letters are better written by people who are involved with kids for several years. I was afraid to mess up his future. I promised myself not to move him again during high school, especially as Princeton High School was one of the best public schools in New Jersey.
At the same time, I got a Visiting Scholar position at Princeton University. Although it didn't pay me any salary, through that position I received university housing, library privileges and an office. I was living on my personal savings and the monthly check my husband was sending me from Israel. My money was running out and I felt completely lost, like so many immigrants. I was new to Princeton; I didn't have friends there; and I was struggling with English. On top of that, I had medical problems, not the least very low energy.
Ingrid Daubechies noticed me at the Princeton math department and approached me. After our conversation, she found some money for me to work on the Math Alive course she was designing. That work was a breath of fresh air. I enjoyed it tremendously, but the part-time salary was not enough. Then I received more help from Ingrid. She appreciated my work on Math Alive a lot, but realized that I needed a different solution. She sacrificed her own interests and started recommending me around. She arranged an interview for me at Telcordia, who offered me a job as a systems engineer.
The decision to accept this job was very painful, because I did not want to leave academia. However, considering that my priority was to keep Alexey in Princeton High School, I didn't feel I had other options. I knew that I couldn't stay much longer at Princeton University and I was aware that getting a University job often requires relocation.
Looking back, I think the reasons behind this decision were more complex than sacrificing my career for my child. If I had known more about social supports for poor families and about other possible research jobs, or if I had been more confident in my research abilities, I might not have left academics.
Alexey triumphed at Princeton High School. The school allowed him to take math courses at Princeton University. He took several, including the course in logic by John Conway and two courses in graph theory by Paul Seymour. Alexey's multi-variable calculus professor complained to me that she couldn't fit her grades into the required curve. If she gave Alexey 100%, the others would have to get less than 20. Luckily, it turned out that because her class was small, she didn't need to bother about making a curve. After three years in Princeton High School, Alexey secured an impressive resume and great recommendation letters and went to MIT to pursue a double major in mathematics and computer science.
Alexey translated from a Russian joke site:
American scientists finally developed a car that runs on water. Unfortunately, at the moment it only runs on water from the Gulf of Mexico.
I've heard about many mathematicians running polymath projects through their blogs. I wasn't planning to do that. It just happened. In this essay, I describe the collaborative effort that was made to solve the following problem that appeared in my blog on July 2009:
Baron Münchhausen has n identical-looking coins weighing 1, 2, …, n grams. The Baron's guests know that he has this set of coins, but do not know which one is which. The Baron knows which coin is which and wants to demonstrate to his guests that he is right. He plans to conduct weighings on a balance scale, so that the guests will be convinced about the weight of every of coin. What is the smallest number of weighings that the Baron must do in order to reveal the weights?
The sequence a(n) of the minimal number of weighings is called the Baron Münchhausen's omni-sequence to distinguish it from the Existential Baron's sequence where he needs the smallest amount of weighings to prove the weight of one coin of his choosing.
In this essay I will describe efforts to calculate a(n). The contributors are: Max Alekseyev, Ilya Bogdanov, Maxim Kalenkov, Konstantin Knop, Joel Lewis and Alexey Radul.
The sequence starts as a(1) = 0, because there is nothing to demonstrate. Next, a(2) = 1, since with only one weighing you can find which coin is lighter.
Next, a(3) = 2. Indeed you can't prove all the coins in one weighing, but in the first weighing you can show that the 1-gram coin is lighter than the 2-gram coin. In the second weighing you can show that the 2-gram coin is lighter than the 3-gram coin. Thus, in two weighings you can establish an order of weights and prove the weight of all three coins.
As you can see in the case of n = 3, you can compare coins in order and prove the weight of all the coins in n − 1 weighings. But this is not at all the optimal number. Let us see why a(4) = 2. In his first weighing the Baron can put the 1- and the 2-gram coins on the left pan of the balance and the 4-gram coin on the right pan. In the future, I will just describe that weighing as 1 + 2 < 4. This way everyone agrees that the coin on the right pan is 4 grams, and the coin that is left out is 3 grams. The only thing that is left to do is to compare the 1-gram and the 2-gram coins in the second weighing.
Later Konstantin Knop sent me a different solution for n=4. His solution provides an interesting example. While looking for solutions, people usually try to have an unbalanced weighing to be "tight". That is, they make it so that the heavier cup is exactly 1 gram heavier than the lighter cup. If you are trying to prove one coin in one weighing, "tightness" is a requirement. But it is not necessary when you have several weighings. Here is the first weighing in Konstantin's solution: 1 + 3 = 4; and his second second weighing is: 1 + 2 < 3 + 4. We see that the second weighing has a weight difference of four between pans.
Next, a(5) = 2. We can have the first weighing the same as before: 1 + 2 < 4, and the second weighing: 1 + 4 = 5. The second weighing confirms that the heavy coin on the right pan in the first weighing can't be the heaviest one, thus it has to be the 4-gram coin. After that you can see that every coin is identified.
Next, a(6) = 2. The first weighing, 1 + 2 + 3 = 6, divides all coins into three groups: {1,2,3}, {4,5} and {6}. We know to which group each coin belongs, but we do not know which coin in the group is which. The second weighing: 1 + 6 < 3 + 5, identifies every coin. Indeed, the only possibility for the left side to weigh less than the right side is when the smallest weighing coin from the first group and 6 are on the left, and the two largest weighing coins from the first two groups are on the right.
When I was writing my essay I suspected that n = 6 is the largest number for which a solution can be established in two weighings, but I didn't have any proof. So I was embarrassed to show my solutions of three weighings n equals 7, 8 and 9.
On the other hand I published the solutions suggested by my son, Alexey Radul, for n = 10 and n = 11. In these cases the theoretical lower bound of log3(n) for a(n) is equal to 3, and finding solutions in three weighings was enough to establish the value of the sequence a(n) for n = 10 and n = 11.
So, a(10) = 3, and here are the weighings. The first weighing is 1 + 2 + 3 + 4 = 10. After this weighing, we can divide the coins into three groups {1,2,3,4}, {5,6,7,8,9} and {10}. The second weighing is 1 + 5 + 10 < 8 + 9. After the second weighing we can divide all coins into groups we know they belong to: {1}, {2,3,4}, {5}, {6,7}, {8,9} and {10}. The last weighing contains the lowest weighing coin from each non-single-coin group on the left and the largest weighing coin on the right, plus, in order to balance them, the coins whose weights we know. The last weighing is 2 + 6 + 8 + 5 = 4 + 7 + 9 + 1.
Similarly, a(11) = 3, and the weighings are: 1 + 2 + 3 + 4 < 11; 1 + 2 + 5 + 11 = 9 + 10; 6 + 9 + 1 + 3 = 8 + 4 + 2 + 5.
After publishing my blog I wrote a letter to the Sequence Fans mailing list asking them to expand the sequence. Max Alekseyev replied with the results of an exhaustive search program he wrote. First of all, he found a counter-intuitive solution for n=6. Namely, the following two weighings: 1 + 3 < 5 and 1 + 2+ 5 < 3 + 6. He also confirmed that it is not possible to identify the coins in two weighings for n=7, n=8 and n=9.
So now I can stop being embarrassed and proudly present my solution for n=7 in three weighings. That is, a(7) = 3 and the first weighing is: 1+2+3 < 7, and it divides all the coins into three groups {1,2,3}, {4,5,6} and 7. The second weighing, 1 + 4 < 6, divides them even further. Now we know the identity of every coin except the group {2,3}, which we can disambiguate with the third weighing: 2 < 3.
In many solutions that I've seen, one of the weighings was very special: every coin on one cup was lighter than every coin on the other cup. I wondered if that was always the case. Konstantin Knop send me a counterexample for n=7. The first weighing is: 1 + 2 + 3 + 5 = 4 + 7. The second is: 1 + 2 + 4 < 3 + 5. The third is: 1 + 3 + 4 = 2 + 6.
Later Max Alekseyev sent me two more special solutions for n=7. The first one contains only equalities: 2 + 5 = 7; 1 + 2 + 4 = 7; 1 + 2 + 3 + 5 = 4 + 7. The second one contains only inequalities: 1 + 3 < 5; 1 + 2 + 5 < 3 + 6; 5 + 6 < 2 + 3 + 7.
Moving to the next index, a(8) = 3 and the first weighing is: 1 + 2 + 3 + 4 + 5 < 7 + 8. The second weighing is: 1 + 2 + 5 < 4 + 6. After that we have identified all coins but two groups {1,2} and {3,4} that can be resolved by 2 + 4 = 6.
Meanwhile my blog received a comment from Konstantin Knop who claimed that he found solutions in three weighings for n in the range between 12 and 17 inclusive and four weighings for n = 53. I had already corresponded with Konstantin and knew that his claims are always well-founded, so I didn't doubt that he had found the solutions.
Later I began to write a paper with Joel Lewis on the upper bound of the omni-sequence, where we prove that a(n) ≤ 2 ⌈log2n⌉. For this paper, we wanted a comprehensive set of examples, so I emailed Konstantin asking him to write up his solutions. He promptly sent me the results and mentioned that he had found the weighings together with Ilya Bogdanov. They used several different ideas in the solutions. First I'll describe their solutions based on ideas we've already seen, namely to compare the lightest coins in the range to the heaviest coins.
Here is the proof that a(13) = 3. The first weighing is: 1 + … + 8 = 11 + 12 + 13, and it identifies the groups {1, 2, 3, 4, 5, 6, 7, 8}, {9, 10} and {11, 12, 13}. The second weighing is: (1 + 2 + 3) + 9 + (11 + 12) = (7 + 8) + 10 + 13, and it divides them further into groups {1, 2, 3}, {4, 5, 6}, {7, 8}, {9}, {10}, {11, 12}, {13}. And the last weighing identifies all the coins: 1 + 4 + 7 + 11 + 9 + 10 = 3 + 6 + 8 + 12 + 13.
Similarly, let us show that a(15) = 3. The first weighing is: 1 + … + 7 < 14 + 15, and it divides the coins into three groups {1, 2, 3, 4, 5, 6, 7}, {8, 9, 10, 11, 12, 13}, and {14, 15}. The second weighing is: (1 + 2 + 3) + 8 + (14 + 15) = (5 + 6 + 7) + (12 + 13), and this divides them further into groups {1, 2, 3}, {4}, {5, 6, 7}, {8}, {9, 10, 11}, {12, 13} and {14, 15}. The third weighing identifies every coin: 1 + 5 + 8 + 9 + 12 + 14 = 3 + 7 + 11 + 13 + 15.
As I mentioned earlier it is not always possible to find the first weighing which will nicely divide the coins into groups. We already discussed an example, n = 5, in which neither of the two weighings divided the coins into groups. Likewise, the same thing happened in the second mysterious solution for n = 6. What these solutions have in common is that the first weighing nearly divides everything nicely. The left pan is almost the set of the lightest coins and the right pan is almost the set of the heaviest coins. But not quite.
That is not our only situation in which the first weighing does not quite divide the coins into groups. For example, here is Konstantin's solution for a(9) = 3. For the first weighing, we put five coins on the left pan and two coins on the right pan. The left pan is lighter. This could happen in three different ways:
The second weighing, 1 + 2 + 3 = 6, in which we took three coins from the left pan and balanced them against one coin – again from the left pan – could only happen in case "C." After the two weighings, the following groups were identified: {1, 2, 3}, {4}, {5, 7}, {6}, {8, 9}. The third weighing, 1 + 4 + 5 + 8 < 3 + 7 + 9, identifies all the coins.
A similar technique is used in the solution that Konstantin sent to us to demonstrate that a(12) = 3. The first weighing is: 1 + 2 + 3 + 4 + 5 + 6 < 10 + 12. The audience which sees the results of the weighings understands that there are three possibilities for the distribution of coins:
The second weighing, (1 + 2 + 3) + (7 + 8) + 10 < (9 + 11 ) + 12, convinces the audience that the left pan must weigh at least 31 if the first weighing was case "A" above (31 = 1 + 2 + 3 + 7 + 8 + 10) or "C" (31 = 1 + 2 + 3 + 6 + 8 + 11), and at least 32 (32 = 1 + 2 + 3 + 7 + 8 + 11) if the first weighing was case "B." At the same time the right pan is not more than 12 + 9 + 11 = 32 for case "A" above, not more than 12 + 9 + 10 = 31 for case "B" and not more than 12 + 9 + 10 = 31 for case "C."
Hence the inequality in the second weighing is only possible when the first weighing was indeed as described by case "A" above. Consequently, the first two weighings together identify groups: {1, 2, 3}, {4, 5, 6}, {7, 8}, {9}, {10}, {11} and {12}. The third weighing, 1 + 4 + 7 + 11 + 12 < 3 + 6 + 8 + 9 + 10, identifies all the coins.
Other cases that Konstantin Knop sent me used a completely different technique. I would like to explain this technique using the mysterious solution for n = 6 found by Max Alekseyev. Suppose we have six coins labeled c1, … c6. The first weighing is: c1 + c3 < c5. The second weighing is: c1 + c2 + c5 < c3 + c6.
Let us prove that these two weighings identify all the coins. Let us replace the two inequalities above with the following: c1 + c3 − c5 ≤ −1, and c1 + c2 + c5 − c3 − c6 ≤ −1. Now we multiply the first inequality by 3 and the second by 2 and sum the results. We get: 5c1 + 2c2 + c3 + 0c4 − c5 − 2c6 ≤ −5. Note that the coefficients for labels are in a decreasing order. By the rearrangement inequality the smallest value the expression 5c1 + 2c2 + c3 + 0c4 − c5 − 2c6 reaches is when the labels on the coins match the indices. This smallest value is −5. Hence, the labels have to match the coins.
The technique that Konstantin and his collaborators are using is to search for appropriate coefficients to multiply the weighings by, rather than searching for the weighings themselves. In lieu of lengthy explanations, I will just list the weighings that he uses together with coefficients to multiply them by for their proof that the weighings differentiate coins.
We will start with showing that a(14) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 < 11 + 13 + 14, and: 1 + 2 + 3 + 8 + 11 + 13 = 7 + 9 + 10 + 12, followed by 1 + 4 + 7 + 10 = 3 + 6 + 13. The coefficients to multiply by are {9, 5, 2}.
Next we will show that a(16) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 8 < 14 + 16, and 1 + 2 + 3 + 7 + 9 + 14 = 8 + 13 + 15, followed by 1 + 4 + 7 + 10 + 13 < 3 + 6 + 12 + 15. The coefficients to multiply by are {11, 5, 2}.
Next we will show that a(17) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 < 15 + 16 + 17, and 1 + 2 + 3 + 8 + 11 + 15 + 16 < 7 + 9 + 10 + 14 + 17, and 1 + 4 + 7 + 8 + 12 + 14 = 3 + 6 + 10 + 11 + 16. The coefficients to multiply by are {11, 5, 2}.
Next we will show that a(53) = 4. The weighings are: (1 + 2 + … + 23) + 25 < 47 + (49 + … + 53), and (1 + … + 9) + 24 + (26+ … + 31) + 47 + (49 + … + 52) < (16 + … + 23) + 25 + (41 + … + 46) + 48 + (51 + 52), and (1 + 2 + 3) + (10 + 11) + (16 + 17 + 18) + 24 + (26 + 27) + (32 + 33 + 34) + (41 + 42 + 43) + 47 + 49 + 53 =(7 + 8 + 9) + 15 + (22 + 23) + 25 + (30 + 31) + (38 + 39) + 40 + (45 + 46) + 48 + (51 + 52), and the last one 1 + 4 + 7 + 10 + 12 + 16 + 19 + 22 + 24 + 28 + 30 + 32 + 35 + 38 + 41 + 45 + 47 + 51 + 53 < 3 + 6 + 9 + 11 + 14 + 18 + 21 + 25 + 27 + 29 + 34 + 37 + 40 + 43 + 48 + 49 + 50 + 52. The coefficients to multiply by are {43, 15, 5, 2}.
When I was working on the paper with Joel Lewis I re-established my email discussions about the Baron's onmi-sequence with Konstantin Knop. At that time Konstantin's colleague, Maxim Kalenkov, got interested in the subject and wrote a computer search program to find other solutions that can be proven with the rearrangement inequality. Thus, we know two more terms of this sequence.
The next known term is a(18) = 3. The weighings are: 1 + 2 + 4 + 5 + 7 + 10 + 12 = 9 + 15 + 17, and 1 + 3 + 4 + 6 + 9 + 11 + 17 = 7 + 12 + 14 + 18, and 2 + 3 + 7 + 8 + 9 + 14 + 15 = 4 + 10 + 11 + 16 + 17. The corresponding coefficients are: {8, 7, 5}.
Similarly, a(19) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 7 + 8 + 10 + 13 = 16 + 18 + 19, and 1 + 2 + 3 + 6 + 9 + 11 + 16 = 8 + 10 + 13 + 17, and 1 + 4 + 6 + 8 + 12 + 18 = 3 + 7 + 11 + 13 + 15. The coefficients are {12, 7, 3}.
Maxim Kalenkov continued his search. He didn't find any new solutions in three weighings, but he found a lot of solutions in four weighings, namely for numbers from 20 to 58. Below are his solutions, with multiplier coefficients in front of every weighing:
a(20) ≤ 4
18: 1+2+3+4+5+10+14+16+18 = 6+7+11+12+17+20
19: 1+2+4+5+12+15+17+19 < 6+8+10+14+18+20
21: 2+6+11+17+18 = 4+9+10+15+16
26: 1+6+7+8+9+10+20 = 2+5+17+18+19
a(21) ≤ 4
18: 3+5+6+11+15+17+19 = 7+8+9+13+18+21
19: 4+6+9+13+16+18+20 = 1+7+11+12+15+19+21
21: 1+4+5+7+12+18+19 = 3+9+10+11+16+17
26: 1+2+3+7+8+9+10+11+21 = 4+5+6+18+19+20
a(22) ≤ 4
18: 3+5+6+12+15+17+20 = 7+8+10+13+18+22
19: 1+2+4+10+13+16+18+21 = 7+9+12+15+20+22
21: 1+6+7+18+19+20 = 2+3+10+11+12+16+17
26: 2+3+7+8+9+10+11+12+22 = 6+18+19+20+21
a(22) ≤ 4
18: 1+2+5+6+12+16+18+22 < 3+7+8+10+13+19+23
19: 1+3+4+6+10+17+19+21 = 7+9+12+14+16+23
21: 3+7+13+14+19+20 = 2+6+10+11+12+17+18
26: 2+7+8+9+10+11+12+23 = 19+20+21+22
a(24) ≤ 4
18: 1+3+6+12+17+19+21+23 < 2+4+7+8+10+13+15+20+24
19: 1+2+4+5+6+10+15+18+20+22 < 7+9+12+14+17+21+24
21: 4+7+13+14+20+21 = 3+6+10+11+12+18+19
26: 2+3+7+8+9+10+11+12+24 = 20+21+22+23
a(25) ≤ 4
18: 1+2+3+5+6+8+9+14+17+19+22 < 10+12+16+20+24+25
19: 2+6+7+9+12+16+18+20+23 = 10+11+14+15+17+22+24
21: 1+2+4+7+8+10+15+20+21+22 = 3+6+12+13+14+18+19+25
26: 3+10+11+12+13+14+24+25 = 2+7+8+9+20+21+22+23
a(26) ≤ 4
18: 1+2+3+5+6+8+9+14+18+20+23 = 10+12+15+21+25+26
19: 2+3+4+6+7+9+12+19+21+24 = 11+14+16+18+23+25
21: 7+8+15+16+21+22+23 = 2+6+12+13+14+19+20+26
26: 1+2+10+11+12+13+14+25+26 = 7+8+9+21+22+23+24
a(27) ≤ 4
18: 1+3+4+6+7+8+9+14+19+21+23+25 < 10+11+13+15+17+22+26+27
19: 1+2+3+5+7+9+13+17+20+22+24 < 4+10+12+14+16+19+23+26
21: 2+3+4+8+10+15+16+22+23 = 1+7+13+14+20+21+27
26: 1+10+11+12+13+14+26+27 = 3+8+9+22+23+24+25
a(28) = 4
18: 3+6+8+9+10+15+19+21+24+26 = 1+5+11+13+16+18+22+27+28
19: 1+4+5+7+9+10+13+18+20+22+25 = 3+6+11+12+15+17+19+24+27
21: 5+6+11+16+17+22+23+24 = 4+9+13+14+15+20+21+28
26: 1+2+3+4+11+12+13+14+15+27+28 = 10+22+23+24+25+26
a(29) = 4
18: 1+3+5+6+7+9+10+16+20+22+25+27 < 11+12+14+17+18+23+28+29
19: 4+8+10+14+18+21+23+26 = 2+3+6+11+13+16+20+25+28
21: 1+2+6+8+9+11+17+23+24+25 = 4+5+14+15+16+21+22+29
26: 2+3+4+5+11+12+13+14+15+16+28+29 = 8+9+10+23+24+25+26+27
a(30) = 4
18: 2+8+10+16+21+23+26+27+28 = 5+11+12+14+17+19+24+29+30
19: 2+4+5+7+9+14+19+22+24+28 = 11+13+16+18+21+26+29
21: 1+5+6+9+10+11+17+18+24+25+26 = 4+14+15+16+22+23+28+30
26: 1+3+4+11+12+13+14+15+16+29+30 < 9+10+24+25+26+27+28
a(31) = 4
18: 1+2+6+9+10+16+21+23+26+28+29 < 3+4+7+11+12+14+17+19+24+30+31
19: 1+2+4+7+14+19+22+24+27+29 < 6+9+11+13+16+18+21+26+30
21: 2+3+7+8+9+11+17+18+24+25+26 = 14+15+16+22+23+29+31
26: 1+3+4+5+6+11+12+13+14+15+16+30+31 = 2+24+25+26+27+28+29
a(32) = 4
18: 1+5+8+9+10+12+13+22+24+27+29 = 6+14+16+18+20+25+30+31
19: 4+6+10+11+13+16+20+23+25+28 = 1+2+8+15+19+22+27+30+32
21: 1+2+6+7+8+11+12+18+19+25+26+27 = 4+5+10+16+17+23+24+31+32
26: 1+2+3+4+5+14+15+16+17+30+31+32 < 11+12+13+25+26+27+28+29
a(33) = 4
18: 1+2+6+7+8+9+10+12+13+23+27+29+30 = 3+14+15+17+19+21+25+31+32
19: 1+2+10+11+13+17+21+24+25+28+30 = 4+6+8+14+16+20+23+27+31+33
21: 1+3+4+8+11+12+14+19+20+25+26+27 < 7+10+17+18+24+30+32+33
26: 3+4+5+6+7+14+15+16+17+18+31+32+33 = 11+12+13+25+26+27+28+29+30
a(34) = 4
18: 2+3+4+8+10+12+13+19+24+28+30+31 < 6+14+15+17+20+22+26+32+33
19: 1+2+3+5+6+9+11+13+17+22+25+26+29+31 = 4+8+14+16+19+21+24+28+32+34
21: 1+3+6+7+8+11+12+14+20+21+26+27+28 = 2+5+17+18+19+25+31+33+34
26: 1+2+4+5+14+15+16+17+18+19+32+33+34 = 3+11+12+13+26+27+28+29+30+31
a(35) = 4
18: 1+3+4+6+8+10+11+13+19+24+26+29+31+32 = 14+15+17+20+22+27+33+34+35
19: 1+4+7+9+11+12+17+22+25+27+30+32 = 6+14+16+19+21+24+29+33+35
21: 2+12+13+14+20+21+27+28+29+35 = 4+7+8+11+17+18+19+25+26+32+34
26: 1+2+3+4+5+6+7+8+14+15+16+17+18+19+33+34 = 12+13+27+28+29+30+31+32
a(36) = 4
18: 1+2+3+4+5+9+11+12+14+15+20+25+29+31+32 < 6+16+18+21+23+27+33+34+36
19: 1+2+4+5+6+8+10+12+13+15+18+23+26+27+30+32 < 16+17+20+22+25+29+33+35+36
21: 1+3+5+13+14+16+21+22+27+28+29+36 = 2+8+9+12+18+19+20+26+32+34+35
26: 2+6+7+8+9+16+17+18+19+20+33+34+35 = 5+13+14+15+27+28+29+30+31+32
a(37) = 4
18: 1+2+3+6+8+10+12+13+15+16+21+27+30+32+33 = 4+9+17+19+22+24+28+34+35+37
19: 1+2+3+7+9+11+13+14+16+19+24+26+28+31+33 = 5+6+10+17+18+21+23+30+34+36+37
21: 3+4+5+9+10+14+15+17+22+23+28+29+30+37 = 1+7+8+13+19+20+21+26+27+33+35+36
26: 1+4+5+6+7+8+17+18+19+20+21+34+35+36 = 3+14+15+16+28+29+30+31+32+33
a(38) = 4
18: 2+3+4+7+9+11+13+15+16+21+26+28+31+33+34 = 5+10+17+18+19+22+24+29+35+36+38
19: 1+3+4+8+10+12+13+14+16+19+24+27+29+32+34 = 7+11+17+21+23+26+31+35+37+38
21: 1+2+4+5+10+11+14+15+17+22+23+29+30+31+38 = 8+9+13+19+20+21+27+28+34+36+37
26: 5+6+7+8+9+17+18+19+20+21+35+36+37 = 4+14+15+16+29+30+31+32+33+34
a(39) = 4
18: 1+2+4+7+9+11+13+14+16+22+27+29+32+34+35 = 10+17+18+20+23+25+30+36+38+39
19: 2+3+4+8+10+12+14+15+20+25+28+30+33+35 = 5+7+11+17+19+22+24+27+32+37+38
21: 1+2+3+5+10+11+15+16+17+23+24+30+31+32+38 < 8+9+14+20+21+22+28+29+35+36+37
26: 1+5+6+7+8+9+17+18+19+20+21+22+36+37 = 15+16+30+31+32+33+34+35
a(40) = 4
18: 1+2+3+4+5+8+9+12+14+15+17+18+23+27+29+32+34+35 < 7+10+19+21+24+26+30+36+37+39+40
19: 1+3+4+5+7+10+13+15+16+18+21+26+28+30+33+35 < 6+8+12+20+23+25+27+32+36+38+39
21: 1+5+6+10+11+12+16+17+24+25+30+31+32+39 < 3+9+15+21+22+23+28+29+35+37+38
26: 2+3+6+7+8+9+19+20+21+22+23+36+37+38 = 5+16+17+18+30+31+32+33+34+35
a(41) = 4
18: 1+3+5+6+8+10+12+14+15+17+18+23+28+30+33+35+36 < 7+11+19+21+24+26+31+37+38+40+41
19: 1+2+4+5+6+7+9+11+13+15+16+18+21+26+29+31+34+36 = 8+12+19+20+23+25+28+33+37+39+40
21: 1+4+6+11+12+16+17+19+24+25+31+32+33+40 < 9+10+15+21+22+23+29+30+36+38+39
26: 2+3+7+8+9+10+19+20+21+22+23+37+38+39 = 6+16+17+18+31+32+33+34+35+36
a(42) = 4
18: 1+3+5+6+7+11+13+15+16+18+24+29+31+34+36+37 = 2+8+12+19+20+22+25+27+32+38+40+41
19: 1+2+4+6+7+10+12+14+16+17+18+22+27+30+32+35+37 = 3+13+19+21+24+26+29+34+39+40+42
21: 1+2+3+4+5+7+8+12+13+17+19+25+26+32+33+34+40 = 10+11+16+22+23+24+30+31+37+38+39
26: 2+3+8+9+10+11+19+20+21+22+23+24+38+39 = 7+17+18+32+33+34+35+36+37
a(43) = 4
18: 1+2+5+6+7+10+12+14+16+17+19+20+29+31+34+36+37 = 8+21+23+25+27+32+38+39+41+42
19: 2+4+5+6+7+8+11+15+17+18+20+23+27+30+32+35+37 = 10+14+22+26+29+34+38+40+41+43
21: 1+2+3+7+13+14+18+19+25+26+32+33+34+41 < 5+11+12+17+23+24+30+31+37+39+40
26: 1+3+4+5+8+9+10+11+12+21+22+23+24+38+39+40 < 7+18+19+20+32+33+34+35+36+37
a(44) = 4
18: 1+2+5+6+7+8+10+11+14+16+17+19+20+25+30+32+35+37+38 = 3+12+21+22+23+26+28+33+39+40+42+44
19: 1+2+3+7+8+12+15+17+18+20+23+28+31+33+36+38+44 = 9+10+14+21+25+27+30+35+39+41+42+43
21: 2+3+4+6+8+9+12+13+14+18+19+21+26+27+33+34+35+42 = 11+17+23+24+25+31+32+38+40+41+44
26: 1+3+4+5+9+10+11+21+22+23+24+25+39+40+41 = 8+18+19+20+33+34+35+36+37+38
a(45) = 4
18: 2+4+5+6+11+13+16+17+18+20+21+26+31+33+36+38+39 = 7+9+14+22+24+27+29+34+40+42+43+45
19: 1+2+4+6+9+12+14+18+19+21+24+29+32+34+37+39+45 = 8+11+16+23+26+28+31+36+40+41+42+44
21: 1+2+3+5+7+8+14+15+16+19+20+27+28+34+35+36+42 = 4+12+13+18+24+25+26+32+33+39+41+45
26: 1+3+4+7+8+9+10+11+12+13+22+23+24+25+26+40+41 = 19+20+21+34+35+36+37+38+39
a(46) = 4
18: 1+2+3+5+6+8+9+14+16+17+19+21+22+26+31+33+36+38+39 = 10+12+23+27+29+34+40+41+42+43+45
19: 2+4+6+7+8+9+12+15+18+20+22+29+32+34+37+39+45 = 3+11+14+17+23+24+26+28+31+36+40+42+44
21: 1+3+7+9+10+11+17+20+21+23+27+28+34+35+36+42 = 6+15+16+25+26+32+33+39+41+45+46
26: 1+2+3+4+5+6+10+11+12+13+14+15+16+23+24+25+26+40+41 = 9+20+21+22+34+35+36+37+38+39
a(47) = 4
18: 2+3+5+7+8+9+13+14+16+18+19+21+22+27+32+34+37+39+40 < 11+23+24+28+30+35+41+42+43+44+46
19: 1+3+6+7+8+9+11+17+19+20+22+30+33+35+38+40+46 < 5+10+13+16+23+25+27+29+32+37+41+43+45
21: 1+2+4+5+9+10+15+16+20+21+23+28+29+35+36+37+43 < 7+14+19+26+27+33+34+40+42+46+47
26: 1+2+3+4+5+6+7+10+11+12+13+14+23+24+25+26+27+41+42 < 9+20+21+22+35+36+37+38+39+40
a(48) = 4
18: 3+5+6+7+8+13+17+19+20+22+23+28+33+35+38+40+41 < 9+11+15+24+26+29+31+36+42+44+45+47
19: 1+4+6+8+11+14+15+18+20+21+23+26+31+34+36+39+41+47 < 3+10+13+17+24+25+28+30+33+38+42+43+44+46
21: 1+2+3+7+8+9+10+15+16+17+21+22+24+29+30+36+37+38+44 = 6+14+20+26+27+28+34+35+41+43+47+48
26: 1+2+3+4+5+6+9+10+11+12+13+14+24+25+26+27+28+42+43 = 8+21+22+23+36+37+38+39+40+41
a(49) = 4
18: 2+3+6+7+8+9+10+15+18+20+21+23+24+29+34+38+40+41+49 < 12+16+25+26+30+32+36+42+43+44+45+47
19: 1+3+5+7+9+10+12+14+16+19+21+22+24+32+35+36+39+41+47 < 11+18+25+27+29+31+34+38+42+44+46+49
21: 1+2+3+4+8+10+11+16+17+18+22+23+25+30+31+36+37+38+44 < 7+14+15+21+28+29+35+41+43+47+48+49
26: 1+2+4+5+6+7+11+12+13+14+15+25+26+27+28+29+42+43 = 10+22+23+24+36+37+38+39+40+41
a(50) = 4
18: 1+2+3+4+6+7+9+10+11+15+16+19+20+21+23+24+34+36+39+41+42+50 = 12+17+25+26+28+30+32+37+43+44+45+46+48
19: 2+3+5+7+8+10+11+17+21+22+24+28+32+35+37+40+42+48 = 4+13+15+19+25+27+31+34+39+43+45+47+50
21: 1+3+4+8+9+11+12+13+17+18+19+22+23+25+30+31+37+38+39+45 = 7+16+21+28+29+35+36+42+44+48+49+50
26: 1+2+4+5+6+7+12+13+14+15+16+25+26+27+28+29+43+44 = 11+22+23+24+37+38+39+40+41+42
a(51) = 4
18: 2+3+4+6+8+10+11+15+17+19+20+21+23+24+30+35+37+40+42+43+50 < 12+13+18+25+26+28+31+33+38+44+46+47+49+51
19: 1+3+4+7+9+11+13+16+18+21+22+24+28+33+36+38+41+43+49 < 6+15+19+25+27+30+32+35+40+45+46+48+50
21: 1+2+4+5+6+9+10+11+12+18+19+22+23+25+31+32+38+39+40+46+51 < 16+17+21+28+29+30+36+37+43+44+45+49+50
26: 1+2+3+5+6+7+8+12+13+14+15+16+17+25+26+27+28+29+30+44+45 < 11+22+23+24+38+39+40+41+42+43+51
a(52) = 4
18: 2+5+7+8+10+11+12+17+19+22+25+26+31+35+37+40+42+43+51 = 3+13+15+20+27+28+29+32+38+44+46+47+49+52
19: 1+2+3+6+8+9+11+12+15+18+20+23+24+26+29+36+38+41+43+49 = 5+14+17+22+27+31+33+35+40+45+46+48+51
21: 1+3+4+5+9+10+12+13+14+20+21+22+24+25+27+32+33+38+39+40+46+52 = 8+18+19+29+30+31+36+37+43+44+45+49+50+51
26: 1+2+3+4+5+6+7+8+13+14+15+16+17+18+19+27+28+29+30+31+44+45 = 12+24+25+26+38+39+40+41+42+43+52
a(53) = 4
18: 2+3+4+7+8+9+10+11+12+17+19+21+23+25+26+31+36+38+41+43+44+52 < 5+13+15+27+29+32+34+39+45+46+47+48+50+53
19: 1+3+4+5+9+11+12+15+18+22+23+24+26+29+34+37+39+42+44+50 = 7+14+17+21+27+28+31+33+36+41+45+47+49+52
21: 1+2+4+5+6+7+10+12+13+14+20+21+24+25+27+32+33+39+40+41+47+53 < 9+18+19+23+29+30+31+37+38+44+46+50+51+52
26: 1+2+3+5+6+7+8+9+13+14+15+16+17+18+19+27+28+29+30+31+45+46 = 12+24+25+26+39+40+41+42+43+44+53
a(54) = 4
18: 2+3+4+6+8+9+11+12+13+18+20+23+25+26+28+29+33+37+39+41+42+43+51 = 5+14+16+21+30+31+32+35+44+45+47+48+49+52+54
19: 1+3+4+5+7+9+10+12+13+16+19+21+24+26+27+29+32+35+38+43+49+54 < 6+15+18+23+30+33+34+37+41+44+46+47+51+53
21: 1+2+4+5+6+10+11+13+14+15+21+22+23+27+28+30+34+40+41+47+52+53 < 9+19+20+26+32+33+38+39+43+45+46+49+50+51
26: 1+2+3+5+6+7+8+9+14+15+16+17+18+19+20+30+31+32+33+44+45+46 < 13+27+28+29+40+41+42+43+52+53+54
a(55) = 4
18: 2+3+6+8+9+11+12+13+18+19+22+24+25+27+28+32+37+39+42+44+45+54 < 4+14+16+20+29+31+33+35+40+46+47+49+50+52+55
19: 1+2+3+4+7+9+10+12+13+16+20+23+25+26+28+31+35+38+40+43+45+52 < 6+15+18+22+30+32+34+37+42+46+48+49+51+54
21: 1+3+4+5+6+10+11+13+14+15+20+21+22+26+27+33+34+40+41+42+49+55 = 9+19+25+31+32+38+39+45+47+48+52+53+54
26: 1+2+4+5+6+7+8+9+14+15+16+17+18+19+29+30+31+32+46+47+48 = 13+26+27+28+40+41+42+43+44+45+55
a(56) = 4
18: 2+3+4+7+9+10+12+13+14+18+20+23+25+26+28+34+39+41+44+46+47+55 < 5+15+16+21+29+30+32+35+37+42+48+50+52+53+56
19: 1+3+4+5+8+10+11+13+14+16+19+21+24+26+27+28+32+37+40+42+45+47+52 < 7+18+23+29+31+34+36+39+44+49+51+54+55+56
21: 1+2+4+5+6+7+11+12+14+15+21+22+23+27+29+35+36+42+43+44+53+54 < 10+19+20+26+32+33+34+40+41+47+48+49+52+56
26: 1+2+3+5+6+7+8+9+10+15+16+17+18+19+20+29+30+31+32+33+34+48+49+56 = 14+27+28+42+43+44+45+46+47+53+54+55
a(57) = 4
18: 2+3+4+7+9+10+12+14+18+19+22+24+27+32+35+36+39+43+45+49+51 < 5+13+16+21+26+28+31+34+38+41+42+48+50+53+56
21: 1+3+4+5+8+10+11+14+16+18+20+21+25+29+31+35+38+40+41+46+51+53+57 < 7+15+19+24+26+30+36+37+39+42+45+47+48+52+55+56
25: 1+2+4+5+6+7+11+12+13+15+18+21+23+24+26+29+32+34+37+41+44+45+48+55 = 10+20+22+31+33+36+40+43+47+51+53+54+56+57
33: 1+2+3+5+6+7+8+9+10+13+15+16+17+19+20+22+26+28+30+31+33+36+42+47+56 = 18+29+32+35+41+44+45+46+49+51+55+57
a(58) = 4
17: 2+3+4+8+9+10+12+13+14+21+22+25+26+27+29+30+37+42+43+45+46+47+56 < 5+15+16+20+31+32+33+35+36+41+48+49+51+52+53+55
20: 1+3+4+5+7+10+11+13+14+16+19+20+24+27+28+30+33+36+40+41+44+47+53 = 8+17+21+25+31+37+38+42+45+48+50+51+56+57
21: 1+2+4+5+6+8+11+12+14+15+17+20+23+25+28+29+31+35+38+41+45+51+55+57 < 10+19+22+27+33+34+37+40+43+47+49+50+53+54+56
26: 1+2+3+5+6+7+8+9+10+15+16+17+18+19+21+22+31+32+33+34+37+48+49+50 < 14+28+29+30+41+44+45+46+47+55+57+58
You might ask why this piece is titled George Hart, when the only man in the photo on the left is John Conway. George Hart is related to this picture in three different ways.
First, this picture is of the math department common room at Princeton University. It was taken during a joint event of the WaM and SWIM programs in June, 2009. It shows the Zometool workshop conducted by George Hart that resulted in the construction of the expanded 120-cell, which appears in the photo's foreground.
The second connection to George Hart is that beautiful shiny object under the lights on the far left. The object is the propello-octahedron sculpture that George Hart created out of 150 CDs. The sculpture has been in the common room for many years, and I have always loved it.
Unfortunately, the sculpture was slowly degrading, even losing some of its parts. I visited Princeton in August 2008 and realized that the sculpture was facing a short life expectancy, so I took the picture of it that is below. I couldn't find any angle to shoot the photo that hid the lost pieces. The sculpture survived until my visit in June 2009, as evidenced by the first picture. But unfortunately it wasn't there any more during my last visit in May 2010.
Oops, I almost forgot. I promised you a third way in which George Hart relates to the first picture. He is the one who took it.
My coauthor Konstantin Knop publishes cute math problems in his blog (in Russian). Recently he posted a coin weighing problem that was given at the 2010 Euler math Olympiad in Russia to eighth graders. The author of the problem is Alexander Shapovalov.
Among 100 coins exactly 4 are fake. All genuine coins weigh the same; all fake coins, too. A fake coin is lighter than a genuine coin. How would we find at least one genuine coin using two weighings on a balance scale?
It is conceivable that your two weighings may find more than one genuine coin. The more difficult question that Konstantin and his commentators discuss is the maximum number of genuine coins you can guarantee to identify in two weighings. Konstantin and the others propose 14 as the answer, but do not have a proof yet.
I wonder if one of you can find a bigger number than Konstantin or alternatively a proof that indeed 14 is the largest possible.
You might ask, considering the title of this piece, why I think that coins are scary. No, I am not afraid of coins. It scares me that this problem was given to eighth graders in Russia, because I cannot imagine that it would be given to kids that age in the USA.
By the way, ten eighth grade students in Russia solved this problem during the competition.
I recently wrote two pieces about the puzzle relating to sons born on a Tuesday: A Son born on Tuesday and Sons and Tuesdays. I also posted a beautiful essay on the subject by Peter Winkler: Conditional Probability and "He Said, She Said". Here is the problem:
You run into an old friend. He has two children, but you do not know what their gender is. He says, "I have a son born on a Tuesday." What is the probability that his second child is also a son?
A side note. My son Alexey explained to me that I made an English mistake in the problem in those previous posts. It is better to say "born on a Tuesday" than "born on Tuesday." I apologize.
Despite this error, I was gratified to hear from a number of people who told me that I had converted them from their solution to my solution. To ensure that the conversion is substantial, I've created a new version of the puzzle on which my readers can test out their new-found understanding. Here it is:
You run into an old friend. He has two children, but you do not know what their gender is. He says, "I have a son born on a Tuesday." What is the probability that his second child is born on a Wednesday?
I was terribly shy when I was a teenager. I worked on this problem and overcame it. But when I moved to the US my shyness returned in a strange form. I was fine around Russians but shy around Americans. At first I assumed that it was a language problem.
I became friends with a Russian sexologist and psychotherapist. He pointed out that I never initiated a conversation with Americans and so I realized that my shyness had returned. He prescribed an exercise for me: I had to invite a new American guy to lunch once a week.
Why guys? Maybe because he was a sexologist or maybe because my problems with self-esteem were more pronounced when I was around men. In any case, I decided to do the exercise.
To paint the full picture I need to add some relevant details. At that time I was married, although I didn't wear a ring, and wasn't especially interested in other men. The reason I didn't wear a ring was that Joseph, my husband at the time, did not himself want to wear a ring. As I love symmetry in relationships more than I love rings, I didn't wear one either.
The men I was about to invite to lunch were mere acquaintances, because I had not yet made any American friends. So although I didn't intend to hide it, they may not have realized that I was married.
Two things surprised me in this exercise. First, it was very easy. Most people agreed to do lunch with me.
Second, every man I invited mentioned his girlfriend. This was unexpected. From my experience with Russians, I anticipated that every man would hide his involvement with someone else, even with a wife, at least for some time. At the very least, many Russian men would try to flirt.
The Americans were different. Unclear why I had invited them out, they wanted to be upfront with me from the start, just in case I was interested in them. Since that experience, I admire the way that American men come clean.
I never invited any of these guys out twice: I just needed a supply of new men for my exercise in overcoming my shyness. I wonder if they thought I was put off by their confessions. Perhaps my loss of interest in them after the first lunch confirmed their suspicions that I was attracted to them.
The sexologist's exercise was a success. Today I have no trouble inviting someone to lunch.
John Conway has a T-shirt with his theorem on it. I couldn't miss this picture opportunity and persuaded John to pose for pictures with his back to me. Here is the theorem:
If you continue the sides of a triangle beyond every vertex at the distances equaling to the length of the opposite side, the resulting six points lie on a circle, which is called Conway's circle.
Poor John Conway had to stand with his back to me until I figured out the proof of the theorem and realized which point must be the center of Conway's circle.
Heard somewhere:
Teacher: What's bigger: 22/7 or 3.14?
Student: They are equal.
Teacher: Why do you say that?!
Student: They are both equal to π.
For the last three years I've been coming to the Institute for Advanced Study in Princeton every spring for the Women and Mathematics program. Every year I am assigned to an office in the main building: Fuld Hall.
The problem is that there is a different office that I crave. Every year I go and check on it over in Simonyi Hall, where the Mathematics Department is located. This year I took this photo of the empty name-tag, hoping that one day it will say Tanya Khovanova.
Credit cards often keep your mother's maiden name in their database for security purposes. This so called "security" is based on two assumptions:
Were these assumptions true, only your close relatives would know your mother's maiden name. In reality, if your mother was never married, then your last name is the same as your mother's last name. So, crooks who are trying to steal identities can try to use your last name as your mother's presumed maiden name. Very often they will succeed. Besides, many women do not change their last names. If you have a different last name from your mother, but your mother uses her maiden name, then the bank's security question is not very secure at all.
If you want your identity to be secure you might need to invent a maiden name for your mother. Alternatively, perhaps your parents can tell you a family secret that will help you choose a name that is related to you, but not transparent to the public.
My relative Martin took his wife's last name after their marriage. Before his children apply for credits cards and bank accounts, he needs to explain to them that it is better for them to use his maiden name as their mother's maiden name for banking purposes.
I was driving on MassPike when, for no apparent reason, a car driving in the opposite direction started flashing its headlights. I remembered the Russian tradition of informing the oncoming traffic that the police are nearby. So I adjusted my speed and very soon I saw a police car. I got this warm feeling in my heart because I didn't need to panic or check my speedometer. I mentally thanked that anonymous Russian driver and started wondering why the tradition had not been adopted in the USA. Is it because we are so responsible that we want to punish speeders, or do we think that the police are on our side?
The following problem was sent to me by Joel Lewis.
You have 12 mice, one of which eats faster than all the others. You need to find it. You have a supply of standard cupcakes that you value very much and want to minimize how many of them you have to use. The only way you can find the mouse is to give cupcakes to several groups of mice and see which group is the fastest.
We assume that mice chew at a constant speed and all the mice in one group can attack the cake at the same time. I love this puzzle because I love coin problems. Let me restate the puzzle as a coin problem:
You have 12 coins, one of which is fake and weighs less than all the others. You have a balance scale with multiple pans, that is you can weigh several things at once and order them by weight. You do not care about the total number of weighings as in most classical coin puzzles, instead, this time using a pan is expensive and you want to find the fake coin with as few pan-uses as possible.
Spoiler warning: below I will discuss the solution for n mice.
You can, of course, give a cake to every mouse and see which one finishes first. You can save one cake by giving cakes at the same time to all but one of the mice. If everyone finishes simultaneously, the faster mouse is the unfed one.
It wastes cakes to give them to unequally-sized groups of mice. We can do better by copying the classical way to find a fake coin with the minimum number of weighings. That is, for each test, divide the mice into three groups as evenly as possible and give a cake to each of two equally-sized groups. The number of cakes you use is about 2log3n.
I wouldn't have written this essay if that was the solution. Sometimes you can do even better. For example, you can find the faster mouse out of 12 using only 5 cakes.
First, if you give out k cakes in one test, the test tells you which of k+1 groups the mouse is in. In the worst case, the faster mouse will be in the biggest group, so you should minimize the biggest group. Hence, your groups that get cakes should have ⌈n/(k+1)⌉ mice.
A test with one cake gives no information. I argue that giving out more than three cakes doesn't gain anything. Indeed, suppose we use four cakes in a test. That is, we divide the mice into five groups A, B, C, D and E, of which the first four are the same size. We can simulate the test by two tests in each of which we give out two cakes. In the first test we give cakes to A+B and C+D. If one of the groups is faster, say A+B, then in the second test give cakes to A and B; if not, E has the faster mouse. I leave it as an exercise to simulate a test with more than four cakes.
Thus, in an optimal strategy we can use two or three cakes per test. Also, if you give one test with k − 1 cakes and the next one with m − 1 cakes, you can switch them with the same effect. The largest group after either order of tests will have at most ⌈n/km⌉ mice.
I don't need two tests of three cakes each, which would give me a group of size at least ⌈n/16⌉. I can achieve the same result with three tests of two cakes each, with the faster mouse restricted to a group of size at most ⌈n/27⌉.
That means all my tests use two cakes, except I might use three cakes once. It doesn't matter in what order I conduct the tests, so I can wait until the end to use three cakes. I leave it as an exercise to the reader that the only small number of mice for which we would prefer three cakes is four. From this it follows quickly that for numbers of mice between 3 * 3i + 1 and 4 * 3i, the number of cakes is 2i + 1. For numbers between 4 * 3i + 1 and 3i+2 the answer is 2i + 2.
A week ago I chatted with my son Sergei about memorable math problems. He mentioned problem 5 from USAMO 2007. The problem can be reduced to the following:
Prove that (x7 + 1)/(x + 1) is composite for x = 77n, where n is a non-negative integer.
Perhaps Sergei remembered this problem because as far as he knew, he was the only one in that competition to solve it. That made me curious as to how he solved it. His solution is available as solution 2 at the Art of Problem Solving website. His solution seemed mysterious and impossible to invent on the spot. I became even more curious to understand the thought process underlying his solution.
Here is his recollection:
We need to factor x6 − x5 + x4 − x3 + x2 − x + 1. If such factoring existed it would have been known. Therefore, we need to somehow use the fact that x = 77n. What is the simplest way to factor? We should try to represent the polynomial in question as the difference of squares. Luckily, x is an odd power of 7. We can make it a square if we multiply or divide it by 7 or another odd power of 7. So with a supply of squares on one side, we need to find a match for one of them to build the difference.
Let us simplify the problem and see what happens for (y3 + 1)/(y + 1) for y = 33n, when n = 1. In other words we want to represent 703 as a difference of squares. This can be done: 703 = 282 − 92. Now let us see how we can express 282 and 92 through y which in this case is equal to 27. The first term is (y + 1)2, and the second is 3y.
Now let's go back to 7 and x, and check whether (x + 1)6 − (x6 − x5 + x4 − x3 + x2 − x + 1) is 7x. Oops, no. The difference is 7x5 + 14x4 + 21x3 + 14x2 +7x. On the plus side, it is divisible by 7x which we know is a square. The leftover factor is x4 + 2x3 + 3x2 +2x + 1, which is a square of x2 +x + 1.
The problem is solved, but the mystery remains. The problem can't be generalized to numbers other than 3 and 7.
I decided to take a closer look at the Putnam Competition. I analyzed the results of the three top contenders for the best Putnam teams: Harvard, MIT, Princeton. I looked at the annual number of Putnam Fellows from each of these three schools starting from 1994.
| Year | Harvard | MIT | Princeton |
|---|---|---|---|
| 1994 | 2 | 0 | 1 |
| 1995 | 3 | 0 | 0 |
| 1996 | 2 | 0 | 0 |
| 1997 | 4 | 0 | 0 |
| 1998 | 2 | 0 | 1 |
| 1999 | 2 | 1 | 0 |
| 2000 | 2 | 2 | 0 |
| 2001 | 2 | 1 | 0 |
| 2002 | 2 | 2 | 0 |
| 2003 | 1 | 2 | 1 |
| 2004 | 0 | 3 | 2 |
| 2005 | 2 | 3 | 1 |
| 2006 | 1 | 3 | 0 |
| 2007 | 1 | 2 | 1 |
| 2008 | 1 | 3 | 0 | 2009 | 1 | 2 | 0 |
As you may notice MIT couldn't even generate a Putnam Fellow until 2000, but starting from 2003 MIT consistently had more Putnam Fellows than Harvard or Princeton.
Richard Stanley, the coach of the MIT team, kindly sent me the statistics for the most recent competition, held in 2009.
| Category | Overall | MIT |
|---|---|---|
| Number of participants | 4036 | 116 |
| Mean score | 9.5 | 34.7 |
| Median score | 2 | 31 |
| Geometric mean | 0 | 0 |
| Percent of 0 score | 43.7 | 4.3 |
Furthermore, MIT had 40% in the top 5, 33% in the top 15, 32% in the top 25, and 35% in the top 81. For comparison, in the top 81, MIT had 28 winners — more than the next three schools together: Caltech 11, Harvard 9, Princeton 7.
No comment.
OnlineDegree.net selected the 50 Best Blogs for Math Majors, and I am pleased that Tanya Khovanova's Math Blog is number two. Since they did not explain their criteria, I suspected that it might be according to the number of Google hits. To double check, I Googled "math blog" and once again my blog was number two.
This might be the right moment to acknowledge the others involved with my blog. First, Sue Katz, my writing teacher and editor, corrected the English in most of my posts. Now I do not "do" mistakes in English any more, I make them.
My sons, Alexey and Sergei, are a huge support. Sometimes my poor kids have to listen endlessly to my latest idea, until I am ready to write about it. And then they will even read the final piece, and continue to encourage me.
But the most important motivators are you, my readers. Your comments, your personal emails and your feedback keep me writing.
I am usually disappointed with American math text books. I have had an underwhelming experience with them. Often I open a book and in the next 15 minutes, I find a mistake, a typo, a misguided explanation, sloppiness, a misconception or some other annoyance.
I was pleasantly surprised when I opened the book Introduction to Algebra
Look at problem 7.3.1.
Five chickens can lay 10 eggs in 20 days. How long does it take 18 chickens to lay 100 eggs?
There is nothing wrong with this problem. But the book is accompanied by the Introduction to Algebra Solutions Manual
The number of eggs is jointly proportional to the number of chickens and the amount of time. One chicken lays one egg in 10 days. Hence, 18 chickens will lay 100 eggs in 1000/18 days, which is slightly more than 55 and a half days.
What is wrong with this solution? Richard Rusczyk is human after all.
I like this book for its amazing accuracy and clean explanations. There are also a lot of diverse problems in terms of difficulty and ideas. Richard Rusczyk has good taste. Many of the problems are from different competitions and require inventiveness. I like that there are a lot of challenge problems that go beyond the boring parts of algebra. Also, I like that important points of algebra are chosen wisely and are emphasized.
This book might not be for everyone. It doesn't have pretty pictures. It doesn't have color at all. This is not a flaw for a math book. The book concentrates on ideas and problems, not entertainment. So if you're looking for math entertainment, you'll find it on my blog. For solid study, try Richard Rusczyk's books
I love Raymond Smullyan's books
This happened at the Gathering for Gardner 8. We were introduced and then later that day, the conference participants were treated to a dinner event that included a magic show. In one evening I saw more close-up magic tricks than I had in my whole life. This left me lightheaded, doubting physics and my whole scientific outlook on life.
Afterwards, Raymond Smullyan joined me in the elevator. "Do you want to see a magic trick?" he asked. "I bet I can kiss you without touching you." I was caught off guard. At that moment I believed anything was possible. I agreed to the bet.
He asked me to close my eyes, kissed me on the cheek and laughed, "I lost."
As a writer of books on mathematical puzzles I am often faced with delicate issues of phrasing, none more so than when it comes to questions about conditional probability. Consider the classic "X has two children and at least one is a boy. What is the probability that the other is a boy?"
It is reasonable to interpret this puzzle as asking you "What is the probability that X has two boys, given that at least one of the children is a boy" in which case the answer is unambiguously 1/3—given the usual assumptions about no twins and equal gender frequency.
This puzzle confounds people *legitimately*, however, because most of the ways in which you are likely to find out that X has at least one boy contain an implicit bias which changes the answer. For example, if you happen to meet one of X's children and it's a boy, the answer changes to 1/2.
Suppose the puzzle is phrased this way: X says "I have two children and at least one is a boy." What is the probability that the other is a boy?
Put this way, the puzzle is highly ambiguous. Computer scientists, cryptologists and others who must deal carefully with message-passing know that what counts is not what a person says (even if she is known never to lie) but *under what circumstances would she have said it.*
Here, there is no context and thus no way to know what prompted X to make this statement. Could he instead have said "At least one is a girl"? Could he have said "Both are boys"? Could he have said nothing? If you, the one faced with solving the puzzle, are desperate to disambiguate it, you'd probably have to assume that what really happened was: X (for some reason unconnected with X's identity) was asked whether it was the case that he had at least one son, and, after being warned—by a judge?—that he had to give a yes-or-no answer, said "yes." An unlikely scenario, to say the least, but necessary if you want to claim that the solution to the puzzle is 1/3.
Consider the puzzle presented (according to Alex Bellos) by Gary Foshee at the recent 9th Gathering for Gardner:
I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?
If the puzzle was indeed put exactly this way, and your life depended on defending any particular answer, God help you. You cannot answer without knowing, for example, what the speaker would have said if he had one boy and one girl, and the boy was born on Wednesday. Or if he had two boys, one born on Tuesday and one on Wednesday. Or two girls, both born on Tuesday. Et cetera.
Now, there is nothing mathematically wrong (given the usual assumptions, including X being random) about saying that "The probability that X has two sons, given that at least one of X's two children is a boy born on Tuesday, is 13/27." But if that is to be turned into an unambiguous puzzle attached to a presumed situation, some serious hypothesizing is necessary. For instance: you get on the phone and start calling random people. Each is asked if he or she has two children. If so, is it the case that at least one is a boy born on a Tuesday? And if the answer is again yes, are the children both boys? Theoretically, of the times you reach the third question, the fraction of pollees who say "yes" should tend to 13/27.
Kind of takes the fun out of the puzzle, though, doesn't it? Kudos to Gary for stirring up controversy with a quickie.
I recently discussed the following problem:
You run into an old friend. He has two children, but you do not know their genders. He says, "I have a son born on Tuesday." What is the probability that his second child is also a son?
I had heard this problem at the Gathering for Gardner 9 in a private conversation. My adversary had been convinced that the answer to the problem is 13/27. I came back to Boston from the gathering and wrote my aforementioned essay in which I disagreed with his conclusion.
I will tell you my little secret: when I started writing I substituted Wednesday for Tuesday. Then I checked my sons' birthdays and they were born on Saturday and Tuesday. So I changed my essay back to Tuesday.
After I published it people sent me several links to other articles discussing the same problem, such as those of Keith Devlin and Alex Bellos, both of whom think the answer is 13/27. So I invented a fictional opponent — Jack, and here is my imaginary conversation with him.
Jack: The probability that a father with two sons has a son born on Tuesday is 13/49. The probability that a father with a son and a daughter has a son born on Tuesday is 1/7. A dad with a son and a daughter is encountered twice as often as a dad with just two sons. Hence, we compare 13/49 with 14/49, and the probability of the father having a second son is 13/27.
Me: What if the problem is about Wednesday?
Jack: It doesn't matter. The particular day in question was random. The answer should be the same: 13/27.
Me: Suppose the father says, "I have a son born on *day." He mumbles the day, so you do not hear it exactly.
Jack: Well, as the answer is the same for any day, it shouldn't matter. The probability that his second child will also be a son is still 13/27.
Me: Suppose he says, "I have a son born …". So he might have continued and mentioned the day, he might not have. What is the probability?
Jack: We already decided that it doesn't depend on the day, so it shouldn't matter. The probability is still 13/27.
Me: Suppose he says, "I have a son and I do not remember when he was born." Isn't that the same as just saying, "I have a son." And by your arguments the probability that his second child is also a son is 13/27.
Jack: Hmm.
Me: Do you remember your calculation? If we denote the number of days in a week as d, then the probability of him having a second son is (2d−1)/(4d−1). My point is that this probability depends on the number of days of the week. So, if tomorrow we change a week length to another number his probability of having a son changes. Right?
At this point my imaginary conversation stops and I do not know whether I have convinced Jack or not.
Now let me give you another probability problem, where the answer is 13/27:
You pick a random father of two children and ask him, "Yes or no, do you have a son born on Tuesday?" Let's make a leap and assume that all fathers know the day of the births of their children and that they answer truthfully. If the answer is yes, what is the probability of the father having two sons?
Jack's argument works perfectly in this case.
My homework for the readers is: Explain the difference between these two problems. Why is the second problem well-defined, while the first one is not?
On March 5, 2010 I visited Princeton and had dinner with John Conway at Tiger Noodles. He gave me the second Doomsday lesson right there on a napkin. I described the first Doomsday lesson earlier, in which John taught me to calculate the days of the week for 2009. Now was the time to expand that lesson to any year.
As you can see on the photo of the napkin, John uses his fingers to make calculations. The thumb represents the DoomsDay Difference, the number of days your birthday is ahead of DoomsDay for a given year. To calculate this number you have to go back to my previous post.
The index finger represents the century adjustment. For example, the Doomsday for the year 1900 is Wednesday. Conway remembers Wednesday as We-are-in-this-day. He invented his algorithm in the twentieth century, not to mention that most people who use his algorithm were born in that century. Conway remembers the Doomsday for the year 2000 as Twosday.
The next three fingers help you to calculate the adjustment for a particular year. Every non-leap year has 52 weeks and one day. So the Doomsday moves one day of the week forward in one year. A leap year has one extra day, so the Doomsday moves forward two days. Thus, every four years the Doomsday moves five days forward, and, consequently, every twelve years it moves forward to the next day of the week. This fact helps us to simplify our year adjustment by replacing every dozen of years with one day in the week.
The middle finger counts the number of dozens in the last two digits of your year. It is important to use "dozen" instead of "12" as later we will sum up all the numerals, and the word "dozen" will remind us that we do not need to include it in the sum.
The ring finger represents the remainder of the last two digits of the year modulo 12, and the pinkie finger represents the number of leap years in that remainder.
John made two sample calculations on the napkin. The first one was for his own birthday — December 26, 1937. John was born exactly on Doomsday. I suspect that that is the real reason he called his algorithm the Doomsday Algorithm. The century adjustment is Wednesday. There are 3 dozens in 37, with the remainder 1 and 0 leap years in the remainder. When we add four more days to Wednesday, we get Sunday. So John Conway was born on Sunday.
The second napkin example was the day we had dinner: March 5, 2010. March 5 is 5 days ahead of the Doomsday. The century adjustment is Twosday, plus 0 dozens, 10 years in the remainder and 2 leap years in the remainder. 5 + 0 + 10 + 2 equals 3 modulo 7. Hence, we add three days to Tuesday, demonstrating that we dined out together on Friday. But then, we already knew that.
Do you know that some Russian letters are shaped exactly as some letters in the English alphabet? The shapes are the same, but the sounds of the letters are not. My Russian last name can be completely spelled using English letters: XOBAHOBA.
The adequate translation of my last name into English is Hovanova. You might ask where the first "K" came from. For many years French was considered the language of diplomacy and the USSR used French as an official language for traveling documents.
But "H" in French is silent and "Hovanova" would have been pronounced as "Ovanova." To prevent that, Russians used "kh" for the "h" sound.
Now to my first name. I was born Tatyana, for which Tanya is a nickname. Back in Russia, Tanya is used for children and students and Tatyana for adults and teachers. As I was a student throughout my 30 years of life in Russia, I was always Tanya. When I moved to the US, I decided to keep using Tanya, which I much preferred to Tatyana.
A psychiatrist might think that I wanted to be a student forever or refused to grow up. Or I could be accused of being lazy, as Tanya is shorter. In reality, I was just trying to be considerate. Tanya is easier to write and to spell for Americans. Anyway, I already had enough problems spelling out my last name in this country.
Now that more information is getting translated from Russian into English, I keep stumbling on references to me as to Hovanova or Tatyana. For example, the IMO official website used Russian sources to come up with the names of the Russian participants. They then translated the names directly into English, instead of going through French. As a result, on their website I am Tatyana Hovanova. This is not unique to me: many Russian names on the IMO website differ from those peoples' passport names.
By the way, if you Google my last name you will encounter other Khovanovas. Khovanova is not a particularly unusual name. Only one of the Khovanovas that came up in my search results is a close relative. Elizabeth Khovanova is my father's second wife and a dear friend. She is also an accomplished geneticist.
Khovanova is used only for females in Russia. The male equivalent is Khovanov. Surely you have heard of my half-brother Mikhail Khovanov and his homologies.
I gave you the Wise Men Without Hats puzzle in one of my previous posts:
A sultan decides to check how wise his two wise men are. The sultan chooses a cell on a chessboard and shows it to the first wise man. In addition, each cell on the chessboard either contains a pebble or is empty. The first wise man has to decide whether to remove one pebble or to add one pebble to an empty cell. Next, the second wise man must look at the board and guess which cell was chosen by the sultan. The two wise men are permitted to agree on their strategy beforehand. What strategy can they find to ensure that the second wise man will always guess the chosen cell?
My readers solved it. The solution is the following. Let us assign a number between 0 and 63 to every cell of the board. The second wise man takes numbers corresponding to cells with pebbles, converts them to binary and XORs the result. The answer is the cell number that he is seeking. The first wise man can always add or remove a pebble to make the XORing operation of the remaining pebbles produce any given number from 0 to 63.
This solution only works for boards that have a power of two as the number of cells.
Let's look at the solution more closely. Let us create a graph corresponding to this problem. The vertices of the graph will correspond to the positions of pebbles. That means vertices are in one-to-one correspondence with the subsets of the set of 64 elements. Let us connect two vertices if we can get from one position to another by removing or adding a pebble. That means vertices are connected if two corresponding sets differ by exactly one element. We can see that the resulting graph is regular and each vertex is connected to exactly 64 other vertices.
Let us assign one out of 64 colors to each cell of the chessboard. The second wise man can guess the cell by looking at the chessboard. From this we can conclude that there is a bijection from the vertices of the graph to chessboard cells. In other words, we can color the graph in 64 colors. The existence of the strategy for wise men means that we can color the graph in such a way that each vertex is connected to the vertices of all colors.
As each vertex in our graph has exactly 64 neighbors, the graph has the following property: It can be colored in 64 colors in such a way that every vertex is connected to exactly one vertex of every color.
As soon as I realized that there is such a graph-theoretical object, I started to run around MIT asking everyone if such objects were studied or have a name.
It appears that indeed such an object has a name. A graph that can be colored into k colors in such a way that every vertex has exactly one neighbor of every color is called a rainbow graph.
Andrew Woldar discusses properties of such graphs in his paper. In particular, rainbow graphs are matching graphs. Indeed, every vertex is connected to exactly one vertex of the same color. Hence there is a natural pairing on vertices. From here, we can conclude that the smallest size of a rainbow graph is 2k.
Several MIT students liked the wise men problem and the associated graph object so much that they decided to study them. Hwanchul Yoo, SuHo Oh, and Taedong Yun enumerated all rainbow graphs of size 2k. The number of non-isomorphic rainbow graphs of size 2k equals mitthe number of switching classes of graphs with k vertices. The corresponding sequence A002854 starts as: 1, 1, 2, 3, 7, 16, 54. The paper is soon to appear. It is titled "Rainbow Graphs and Switching Classes."
Just received from Victor Gutenmacher:
Fibonacci salad: For today's salad, mix yesterday's leftover salad with that of the day before.
New additions to my trick problems collection:
* * *
It takes 12 minutes to saw a log into 3 parts. How much time will it take to saw it into 4 parts?
* * *
The Davidsons have five sons. Each son has one sister. How many children are there in the family?
* * *
A caterpillar wants to see the world and decides to climb a 12-meter pole. It starts every morning and climbs 4 meters in half a day. Then it falls asleep for the second half of the day during which it slips 3 meters down. How much time will it take the caterpillar to reach the top?
First Name: David
Last Name: (hidden for privacy protection)
Year of Birth: (hidden for privacy protection)
email: buchanan1985@gmail.com
I remember a math problem from my childhood: divide an L-shaped triomino into four congruent parts. The answer is in the picture on the left. Such division is quite appropriately called a reptile (repetitive tiling). Solomon Golomb invented the name many years ago. He wasn't aware that his definition would end up creating Googling problems, for when you search for such a mathematical object you will stumble upon a lot of amphibians.
Similarly, you can divide the same shape into 9 congruent pieces (see the figure on the left).
Suppose you want to divide a shape into pieces that are similar, but not necessarily of the same size. Such tiling is called an irreptile (irregular reptile).
At the Gathering for Gardner 9 I listened to Carolyn Yackel's talk about the L-reptiles and L-irreptiles. One of the ways to create an irreptile is to start with a reptile, then to make a sub-tiling of one of the existing tiles. This procedure can be repeated many times.
Carolyn brought a ceramic table to the Gathering for Gardner. This table is made of two L-shapes. Both shapes are irreptiles, created by this procedure. In one part of the table she started with a 9-reptile, and in the other with a 4-reptile. She sent me this picture of her table to use in this essay.
After her talk I started wondering how many tiles can an L-irreptile be comprised of. We start with one piece: the L-shape itself. If we divide a tile into four smaller tiles we add three more pieces. If we divide it into nine tiles we add eight more pieces. We can mix sub-dividing into four and nine tiles. The total number of tiles that an L-shape can be comprised of by this procedure is all the numbers you can get from 1 by adding three or eight. The sequence is 1, 4, 7, 9, 10, 12, 13, 15, 16, 17 and so on. Starting from 15 we get all the consecutive numbers.
The numbers that are not represented in the above sequence are 2, 3, 5, 6, 8, 11 and 14. Can we divide an L-shape into such numbers of tiles? Benoît Jubin reminded me that there is an L-reptile with six pieces.
Consequently, we can add 5 more pieces to any L-irreptile. Thus, there exists an L-irreptile made out of 11 (1+5+5) and out of 14 (1+8+5) pieces. The numbers that are left are 2, 3, 5 and 8.
While I was discussing L-irreptiles with fans of sequences, David Wilson suggested a conjecture.
David Wilson's conjecture. If there is an L-irreptile, there is a corresponding square-irreptile with similarly-sized pieces.
If this conjecture is true, then we can see that L-irreptiles with 2, 3 or 5 pieces can't exist as corresponding square-irreptiles do not exist.
For example, to prove that 2 or 3 square-irreptiles can't exist, you need to notice that each corner of the square we are trying to tile should belong to a different small tile.
The question of the existence of an 8-irreptile of the L-shape is more interesting and challenging. The square 8-irreptile exists. If you can prove that the L-shape 8-irreptile doesn't exist, then you will automatically prove that the converse to Wilson's conjecture is not true.
Only at MIT. Room 4-231.
Let n coins weighing 1, 2, … n be given. Baron Münchhausen knows which coin weighs how much, but his audience does not. Define a(n) to be the minimum number of weighings the Baron must conduct on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins.In the paper Baron Münchhausen's Sequence, three of us completely described the Baron's sequence. In particular, we proved that a(n) ≤ 2. Here we would like to outline another proof idea, which is interesting in part because it touches the Riemann hypothesis.
We denote the total weight of coins in some set A as |A|.
Lemma. Numbers n that can be represented as Ti + Tj + Tk = 3n, where i ≤ j < k, such that there is a subset A of coins from j + 1 to k such that n = Tj + |A|, can be done in two weighings.
Proof. Suppose Ti + Tj + Tk = 3n and there is a subset A of coins from j + 1 to k such that n = Tj + |A|. We propose the two weighings
[1…j] + A = n
and
[1…i] + B = n + A,
where B is the complement of A in {j + 1, j + 2, … , k}.
If we sum up twice the first weighing with the second weighing we get
3[1…i] + 2[(i + 1)…j] + 2A + B = 3n + A.
In other words, three times the weight of the coins that were on the left side in both weighings, plus twice the weight of the coins that were on the left side in only the first weighing, plus the weight of the coins that were moved from the left cup to the right cup plus the weight of the coins on the left cup in only the second weighing equals three times the weight of the coin on the right cup in both weighings. Hence three times the weight of the coin on the right cup in both weighings can't be less than the weight of the k other coins that participated plus the weight of the j coins that were on the left cup in the first weighing and weren't moved to the right cup, plus the weight of the i coins that were one the left cup in both the first and the second weighing. But because Ti + Tj + Tk = 3n, then 3n is the smallest possible weight of any set of i plus j plus k coins, the coin on the right cup in both weighings has to be the n-coin.
We checked that any number up to 600,000 except 20 can be represented so as to satisfy the Lemma. To show how to solve 20 coins in two weighings is easy, and, as usual, is left as an exercise for the reader. Next, we want to look at the following lemma.
Lemma. Given a set of consecutive numbers {(j + 1), … , k}, if k > 2j + 2, then it is possible to find a subset in the set that sums up to any number in the range from j + 1 to (j + k + 1)(k – j)/2 – j – 1.
We won't prove the lemma, but it means that if k is about twice larger than j, then we have a lot of flexibility for building our set A in the weighing above. For moderately large n (where 600000 >> "moderately large"), it is not hard to prove that this flexibility is sufficient.
Now the question becomes: can we find such a decomposition into triangular numbers? It is enough to find a representation Ti + Tj + Tk = 3n, where Tk is at least 81% of 3n.
We know that decompositions into triangular numbers are associated with decompositions into squares. Namely, if Ti + Tj + Tk = 3n, then (2i + 1)2 + (2j + 1)2 + (2k + 1)2 = 24n + 3. If the largest square is at least 81% of 24n + 3, then the largest triangular number in the decomposition of 3n is at least 81%.
There is a theorem (W. Duke, Hyperbolic distribution problems and half-integral weight Maass forms, in Inventiones Math 92 (1988) p.73-90) that states that in the limit the decompositions of numbers into three squares are equidistributed. That is, if we take some region on the unit sphere x2 + y2 + z2 = 1 (for example, the region |z| > 0.8) and view decompositions of 24n + 3 into squares as points on the sphere x2 + y2 + z2 = 24n + 3, then, as n grows, decompositions whose projections fall into our chosen region are guaranteed to appear.
This theorem is great, because it tells us that for large enough n we will always be able to find a decomposition of 24n + 3 into triangle numbers where one of the triangle numbers will be much bigger than the others, and it will be possible to prove the weight of the n coin in two weighings. Unfortunately, this summary, as stated, does not tell us how large that n needs to be. So we need some exact estimates.
The number of decompositions of m into sums of three squares is about the square root of m. More precisely, it is possible to compute a number N, such that for any number m > N, with at most one exception, the number of decompositions is at least Cm1/2−1/30, where C is a known constant.
The more specific statement of Duke's theorem is that if the number of solutions to the quadratic x2 + y2 + z2 = 24n + 3 is large, for a computable value of "large", then the solutions are equidistributed. More precisely, let us denote 3n by m and fix an area Ω on the unit sphere. Then the number of solutions (x, y, z) such that the unit vector (x, y, z)/|(x, y, z)| belongs to Ω is
1/(4π) Ωh(8m+3) + E(m),
where h(8m+3) is the total number of solutions of x2 + y2 + z2 = 24n + 3, and E(m) is an error term, which starting from some number satisfies the inequality: E(m) ≤ 1000m1/2-1/7.
That's pretty good, because combining these two lets us, at least in principle, actually calculate an N such that for all n > N except maybe one a(n) = 2. After that we hoped to write a program to exhaustively search smaller numbers by computer.
This situation is still somewhat annoying, because that possible exception must then be propagated into the proof, and if we are not careful, possibly into the final theorem. ("No matter how many coins the Baron has, he can prove the weight of one in at most two weighings, except maybe one number of coins, and we don't know which...") This is where the Riemann Hypothesis comes in. If the Riemann Hypothesis is true, then that exception isn't there, and all is sunlight and flowers.
The beauty of the Baron's puzzle is such that we actually do not need the Riemann hypothesis. As we can use unbalanced weighings, it is enough to find a good decomposition for one out of the four numbers 3n, 3n-1, 3n-2, or 3n-3.
Instead of finding all these exact estimates we found a different elementary proof of our theorem. But we are excited that methods that are used in very advanced number theory can be used to solve a simple math problem that can be described to middle school children.
It would be great if someone decided to finish this proof.
I am used to thinking that a "woman in numbers" means a female number theorist. But not anymore. I just discovered drawings by Svetlana Bogatyr. From now on the expression a "woman in numbers" will convey an additional meaning to me.
I am grateful to Svetlana for permitting me to post several of her drawings. The "Mature Woman" is on the left. "Eurydice", "Girl in Scarf" and "Holland Woman " are below.
Enjoy.
I just invented a two-player game. To start, you have an empty n by n board. When it's your turn you must write an integer between 1 and n into an empty cell on the board. Your integer has to differ from the integers that are already present in the same row or column. If you finish filling up the board, you will get a Latin square and the game will be a tie. The person who doesn't have a move loses. What is the best strategy?
Let's see what happens if n is 2. The first player puts any number in one of the four corners of the 2 by 2 board. The second player wins by placing a different number in the opposite corner.
I played this game with my son Sergei Bernstein on a 3 by 3 board. We discovered that the first player can always win. Since this game is so much fun, I'll leave it to the reader to play it and to find the winning strategy for the first player.
Can you analyze bigger boards? Remember that this game has many symmetries. You can permute rows and columns. Also, you can permute numbers.
While we were playing Sergei invented two theorems.
Sergei's theorem 1. If n is odd the first player can guarantee a tie.
Proof. In the first move the first player writes (n+1)/2 in the center cell. If the second player puts number x in any cell the first player puts number n+1−x into the cell that is rotationally symmetric to the second player's cell with respect to the center. With this strategy the first player will always have a legal move.
Sergei's theorem 2. If n is even the second player can guarantee a tie.
Proof. If the first player puts number x in any cell the second player puts number n+1−x into the cell that is vertically symmetric to the first player's cell with respect to the vertical line of symmetry of the board. With this strategy the second player will always have a legal move.
As you play the game, let me know if you develop any theorems of your own.
Suppose you meet a friend who you know for sure has two children, and he says: "I have a son born on Tuesday." What is the probability that the second child of this man is also a son?
People argue about this problem a lot. Although I've discussed similar problems in the past, this particular problem has several interesting twists. See if you can identify them.
First, let us agree on some basic assumptions:
Now let us consider the first scenario. A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: "I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun." If his statement is, "I have a son born on Tuesday," what is the probability that the second child is also a son?
The probability that a father of two daughters will make such a statement is zero. The probability that a father of differently-gendered children will produce such a statement is 1/14. Indeed, with a probability of 1/2 the son is chosen over the daughter and with a probability of 1/7 Tuesday is the birthday.
The probability that a father of two sons will make this statement is 1/7. Among the fathers with two children, there are twice as many who have a son and a daughter than fathers who have two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 1/2 for the second child to also be a son.
Now let us consider the second scenario. A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son's day of birth. If he has two sons, he has to choose one at random. His statement will be, "I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun." If his statement is, "I have a son born on Tuesday," what is the probability that the second child is also a son?
The probability that a father of differently-gendered children will produce such a statement is 1/7. If he has two sons, the probability will likewise be 1/7. Among the fathers with two children, twice as many have a son and a daughter as have two sons. Plugging these numbers into the formula for calculating the conditional probability gives us a probability of 1/3 for the second child to also be a son.
Now let us consider the third scenario. A father of two children is picked at random. If he doesn't have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, "I have a son born on Tuesday." What is the probability that the second child is also a son?
The probability that a father of two daughters will have a son born on Tuesday is zero. The probability that a father of differently-gendered children will have a son who is born on Tuesday is 1/7. The probability that a father of two sons will have a son born on Tuesday is 13/49. Among the fathers with two children, twice as many have a son and a daughter than two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 13/27 for the second child to also be a son.
Now let's go back to the original problem. Suppose you meet your friend who you know has two children and he tells you, "I have a son born on Tuesday." What is the probability that the second child is also a son?
What puzzles me is that I've never run into a similar problem about daughters or mothers. I've discussed this math problem about these probabilities with many people many times. But I keep stumbling upon men who passionately defend their wrong solution. When I dig into why their solution is wrong, it appears that they implicitly assume that if a man has a daughter and a son, he won't bother talking about his daughter's birthday at all.
I've seen this so often that I wonder if this is a mathematical mistake or a reflection of their bias.
How to solve the original problem? The problem is under-defined. The solution depends on the reason the father only mentions one child, or the Tuesday.
The funny part of this story is that I, Tanya Khovanova, have two children. And the following statement is true: "I have a son born on Tuesday." What is the probability that my second child is a son?
How many different months' lengths are possible?
For "simplicity" let's stick to the Gregorian calendar.
Lennart Green is an amazing magician who performs card tricks. He is so good that the judges at the competition of the International Federation of Magic Societies didn't believe his tricks. They assumed that he used the help of stooges, and unfairly disqualified him. At the next competition he used a judge to assist him and won first place in the cards category.
I saw his performance twice. Both times he brought a woman to the stage, but each time it was a different woman. It was clear that he talked to each woman before the performance, presumably asking her permission. Furthermore, during his performances, both of the women looked slightly bored, implying that it might be not their first time. My first impression was that the women were a part of the act. I was fooled just as the judges had been fooled.
What can I say? Lennart Green isn't skilled at picking up the right women. Watch his performance at TED, and remember that he proved that the assistants were clueless.
When the Abel Prize was announced in 2001, I got very excited and started wondering who would be the first person to get it. I asked my friends and colleagues who they thought was the greatest mathematician alive. I got the same answer from every person I asked: Alexander Grothendieck. Well, Alexander Grothendieck is not the easiest kind of person to give a prize to, since he rejected the mathematical community and lives in seclusion.
Years later I told this story to my friend Ingrid Daubechies. She pointed out to me that my spontaneous poll was extremely biased. Indeed, I was asking only Russian mathematicians living abroad who belonged to "Gelfand's school." Even so, the unanimity of those responses continues to amaze me.
Now several years have passed and it does not seem that Alexander Grothendieck will be awarded the Prize. Sadly, my advisor Israel Gelfand died without getting the Prize either. I am sure I am biased with respect to Gelfand. He was extremely famous in Soviet Russia, although less well-known outside, which may have affected the decision of the Abel's committee.
I decided to assign some non-subjective numbers to the fame of Gelfand and Grothendieck. On Pi Day, March 14, 2010, I checked the number of Google hits for these two men. All the Google hits in the rest of this essay were obtained on the same day, using only the full names inside quotation marks.
Google hits do not give us a scientific measurement. If the name is very common, the results will be inflated because they will include hits on other people. On the other hand, if a person has different spellings of their name, the results may be diminished. Also, people who worked in countries with a different alphabet are at a big disadvantage. I tried the Google hits for the complete Russian spelling of Gelfand: "Израиль Моисеевич Гельфанд" and got an impressive 137,000.
Now I want to compare these numbers to the Abel Prize winners' hits. Here we have another problem. As soon as a person gets a prize, s/he becomes more famous and the number of hits increases. It would be interesting to collect the hits before the prize winner is announced and then to compare that number to the results after the prize announcement and see how much it increases. For this endeavor, the researcher needs to know who the winner is in advance or to collect the data for all the likely candidates.
John Thompson is way beyond everyone else's range because he shares his name with a famous basketball coach. But my point is that Gelfand and Grothendieck could have been perfect additions to this list.
I have this fun book at home written by Clifford Pickover and titled Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning
The most puzzling thing about this list is that there is no overlap with the Abel Prize winners. Here is the list with the corresponding Google hits.
Since there are other great mathematicians with a lot of hits, I started trying random names. In the list below, I didn't include mathematicians who had someone else appear on the first results page of my search. For example, there exists a film director named Richard Stanley. So here are my relatively "clean" results.
If prizes were awarded by hits, even when the search is polluted by other people with the same name, then the first five to receive them would have been:
If we had included other languages, then Gelfand might have made the top five with his 48,000 English-language hits plus 137,000 Russian hits.
This may not be the most scientific way to select the greatest living mathematician. That's why I'm asking you to tell me, in the comments section, who you would vote for.
I got this problem from my friend, a middle-school math teacher, Tatyana Finkelstein.
We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weight the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale.
Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.
I would like to add an extra twist to the problem above. It is conceivable that there might be several different strategies for finding the direction in which the weight of the fake coin deviates from the real coins. In this case it is better to choose a strategy that can redeem as many coins as possible — that is, to identify the maximum number of coins as real.
The number of coins you identify as real depends on the outcomes of your weighings. Then what is the precise definition of the best strategy?
Let us call a strategy k-redeem if after the weighings you are guaranteed to demonstrate that k coins are real, but you are not guaranteed to demonstrate that k+1 coins are real. Your task is to analyze two-weighing strategies and choose the most profitable one — the strategy that guarantees to redeem the largest possible number of coins, that is, a k-redeem strategy for the largest k.
Sara was born in Boston on February 29, 2008 at 11:00 am. Her parents were quite upset that their calendar-challenged daughter would only be able to celebrate her birthday once in four years. Luckily, science can help Sara's parents. How? Sara can celebrate her birthday every year at the moment when the Earth passes the same point on its orbit around the Sun as when Sara was born.
Assuming that Sara lives her entire life in Boston and that the daylight savings time is not moved earlier into February, your task is to calculate the schedule of Sara's birthday celebrations for 100 years starting from her birth. To simplify your homework, you can approximate one year as 365 days and 6 hours.
Joseph DeVincentis heard my prayers and created an index for MIT mystery hunt puzzles. He created it not because I requested it, but rather because he was on the writing team this year and they needed it. Anyway, finally there is an index.
I have to warn you, though, that this index was created for people who have already solved the puzzles, so the index contains hints for many of the problems and, on rare occasions, solutions.
Now I will do the math index for this year, and I promise that I will avoid big hints.
One fine day in January 2010, John H. Conway shared with me his recipe for success.
1. Work at several problems at a time. If you only work on one problem and get stuck, you might get depressed. It is nice to have an easier back-up problem. The back-up problem will work as an anti-depressant and will allow you to go back to your difficult problem in a better mood. John told me that for him the best approach is to juggle six problems at a time.
2. Pick your problems with specific goals in mind. The problems you work on shouldn't be picked at random. They should balance each other. Here is the list of projects he suggests you have:
3. Enjoy your life. Important problems should never interfere with having fun. When John Conway referred to having fun, I thought that he was only talking about mathematics. On second thought, I'm not so sure.
* * *
Birthdays are beneficial for your health. A new breakthrough statistical study unequivocally proved that the more birthdays one has the longer one lives.
* * *
We know through Erdös that "a mathematician is a device for turning coffee into theorems". It thus follows by duality that a comathematician is a device for turning cotheorems into ffee.
* * *
- What do you do when you see a beautiful girl?
- I download her.
* * *
Programmers wear red T-shirts to match the color of their eyes.
* * *
We invented the decimal system, because humans have ten fingers on their hands; and 32-bit computers, because humans have 32 teeth in their mouths.
* * *
A general shows off a new tank and boasts:
- You see a tank supplied with the most modern computer technology.
- What is the speed of its computer?
- The same as the speed of the tank, of course.
A sultan decides to check how wise his two wise men are. The sultan chooses a cell on a chessboard and shows it to the first wise man. In addition, each cell on the chessboard either contains a rock or is empty. The first wise man has to decide whether to remove one rock or to add one rock to an empty cell. Next, the second wise man must look at the board and guess which cell was chosen by the sultan. The two wise men are permitted to agree on the strategy beforehand. What strategy can they find to ensure that the second wise man will always guess the chosen cell?
If you are stuck, there are many approaches to try. You can attempt to solve the puzzle for a board of size 1 by 2, or for a board of size 1 by 3. Some might find it easier to solve a version in which you are allowed to have a multiple number of rocks on a cell, and the first wise man is permitted to add a rock to a cell that already contains rocks.
This year I am again on the organizing committee of the Women and Mathematics program at Princeton's Institute for Advanced Study. Our subject is "p-adic Langlands Program." It is a fashionable, advanced and very influential program connecting number theory and representation theory.
We invite undergraduate students, graduate students and post-docs to apply. In 2009-2010 the Institute has been running a special year in Analytic Number Theory. That has brought many number theorists to the institute already, so there will be a lot of people to talk to.
Last year I promised to hold a math party during the program. But I had to cancel it due to a scheduling conflict with George Hart's ZomeTool Workshop. I am planning a party this year. Either way, we'll have fun.
If you want to learn about the Langlands program, to spent time on the beautiful grounds of the Institute, to eat in one of the best cafeterias around, and to make new friends with other women interested in number theory, then please apply. The application deadline is February 20.
Suppose you want to increase your chances of winning the lottery jackpot by pooling money with a group of coworkers. There are several issues you should keep in mind.
When you pool the money and you hit the jackpot, the money has to be split. If you bought 10,000 tickets and the jackpot that you win is $100 million, then each ticket is entitled to a mere $10,000. Your chances of hitting the jackpot in the first place are 1 in 17,500 and you're not going to get rich off what you win.
Perhaps you'd be satisfied with a small profit. However, as I calculated in my previous piece on the subject, even if you include the jackpot in the calculation of the expected return, the Mega Millions game never had, and probably never will have a positive return.
Despite this fact, people continue to pool money in the hopes of winning big. However, there are more problems in doing this than just its non-profitability.
Consider a scenario. Your coworkers collected $1,000 to buy 1,000 lottery tickets. You give the money to Jerry who buys the tickets. Jerry can go to a store and buy 1,005 tickets. After the lottery he checks the tickets, takes the best five for himself and comes back to work with 1,000 disappointing tickets.
It is more likely that Jerry is cheating or that he will lose the tickets than it is that your group will win the jackpot. But there is a probabilistic way to check Jerry's integrity. According to the odds, every 40th ticket in Mega Millions wins something. Out of 1,000 tickets that Jerry bought, you should have about 25 that win something. If Jerry systematically brings back tickets that win less often than expected, you should replace Jerry with someone else.
There are methods to protect your group against cheating. For example, you can ask another person to join Jerry in purchasing the tickets, which they then seal in an envelope that they both sign.
Alternatively, you yourself could be the person responsible for buying 1,000 tickets. How would you protect yourself from suspicion of cheating? The same way as I mentioned above: bring along some witnesses and have everyone sign the sealed envelope.
The most reliable way to prevent Jerry from cheating is to have him write down all the ticket numbers and send this information to everyone before the drawing. This way he can't replace one ticket with another. But this is a lot of work for tickets that are usually worth less than the money you collected to buy them.
But there are other kinds of dangers if you use this supposedly reliable method. If you bought a lot of tickets the probability of winning a big payoff increases. Suppose Jerry publicly locks the envelope in a desk drawer in his office. If one ticket wins $10,000, and everyone knows all the ticket combinations, suddenly Jerry's desk drawer becomes a very unsafe place to keep the tickets.
Scams are not your only worry. You shouldn't buy the same combination twice — whether picking randomly or not. You really do not want to waste a ticket and end up sharing the jackpot with yourself.
You cannot change the odds of hitting the jackpot, but you can change the odds of sharing it with others. Indeed, there are people who do not buy random combinations, but rather pick their favorite numbers, like birthdays. You can reduce the probability of sharing the jackpot if you choose the combinations for your tickets wisely, by picking numbers that other people are unlikely to pick.
Still want to try the lottery? If you feel a need to throw your money away, instead of buying lottery tickets, feel free to donate to this blog.
In one of my previous pieces, I discussed returns on the Mega Millions lottery game, assuming that you buy a small number of tickets. In such a case winning the jackpot has zero probability. So I argued that if you want to estimate the profitability of the lottery as an investment, you have to remove the jackpot money from the calculation.
Today I will discuss what the formal expected return is. That is, I will include the jackpot money in the calculation. Since I argued against including the jackpot in my last article, you might wonder why I've then turned around to look into this.
I think this mathematical exercise will be fun. Besides, on a practical note, it is useful to know when the formal expected return is more than 100%, because then it might make sense to pool money with other people. Keep in mind though that if you want a chance to hit the jackpot, the total number of tickets you buy must be really big. For example, even if you manage to pool $10,000 for tickets, your probability of winning the jackpot in Mega Millions is only one in 17,500 — still minuscule.
If you buy only one ticket, you'll lose. If you manage to pool a lot of money and the probability of the jackpot becomes noticeable, that is, non-zero, could the jackpot be large enough that the lottery becomes a good investment?
For this calculation, I'm still assuming that you buy a relatively small number of tickets. If you buy millions of tickets the calculation is slightly different, and I will write about that later.
You might think that when the jackpot is bigger than the odds, it makes sense to play. I am discussing the Mega Millions game, where the odds of winning the jackpot are one in 175 million. So if the jackpot is more than $175 million, then it is profitable to play. Right?
Wrong. As I mentioned in my previous piece, after reducing for taxes, you get about 16% of your money back through smaller payouts. Hence, you need to recover the other 84% through the jackpot. So the jackpot should be more than 175*.84 = 147 million dollars. This sounds even better. Right?
Wrong. No one receives the jackpot. Winners can chose to immediately receive the lump sum, which equals the money lottery organizers have actually set aside for it. Alternatively, the lottery organizers can invest the lump sum and give winners a yearly distribution over many years, the total of which will equal the jackpot.
Suppose for simplicity the lump sum is half of the jackpot. That means we need the jackpot to be $294 million ($147 x 2). Right?
Oops. As usual, we forgot about taxes. To exacerbate your pain, I have to add that the winnings are taxable. Suppose you have to pay 30% from the jackpot. That means the jackpot needs to be $424 ($294/0.7) million in order to justify pooling money. OK?
We haven't seen jackpots that big yet. But neither have we finished the calculation. There is a probability that you might have to share the jackpot with other winners. To calculate this probability, we need to calculate the number of tickets sold. That means, your expected return depends not only on the size of the jackpot, but also on the number of these tickets.
But even if you know the number of tickets sold, we cannot calculate the expected returns precisely because people don't always buy tickets with random combinations, but often pick their own numbers.
When the jackpot is large people start buying tons of tickets, so we can expect that many of them buy quick-picks. Let us assume for now that the vast majority of people do not choose their own numbers, but buy tickets at random. Suppose 200 million tickets were sold. That is a very big number. Last time that many tickets were sold was when the jackpot was $390 million in March 2007. By the way, that was the largest jackpot ever.
In order to finish the calculation, we need to establish the probability of several winners, given that 200 million random tickets were sold:
| Number of winners | Probability |
|---|---|
| No winner | 0.3204 |
| One winner | 0.3647 |
| Two winners | 0.2075 |
| Three winners | 0.0787 |
| Four winners | 0.0224 |
| Five winners | 0.0051 |
| Six winners | 0.0010 |
From here we can calculate the adjustment coefficient, that is, the proportion of money you are expected to get from the jackpot given that there are 200 million players in the game. The coefficient is calculated from the table above as (0.3647 + 1/2*0.2075 + 1/3*0.0787 +1/4*0.0224 + 1/5*0.0051 + 1/6*0.0010)/(1 - 0.3204), and is equal to 0.7379. We need to divide our previous figure of $424 million by the adjustment coefficient. The result is $575 million.
Given that a $390 million jackpot attracted more than $200 million in tickets, we can expect that the $575 million jackpot will make people completely crazy and attract even more money. So I do not anticipate that the Mega Millions game will ever have a positive formal expected gain. My conclusion is that not only is there no financial sense in buying a single lottery ticket, but also none in pooling money.
Of course, you can buy tickets for non-financial reasons, like pumping up your adrenaline. In any case, I showed you the method to calculate your expected return, or, more appropriately, your expected loss.
Here is a fun math activity I use with my students, after I teach them to play the game of set. To other teachers — feel free to clone this idea.
First, I ask the students if they know what a magic square is. They usually do know that a magic square is a three-by-three square of distinct digits, so that every row, column and diagonal has the same sum. Then I ask them what a magic set square might be. Often they guess correctly that it is a three-by-three square made of set cards, so that every row, column and diagonal form a set. Once that's established, I have them build magic set squares.
While they're building them, I ask a lot of questions, from how many cards there should be in the deck to how many different sets there are.
Once the squares are built, I ask them what a magic set cube might be. Their next task is to use their magic set squares as the bottom layer in building magic set cubes. In order to see all the cards in the cube, I instruct them to arrange the layers (bottom, middle and top) side-by-side.
As they're working on their cubes, I continue quizzing them. How many main diagonals does a cube have? Once they confirm that the answer is four, I ask them to show me those diagonals in their magic set cubes and check that they are sets. I might also ask them how many different magic set squares should be inside a magic cube. This is a theoretical math question they need to answer before finding them in their own model. Next they need to identify the different sets that form lines inside their cubes.
At this point, some students guess my next request: to construct a magic set hypercube.
After students build their hypercubes, they never want to destroy them. They like comparing the different hypercubes and often take photos of them. If there's still time left, I can continue in several directions. For example, they can count the main diagonals of the hypercube and find them in their models. Alternatively, they can find a "no set" — the largest possible set of cards inside a magic set hypercube that doesn't contain a set.
Math is usually about thinking, but this is one activity the students can do with their hands. And that adds another layer of magic.
I promised to discuss how to improve the accuracy of your guessing at AMC 10/12, or other tests for that matter. There are two types of guessing. First, meta-guessing is when you do not look at the problem, but rather guess from just looking at the choices. Second, in the one I refer to as an educated-guess you do look at the problem. Instead of completely solving it, you try to deduce some information from the problem that will help you eliminate some of the choices. In this essay, I discuss both methods.
But first let me emphasize that solving problems is a more important skill than meta-guessing from a list of choices. Solving problems not only teaches you to think mathematically, but also increases your brain power. Spending time improving your meta-guessing skills can help you at multiple-choice tests and may give you insight into how test designers think, but this will not increase your math knowledge in the long run.
On the other hand, educated-guessing is a very useful skill to obtain. Not only can it improve your score during a test, the same methods can be applied to speed up the process of re-checking your answers before handing in the test. This skill will also come in handy when you start your research. Some problems in research are so difficult that even minor progress in estimating or describing your answer is beneficial.
Before discussing particular methods, let me remind you that AMC 10/12 is a multiple-choice competition with five choices for each question. The correct answer brings you 6 points. A wrong answer brings you 0 points, and not answering brings you 1.5 points. So if you randomly guess one of the five choices, your expected average score is 1.2 points, which is 0.3 less than your score for an unanswered question. Thus guessing is unprofitable on average.
However, if you can eliminate one choice, your expected average score becomes 1.5 points. In this case guessing doesn't bring you points on average, but it does create some randomness in your results. For strategic reasons, you might prefer guessing, as I discussed in my earlier piece.
If you can eliminate two wrong choices, then guessing becomes profitable. A random guess out of three possibilities brings you 2 points, a better result than 1.5 points for an unanswered question. Even more, if you can eliminate three choices, then guessing will increase your score by 3 points on average.
Now that we've covered the benefits of excluding choices before guessing, I would like to discuss how to exclude choices by just looking at them. Let us take one of the problems from the 2002 AMC10B. Here are the choices: (-2,1), (-1,2), (1,-2), (2,-1), (4,4). The pair (4,4) is a clear outlier. I suggest that an outlier can't be a correct choice. If (4,4) were the correct answer, then it would have been enough, instead of solving the problem, to use some intermediate arguments to choose it. For example, if you can argue that both numbers in the answer must be at least 2, or must be positive or be even, then you can get the correct answer without solving the problem. Any problem for which you can easily pick the correct answer without solving it is an unacceptably poor problem design. Thus, (4,4) can't be the correct answer, and should be eliminated during guessing.
Let us look at a 2002 AMC10A problem with the following choices: 4/9, 2/3, 5/6, 3/2, 9/4. Test designers want to create choices that are plausible. They try to anticipate possible mistakes. In this set of choices, we can deduce that one of the mistakes that they anticipate is that students will confuse a number with its inverse. In this case 5/6 can't be the correct answer. Otherwise, 6/5 would have been included as a choice. In another similar example from the 2000 AMC10 with choices -2, -1/2, 1/3, 1/2, 2, the designers probably hope that students will confuse numbers with their inverses and negations. Hence, we can exclude 1/3.
Sometimes the outlier might hint at the correct answer. Suppose you have the following list of choices: 2, 1/2π, π, 2π, 4π. The number 2 is an outlier here. Probably, the problem designers were contemplating that students might forget to multiply by π. In this case the likely correct answer would be 2π, because only from this answer can you get 2 by forgetting to multiply by π.
As an exercise, try to eliminate the wrong choices from the following set from a problem given at 2000 AMC10: 1/(2m+1), m, 1-m, 1/4m, 1/8m2.
AMC designers know all of these guessing tricks, so they attempt to confuse the competitors from time to time by going against common sense. For example, in a 2002 AMC10A problem the choices were: -5, -10/3, -7/3, 5/3, 5. I would argue that -7/3 is a clear outlier because all the others are divisible by 5. Furthermore, there is no point in including so many answers with 3 in the denominator unless there is a 3 in the denominator of the correct answer. So I would suggest that one of -10/3 and 5/3 must be the answer. My choice would be 5/3, as there is a choice of 5 which I would assume is there for students who forgot to divide by 3. As I said the designers are smart and the actual answer is -10/3. They would have tricked me on this problem.
One of the best ways to design a multiple choice question is to have an arithmetic progression as a list of choices. There is no good way to eliminate some choices from 112, 113, 114, 115, 116. Unfortunately for people who want to get an advantage by guessing, many of the problems at AMC have an arithmetic progression as their choices.
Now that we've seen the methods of meta-guessing, let's look at how to make an educated guess. Let us look at problem 1 in 2003 AMC10A.
What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?
Without calculations we know that the answer must be odd. Thus, we can immediately exclude three out of five choices from the given choices of 0, 1, 2, 2003, 4006. Parity consideration is a powerful tool in eliminating wrong answers. Almost always you can decide the parity of the answer much faster than you can calculate the answer. Similar to using parity, you can use divisibility by other numbers to have a fast elimination. Here is a problem from 2000 AMC10:
Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?
The first four numbers entered must be divisible by 4. The total of the given numbers is divisible by 4. Hence, the last number must also be divisible by 4. This reasoning eliminates three out of five choices.
Another powerful method is a rough estimate. Let us look at the next problem in 2003 AMC10A.
Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $2366, how many members are in the League?
If we notice that the cost per person is more than $25, we can conclude that there are less than a hundred members in the League. Given the choices of 77, 91, 143, 182, and 286, we immediately can eliminate three of them.
Another method is to use any partial knowledge that you may have. Consider this problem from 2003 AMC10A:
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?
You might remember that there is a formula for this. Even if you do not remember the exact formula, you might still have a vague memory that the answer must be a binomial coefficient that somehow uses the number of cookies and the number of flavors. Looking at the choices — 22, 25, 27, 28, 29 — you can see that the only choice that appears in the first 10 rows of the Pascal's triangle is 28. So you should go with 28.
It is easy to talk about easy problems; let us see what we can do about difficult ones. Consider the last problem on 2003 AMC10A:
Let n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q+r divisible by 11?
The choices are 8180, 8181, 8182, 9000, 9090. They can be naturally split into two groups: three choices below 9000 and the rest. By my rules of removing outliers the group of numbers below 9000 seems the more promising group. But I would like to discuss how to approximate the answer. There is no reason to believe that there is much correlation between remainders by 100 and divisibility by 11. There is a total of 90,000 5-digit numbers; among those numbers, approximately 90,000/11 = 8182 is divisible by 11, so we should go with the group of answers close to 8182.
Another way of thinking about this problem is the following. There are 900 different quotients by 100 to which we add numbers between 0 and 99. Thus for every quotient our sums are a set of 100 consecutive numbers. Out of 100 consecutive numbers usually 9, and rarely 10, are divisible by 11. Hence, the answer has to be less than 9000.
Sometimes methods you use for guessing can bring you the answer. Here is a problem from 2001 AMC12:
What is the product of all odd positive integers less than 10000? (A) 10000!/5000!2, (B) 10000!/25000, (C) 9999!/25000, (D) 10000!/(250005000!), (E) 5000!/25000.
For a rough estimate, I would take a prime number and see in what power it belongs to the answer. It's simplest to consider a prime number p that is slightly below 5000. Then p should appear as a factor in the product of all odd positive integers below 10000 exactly once. Now let us look at the choices. Number p appears in 5000! once and in 10000! twice (as p and 2p). Hence, it appears in (A) zero times, and twice each in (B) and (C). We also can rule out (E) as the product of odd numbers below 10000 must be divisible by primes between 5000 and 10000, but 5000! doesn't contain such primes. Thus the answer must be (D).
The method I just described won't produce the formula. But the ideas in this method allow you to eliminate all the choices except the right one. Moreover, this method provides you with a sanity check after you derive the formula. It also helps to build your mathematical intuition.
I hope that you will find my essays about AMC useful. And good luck on February 9!
I teach students to solve math problems by appreciating the big picture, or by noticing the problem's inner symmetries, or through a deep understanding of the problem. In the long run, one thing leads to another: such training structures their minds so that they are better at understanding mathematics and, as a consequence, they perform well at math competitions.
That is why, when AMC is still far away, I do not give my students a lot of AMC problems; rather, I pick problems that contain useful ideas. When I do give AMC problems, I remove the multiple choices, so they understand the problems completely, instead of looking for shortcuts. For example, this problem from AHSME 1999 is a useful problem with or without choices.
What is the largest number of acute angles that a hexagon can have?
As AMC approaches, we start discussing how to solve problems given multiple choices. Training students for AMC is noticeably different from teaching mathematics. For example, some problems are very specific to AMC. They might not even exist without choices. Consider this problem from the 2001 AMC12:
A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?
Here we really need the choices in order to pick one complex number, such that its real part is an integer or a half integer and, in addition, the product of the number with its conjugate produces an integer.
Sometimes the choices distract from solving the problem. For example, in the following problem from the 2005 AMC12, having choices might tempt students to try to eliminate them one by one:
The sum of four two-digit numbers is 221. None of the eight digits is 0 and no two of them are same. Which of the following is not included among the eight digits?
Without the choices, students might start considering divisibility by 9 right away.
On some occasions, the choices given for the problems at AMC make the problem more interesting. Here is an example from the 2000 AMC10:
Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
The choices are 21, 60, 119, 180 and 231. We can immediately see that the answer must be odd. Because the span of the three remaining choices is so wide, we suspect that we can eliminate the smallest and the largest. Trying for 5 and 7 — the two smallest primes in the range — we can eliminate 21. Similarly, checking the two largest primes in the range, we can eliminate 231. This leaves us with the answer: 119. If the choices were different, we might have lost the interplay between the solution and the list of choices. Then, solving the problem would have been slower and more boring. There are ten pairs of prime numbers to check. And we would need on average to check five of them until we stumbled on the correct choice.
In other cases having multiple choices makes the problem more boring and less educational. Here is another problem from the same competition.
Two non-zero real numbers, a and b, satisfy ab = a – b. Find a possible value of a/b + b/a – ab.
Solving this problem without choices can teach students some clever tricks that people use when playing with expressions. Indeed, when we collect a/b + b/a into one fraction (a2 + b2)/ab, we might remember that a2 + b2 is very close to (a – b)2, and see from here that a/b + b/a – 2 is (a – b)2/ab, which, given the initial condition, equals ab. Thus, we can get the answer: 2.
On the other hand, if you look at the multiple choices first: -2, -1/2, 1/3, 1/2, 2, you might correctly assume that the answer is a number. Thus, the fastest way to solve it is to find an example. If a = 1, then b must be 1/2, and the answer must be 2. This solution doesn't teach us anything new or interesting.
My next example from the 2002 AMC10 is similar to the previous one. The difference is that the solution with multiple choices is even more boring, while the solution without these choices is more interesting and beautiful.
Let a, b, and c be real numbers such that a – 7b + 8c = 4 and 8a + 4b – c = 7. Then a2 – b2 + c2 is: 0, 1, 4, 7, or 8.
Given the choices, we see that the answer is a number. Hence we need to find any solution for the system or equations: a – 7b + 8c = 4 and 8a + 4b – c = 7. For example, if we let c = 0, we have two linear equations and two variables a and b that can be solved by a straightforward computation. Then we plug the solution into a2 – b2 + c2.
Without knowing that the result is a number, we need to look at the symmetries of our two initial equations. We might discover a new rule:
If we have two expressions ax + by and bx – ay, where a and b are switched between variables and there is a change in sign, it is a good idea to square each of them and sum them up, because the result is very simple: (a2 + b2)(x2 + y2).
Hence, in our initial problem we need to move the term with b in our two linear equations to the right; then square them and sum up the results. This way we may get a very simple expression. And indeed, this trick leads to a solution and this solution provides insight into working with algebraic expressions.
This is the perfect problem to linger over, assuming you're not in the middle of a timed competition. It might make you wonder for which parameters this problem works. You might discover a new theorem that allows you to create a very similar problem from any number that can be represented as a sum of two non-trivial squares in two different ways.
To prepare my students for AMC I need to teach them tricks that are not useful at USAMO, or in mathematics in general for that matter. Many tricks distract from new ideas or from understanding the problem. All they give us is speed.
This bothers me, but to pacify myself, I keep in mind that most of my students will not become mathematicians and it might be useful in their lives to be able to make split-second decisions among a small number of choices.
However, it seems like Americans have the opposite problem: we make quick decisions without thinking. I'm concerned that training for multiple choice tests and AMC competitions aggravates this problem.
What object is missing?
Should you try to guess an answer to a multiple-choice problem during a test? How many problems should you try to guess? I will talk about the art of boosting your guessing accuracy in a later essay. Now I would like to discuss whether it makes sense to pick a random answer for a problem at AMC 10.
Let me remind you that each of the 25 problems on the AMC 10 test provides five choices. A correct answer brings you 6 points, a wrong answer 0 points and not answering at all gives you 1.5 points. So guessing makes the expected average per problem to be 1.2 points. That is, on average you lose 0.3 points when guessing. However, if you are lucky, guessing will gain you 4.5 points per problem, and if you are unlucky, it will lose you 1.5 points per problem.
So we see that on average guessing is unprofitable. But there are situations in which you have nothing to lose if you get a smaller score and a lot to gain if you get a better score. Usually the goal of a competitor at AMC 10 is to get to AIME. For that to happen, you need to get 120 points or be in the highest one percentile of all competitors. This rule complicates my calculations. So I decided to simplify it and say that your goal is to get 120 points and then see what mathematical results I can get out of that simplification.
First, suppose you are so accurate that you never make mistakes. If you have solved 20 problems, then your score without guessing is 127.5. If you start guessing and guess wrongly for all of the last five questions, you still have your desired 120 points. In this instance it doesn't matter whether you guess or not.
Suppose on the other hand that you are still accurate, but less powerful. You have only solved 15 problems, so your score without guessing is 105. Now you must be strategic. Your only chance to get to your goal of 120 points is to guess. Suppose you randomly guess the answers for the 10 problems you didn't solve. To make it to 120, you need to guess correctly at least five out of the ten remaining problems. The probability of doing so is 3%. Here is a table of your probability of making 120 points if you solve correctly n problems and guess the other problems.
| n | Probability |
|---|---|
| 20 | 1.0 |
| 19 | 0.74 |
| 18 | 0.42 |
| 17 | 0.20 |
| 16 | 0.09 |
| 15 | 0.03 |
| 14 | 0.01 |
| 13 | 0.004 |
| 12 | 0.001 |
We can see that if you solved a small number of problems, then the probability of getting 120 points is minuscule; but as the number of problems you solved increases, so does the probability of getting 120 points by guessing.
The interesting part is that if you have solved 19 problems, you are guaranteed to get to AIME without guessing. On the other hand, if you start guessing and all your guesses are wrong, you will not pass the 120 mark. The probability of having all six problems wrong is a not insignificant 26%. In conclusion, if you are an accurate solver and want to have 120 points, it is beneficial to guess the remaining problems if you solved fewer than 19 problems. It doesn't matter much if you solved fewer than 10 problems or more than 19. But you shouldn't guess if you solved exactly 19.
If you are not 100% accurate things get more complicated and more interesting. To decide about guessing, it is crucial to have a good estimate of how many mistakes you usually make. Let's say that you usually have two problems wrong per AMC test. Suppose you gave answers to 20 problems at AMC. What's next? Let us estimate your score. Out of your 20 answers you are expected to get 18*6 points for them plus 5*1.5 points for the problems you didn't answer. Your expected score is 115.5. You are almost there. You definitely should guess. But does it matter how many questions you are trying to guess?
The correct answer to one question increases your score by 4.5 with probability 0.2, and the wrong answer decreases your score by 1.5 with probability 0.8. One increase by 4.5 is enough for you goal of 120 points. So if you guess one question, with probability 0.2, you get 120 points. If you guess two questions, then your outcome is as follows: you increase your score by 9 points with probability 0.04; you increase your score by 3 points with probability 0.32; and you decrease your score by 3 points with probability 0.64. As you need at least a 4.5 increase in points, it is not enough to guess one question out of two. You actually need to guess correctly on both of them. The probability of this happening is 0.04. It is interesting, but you have a much greater chance to get to your goal if you guess just one question than if you guess two. Overall, here is the table of probabilities to get to 120 points where m is the number of questions you are guessing.
| m | Probability |
|---|---|
| 0 | 0 |
| 1 | 0.20 |
| 2 | 0.04 |
| 3 | 0.10 |
| 4 | 0.18 |
| 5 | 0.26 |
Your best chances are to guess all the remaining questions.
By the end of the test you know how many questions you answered, but you don't know how many errors you made. The table below tells you what you need in order to get 120 points. Here is how you read the table: The number of problems you solved is in the first column. If you are sure that the number of mistakes is not more than the number in the second column, you can relax as you made at least 120 points. The last column gives the score.
| Answered problems | Mistakes you can afford | Score |
|---|---|---|
| 19 | 0 | 123.0 |
| 20 | 1 | 121.5 |
| 21 | 2 | 120.0 |
| 22 | 2 | 124.5 |
| 23 | 3 | 123.0 |
| 24 | 4 | 121.5 |
| 25 | 5 | 120.0 |
If the number of mistakes you made is one more than in the corresponding row of the table, you should start guessing in order to try to get 120 points. Keep in mind that there is a risk: if you are not sure how many problems you solved already and start guessing, you might ruin your achievement of 120 points.
In the next table I show how many questions you can guess without the risk of going below 120 points. The word "all" means that it is safe to guess all the remaining questions.
| Answered problems | Mistakes you can afford | Score | Non-risky guesses |
|---|---|---|---|
| 19 | 0 | 123.0 | 2 |
| 20 | 1 | 121.5 | 1 |
| 21 | 2 | 120.0 | 0 |
| 22 | 2 | 124.5 | all |
| 23 | 3 | 123.0 | all |
| 24 | 4 | 121.5 | all |
You can see that if your goal is to get 120 points, your dividing line is answering 22 questions. If you solved 22 questions or more, there is no risk in guessing. Namely, if you have already achieved more than 120 points, guessing will not take you below that. But if you made more errors than are in the table, then guessing might be beneficial. Hence, you should always guess in this case — you have nothing to lose.
Now I would like to show you my calculations for a situation in which you are close to 120 points and need to determine the optimum number of questions to guess. The first column is the number of answered questions. The second column is the number of mistakes. The third column is your expected score without guessing. The fourth column is the optimum number of questions you should guess. And the last column lists your chances to get 120 points if you guess the number of questions in the fourth column.
| Answered problems | Mistakes | Score | Questions to guess | Probability of success |
|---|---|---|---|---|
| 17 | 0 | 114.0 | 8 | 0.20 |
| 18 | 0 | 118.5 | 7 | 0.42 |
| 18 | 1 | 112.5 | 7 | 0.15 |
| 19 | 1 | 117.0 | 2 | 0.36 |
| 19 | 2 | 111.0 | 6 | 0.10 |
| 20 | 2 | 115.5 | 5 | 0.26 |
| 21 | 3 | 114.0 | 4 | 0.18 |
| 22 | 3 | 118.5 | 3 | 0.49 |
| 22 | 4 | 112.5 | 3 | 0.10 |
| 23 | 4 | 117.0 | 2 | 0.36 |
| 23 | 5 | 111.0 | 2 | 0.04 |
| 24 | 5 | 115.5 | 1 | 0.20 |
You can see that almost always if you are behind your goal, you should try to guess all of the remaining questions, with one exception: if you answered 19 questions and one of them is wrong. In this case you should guess exactly two questions — not all that remain.
Keep in mind that all these calculations are very interesting, but don't necessarily apply directly to AMC 10, because I simplified assumptions about your goals. It may not be directly applicable, but I hope I have expanded your perspective about how you can use math to help you understand how better to succeed at math tests and how to design your strategy.
I plan to teach you how to guess more profitably, and this skill will also advance your perspective.
Lottery is a tax on people bad at math.
In this article I calculate how bad the lottery is as an investment, using Mega Millions as an example. To play the game, a player pays $1.00 and picks five numbers from 1 to 56 (white balls) and one additional number from 1 to 46 (the Mega Ball number, a yellow ball).
During the drawing, five white balls out of 56 are picked randomly, and, likewise, one yellow ball out of 46 is also picked independently at random. The winnings depend on how many numbers out of the ones that a player picks coincide with the numbers on the balls that have been drawn.
So what is your expected gain if you buy a ticket? We know that only half of the money goes to payouts. Can you conclude that your return is 50%?
The answer is no. The mathematical expectation of every game is different. It depends on the jackpot and the number of players. The more players, the bigger is the probability that the jackpot will be split.
Every Mega Millions playslip has odds printed on the back side. The odds of hitting the jackpot are 1 in 175,711,536. This number is easy to calculate: it is (56 choose 5) times 46.
How much is 175,711,536? Let's try a comparison. The government estimates that in the US we have 1.3 deaths per 100 million vehicle miles. If you drive one mile to buy a ticket and one mile back, your probability to die is 2.6/100,000,000. The probability of dying in a car accident while you drive one mile to buy a lottery ticket is five times higher than the probability of winning the jackpot.
Suppose you buy 100 tickets twice a week. That is, you spend $10,000 a year. You will need to live for 1,000 years in order to make your chances of winning the jackpot be one out of 10. For all practical purposes, the chance of winning the jackpot are zero.
As the probability of winning the jackpot is zero, we do not need to include it in our estimate of the expected return. If you count all other payouts then you are likely to get back 18 cents for every dollar you invest. You are guaranteed to lose 82% of your money. If you spend $1000 a year on lottery tickets, on average you will lose $820 every year.
If you do not buy a lot of tickets your probability of a big win is close to zero. For example, the probability of winning $250,000 (that is guessing all white balls, and not guessing a yellow ball) by buying one ticket is about 1 in 4 million. The probability of winning $10,000 — the next largest win — is close to 1 in 700,000. If we say that you have no chance at these winnings anyway, then your expected return is even less: it is 10 cents per every dollar you invest.
You might ask what happens if we pool our money together. When a lot of tickets are bought then the probability of winning the jackpot stops being zero. I will write about this topic later. For now this is what I would like you to remember. From every dollar ticket:
I am not at all trying to persuade you not to buy tickets. Lottery tickets have some entertainment value: they allow you to briefly dream about what you would do with those millions of dollars. But I am trying to persuade you not to buy lottery as an investment and not to put more hope into it than it deserves. If you treat lottery tickets as tickets to a movie that is played in your head, you will never buy more than one ticket at a time.
That is it. I advise you not to buy more than one ticket at a time. One ticket will allow you to dream about the expression on your sister's face when she sees your new $5,000,000 mansion, but will not destroy your finances.
Should you allocate time for checking your answers during important tests? I will use AMC 10/12 as an example, but you can adjust the calculations for any other test.
AMC 10/12 is a math competition that asks 25 multiple-choice questions that you need to solve in 75 minutes. You get 6 points for a correct answer, 1.5 points for an unanswered question and 0 points for a wrong answer.
Whether or not to take time to check your answers depends on you and the situation. If you finished your test and have some time left over, then surely you should use the extra time for checking your answers. If you only have three minutes left and your next problems are too complex to be dealt with in that time, then it is logical to use these moments to check back.
Sometimes, though, it isn't worth it to check your answers. If you haven't finished the test, but are a super-accurate person and never make mistakes, then it is better to continue working on the next problem than to waste time checking your correct answers. Also, if you rarely catch your own mistakes anyway, it doesn't make sense to check.
But things are not usually so clear. By the end of the test, most people need to make a decision: continue working through the problems or use the final moments to check the answers? How can you best decide if you should allocate time for checking and, if so, how much time?
The problems in AMC tests increase in difficulty. I suggest that each time you take the test or practice for AMC, take note of two things. How long did it take you to solve the test's last problem and what is the level of your accuracy for it. Suppose you know that at the end of the AMC test you can solve a problem in about 10 minutes and it is correct about 90% of the time. That means that investing the last ten minutes in solving the next problem will give you on average 5.4 points. If you remember that a blank answer gives you 1.5 points, you should realize that solving the last problem increases your score by 3.9 points. If you are very accurate, your score can increase more, but not more than 4.5 points.
Try conducting the following experiment. Take an AMC test from a past year. Do it for 65 minutes — the time of the test minus the time you need for that last problem. Then spend the last 10 minutes checking and correcting your answers. Now let us calculate how profitable that would be. Compare the scores you would have gotten without your corrections and with your corrections. If checking increases your score by more than 3.9 points, it is more profitable to check than to solve the next problem. If you do not make errors when you're trying to make corrections, the rule of thumb is that correcting one mistake is better than solving one problem. Indeed, your score increases by 6 points if you correct a mistake, and by not more than 4.5 points if you solve the next problem.
On all tests that punish wrong answers, correcting an error produces more points than solving a new question.
If you find that checking is profitable, but you can't check all the problems in ten minutes, you should consider allocating more time. Keep in mind, though, that you should adjust the sample calculation above for the last two problems. Remember, the next to the last problem is generally easier than the last problem. So if it takes you ten minutes to solve the last problem in the test, it most probably will take you less than twenty minutes to solve the last two. Also, since the difficulty increases throughout the test, the accuracy of the second to last problem might be better than the accuracy of the last problem. In addition, the first ten minutes that you check may be more productive than the next ten minutes of checking. So if you wonder if you should forgo the last two problems in order to check your earlier work, you have to redo the experiment anew, measuring both how long it takes you to solve those two problems and the benefits of checking.
This discussion can potentially help you to increase your score. However, there are other strategic considerations to weigh when deciding whether or not to check your work. For example, if the number of mistakes in your tests varies and sometimes you are 100% accurate and you are one problem away from your goal to get to AIME, it is more profitable to go for the last problem and hope for the best. I will discuss the strategic considerations for AMC some other time.
This essay is written especially for high school and undergrad math lovers who enjoy problem solving and who plan to major in mathematics. One of the authors, Tanya, often received this advice when she was an undergraduate in Russia: "Problem solving is child's play. You'll have to change your attitude if you plan to succeed in research."
Perhaps that's why some famous problem solvers, even those who won gold medals at IMO, became not-so-famous mathematicians. To help you avoid that fate, we'll discuss the ways in which research is unlike problem solving.
Yes and no. There are many mathematicians who continue problem solving as their form of research. Remember Paul Erdos who used to suggest a lot of problems and even offered money rewards for solutions. Many mathematicians solve problems posed by other people. You might consider Andrew Wiles as the ultimate math problem solver: he proved Fermat's last theorem, which had been open for 400 years. Though he could not have done it without the many theories that had already been generated in the search to find the elusive proof.
You can become a mathematician and continue to look around you for problems to solve. Even though this is still problem solving, the problems will be very different from competition problems, and you will still need to adjust to this type of research.
So, what is the difference between problems that mathematicians solve during competition and the problems they tackle for their research?
Expected answer. In competition problem solving you know there is a solution. Often you know the answer, but you just need to prove it. In research there is no guarantee. You do not know which way it will go. For this reason finding counter-examples and proving that some ideas are wrong is a positive contribution, for it can eliminate some possibilities. So one adjustment is that you might start valuing negative answers.
Difficulty level. Competition problems are designed to be solved in one hour, so you are expected to generate an idea in just minutes. In research the problem might drag on for years, because it is far more difficult. If you get used to the instant gratification of competition problem solving, you might find the lengthy work of research frustrating. It's very important to adjust your expectations so that you won't drop a problem prematurely. You need to measure progress in small intermediate steps and learn to appreciate this different rhythm.
Motivation. Although you miss the euphoria of finding quick solutions, you get a different kind of reward with research. Because no one knows the answer in advance, when you solve the problem, you are the first to do so. You have opened up a new truth.
Time limits. In competitions you have a time limit for every problem. In research you set your time limits yourself. That allows you to put a problem aside and come back later if necessary. In a sense you can think about several problems at the same time.
Your passion. You can choose your problems yourself. Research is much more rewarding if you follow your heart. In competitions you have to spend time on problems you might not like. Here you have an option to choose and pick only the problems that appeal to you. Thus, you become more motivated and as a result more successful.
After solving problems posed by other people, the next step is to pose math problems yourself. As we mentioned before, in research you do not always have a strictly-defined problem. It is a significant adjustment to move from solving already-defined problems to posing the problems yourself.
Generalizations. Often you can generalize from an existing problem to more general cases. For example, if you see a problem for n=3, you can wonder what happens for any n, or for any prime n.
Being on the lookout. Sometimes a situation puzzles you, but you can't formulate a specific problem around that situation. For example, why do most of the terms in the sequence end in 9? Is there a reason for that? Or, you might find that a formula from your integrable systems seminar is similar to a formula from your representation theory class. This might lead you to the essential research question: "What is going on?" You always need to be on the lookout for the right questions.
Value. When you create your own research problems it is crucial to always ask yourself: Is the problem I am creating important? What is the value of this problem? There is no a good reason to create random generalizations of random problems. If the problem you found interests you very much, that is the first sign that it might interest other people; nonetheless, you should still ask yourself how this problem will help advance mathematics.
There are other things to do than solve problems. There are many mathematicians who work differently, who don't solve problems or don't only solve problems. Here are some of the many options mathematicians have:
Building structures. You may not be interested in calculating the answer to a question, but rather in building a new structure or a new theory.
Advancing the language. When you invent new definitions and new notations, you will help to simplify a math language so that the new language will allow you to prove your results and other peoples' results faster and clearer.
Unification. Sometimes you notice two results in two different areas of mathematics with some kind of similarity. Explaining why these results are the same might create a new understanding of things. It is great to unify two different areas of mathematics.
Explaining. Very often proofs are not enough. Why is something true? What's the reason and what's the explanation? It is good to ask yourself a "why" question from time to time, such as, "Why is this proof working?" When you find an answer, it might become easier to understand what to do next and how to generalize your proof.
Directions. Many mathematicians are valued not for the problems they solve or suggest, but for ideas and directions they propose. Finding a new direction for research can generate unexpected opportunities and create tons of math problems on the way. It can be valuable to come up with good conjectures, even if you have no hope of solving them yourself. Two example of this are the Weil conjectures (eventually proved by Deligne) and the Langlands program, which is still incomplete but which has generated a huge amount of important research.
Vision. What is the most general thing that can be proved by this technique? What kinds of improvements and refinements are there? It is good to step back from the problem you solved and meta-think about it.
As you can see, problem solving is just the beginning of all that mathematics can offer you. Mathematicians find these other options very rewarding, so it's worth your while to try these varied aspects of mathematical work to see if you have a taste for other things. If you don't venture beyond problem solving you might miss the full beauty of mathematics.
I could no longer resist: I added a section of physics jokes to my math jokes collection:
* * *
A hydrogen atom says to the bartender, "Hey buddy, have you seen an electron around here? I seem to have lost mine."
"Are you sure you lost it?" the bartender asks.
And the hydrogen atom answers, "I'm positive!"
* * *
Heisenberg gets stopped on the motorway by the police.
Cop: "Do you know how fast you were going sir?"
Heisenberg: "No, but I know exactly where I am."
* * *
A photon checks into a hotel. The bellhop asks him, "Can I help you with your luggage?"
To which the photon replies, "I don't have any. I'm traveling light."
My recent entry, where I asked you to choose the odd one out among these images
was extremely popular. It was republished all around the world and brought my blog as much traffic in one day as I used to get in a month. Not only did I read the many comments I received, I also followed up on other peoples' blogs who reprinted my puzzle — at least those that were in either Russian or English. I also got private emails and had many conversations in person about it. The diversity of answers surprised me, so I would like to share them with you.
As I've said before, I do not think there is a correct answer to this type of question, but I was disappointed by some of the answers. For example, those who simply said, "The green one is the odd one out," made me feel that either they hadn't read the question or hadn't thought about it very much. It's a shame that these people spent more time sharing their opinion with the world than thinking about the problem in the first place.
I wouldn't mind someone arguing that the green one is the odd one out, but in this case an explanation is in order. Many people did offer explanations. Some told me that we perceive the color difference stronger than all other parameters I used, and the green figure pops out of the picture more than anything else. In fact, I personally perceive color difference the strongest among all the parameters, but since there are people who are color blind, I would disregard my feelings for color as being subjective.
You can create a whole research project out of this puzzle. For example, you can run an experiment: Ask the question, but flash the images above very fast, so there is no time for analysis — only time to guess. This allows us to check which figure is the first one that people perceive as different. Or you can vary the width of the frame and see how the perception changes.
Color was not the only parameter among those I chose — shape, color, size and the existence of a frame — that people thought was more prominent. My readers weighed these parameters unequally, so each argued the primary importance of the parameter they most emphasized. For example, one of my friends argued that:
The second figure should be the odd one out as, first, it is the only one without a frame, and, second, it is the only one comprised of one color rather than two. So it differs by two features, as others differ only by one feature.
A figure having one color is the consequence of not having a frame, so this particular friend of mine inflated the importance of not having a frame.
However, I can interpret any feature as two features. For example, I can say that the circle is the odd one out because not only is it a different figure, but it also doesn't have any angles. Similarly, the last one is the smallest one and the border width is in a different proportion to its diameter.
On a lighter side, there were many funny answers to the puzzle:
For the which-is-the-odd-one-out questions, the designer of the question is usually expecting a particular answer. So here's the answer I expected:
There is only one green figure. Wait a minute, there is only one circle. Hmm, there is only one without a frame and there's only one small figure. I see! The first one is the only figure that is not the odd one, that doesn't have a special property, so the first must be the odd one out. This is cool!
And the majority of the answers were exactly as I expected.
Since this is a philosophical problem, some of the responses took it to a different level. One interesting answer went like this:
All right, the last four figures have special features; the first figure is special because it is normal. Hence, every figure is special and there are no odd ones here.
I like this answer as the author of it equated regular features with a meta-feature, and it is a valid choice. This answer prompted me to write another blog entry with a picture where I purposefully tried to not have an odd one out:
Though I wrote that the purpose of this second set of images is to show an example where there is no odd one out, my commentators still argued about which one was the odd one out here.
Finally, I would like to quote Will's comment to my first set of images:
The prevailing opinion is that the first is least unique and is therefore the oddest. But it is the mean and the others are one deviation from it. Can the mean be the statistical anomaly?
And Cedric replied to Will:
Yes, I think the mean can be a statistical anomaly. The average person has roughly one testicle and one ovary. But a person with these characteristics would certainly be an anomaly.
I started my blog about two year ago. I have written about 200 entries. According to my traffic reports, these have been the top ten most popular entries:
My most popular category was Math Humor.
Israel Gelfand's memorial is being held at Rutgers on December 6, 2009. I was invited as Gelfand's student.
My relationship with Gelfand was complicated: sometimes it was very painful and sometimes it was very rewarding. I was planning to attend the memorial to help me forget the pain and to acknowledge the good parts.
I believe that my relationship with Gelfand was utterly unique. You see, I was married three times, and all three times to students of Gelfand.
Now that I know that I can't make it to the memorial, I can't stop wondering how many single male students of Gelfand will be there.
I've translated two problems from the 2009 Moscow Math Olympiad. In both of them our characters are genetically engineered octopuses. The ones with an even number of arms always tell the truth; the ones with an odd number of arms always lie. In the first problem (for sixth graders) four octopuses had a chat:
- "I have 8 arms," the green octopus bragged to the blue one. "You have only 6!"
- "It is I who has 8 arms," countered the blue octopus. "You have only 7!"
- "The blue one really has 8 arms," the red octopus said, confirming the blue one's claim. He went on to boast, "I have 9 arms!"
- "None of you have 8 arms," interjected the striped octopus. "Only I have 8 arms!"
Who has exactly 8 arms?
Not only do octopuses lie or tell the truth according to the parity of the number of their arms, it turns out that the underwater world is so discriminatory that only octopuses with six, seven or eight arms are allowed to serve Neptune. In the next problem (for seventh graders), four octopuses who worked as guards at Neptune's palace were conversing:
- The blue one said, "All together we have 28 arms."
- The green one said, "All together we have 27 arms."
- The yellow one said, "All together we have 26 arms."
- The red one said, "All together we have 25 arms."
How many arms does each of them have?
My students enjoyed the octopuses, so I decided to invent some octopus problems of my own. In the first problem, the guards from the night shift at Neptune's palace were bored, and they started to argue:
- The magenta one said, "All together we have 31 arms."
- The cyan one said, "No, we do not."
- The brown one said, "The beige one has six arms."
- The beige one said, "You, brown, are lying."
Who is lying and who is telling the truth?
In the next problem the last shift of guards at the palace has nothing better to do than count their arms:
- The pink one said, "Gray and I have 15 arms together."
- The gray one said, "Lavender and I have 14 arms together."
- The lavender one said, "Turquoise and I have 14 arms together."
- The turquoise one said, "Pink and I have 15 arms together."
What number of arms does each one have?
An ancient Russian joke:
Patient: Doctor, is there a medicine I can use to prevent my girlfriends from become pregnant?
Doctor: Kefir.
Patient: Should I drink it before or after sex?
Doctor: Instead of.
I have a more pleasurable suggestion than drinking kefir: date postmenopausal women. There are many other reasons why men enjoy dating older women, but since my blog is about mathematics, I would like to dig into some relevant numbers.
We know that boys are born more often than girls, and men die earlier than women. Somewhere around age 30 the proportion in population switches from more boys to more girls. And it gets more skewed with age. So there's a deficit of older men. In addition, a big part of the population is married, making the disproportions in singles group more pronounced. So I decided to look at the numbers to see how misshaped the dating scene is.
This 2008 data comes from the U.S. government census website's table "Marital Status of the Population by Sex and Age: 2008. (Numbers in thousands. Civilian non-institutionalized population.)" To calculate the number of singles, I summed up the widowed, divorced and never married columns.
| Age Group | Single Male | Single Female | Ratio M/F |
|---|---|---|---|
| Total | 44,707 | 51,293 | 0.87 |
| 15 to 17 years | 6,729 | 6,513 | 1.03 |
| 18 to 24 years | 13,074 | 11,848 | 1.10 |
| 25 to 29 years | 6,639 | 5,224 | 1.27 |
| 30 to 34 years | 3,901 | 3,343 | 1.17 |
| 35 to 39 years | 3,354 | 2,965 | 1.13 |
| 40 to 44 years | 3,410 | 3,270 | 1.04 |
| 45 to 49 years | 3,476 | 3,591 | 0.97 |
| 50 to 54 years | 2,979 | 3,385 | 0.88 |
| 55 to 59 years | 2,309 | 3,123 | 0.74 |
| 60 to 64 years | 1,552 | 2,746 | 0.57 |
| 65 to 69 years | 1,082 | 2,423 | 0.47 |
| 70 to 74 years | 787 | 2,162 | 0.36 |
| 75 to 79 years | 790 | 2,391 | 0.33 |
| 80 to 84 years | 685 | 2,430 | 0.28 |
| 85 years and over | 669 | 2,391 | 0.28 |
These data alone cannot explain the dating situation. For example, I have no way of knowing what proportion of each gender isn't interested in dating the opposite sex, or even in dating altogether. But the trend is quite clear: the proportion of men in younger categories is much higher. That implies that there is less competition for older women. So those young men who are open to dating much older women might have more options and those options might be more interesting.
I just turned 50 and plan to return to dating again. Looking at the data I see that there are 11 million single men older than me and 34 million who are younger than me. If I were to pick a single man randomly, I am three times more likely to end up with a younger man.
Supposedly we live in a free society, where people can do what they want as long as they do not harm anyone else. Still our society often disapproves of women dating much younger men. Consider this definition from Wikipedia:
"Cougar — a woman over 40 who sexually pursues a much younger men."
This derogatory term portrays such women as predatory. Not only is there nothing wrong with women dating younger men, but it makes no sense for older women to ignore the imbalance of the dating scene and be closed to relationships with much younger men. After all, the demographics are also affected by the fact that women live longer, probably because of their healthy life style, non-risky behavior and positive attitude to life.
Can someone explain to me again why sane, healthy, non-risky women with positive attitudes to life are called "cougars"?
The first episode of Numb3rs: Season Six
Suppose you are in a prison and the guard says to you, "You will be hanged tomorrow at noon and it will be a surprise." You presume that you can't be surprised since they already told you, so there is a contradiction in what they've said. Therefore, you conclude that they can't hang you and you relax. Next day at noon the guard comes for you, to take you to be hanged, and you are utterly surprised. Oops.
What I do not like about this paradox is that it assumes that you do not know about the paradox. I, on the other hand, imagine that you, my reader, are logical and intelligent. So the moment the guard tells you that you will be hanged tomorrow at noon and it will be a surprise, you realize that the situation depends on what you decide to believe in now. If you decide that you won't be hanged tomorrow, then you will have a relatively relaxing day today and you will be caught by surprise tomorrow and die. If you decide that you will die tomorrow, then you will have a nerve-wracking day today, but the guard may release you, to save his honor, since you won't be surprised.
The original hangman's paradox in which the guard tells you that you will be hanged on a weekday the following week and that you will be caught by surprise, also assumes that you are not aware of the paradox. If you are aware of the paradox, you know that usually guards in this paradox come for you on Wednesday, so you can prepare yourself. Actually, to guarantee your survival, if not your feeling of moral superiority, you can daily persuade yourself to belief that you will be hanged at noon the next day. This way, you will never be caught by surprise. If you are a person who can control your own beliefs, you may be able to save your life.
The book An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica is targeted at people preparing for USAMO and IMO. It contains a lot of problems on Diophantine equations from math Olympiads used in various math Olympiads all over the world.
The first chapter discusses several methods for solving Diophantine equations: decomposition, using inequalities, using parameters, modular arithmetic, induction, infinite descent, and other miscellaneous ideas. Each sub-chapter starts with a short description of the method, accompanied by several solutions to sample problems. At the end of each sub-chapter there are a plethora of exercise problems.
The second and the third chapters are more theoretical. The former discusses some classical equations and the latter looks at Pell's equation. These two chapters also contain problems, but the bulk of the chapters is devoted to basic theory that is essential to an understanding of Diophantine equations.
For those who are training for the Olympiads, this is an important book to own, not only because there are few other books on the subject, but because it provides so many useful problems.
I've long complained that most training books for math competitions leave out any discussion of how we choose a method by just looking at a problem. Andreescu and Andrica didn't fill that gap with this book.
Perhaps in their next book they will point out clues that indicate that a particular problem might be solved by the parametric method. And explain which types of problems are best solved with induction. Let them challenge students to find those clues in a problem that help us to judge which method might be most promising, instead of randomly trying one method after another. Let me give you a sample problem from the book, which originated at the Balkan Mathematical Olympiad:
Prove that the equation x5 – y2 = 4 has no solutions in integers.
The solution is to take the equation modulo 11, and see that it is impossible.
Is there a reason to start with the modular arithmetic method and not with other methods? If we use modular arithmetic, do we recognize why it's best to start with 11? I'm convinced that this problem has sufficient clues to suggest starting with checking this equation modulo 11.
I wonder if you, my readers, agree with me. If so, can you explain which hints in the problem lead to taking the equation modulo 11? I believe it should be a part of competition training to learn to identify clues that suggest that one direction might be preferable to the others.
I love Harvard-MIT Math Tournaments. I like the mini-events, especially when I learn a new game. I also like the guts round, where I enjoy the adrenaline rush of watching the progress in real time. I also like the fact that I know many of the kids from different teams: my current students, my former students, the members of my club, my Sergei's friends.
The problems for the competitions are designed by undergraduate students at MIT and Harvard. Kudos to them. Still, I was somewhat disappointed with the November 2009 problems. Most problems are variations of standard problems with different parameters. It is not easy to design a problem, but I was hoping for something fresh.
My favorite problem from the HMNT 2009 tournament was in the theme round:
There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:Who are the liars?
- Alan: "All of us are truth-tellers."
- Bob: "No, only Alan and I are truth-tellers."
- Casey: "You are both liars."
- Dan: "If Casey is a truth-teller, then Eric is too."
- Eric: "An odd number of us are liars."
My second favorite problem was in the guts round:
Six men and their wives are sitting at a round table with 12 seats. These men and women are very jealous — no man will allow his wife to sit next to any man except for himself, and no woman will allow her husband to sit next to any woman except for herself. In how many distinct ways can these 12 people be seated such that these conditions are satisfied? (Rotations of a valid seating are considered distinct.)
This was the funniest problem:
You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it by cutting off its two claws and six legs and attacking its weak point for massive damage. You cannot cut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak point until you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab? (Note that the legs are distinguishable, as are the claws.)
It is difficult to arrange so many problems for four rounds without mistakes. The error in the following problem is not a typo and it bothers me that no one caught it:
Pick a random digit in the decimal expansion of 1/99999. What is the probability that it is 0?
Hey, there is no uniform distribution on an infinite set of integers: picking a random digit is not defined.
Last month I gave my students a problem from Raymond Smullyan's book The Riddle of Scheherazade
A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. How many of them can each say that it is the same color as another one of Hassan's horses?
Half of my students failed to notice the trick and gave the wrong answer. Recently I gave them the continuation of the problem from the same book:
A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. Assuming now that Hassan's horses can talk, how many of them can each say that it is the same color as another one of Hassan's horses?
This time the majority of my students didn't notice the trick. This motivated me to continue playing jokes with them. Unfortunately though, Raymond Smullyan had only two problems about Hassan's horses, so I have to invent the next one myself. Here is what I plan to give my students next time:
A certain sheik named Hassan has eight horses. Four of them are white, three are black, and one is brown. Assuming now that Hassan's horses can talk and always tell the truth, how many of them will say that it is the same color as another one of Hassan's horses?
Feel free to continue the series.
My recent blog puzzle where my readers had to choose the odd one out became extremely popular and was republished in many blogs around the world. Some commentators decided that my posting was a joke and an example where the odd one out didn't exist. I have to disappoint them: as a protest against find-the-odd-one-out questions and to illustrate that sometimes there is no good choice for the odd one out I would have chosen a different picture:
Can you find the odd one out?
Inspired by Michael Huber, who in his new book Mythematics
The day was passing towards sunset when the Company finally caught a long-awaited gleam of water, from which sparkled flickers of sunlight. As they quietly drew nearer, they laid their eyes on the next obstacle — a river that they had to transverse. The Company was footsore and tired and the hobbits were starving. But they couldn't rest yet. They needed to collect materials with which to construct their raft before it became too dark. By nightfall they managed to build a tiny raft, and eagerly started their supper.
They couldn't wait until dawn to build more rafts, for they needed to cross the river now. So while they rested, Aragorn smoked his pipe and began to contrive a plan.
Aragorn was in charge and there were eight of them. The four hobbits — Frodo, Sam, Merry and Pippin — were not very useful in battle. However, the four strong fighters — Aragorn, Gimli, Legolas and Boromir, who were sworn to protect the ring-bearer Frodo — were the best in the land.
The small raft they had built would not hold a lot of weight. Aragorn and Boromir were the heaviest. Gimli was short, but together with his armor he weighed as much as either Aragorn or Boromir. Each one of these three heaviest warriors was close to the raft's maximum capacity, so they had to each be alone on the raft while crossing the river. Among the strong fighters, only Legolas was able to cross the river with a hobbit. The raft could also accommodate two hobbits.
Weight was not Aragorn's only consideration: the current was dangerously fast. All the strong men could row, but among the hobbits, only Sam was strong enough to row against such a swift current.
Aragorn also worried about the orcs, who were roaming on both sides of the river. He didn't want to leave any hobbit(s) alone on a riverside, without the safeguard of a strong fighter. Because he was the ring-bearer, Frodo needed extra protection. Aragorn wanted Frodo to be accompanied by at least two strong men. But lately Boromir had become restless when he was around the ring and Aragorn couldn't count on him to look after Frodo. That is, while on the riverside, Frodo's protection had to come from two out of the three remaining strong men: Aragorn, Legolas and Gimli.
Can you help Aragorn design a plan to cross the river?
In a Turing test a human judge on one end of an interface interacts with either a computer or another human through this interface. If the judge can't differentiate a machine from a human, then the computer is said to pass the test. One big goal of folks working in Artificial Intelligence is to build a computer that, when subjected to this test, is indistinguishable from a human.
However, while some people are working hard trying to build programs that can pass as humans, other people are working hard inventing tests that can differentiate between humans and those programs. Such tests are sometimes called Reverse Turing tests. As computer science progresses, the programs that are pretending to be humans as well as reverse tests are becoming increasingly complex.
For example, banks frequently want to prevent malicious computer programs from trying to log into their customers' accounts. As a nice touch the judges are computers in this case. There are different methods designed to confirm that a human is trying to log in. In one of them a picture of a word, called CAPTCHA is presented on a screen, and the program requires that this word be typed in.
I wanted a CAPTCHA with words "Turing Test" in it for this posting. I looked online trying to find a way to do it. I couldn't. There is a ton of software that can produce random CAPTCHAs from a dictionary but nothing could do a particular word. Finally, rather than looking for software, I found a human, a kind gentlemen named Leonid Grinberg who with some GIMP help manually implemented a self-referencing " symbolizing the race between computers and tests.
As text recognition software becomes better and better, these CAPTCHAs become more and more difficult to read by a human. The last time I tried to login, I was only able to type the right word on my fourth try. Very soon computers will be better than humans at parsing CAPTCHAs. Humans are loosing the race on visual methods like this one.
Here's another example. Some malicious software can recognize and capture email addresses on webpages to use for spam. While we don't want them to recognize email addresses, we do want people to be able to do so. Thus we need a way to present email addresses as a reverse Turing test.
The standard safety recommendation is to avoid writing out the full and exact email address. Here's an example: billgates AT gmail DOT com. Actually, I think computers are so smart nowadays that they can learn this trick. Another idea is to show a picture of your email address instead of using characters. Here we return to the image idea, which most computers can nowadays recognize.
Another idea of how to hide an email address is to give simple clues, which point to characters in the email address. For example, if you have "4" in your address, you might say that the character is the sum of two and two. I already invented a version of my email address in which each letter of my username is an answer to a simple question. Unfortunately, I think that the question answering systems like Start, as well as its huge new competitor Wolfram Alpha, will learn to answer these questions very soon. I can construct more sophisticated questions, but that would require my readers to spend more time to figure it out including going back to school for a calculus class.
So, recently, I've come up with a new idea. I made the description of my email simpler, but the paragraph describing my email didn't contain all the necessary information:
I have an email account with Yahoo. My account name consists of seven lower case letters: five letters of my first name concatenated with the first two letters of my last name.
People who want to contact me can easily find my name in the title of my webpage or in my url, but I hoped it would take the evil computers some time to figure out what to look for, where it's located and how to turn it into an address.
The day after I changed my contact web page, I went to my math coaching work at AMSA. During my break, I wanted to unwind by solving a light up puzzle, but it appeared that the new security system at AMSA forbids Internet access to all gaming sites. Thus, being still wound up I decided to do some work and went to my personal page for some materials. I was blocked again. The software politely informed me that access to personal websites was not permitted either. Oops. If a computer can understand that it is a personal website, it probably can figure out the name of the corresponding person. Oops-Oops-Oops. I am loosing the race against computers again. My recent idea to protect my email address from spam lasted one day until my first reality check.
In my days of competing in math, I met guys who could solve any geometry problem by using coordinates: first they would assign variables to represent coordinates of different points, then they would write and solve a set of equations. It seemed so boring. Besides, this approach doesn't provide us with any new insight into geometry.
I find geometric solutions to geometry problems much more interesting than algebraic solutions. The geometric solutions that use geometric transformations are often the shortest and the most beautiful.
I.M. Yaglom wrote a great trilogy called The Geometric Transformations. The first book of this trilogy
I think geometry is the weakest link for the USA math team. So we have to borrow the best geometry books from other countries. This trilogy was translated from Russian and Russians are known for their strong tradition of excellence in teaching geometry.
Below you can find sample problems from Geometric Transformations 1
Problem 1. Let A be a point outside a circle S. Using only a straightedge, draw the tangents from A to S.
Problem 2. At what point should a bridge be built across a river separating two towns A and B (see figure) in order that the path connecting the towns be as short as possible? The banks of the river are assumed to be parallel straight lines, and the bridge is assumed to be perpendicular to the river.
Problem 3. Suppose you have two lines drawn on a piece of paper. The intersection point A of the two lines is unreachable: it is outside the paper. Using a ruler and a compass, draw a line through a given point M such that, were the paper bigger, point A would belong to the continuation of the line.
I added my favorite webcomics from "Not So Humple Pi" to my collection of funny math pictures.
Suppose you got accepted to the college of your dreams, say MIT. If you are so poor that MIT gives you a full financial package or you are so rich that the cost is not an issue, then you might throw a party. Everyone else, however, needs to wait for the financial package letter from MIT. The dream depends on the willingness and the ability of the parents to pay.
Suppose your father looks at the bill in shock. Then he takes you for a walk and tells you to forget about MIT and go to the state college, as he can't pay the requested amount.
If you know for sure that your father has the money, what is the first question that you should ask him? The first question should be: "Are you still married to my mother?" If you are not completely clueless, you ought to know the answer to this question already. The family status of your parents may be the deciding factor in whether or not you can get your father to pay.
If your parents are divorced, your college expenses might be covered by their divorce agreement. In this case, there would be a legal document designating how your parents need to pay. If your father refuses to pay, your mother can use the divorce agreement to threaten your father with a complaint. The threat might be enough. If it is not, the court will probably force the reluctant father to pay according to the divorce agreement. So if your parents are divorced, it might be a good idea for you to scrutinize their divorce agreement.
Even if your parents' lawyers neglected to include college expenses in the divorce agreement, you might still be able to finance your college education. Your mother, for example, might sue your father for college expenses.
I wonder what happens if the divorce agreement covers your college expenses, but neither parent wants to pay. I'm curious whether or not it is possible for the child to sue the parents based on the agreement he/she is not a party to. If any reader knows the answer, I'd appreciate hearing from you.
If your parents are together, there is no divorce agreement to protect your interests. It seems that legally the situation favors the children of divorced parents. If your parents do not love each other and have stayed in their marriage for your sake, it might be to your financial advantage to persuade them to divorce well before you need to go to college. Do not disregard reminding their lawyers to include college expenses in the agreement.
Israel Gelfand was my scientific adviser from the time I was 15. This is the story of how Gelfand helped me, when at 20 I was an undergrad at Moscow State University. At that time, I was married to Sasha (Alexander) Goncharov, who was also Gelfand's student.
Sasha was more driven by mathematics than I. I had a lot of different interests: I wanted to hang out with friends, go to movies and read books. Sasha only wanted to do mathematics. His only other obsession was with what our colleagues (including me) were doing mathematically. So he was constantly asking me about the math problems I was thinking about.
For example, I was sitting at my side of the desk working, and he asked me to tell him about my problem. A few minutes later, I was forced to interrupt my work to go grocery shopping, because the household chores fell to me. As soon as I returned with bread and milk, Sasha excitedly told me the solution to my problem. It made me feel stupid, as if I should have solved it while I was waiting in the line for bread and milk. That feeling blocked out all the other feelings I should have been noticing, such as frustration and annoyance with Sasha.
Without his interference, I would have happily solved the problem myself. I was about to start my serious research, but I worried that I'd end up as a supplier of new problems for his papers.
You might wonder why I didn't stop sharing my math with Sasha. But at that time, I wasn't very in touch with my feelings and I prided myself on being a logical person. The idea that a husband and wife would discuss their work together seemed logical. Besides, even though I wasn't particularly interested, Sasha was always ready to tell me about his math problems. It seemed important for me to be fair and to reciprocate. So I was stuck in a situation I didn't know how to resolve.
I never confided this issue to any math colleagues. I never mentioned it to Gelfand — mostly because I was too scared of him to initiate any conversation. Besides, Gelfand delegated most of his responsibilities to others, because he was quite famous and busy. For example, all official paperwork related to his adviser role was done by Alexandre Kirillov. With me avoiding Gelfand and Gelfand being busy, we almost never spoke one-on-one.
You can understand my surprise when one day Gelfand approached Sasha and me to have a chat. He told us that we were about to start our own research, and while he permitted me to ask Sasha about what he was doing, he would not allow Sasha to interfere with my research.
Gelfand was a great judge of character. Without anyone telling him, he perceived what was going on in our marriage and gave me an excuse to stop Sasha's prying. It was an appreciated gift.
I am strongly opposed to questions of the type "which is the odd one out" during IQ tests. On the other hand, I do not mind them in different settings, especially when they are fun. Inspired by Martin Gardner, I spent a lot of time drawing this picture, and now I have to share it with the world. So, which is the odd one out?
I love the MIT Mystery Hunt. I like the adrenalin rush when solving problems under pressure. Plus, I like the togetherness of doing problems with other people. During the hunt I usually do not have time to look at all the puzzles: some of them are solved by my teammates while I'm sleeping and others are solved before I get to see them.
I've never tried to go back and check out the puzzles I missed nor the puzzles from the previous hunts, probably because without the goal of winning and without my team, I might find them boring. Often the solving process involves tedious Internet browsing to identify the images of different people or objects. I would only be motivated if the puzzle were related to something I am very interested in, such as Ballroom dancing. But I'm not thrilled at the thought of browsing through all the problems in order to find one that is relevant to the Tango.
In short, I need an index to the puzzles. For example, it would be nice to direct the lovers of square dancing to the Do Sa Do puzzle, or fans of Star Trek to the Alien Species puzzle. I hope that nobody blames me for hinting that those aliens are from Star Trek. I'm convinced that Trekkies who only want to solve Star Trek-related puzzles would immediately recognize them anyway. I do believe that I am not revealing too much by saying that the Facebook puzzle will appeal to the aficionados of the television show "24".
It would be extremely useful to humanity to at least mark the MIT Mystery Hunt puzzles that are self-consistent, and do not require activities. For example, some of the puzzles involve interaction with headquarters, so you can't solve them after the hunt. Some of the puzzles might expire, as for example the puzzle with pictures of different announcements in the infinite corridor.
Unfortunately, such an index doesn't exist, and I do not have the time or expertise to create one myself. But I can fill this void at least partially by presenting a guide to math puzzles from the previous four hunts. I can't promise that my guide is complete, as navigating the MIT Mystery Hunt website is very tiresome.
Before going into the math puzzles, I would like to list Sergei's favorite type of puzzle: Duck Konundrums. The first Duck Konundrum puzzle appeared in 2000. It was created by Dan Katz, which is why his initials are in the title. One really needs to follow the instructions for this puzzle. This is very unusual as traditionally hunt puzzles do not have instructions at all. Do not be relieved: the instructions are really very complicated. The next Duck Konundrum puzzle appeared in 2002 and was considered to be even more amusing. People loved it, and this type of puzzle became a tradition in subsequent hunts. Here is my list of Duck Konundrums:
Many Mystery Hunt puzzles appeal to mathematicians. I have to warn you though. Puzzles often are divided into two stages. In the first stage, you need to solve a puzzle, like solving sudoku, a crossword or finding all the wedding dates of the people in the pictures. The second stage requires you to produce a word or a phrase that is the answer to the puzzle. The second stage might be as simple as listing the people in chronological order of their wedding dates and then taking the first letters of their last names. This second stage could also be quite difficult. Depending on your tastes one stage of the puzzle might be much more rewarding than the other. If you love solving sudokus, you might find it more fun to just stop with that solution, instead of continuing on to the second stage.
2006
2007
2008
2009
It would also be nice to have some ratings for puzzles. I am not sure how to persuade the webmasters of the MIT Mystery Hunt page to do the index and the rating. Feel free to send them an encouraging email.
In the book "Mythematics: Solving the Twelve Labors of Hercules
The problems in Mythematics are quite advanced. They range in topic from algebra, geometry and probability to differential equations and integral calculus. Plus, as a reward for helping Hercules, Huber gives you variations on Sudoku puzzles.
Solving some nice math problems might not be the only reason for people to buy this book. Here are some other reasons:
I like Huber's approach. Future possibilities for more books are endless. Let's write new math problems based on Harry Potter, Batman, the Bible or, maybe, The Joy of Sex
Konstantin Knop sent me the following coins puzzle, which was created by Alexander Shapovalov and first appeared at the Regional round of the all-Russian math Olympiad in 2000.
Baron Münchhausen has 8 identical-looking coins weighing 1, 2, …, 8 grams. The Baron knows which coin is which and wants to demonstrate to his guests that he is right. He plans to conduct one weighing on a balance scale, so that the guests will be convinced about the weight of one of the coins. Can the Baron do this?
This being a sequence-lover blog, we want to create a sequence out of this puzzle. The sequence is the following: Let the Baron initially have n identical-looking coins that weigh exactly 1, 2, …, n grams. Then a(n) is the minimum number of weighings on a balance scale that the Baron needs in order to convince his guests about the weight of one of those coins.
The original puzzle can be restated as asking whether a(8) = 1. The sequence is defined starting from index 1 and the first several terms are easy to calculate: 0, 1, 1, 1, 2, 1, 1, 1. Can you continue this sequence?
Let's look at where ones occur in this sequence:
Theorem. If the weight of a coin can be confirmed with one weighing, then one cup of that weighing must contain all the coins with weights from 1 to some i, and the other cup must contain all the coins with weights from some j to n. Furthermore, either the scale must balance, or the cup containing the 1-gram coin must be lighter.
Proof. What does it mean for the Baron to convince his guests about the weight of some coin with one weighing? From the perspective of the guests, a weighing is a number of coins in one cup, a number of coins in the other cup, and a number of coins not on the scale, together with the result the scale shows (one or the other cup heavier, or both the same weight). For the guests to be convinced of the weight of some particular coin, it must therefore be the case that all possible arrangements of coin weights consistent with that data agree on the weight of the coin in question. Our proof strategy, therefore, is to look for ways to alter a given arrangement of coin weights so as to change the weight given to the coin whose weight is being demonstrated, thus arriving at a contradiction.
First, obviously, the coin whose weight k the Baron is trying to confirm has to be alone in its group: either alone on some cup or the only coin not on the scale. After that we can divide the proof of the theorem into several cases.
Case 1. The coin k is on a cup and the scale is balanced. Then we want to show two things: k = n, and the coins on the other cup weigh 1, 2, …, i grams for some i. For the first part, observe that if k < n, then the coin with weight k+1 must not be on the scale (otherwise it would overbalance coin k). Therefore, we can substitute coin k+1 for coin k, and substitute a coin one gram heavier for the heaviest coin that was on the other cup, and produce thereby a different arrangement with the same observable characteristics but a different weight for the coin the Baron claims has weight k.
To prove the second part, suppose the contrary. Then it is possible to substitute a coin 1 gram lighter for one of the coins on the other cup. Now, if coin k-1 is not on the scale, we can also substitute k-1 for k, and again produce a different arrangement with the same observable characteristics but a different weight for the coin labeled k. On the other hand, if k-1 is on the scale but k-2 is not, then we can substitute k-2 for k-1 and then k-1 for k and the weighing is again unconvincing. Finally, if both k-1 and k-2 are on the scale, and yet they balance k, then k=3 and the theorem holds.
Consequently, k = n = 1 + 2 + … + i is a triangular number.
Case 2. The coin k is on the lighter cup of the scale. Then: first, k = 1, because otherwise we could swap k and the 1-gram coin, making the light cup lighter and the heavy cup heavier or unaffected; second, the 2-gram coin is on the heavy cup and is the only coin on the heavy cup, because otherwise we could swap k with the 2-gram coin and not change the weights by enough to affect the imbalance; and finally n = 2 because otherwise we could change the weighing 1 < 2 into 2 < 3.
Thus the theorem holds, and the only example of this case is k = 1, n = 2.
Case 3. The coin k is on the heavier cup of the scale. Then k = n and the lighter cup consists of some collection of the lightest available coins, by the same argument as Case 1 (but even easier, because there is no need to maintain the balance). Furthermore, k must weigh exactly 1 gram more than the lighter cup, because otherwise, k-1 is not on the lighter cup and can be substituted for k, making the weighing unconvincing.
Consequently, k = n = (1 + 2 + … + i) + 1 is one more than a triangluar number.
Case 4. The coin k is not on a cup and the scale is not balanced. Then, since k must be off the scale by itself, all the other coins must be on one cup or the other. Furthermore, all coins heavier than k must be on the heavier cup, because otherwise we could make the lighter cup even lighter by substituting k for one of those coins. Likewise, all coins lighter than k must be on the lighter cup, because otherwise we could make the heavier cup even heavier by substituting k for one of those coins. So the theorem holds; and furthermore, the cups must again differ in weight by exactly 1 gram, because otherwise we could swap k with either k-1 or k+1 without changing the weights enough to affect the result on the scale.
Consequently, the weight of the lighter cup is k(k-1)/2, the weight of the heavier cup is 1 + k(k-1)/2. Thus the total weight of all the coins is n(n+1)/2 = k2+1. In other words, case 4 is possible iff n is the index of a triangular number that is one greater than a square.
Case 5. The coin k is not on a cup and the scale is balanced. This case is hairier than all the others combined, so we will take it slowly (noting first that all the coins besides k must be on some cup).
Lemma 1. The two coins k-1 and k-2 must be on the same cup, if they exist (that is, if k > 2). Likewise k-2 and k-4; k+1 and k+2; and k+2 and k+4.
Proof. Suppose they're not. Then we can rotate k, k-1, and k-2, that is, put k on the cup with k-1, put k-1 on the cup with k-2, and take k-2 off the scale. This makes both cups heavier by one gram, producing a weighing with the same outward characteristics as the one we started with, but a different coin off the scale. The same argument applies to the other three pairs of coins we are interested in, mutatis mutandis.
Lemma 2. The four coins k-1, k-2, k-3 and k-4 must be on the same cup if they exist (that is, if k ≥ 5).
Proof. By Lemma 1, the three coins k-1, k-2, and k-4 must be on the same cup. Suppose coin k-3 is on the other cup. Then we can swap k-1 with k-3 and k with k-4. Each cup becomes heavier by 2 grams without changing the number of coins present, the balance is maintained, and the Baron's guests are not convinced.
Lemma 3. If coin k-4 exists, that is if k ≥ 5, all coins lighter than k must be on the same cup.
Proof. By Lemma 2, the four coins k-1, k-2, k-3 and k-4 must be on the same cup. Suppose some lighter coin is on the other cup. Call the heaviest such coin c. Then, by choice of c, the coin with weight c+1 is on the same cup as the cluster k-1, &hellip, k-4, and is distinct from coin k-2 (because c is on a different cup from k-3). We can therefore swap c with c+1 and swap k with k-2. This increases the weight on both cups by 1 gram without changing how many coins are on each, but moves k onto the scale. The Baron's guests are again unconvinced.
Lemma 4. The theorem is true for k ≥ 5.
Proof. By Lemma 3, all coins lighter than k must be on the same cup. Further, if a coin with weight k+4 exists, then by the symmetric version of Lemma 3, all coins heavier than k must also be on the same cup. They must be on the other cup from the coins lighter than k because otherwise the scale wouldn't balance, and the theorem is true.
If no coin with weight k+4 exists, that is, if n ≤ k+3, how can the theorem be false? All the coins lighter than k must be on one cup, and their total weight is k(k-1)/2. Further, in order to falsify the theorem, at least one of the coins heavier than k must also be on that same cup. So the minimum weight of that cup is now k(k-1)/2 + k+1. But we only have at most two coins for the other cup, whose total weight is at most k+2 + k+3 = 2k + 5. For the scale to even have a chance of balancing, we must have
k(k-1)/2 + k+1 &le 2k + 5 ⇔ k(k-1)/2 ≤ k + 4 ⇔ k(k-1) ≤ 2k + 8 ⇔ k2 - 3k - 8 ≤ 0.
Finding the largest root of that quadratic we see that k < 5.
So for k ≥ 5, the collection of all coins lighter than k is heavy enough that either one needs all the coins heavier than k to balance them, or there are enough coins heavier than k that the theorem is true by symmetric application of Lemma 3.
Completion of Case 5. It remains to check the case for k < 5. If n > k+3, then coin k+4 exists. If so, all the coins heaver than k must be on the same cup. Furthermore, since k is so small, they will together weigh more than half the available weight, so the scale will be unbalanceable. So k < 5 and n ≤ k+3 &le 7.
For lack of any better creativity, we will tackle the remaining portion of the problem by complete enumeration of the possible cases, except for the one observation that, to balance the scale with just the coin k off it, the total weight of the remaining coins, that is, n(n+1)/2 - k must be even. This observation cuts our remaining work in half. Now to it.
Case 5. Seven Coins. n = 7. Then 5 > k ≥ n - 3 = 4, so k = 4. Then the weight on each cup must be 12. One of the cups must contain the 7 coin, and no cup can contain the 4 coin, so the only two weighings the Baron could try are 7 + 5 = 1 + 2 + 3 + 6, and 7 + 3 + 2 = 1 + 5 + 6. But the first of those is unconvincing because k+1 = 5 is not on the same cup as k+2 = 6, and the second because it has the same shape as 7 + 3 + 1 = 2 + 4 + 5 (leaving out the 6-gram coin instead of the asserted 4-gram coin).
Case 5. Six Coins. n = 6. Then 5 > k ≥ n - 3 = 3, and n(n+1)/2 = 21 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 9. The 6-gram coin has to be on a cup and the 3-gram coin is by presumption out, so the Baron's only chance is the weighing 6 + 2 + 1 = 4 + 5, but that doesn't convince his skeptical guests because it looks too much like the weighing 1 + 3 + 4 = 6 + 2.
Case 5. Five Coins. n = 5. Then 5 > k ≥ n - 3 = 2, and n(n+1)/2 = 15 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 6. The only way to do that is the weighing 5 + 1 = 2 + 4, which does not convince the Baron's guests because it looks too much like 1 + 4 = 2 + 3.
Case 5. Four Coins. n = 4. Then the only way to balance a scale using all but one coin is to put two coins on one cup and one on the other. The only two such weighings that balance are 1 + 2 = 3 and 1 + 3 = 4, but they leave different coins off the scale.
The remaining cases, n < 4, are even easier. That concludes the proof of Case 5.
Consequently, by the argument similar to the one in case 4 we can show that the number of coins in case 5 must be the index of a square triangular number.
This concludes the proof of the theorem.
Now we can describe all possible numbers of coins that allow the Baron to confirm a coin in one weighing, or, in other words, the indices of ones in the sequence a(n). The following list corresponds to the five cases above:
If we have four coins, then the same weighing 1+2 < 4 identifies two coins: the coin that weighs three grams and is not in a cup and the coin weighing four grams that is in a cup. The other case like this is for two coins. Comparing them to each other we can identify each of them. It is clear that there are no other cases like this. Indeed, for the same weighing to identify two different coins, it must be the n-gram on the cup, and the n-1 coin off the scale. From here we can see that n can't be big.
As usual we want to give something to think about to our readers. We have given you the list of sequences describing all the numbers for which the Baron can prove the weight of one coin in one weighing. Does there exist a number greater than four that belongs to two of these sequences? In other words, does there exist a total number of coins such that the Baron can have two different one-weighing proofs for two different coins?
To conclude this essay we would like to note that the puzzle we are discussing is related to the puzzle in one of Tanya's previous posts:
You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same, except for their labels. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?
The latter puzzle appeared at the last round of Moscow math Olympiad in 1991. The author of this problem was Sergey Tokarev.
I would like to tell you a story from my childhood and how I started on my math path.
When I was in elementary and middle school, I was very good with mathematics. Actually, I was by far the best math student in my class and my math teacher didn't know what to do with me. Our algebra book had 2,000 problems and was intended to cover three years of study. But I worked out those problems, one after another, whenever I had a free minute in my math class. As I result, I got way ahead of everyone else.
One day a new boy named Lenya Kostyukov joined our class. He had extraordinarily long eye-lashes that covered his eyes, and all the girls envied him. He was a nice smart kid, but other than his lashes, I didn't notice him very much. After a year or two, he announced that he was leaving our school, because he had been accepted to a math school for gifted children.
"Why is he going to a math school? I am the math star here. Why aren't I going to a math school?" I knew about math schools, and I knew that I was good at math; I just never made the connection. I never felt that I was supposed to apply. Despite enjoying my reputation, I just passively went with the flow. Lenya figuratively kicked me in the butt. If he can, why can't I?
So I applied to the same school on the last permissible day and was accepted. It turned out that I accidentally went to the room where they were giving the test for a grade higher than mine. I passed it with flying colors. My parents, though, were scared of a long commute and didn't really want me to go so far away. They found a different math school closer to home, and used my extraordinary results to convince that school to accept me, even though their application date had passed.
For many years I continued to be a very passive person. Applying to a math school was the single big step I took for myself, but it was a defining step. I am grateful to Lenya for that. Or more likely to his parents, who were actively looking around trying to find the best place for their gifted son, and as a byproduct found a place for me. Once I was on the path of mathematics, I had the guidance of teachers and supervisors, for better or for worse, which allowed me to continue to be passive.
I have described my defining moment to you, but I don't want to leave you in the dark about Lenya's fate. Here's what happened to him.
As I mentioned, Leonid (Lenya's formal name) and I ended up in different math schools, so I lost track of him. Four years later I went to study math at Moscow State University and stumbled upon a guy with very long eye-lashes. We recognized each other immediately and eventually became friends.
He was doing logic and was very good at it. He was recommended for graduate school. But by that time our MSU administration noticed his lashes too. The lashes were obviously very suspicious; they hinted at the existence of non-Russian blood in his veins. As it was the period of brutal anti-semitism at MSU, they didn't allow him to go to graduate school.
Leonid Kostyukov dropped mathematics and became a famous writer.
Here is a strange puzzle that was inspired by the palindrome problem. Suppose you have a sequence of words in some alphabet with the initial term a and all the other terms b: a, b, b, b, b, etc. Suppose this sequence generates palindromes every time you concatenate the first several terms, not counting the first term itself. So, ab, abb, abbb, and so on — are all palindromes. We call words b "papaya" words, when a exists, such that a and b generate the sequence with this palindrome property. Can you describe papayas?
Theorem. The word b is a papaya word if and only if b is a substring of Reverse(b)Reverse(b).
Proof. After we have added b so many times that the initial part a is much smaller than half of the concatenated string, the middle part of the concatenated string would consist of several words copies of the word b. The middle of the reverse string consists of several concatenations of Reverse(b). So the word b must be a substring of Reverse(b)Reverse(b). On the other hand, suppose b is a substring of Reverse(b)Reverse(b). Then Reverse(b)Reverse(b) is of the form xby, and we can choose a = y.
Theorem. Papaya words are either palindromes or concatenations of two palindromes.
Proof. Suppose our word consists of two palindromes cd. Then the reverse of it is dc and its double is dcdc. The word cd is a substring of dcdc, thus according to the first theorem, cd is a papaya word. Let's do the other direction. Suppose the word b is a substring of Reverse(b)Reverse(b). Then Reverse(b)Reverse(b) is of the form xby. Then b = yx, and Reverse(b) = xy, which equals Reverse(x)Reverse(y). From here, Reverse(x) = x and Reverse(y) = y. If x or y has zero length, then our word is a palindrome. QED.
Hey, do you already know why we call these words papayas?
Just for fun we would like to study the structure of papaya words. Any one-character or two-character word is a papaya word, so the patterns are: a, aa, ab. For three-character words there are four patterns: aaa, aab, aba, abb. For four-character words there are 10 patterns: aaaa, aaab, aaba, abaa, aabb, abab, abba, abbb, abac, abcb. In this manner we get the sequence of the number of n-character papaya patterns: 1, 2, 4, 10, 21, 50 etc, which is sequence A165137 in the OEIS. This sequence depends on the number of letters in your alphabet. But the first n terms of these sequences are the same for all alphabets that have at least n letters.
Let us assume that we are working with an infinite alphabet. The complementary sequence would be the number of patterns for non-papaya words. The total number of patterns is described by sequence A000110 — Bell numbers: the number of word structures of length n using an infinite alphabet. So the beginning of this complementary sequence A165610 is: 0, 0, 1, 5, 31, 153, etc. The list of corresponding patterns is abc, aabc, abbc, abcc, abca, abcd, etc.
Historically, we first invented the corresponding sequence for numbers, not for words. We call a number a papaya number if it is a palindrome or a concatenation of two palindromes. If we use numbers instead of words in the problem, we need to carefully look at what happens if we encounter initial zeroes. Let's take the papaya number 2200100, and see if we can find a number a, such that adding 22010 repeatedly to this sequence starting with a will always generate a palindrome. The number a must be 00100. But this is not a number. We have two choices: to say that we are working with strings of digits, or to allow several numbers to start the sequence before we add b repetitively and before getting to palindromes. In the latter case our sequence can start 0, 0, 100, 22010, 22010, and so on.
As we mentioned before, the number of patterns of papaya numbers will start the same as the number of patterns of papaya words. Later the sequence of patterns of numbers A165136 will be smaller than the corresponding sequence for words. As the sequence of Bell numbers is much more famous than the sequence A164864 of patterns of numbers, we expect the papaya patterns sequence corresponding to the infinite alphabet to be more interesting and important than the sequence of papaya patterns for numbers.
Though papaya numbers might be less important than papaya words with an infinite alphabet, they have an advantage in that we can generate more sequences with them. For example we can calculate the number of positive papaya number with n digits, as in the sequence A165135: 9, 90, 252, 1872, 4464, 29250, etc. And we can also calculate the sequence A165611 of n-digit non-papaya numbers: 0, 0, 648, 7128, 85536 etc.
This geometric problem was given to me by Arkady Berenstein:
There are n points on the plane, but not all of them are on one line. Prove that a line exists that passes through exactly two points of this set.
Arkady gave me a beautiful solution to this problem. First, draw a line through each pair of points. Suppose you calculate all the distances from each point to all the lines that the point doesn't belong to. Pick the smallest distance. The corresponding line will be the one with two points. To finish the solution you need to fill in the details. That process is usually left to the reader.
I suspect that there might also be a solution using linear algebra. Can you find one?
I would like to reformulate this problem without using geometry. Suppose there is a set of n elements. Let's call a family of subsets line-like if any two distinct subsets of this family can have as an intersection not more than one element. Then the geometry problem above has a set-theoretical analogue:
You have a set of n elements and a line-like family of subsets of this set such that any two elements of the set belong to a subset from this family, and that the family doesn't contain the whole set. Is it true that there always exists a subset in this family consisting of two elements?
Usually I give such problems as homework for the reader, but this time I decided to change my habit, so I'm including the picture which contains the solution of this problem by my son Alexey Radul.
Conclusion: geometry is important.
I remember this question from my childhood:
Why is the South Pole colder than the North Pole?
Indeed, the average winter temperature at the North Pole of -34°C is the same as the temperature at the South Pole at the beginning and end of its summer. The South Pole is only warmer than the North Pole 40 days per year. So the South Pole is a much, much colder place. According to Wikipedia there are three major reasons for this:
I remember when I was a child my father gave me a completely different explanation.
The Earth's orbit is not a circle, but rather an ellipse. According to Kepler's second law: "A line joining a planet and the sun sweeps out equal areas during equal intervals of time." This means that the earth has a slower angle motion around the aphelion — in its furthest point — than around the perihelion — in its closest point to the Sun. Consequently, the summer is longer than the winter for the North Pole, whereas the opposite is true for the South Pole.
Something in my father's explanation bothered me. Now I understand what: though the summer is longer at the North Pole, it should get less sunshine as the North Pole is further away from the Sun than the South Pole during its summer. So the effects might cancel each other out. In any case, as the earth's orbit is almost circular, the contribution of the shape of the orbit should be minor, compared to the effects of elevation, the water underneath and continentality.
On the other hand, it is possible that my father wasn't talking about the poles, but rather about the difference in hemispheres. I wonder if someone can calculate if there is a difference in the amount of sunlight the poles get due to the fact that the Earth's orbit is not circular. Is the temperature different for the places that are equidistant from the equator, and have similar elevation and continentality, but which are located in different hemispheres?
I remember a funny article explaining why the northern hemisphere has more land. They said that continents drifted into the northern hemisphere because they wanted a nicer climate.
I bought the book "Heard on The Street: Quantitative Questions from Wall Street Job Interviews" by Timothy Falcon Crack
The book has problems in logic, probability, statistics and finance, as well as a very useful chapter of general interview questions. If you're interested in buying this book, I should mention that some questions require calculus and knowledge of financial terms.
I love the author's taste in problems, and here are some sample questions from the book.
Question 2.7: How many degrees (if any) are there in the angle between the hour and minute hands of a clock when the time is a quarter past three?
Question 5.1.14: Welcome to your interview. Sit in this chair. Excuse me while I tie your arms and legs to the chair. Thank you. Now we are going to play "Russian roulette." I have a revolver with six empty chambers. Watch me as I load the weapon with two contiguous rounds (i.e., two bullets side-by-side in the cylindrical barrel). Watch me as I spin the barrel. I am putting the gun against your head. Close your eyes while I pull the trigger. This is your lucky day: you are still alive! Our game differs from regular Russian roulette because I am not going to add any bullets to the barrel before we continue, and I am not going to give you the gun.
My question for you: I am going to shoot at you once more before we talk about your résumé. Do you want me to spin the barrel once more, or should I just shoot?
Question 6.1.16: Tell me something you tried but ended up quitting on.
I can tell you what I would have answered to the last question: I tried smoking, but ended up quitting.
Mathematically we can describe a marriage by a graph. People are vertices and two spouses are connected by an edge.
Mathematical models tend to oversimplify life, so let us assume that a person can only be one of two genders. Therefore, the vertices of a graph are colored in two colors: pink and blue. In this article I explore the graph theory of different types of marriages.
A monogamous couple is represented as a complete K2 graph: two vertices connected by an edge. The graph is bipartite, no matter how you color it. But actually our vertices are already colored from the start. If we are considering traditional marriage, one vertex is pink and the other is blue.
Historically, the second most common type of marriage is polygyny, in which one man has several wives. Less common in history, but a mathematically equivalent type, is polyandry, in which one woman has several husbands. Both these types of marriages emphasize inequality, as husbands and wives have completely different sets of rights.
From a mathematical point of view, polygyny and polyandry are described by star graphs. Star graphs are bipartite graphs and the natural coloring is the one that proves bipartiteness.
The final type of marriage is polygynandry, which refers to a group marriage, where more than one man and more than one woman create a family. Everyone can have sex with everyone else of the other gender. Mathematically this type of marriage corresponds to a complete bipartite graph Kn,m. Actually, in this case I can imagine that a particular pair of people of different genders wouldn't like each other and might not consummate their marriage. So this graph is not necessarily complete.
How can same-sex marriages change the graph theory of marriages? As a graph, a monogamous same-sex marriage is the same bipartite K2 graph as a heterosexual marriage. It will just be less colorful, as both vertices will be of the same color.
But what happens if we add the same-sex idea to polygamous marriages?
Suppose a homosexual man wants to live with several spouses at the same time. What name can we give to a family unit of more than two homosexual men? Homopolygamy? Their marriage graph will be a star graph in which all the vertices are of the same color.
If a man can have several spouses, what about his spouses? Can they form multiple marriages too? If only one person is allowed to engage in several marriages, then we will see inequality within the same gender. If any spouse is allowed to form other marriages, then we will have a situation in which several men are all spouses to each other. So mathematically we will see complete graphs with more than two vertices to represent a marriage. If two people in a group do not like each other and do not want to be married, then the corresponding graph doesn't need to be complete.
By symmetry we can describe a marriage of several women, and mathematically it will be similar to a marriage of several men.
Another interesting aspect is the idea of mixed types of marriages involved in polygamy. Suppose a husband has several wives. Some of them might get bored waiting for his attention, and start spending so much time with each other that they end up developing feelings for each other. Suppose two wives of the same man decide to marry each other. What name would we give to this type of marriage? I am afraid that we do not have enough words to cover all the potential situations.
Suppose we have a heterosexual married couple and the man decides to bring another woman into their house. Thus the transition from a traditional marriage to a polygyny is created. If they got along so well that the first wife decides to marry the second wife, this would require a transition to a new type of marriage. Oh, I see that my essay just went in another direction — how different types of marriages might evolve into each other. For now, I'll leave this for future research.
Talking about different directions. I recently wrote a piece about condoms. Now I have a new generalization for the classic condom puzzle. Suppose we have a mixed-type marriage defined by a graph. Suppose tonight every couple of people corresponding to the edge of this graph wants to have sex with each other. What is the smallest number of condoms they can use? In my condom essay, I didn't define the condom usage for the sex of two women. I will leave it to your imagination and definition.
What follows is the story of a pair of integer sequences, which started life as a Google interview puzzle back in the previous century when VHS video tapes were in use:
Suppose you are buying VHS tapes and want to label them using the stickers that came in the package. You want to number the tapes consecutively starting from 1 and the stickers that come with each package are exactly one of each digit ["0"..."9"]. For your first tape you use only the digit "1", and save all the other digit stickers for later tapes. The next time you will need a digit "1" will be for tape number 10. By this time you will have several unused "1" stickers. What is the next tape number such that after labeling the tape with that number you will not have any "1" stickers remaining?
We can formalize this puzzle in the following way:
Consider a function f which, for a given whole number x, returns the number of times that the digit 1 is needed to write all of the numbers between one and x. For example, f(13) = 6. Notice that f(1) = 1. What is the next largest x such that f(x) = x?
Thus f(x) is the number of "1" stickers needed to label all the tapes up to tape x. When f(x) = x then we have used all of the "1" stickers in labeling the first x tapes.
Let's consider any non-zero digit. In the single and double-digit numbers, there are ten of each digit in the ones column, and ten of each digit in the tens column, so 20 altogether. Early on, the tape number is ahead of the digit count. By the time we get to 20-digit numbers, though, there should be, on average, two of any single non-zero digit per number. Thus, the number of times that any digit is used should eventually catch up with the tape numbers.
Encouraged by assurance of reaching our goal somewhere, we might continue our estimate. In the up-to three-digit numbers there are 300 of each non-zero digit; in the numbers below 10,000 there are 4000; then 50,000, and so on up to 1010, where f(x) and x must (almost) meet. In particular, there are 10,000,000,000 counts for any non-zero digit in the numbers below 10,000,000,000. Hence, were the puzzle asking about any of the digits 2–9, then ten billion could have been an easy answer or, at least a limit on how far we need to search.
Sadly, there is a 1 in the decimal representation of ten billion (and a few zeroes), so we require 1010+1 of the digit "1" to write the numbers [1…1010]. Thus, f(1010) ≠ 1010, so 1010 cannot be the answer to the original puzzle. Thus stymied, Gregory wrote a program to find the solution to the original Google's puzzle. And the answer turned out to be 199,981.
Gregory was overstymied, so he actually wrote a program to solve the puzzle for any non-zero digit. He calculated the beginning of the sequence a(n), where a(n) is the smallest number x > 1 such that the decimal representation of n appears as a substring of the decimal representations of the numbers [1…x] exactly x times.
We already know that a(1) is 199,981. The sequence continues as follows: 199,981, 28,263,827, 372,599,983, 499,999,984, 10,000,000,000, 10,000,000,000, 9,465,000,000, 9,465,000,000, 10,000,000,000, ….
Did you expect this sequence to be increasing? You could have, because smaller numbers tend to contain smaller digits than larger numbers. Then why is the sequence not increasing? As Gregory failed to find a value for the digit 5 below ten billion, he noticed that it's fairly easy to imagine a scenario where you have one less than the number you need, and then the next value has more than you need for equality, and then you equalize again later. In response, Alexey Radul, a friend of one author and a son of the other, suggested a related sequence:
Let a(n) be the smallest number x > 1 such that the decimal representation of n appears as a substring of the decimal representations of the numbers [1…x] more than x times.
The key difference being "more than" rather than "exactly". Starting at 1 then, here are the first nine terms of each sequence:
| n | = | > |
|---|---|---|
| 1 | 199,981 | 199,991 |
| 2 | 28,263,827 | 28,263,828 |
| 3 | 371,599,983 | 371,599,993 |
| 4 | 499,999,984 | 499,999,994 |
| 5 | 10,000,000,000 | 5,555,555,555 |
| 6 | 10,000,000,000 | 6,666,666,666 |
| 7 | 9,465,000,000 | 7,777,777,777 |
| 8 | 9,465,000,000 | 8,888,888,888 |
| 9 | 10,000,000,000 | 9,999,999,999 |
Some of these pairs are interesting in their own right. Notice that 199,991 is ten more than the previously found 199,981. For all the numbers in between, the initial equality holds. Likewise for n=3, each of the numbers between 371,599,983 and 371,599,993 has exactly one three. Hence, the increase in a number by one is the same as the increase in the count of threes. A similar situation holds for n=4.
Gregory has submitted these two new sequences to the Online Encyclopedia of Integer Sequences, as they turned out not to be there yet, and they can be found using the identifiers A163500 for the "=" sequence and A164321 for the ">" sequence. It's not surprising that the values matching this relaxed second condition are more well behaved than those with equality. Do you think the second sequence is always increasing? Wait a minute, let's check that sequence for zero.
In counting zeroes, it is important to remember that we start with one, not zero. In this case the smallest number x such that x is less than or equal to the number of 0s in the decimal representations of [1 … x] is 100,559,404,366. But what is the corresponding number for the "=" sequence? It appears that no such number exists. The challenge of accurately proving it, as they say, is left as an exercise to the reader.
There is no reason that we should be constrained to single digits. The formal statement of the problem provides an obvious generalization, where we consider substrings of each of the numbers [1 … x] rather than digits in those numbers. We should note that we count every occurrence of a substring separately. Thus 11 will be counted twice as a substring of 1113.
We can prove that the "more than" sequence is increasing after the first term. Indeed, for two integers i and j, if i is less than j, then for every occurrence of j, by replacing j with i we get a smaller number with an occurrence of i.
Inspired, Tanya wrote another fancier and faster program to find
values of this sequence for two-digit numbers. Here is the smallest
number x for which the number of "10"s as substrings of the
numbers [1…x] is more than or equal to x. And by a
lucky strike the equality holds. The number is:
109,
The value of a≥(11) might seem like a miracle:
119,
Sadly, a≥(12) is not so pretty:
1,296,624,070,230,872,
We cannot leave off without at least mentioning that the sequence function should next take one more parameter: the base of representation.
We have found only the first few members of these new sequences, and there are many related sequences to be catalogued. We would love to hear tales from your explorations. Enjoy the sequence hunt!
Peter Winkler gave a talk at MIT last fall and, as is customary, the audience was invited to join him for dinner afterwards at a local restaurant. I was eager to dine with Peter because
he is a puzzle collector
Peter is now a columnist at ACM communications. His column is called "Puzzled" and it is featured as part of the section titled "Last Byte." Writing these paragraphs has made me so hungry that I need to go grab something to bite.
Okay, I am back and here is the pizza puzzle.
Alice and Bob ordered a pizza. The pizza is cut into several radial pieces. Both Alice and Bob are greedy and well-mannered at the same time. They each want to get as much pizza as possible for themselves, but they don't want to be obvious about it. They take pieces in turn, starting with Alice. Because they are well-mannered and not obvious, when it is their turn, they only take a piece that is adjacent to the pieces already taken. Throughout the process of consuming this pizza, the untaken pieces are contiguous.
The question is: Is it possible to cut the pizza in such a way that although Alice starts, Bob can guarantee himself more than half?
If you want to think about this puzzle on your own, now would be a good time to pause. Why? Because in the next paragraph, I will give you some hints about how to approach this question.
If the number of pieces is even, then Alice can't lose. She can number the pieces around the circle consecutively, decide whether all the odd pieces or all the even pieces make up a bigger chunk, and then follow the parity.
But what happens if there are an odd number of pieces? Alice has an advantage, for she will get more pieces. But is that enough to guarantee that she will get at least half? Suppose she starts by taking a minuscule piece. Then Bob can number all of the leftover pieces in order and decide if he prefers the even-numbered group or the odd-numbered group. He is in control now, so he can guarantee himself the bigger part of the leftover pieces.
However, that might not be good enough for Bob to win. For example, if there is a very big piece, one that is bigger than half of the pizza, then in the first move Alice wins.
On the other hand, would Alice necessarily win by starting with the biggest piece?
Suppose the biggest piece is significantly less than half. Would Bob have a chance? To his advantage, he does have a lot of control. He can choose the parity of the pieces he wants at the beginning, and he can also switch this parity later, depending on what Alice does. Does he control the situation enough that it would be possible to cut the pizza in his favor?
I have to add that if you can find a solution with N pieces, then you can easily build a solution with N+2 pieces. Suppose you start with a pizza cut into N pieces, such that Bob will win. If you add two adjacent pieces of zero value to this pizza, you will get a pizza cut into N+2 pieces, such that Bob can still win. Indeed, Bob will follow the strategy he used with the N-pieces pizza, except that each time Alice takes one of the new pieces, Bob takes the other.
Can you find a way to cut a pizza so that Bob can guarantee himself more than half?
I wrote a story a while ago about how a clerk at my previous job mistyped my resignation date, substituting January 2007 for my real date, January 2008. As a result, my medical insurance provider decided that I wasn't covered in 2007, and requested that my doctor return the money he had already received.
After several phone calls my medical insurance was reinstated, but I kept receiving bills from my doctor. When I called my insurance, they assured me that everything was fine and that they had paid my doctor. However, my doctor continued to send me bills.
After half a year of phone calls back and forth, someone finally explained to me what was going on. My insurance company had initially requested the money back. The money was never returned to them, because my doctor's office would not pay them a penny until I had paid the doctor first. In my doctor's database, my visits were marked as unpaid.
When the problem was cleared up, the insurance company stopped requesting that the doctor pay them back. But the computer at my doctor's office didn't understand that stop-the-request command. It didn't know what stopping the request meant.
The computers were talking different languages and I was caught in the middle.
My son, Alexey Radul, made his PhD thesis available to the public.
I was amazed at how much he had invested in the thesis. I assumed that the main goal for a dissertation in computer science is to write a ground-breaking code and that the accompanying text is just a formality. However, this is not the case with my son's thesis. I am not fully qualified to appreciate the "ground-breakness" of his code, but his thesis text is just wonderful.
Alexey decided that he wanted to make his thesis accessible to a wide audience. He had to make a lot of choices and decisions while he was designing and coding his prototype, and in his thesis he devoted a lot of effort to explaining this process. He also tried to entertain: I certainly had fun trying to solve an evil sudoku puzzle on page 87, that turned out not to have a unique solution.
In addition to everything else, Alexey is an amazing writer.
I am a proud mother and as such I am biased. So I'll let his thesis speak for itself. Below is the table of contents accompanied by some quotes:
"Revolution is at hand. The revolution everyone talks about is, of course, the parallel hardware revolution; less noticed but maybe more important, a paradigm shift is also brewing in the structure of programming languages. Perhaps spurred by changes in hardware architecture, we are reaching away from the old, strictly time-bound expression-evaluation paradigm that has carried us so far, and looking for new means of expressiveness not constrained by over-rigid notions of computational time."
"Fortunately, there is a common theme in all these efforts to escape temporal tyranny. The commonality is to organize computation as a network of interconnected machines of some kind, each of which is free to run when it pleases, propagating information around the network as proves possible. The consequence of this freedom is that the structure of the aggregate does not impose an order of time. Instead the implementation, be it a constraint solver, or a logic programming system, or a functional reactive system, or what have you is free to attend to each conceptual machine as it pleases, and allow the order of operations to be determined by the needs of the solution of the problem at hand, rather then the structure of the problem's description."
"Every human harbors mutually inconsistent beliefs: an intelligent person may be committed to the scientific method, and yet have a strong attachment to some superstitious or ritual practices. A person may have a strong belief in the sanctity of all human life, yet also believe that capital punishment is sometimes justified. If we were really logicians this kind of inconsistency would be fatal, because were we to simultaneously believe both propositions P and NOT P then we would have to believe all propositions! Somehow we manage to keep inconsistent beliefs from inhibiting all useful thought. Our personal belief systems appear to be locally consistent, in that there are no contradictions apparent. If we observe inconsistencies we do not crash—we chuckle!"
"This is power. By generalizing propagation to deal with arbitrary partial information structures, we are able to use it with structures that encode the state of the search as well as the usual domain information. We are consequently able to invert the flow of control between search and propagation: instead of the search being on top and calling the propagation when it needs it, the propagation is on top, and bits of search happen as contradictions are discovered. Even better, the structures that track the search are independent modules that just compose with the structures that track the domain information."
"A shift such as from evaluation to propagation is transformative. You have followed me, gentle reader, through 137 pages of discussions, examples, implementations, technicalities, consequences and open problems attendant upon that transformation; sit back now and reflect with me, amid figurative pipe smoke, upon the deepest problems of computation, and the new way they can be seen after one's eyes have been transformed."
"The "concurrency problem" is a bogeyman of the field of computer science that has reached the point of being used to frighten children. The problem is usually stated equivalently to "How do we make computer languages that effectively describe concurrent systems?", where "effectively" is taken to mean "without tripping over our own coattails". This problem statement contains a hidden assumption. Indeed, the concurrency itself is not difficult in the least—the problem comes from trying to maintain the illusion that the events of a concurrent system occur in a particular chronological order."
"Time is nature's way of keeping everything from happening at once; space is nature's way of keeping everything from happening in the same place ( with apologies to Woody Allen, Albert Einstein, and John Archibald Wheeler, to whom variations of this quote are variously attributed)."
"I'm meticulous. I like to dot every i and cross every t. The main text had glossed over a number of details in the interest of space and clarity, so those dots and cross-bars are collected here."
You have been warned. You are allowed to read this if you are 17 or over. Otherwise, you can ask your parents to read this to you. Here is a famous old condom puzzle in the version I heard when I was a teenager myself:
A man hires three prostitutes and wants to have sex with all three of them. They all might have different sexually transmitted diseases and they all want to use condoms. Unluckily, they have only two condoms. Plus, they are in the forest and can't buy new condoms. Can the man have sex with all three of the women without danger to any of the four?
Everyone in this problem is so health-conscious, that it might not be such a dirty problem after all. I leave you with the fun challenge of figuring this problem out.
Another fun variation of this problem is when you have two men and two women and two condoms. Every woman wants to have sex with every man. How can they do that?
If you are a teacher and want to use these great puzzles for younger students, you can follow the example of MathWorld and pretend that it's a glove problem between doctors and patients.
Recently my younger son and his MIT friends invented another variation of this problem:
Suppose three gay men all want to have sex with each other and every pair among them wants to do two penetrative sexual acts, switching roles. They want to avoid contaminating each other, and in addition, each man also does not want to cross-contaminate himself from either region to the other. How can they do that using exactly three condoms?
Let me remind you that they plan to perform six sexual acts altogether, meaning that six condoms would be enough. On the other hand, each of them needs two condom surfaces, so they can't do it with less than three condoms. My son showed to me his solution, but I will postpone its publication.
Of course, you can say that this is a glove problem about three surgeons operating on each other.
In addition, you can generalize it to any number of gay men. Here is my solution for four men and four condoms, where letters denote people, numbers denote condoms, and the order of people represents roles: A12D, B32D, C2D, A14C, B34C, D4C, A1B, B3A, C21B, D41B, C23A, D43A.
Can you solve the problem for five or more people?
I used to receive emails requesting link exchanges with other websites. They promised to increase my page rank by creating additional hyperlinks to my pages. I ignored them. If they thought my website was good, why did they need my reciprocity to link to me? Besides, their websites didn't have anything to do with mathematics; they were the sites of dental services or Honda dealers.
I have resisted the temptation so far. The links that I have on my websites are to sites that I recommend. Sometimes I wonder through other people blogrolls and add good links to my blog.
At other times a blog roll exchange happens: I have Google Analytics installed on my sites. From time to time I examine my traffic. When I see a new traffic flow from a particular website, I check that site out. If I like it, I add it to my blogroll.
I wouldn't mind people writing to inform me that they have a link to my website and asking me if I'd like to reciprocate. But this doesn't happen. Instead, strangers write to me offering to put up a link to my website on the condition that I put a link to them. I do not like this imposition.
Recently I received a request for a blog review exchange. I went to that blog and found that all of its postings were reviews of other people's blogs, presumably those who had agreed on this kind of exchange. I checked out several of those other blogs and I didn't find any of them very interesting.
I missed this opportunity to receive that blog review, but on the other hand, if I start linking to random crap, I might lose the respect of my readers.
My previous paragraph reminded me of a Russian joke:
I wonder how a person whose website comes up first in a Google search for "random crap" feels.
Russians assume that such a person will be embarrassed. They do not understand Americans who welcome negative publicity, and purposefully would name their website randomcraponline.com.
I enjoyed a recent discussion on the sequences fans mailing list. David Wilson suggested an idea for hiding email addresses on webpages from bots: change your email slightly and explain how to change it back.
For example, if you want to contact me, you should reverse my login name in the following email address:
hkaynat@yahoo.com
Or, remove all the digits from the following email address:
1ta2nya3kh4@yahoo.com
Here is a palindrome problem by Nazar Agakhanov from the All-Russian Mathematical Olympiad, 1996:
Can the number obtained by writing the numbers from 1 to n, one after another, be a palindrome?
Of course, we should assume that n > 1.
When I give this problem to my friends, they immediately answer, "No, of course not." The reasoning is simple. The last 9 digits of this palindrome should be 987654321 — all different digits. The earliest you can get these digits at the end of your string 1 to n, is when your number n is actually 987654321. By that time your string of digits is really huge. If we take a random number with 2k digits, then the probability of it being a palindrome is 10(-k). There is no reason to think that writing consecutive integers increases the probability of finding a palindrome. So the probability of hitting a palindrome is so low, that you can safely answer, "No, of course not."
After confidently saying "no", my friends usually stop thinking about the problem altogether. Only my friend David Bernstein suggested a simple solution for this problem. You can try to find the proof, but I do not insist that you do. I am about to give you many other fun problems to solve, and you can choose which ones you want to think about.
Naturally, we can replace the sequence of natural numbers in the Agakhanov's problem with any sequence. So, one problem becomes 160,000 different problems when you plug all current sequences from the Online Encyclopedia of Integer Sequences into it.
For some periodic sequences every concatenation you create is a palindrome. For example, for the sequence of all zeroes, every set of the first n terms is a palindrome. More interestingly, you can think of an increasing sequence such that any first n terms comprise a palindrome, as we see in the sequence of repunits: 1, 11, 111, 1111, etc. Can you think of other sequences like this?
For some sequences, not every concatenation you create is a palindrome, but you can obtain an infinite number of palindromes. One example, suggested by Sergei Bernstein, is the sequence a(n) of the last digit of the greatest power of 2 that divides n. Can you think of other sequences like that?
For some sequences only the initial term itself is a palindrome. Beyond that you can't obtain a palindrome by concatenating the first n terms. For a few of those sequences, this fact is easy to prove. Take, for example, the sequence of powers of ten, or the sequence of squares, or the sequence of prime numbers. Can you think of other similar sequences?
There are many sequences that do not produce a palindrome beyond the first term, even if you try many times. I suspect that they do not, in fact, ever produce a palindrome, but I have no clue how to prove that. I have in mind the sequence of the digits of π. Can you suggest other sequences like that?
Instead of solving the initial problem, I gave you a variety of other problems. My last challenge for you is to find other sequences that can replace the sequence of natural numbers in the Agakhanov's problem so that the problem becomes solvable and the solution is interesting.
The PhD program in Russia was limited to exactly three years. My son Alexey was born right after I started it, and I was distracted from my research for quite some time. At that time, my mom, who lived with us, reached her retirement age of 55. Her retirement would have been supported by the government and her pension would have been almost equal to her salary. So I begged my mom to retire and help me with my son Alexey. I couldn't understand why she wouldn't stop working, so I kept pressing.
At the same time, I was frantically trying to find a place for Alexey in a day care center. I finally was successful, but it backfired. Alexey started getting sick all the time. Daycare was overcrowded, with 30 kids to every adult. Workers didn't have time to attend to every kid. Day care workers were so tired that they were relieved when a few kids stayed home sick.
After Alexey was hospitalized for two weeks with severe dysentery, my mother gave up and retired. It was one year before the end of my graduate school. In that year I wrote four papers and my thesis, and I got my PhD.
Now that I am fifty, I understand that my mother really did love her job. Being older and wiser, I recognize what a sacrifice my mother made in retiring in order to look after a grandchild.
Mom, sorry for being so pushy back then and thank you so much for all that you did for us.
This puzzle was brought to me by Leonid Grinberg.
A frog needs to jump across the street. The time is discrete, and at each successive moment the frog considers whether to jump or not. Unfortunately, the frog has crappy eyesight. He knows there are dangerous cars out there, but he can't see them. If a car appears at the same moment that he decides to jump, he will die.
The adversary sends cars, hoping to kill the frog. The adversary knows the frog's algorithm, but can use only a finite number of cars. The frog wants to maximize his chances of survival with his algorithm. The frog is allowed to use a random number generator that the adversary can't predict. Can you suggest an algorithm for the frog to cross the street and survive with a probability of at least 1 − ε?
My guest blogger is Levy Ulanovsky, a maven of physics puzzles. Here is one of his favorite puzzles:
There are n points in 3-dimensional space. Every point is connected to every other point by a wire of resistance R. What is the resistance between any two of these points?
In 2007 Alexander Shapovalov suggested a very interesting coin problem. Here is the kindergarten version:
You present 100 identical coins to a judge, who knows that among them are either one or two fake coins. All the real coins weigh the same and all the fake coins weigh the same, but the fake coins are lighter than the real ones.
You yourself know that there are exactly two fake coins and you know which ones they are. Can you use a balance scale to convince the judge that there are exactly two fake coins without revealing any information about any particular coin?
To solve this problem, divide the coins into two piles of 50 so that each pile contains exactly one fake coin. Put the piles in the different cups of the scale. The scale will balance, which means that you can't have the total of exactly one fake coin. Moreover, this proves that each group contains exactly one fake coin. But for any particular coin, the judge won't have a clue whether it is real or fake.
The puzzle is solved, and though you do not reveal any information about a particular coin, you still give out some information. I would like to introduce the notion of a revealing coefficient. The revealing coefficient is a portion of information you reveal, in addition to proving that there are exactly two fake coins. Before you weighed them all, any two coins out of 100 could have been the two fakes, so the number of equally probable possibilities was 100 choose 2, which is 5050. After you've weighed them, it became clear that there was one fake in each pile, so the number of possibilities was reduced to 2500. The revealing coefficient shows the portion by which your possibilities have been reduced. In our case, it is (5050 − 2500)/5050, slightly more than one half.
Now that I've explained the kindergarten version, it's time for you to try the elementary version. This problem is the same as above, except that this time you have 99 coins, instead of 100.
Hopefully you've finished that warm-up problem and we can move on to the original Shapovalov's problem, which was designed for high schoolers.
A judge is presented with 100 coins that look the same, knowing that there are two or three fake coins among them. All the real coins weigh the same and all the fake coins weigh the same, but the fake coins are lighter than the real ones.
You yourself know that there are exactly three fake coins and you know which ones they are. Can you use a balance scale to convince the judge that there are exactly three fake coins, without revealing any information about any particular coin?
If you are lazy and do not want to solve this problem, but not too lazy to learn Russian, you can find several solutions to this problem in Russian in an essay by Konstantin Knop.
Your challenge is to solve the original Shapovalov puzzle, and for each solution to calculate the revealing coefficient. The best solution will be the one with the smallest revealing coefficient.
My guest blogger is David Wilson, a fellow fan of sequences. It is a nice exercise to understand how this graph works. When you do, you will discover that you can use this graph to calculate the remainders of numbers modulo 7. Back to David Wilson:
I have attached a picture of a graph.
Write down a number n. Start at the small white node at the bottom of the graph. For each digit d in n, follow d black arrows in a succession, and as you move from one digit to the next, follow 1 white arrow.
For example, if n = 325, follow 3 black arrows, then 1 white arrow, then 2 black arrows, then 1 white arrow, and finally 5 black arrows.
If you end up back at the white node, n is divisible by 7.
Nothing earth-shattering, but I was pleased that the graph was planar.
My son Alexey Radul defended his PhD thesis on Propagation Networks on August 4. Here is the abstract.
I propose a shift in the foundations of computation. Modern programming systems are not expressive enough. The traditional image of a single computer that has global effects on a large memory is too restrictive. The propagation paradigm replaces this with computing by networks of local, independent, stateless machines interconnected with stateful storage cells. It offers great flexibility and expressive power, and has therefore been much studied, but has not yet been tamed for general-purpose computation. The novel insight that should finally permit computing with general-purpose propagation is that a cell should not be seen as storing a value, but as accumulating information about a value.
Various forms of the general idea of propagation have been used with great success for various special purposes; perhaps the most immediate example is constraint propagation in constraint satisfaction systems. This success is evidence both that traditional linear computation is not expressive enough, and that propagation is more expressive. These special-purpose systems, however, are all complex and all different, and neither compose well, nor interoperate well, nor generalize well. A foundational layer is missing.
I present the design of a general-purpose propagation system. I illustrate on several examples how the resulting prototype supports arbitrary computation; recovers the expressivity benefits that have been derived from special-purpose propagation systems in a single general-purpose framework, allowing them to compose and interoperate; and offers further expressive power beyond what we have known in the past.
Warning: this essay contains solutions to math problems.
Here is a famous hat puzzle:
A king decides to give 100 of his wise men a test. If together they pass, they can go free. Otherwise, the king will execute all of them. The test goes as follows: the wise men stand in a line one after another, all facing in the same direction. The king puts either a black or a white hat on each wise man. The wise men can only see the colors of the hats in front of them. In any order they want, each one guesses the color of the hat on his own head. Other than that, the wise men cannot speak. To pass, no more than one of them may guess incorrectly. Given that they have time to agree on a strategy beforehand, how can they assure that they will survive?
Instead of discussing the puzzle above, I'd like to look at a different version. It is an infinite variation of the puzzle that my son Sergei brought back from the Canada/USA Mathcamp last year.
The king has a countable number of wise men. The line starts from the left and is infinite in the right direction. The wise men are all facing to the right and they see the infinite tail of the line. Again, the king places either a black or white hat on each head and they can only say one of two words: black or white. Will they be able to devise a strategy beforehand that ensures that not more than one person makes a mistake?
Oh, I forgot to mention: you are allowed to use the axiom of choice.
Here is the solution. You can build an equivalence relation on the possible placements of hats. To be equivalent, two ways of placing the hats should have the same tail. In other words, there is a person such that both hat arrangements to his right are the same. By the axiom of choice you can pick a representative in any equivalence class. The first wise man looks at all the other hats and calculates in how many places the tail differs from the representative of the class they picked. This is a finite number, and by stating one color or the other, he signals the parity of that number. After that, all the wise men say their colors from left to right. Everyone sees the tail and everyone hears the color choices of the people behind. So every wise man can reconstruct the color of his hat with this information. Only the first person may potentially be mistaken.
Many things about this solution bother me. Where is this country that can fit an infinite number of people? What kind of humans can see into infinity? How much time will this procedure take?
Aside from the practical matters, there are mathematical matters that bother me, too. By the axiom of choice you can pick an element in every class. The problem is that all of the wise men have to pick the same element. The axiom of choice claims the existence of a choice function, which picks an element in each set. So the function exists, but can we distribute this function to many wise men? Remember, they need to agree on this function the night before.
We already implicitly assumed that our wise men have a lot of magical abilities. So we can add to those the ability to go through all the possible tails and memorize the representatives for all the tails in one evening.
But still, I am very curious to know what follows from the axiom of choice. Tell me what you think: does the axiom of choice imply that we can distribute the choice function, or do we need a new axiom? In your opinion, will these wise men live?
I do not know why I like the television show America's Got Talent. Sometimes I picture myself on a stage doing what I love to do the most: entertaining people with mathematics. But it wouldn't really work on the stage of America's Got Talent. The audience makes its judgment in the first five seconds of a performance. There is no way I can teach a new math idea in five seconds.
Back to the show. I especially like the auditions. I noticed a strange correlation between what people say before their performance and what happens on the stage. In short, if a person brags that he/she has the greatest talent and that the judges will be blown away, the performance is likely to be pathetic.
My first thought was that the producers were editing it this way in order to boost the drama of the show. Now I wonder if it could be something else. Perhaps people who do not have much talent need to build up their confidence to appear on the show. And, vice versa, people who have talent can afford to be modest.
I didn't see the same correlation when I watched Britain's Got Talent. Could this tendency be a part of our American culture? After all, the message that confidence is all we need to succeed permeates the whole culture.
A pre-stage interview with one of the contestants on the show was especially telling. She said, "I could be the next greatest act in America, because I have the courage, the self-esteem, the confidence, the faith and hope and belief in myself." Talent wasn't mentioned at all.
Yesterday I had a nightmare. I was on the stage of America's Got Talent and Piers Morgan, my favorite judge, was questioning me:
Piers: Do you have a talent people will pay for?
Me: Yes, I do.
Piers: What is it?
Me: I sing so badly people will pay me to stop.
Do you know that numbers have destinies? Well, to have a destiny, a number needs to have a life, or in mathematical terms, destinies are defined with respect to an operation or a function.
I know the term "destiny" from John Conway, the creator of the Game of Life. It would be harmonious to assume that he suggested this term.
Case 1. SOD. Suppose our function is the sum of digits of a number, denoted as SOD. Then the trajectory of a number k is the sequence a(n), such that a(0) = k and a(n+1) = SOD(a(n)).
Two numbers have the same destiny with respect to SOD if the tails of their trajectories coincide. Suppose a(n) and b(n) are two trajectories. Then the numbers a(0) and b(0) on which these trajectories are build have the same destiny if there exist N and M, such that for any j, a(N+j) = b(M+j). In particular, all numbers in the same trajectory a(n) have the same destiny.
In the above example of SOD, any trajectory of a positive integer ends with a one-digit non-zero number repeating many times. It follows that all the natural numbers have 9 different destinies with respect to SOD, which only depend on the remainder of the number modulo 9.
Given an operation, we can build another sequence that is called "the first occurrence of a new destiny". This is the sequence of numbers c(n) such that c(n) is the smallest number with its destiny. For the SOD operation, the sequence of the first occurrences of new destinies is finite and is equal to: 1, 2, 3, 4, 5, 6, 7, 8, 9.
Case 2. The next prime. Let us consider a different example. Let the function f(n) be the next prime after n. Then wherever we start, the tail of the trajectory with respect to the function f(n) is a sequence of consecutive prime numbers. Therefore, with respect to this function all integers have the same destiny.
Case 3. Reverse. Suppose f(n) is a reverse of n. If a number is a palindrome, then its trajectory is a one-cycle consisting of that number. If a number is not a palindrome, then its trajectory is a two-cycle consisting of a number and its reverse. The first appearance of a new destiny is sequence A131058 — a list of numbers n whose reverse is not less than n. In this case, instead of studying destinies, it might be more interesting to study types of destinies. For this operation we have two types: one-cycles for palindromic integers and two-cycles for non-palindromic integers.
Case 4. The sum of proper divisors. For the next example, let f(n) denote the sum of proper divisors. Let's look at the trajectory of 15: it is 15, 9, 4, 3, 1, 0. The sum of proper divisors of zero is not-defined or is equal to infinity, whichever you prefer. So, let us say the trajectory of 15 is finite, and ends with 1, 0. This situation makes the definition of destinies more complicated, but it is appropriate to say that finite sequences have the same destinies if they end with the same number. For our example of sums of proper divisors all finite sequences end with 0. Thus all the numbers whose trajectories are finite have the same destinies. The sequence of new destinies starts 1, 6, 28, 220, …; and we do not know what the next number is because for 276 we do not know the behavior of its trajectory. Even when we know what kind of life the number is living the destinies are not always clear.
Case 5. TITO. The next interesting example is the TITO operation. TITO is an abbreviation of "Turning Inside, Turning Outside". By definition, to calculate TITO(n) you need to reverse the prime factors of n, multiply them back together (with multiplicities) and reverse the result. For example, to calculate TITO(68), we first find prime factors of 68, which are 2, 2 and 17. We reverse them and multiply: 2*2*71 = 284. Then we reverse the result: TITO(68) = 482.
It is easy to see that prime numbers are among the fixed points of the TITO operation. That means all prime numbers have different destinies of the same type: they end with a one-cycle. There are numbers other than prime that have one-cycle destinies. For example, a palindrome that is a product of palindromic primes is a fixed point of the TITO operation. There are other cases too: for example 26 is a fixed point, but is neither prime nor palindromic. There are numbers that have one-cycle destinies, but are not the fixed points of TITO operation themselves. For example, the trajectory of 49 starts with the following sequence: 49, 94, 841, 4648, 8212, 80041, 415003, 282145, 54796, 849667, 3652951, 35429422, 2941879, 27075955, 5275759, 5275759, 5275759. ….
There are numbers that generate two-cycles. For example, 12 and 156. For numbers n that have only palindromic factors, TITO(n) is equal to the reverse of n. If n is not a palindrome and the reverse of n has only palindromic factors, then the trajectory of n is a two-cycle. Not all two-cycles are like that. Take for example 291 which is a product of primes 3 and 97. Thus, TITO(291) is the reverse of 3*79, which is equal to 732. On the other hand, 732 = 2*2*3*61. Hence, TITO(732) = reverse(2*2*3*16) = 291.
There are longer cycles too. Take for example 15, which generates a cycle of length 7: 15, 51, 312, 447, 3282, 744, 213, 15, …. Also, for some numbers we do not know if there is a cycle. The smallest number for which I myself do not know whether the trajectory tends to infinity or collapses into a cycle is 78.
For the TITO operation we might be more interested in types of destinies. Personally, I am interested not only in the types of destinies, but also in sets of numbers that have the same destinies for the same reasons. For example, I am interested in dividing numbers of one-cycle TITO destinies into three groups: primes, palindromes with palindromic primes and other cases.
Case 6. RATS. I kept the RATS operation for dessert as, in my opinion, it is the most interesting operation with respect to destinies. RATS is the abbreviation of Reverse Add Then Sort. For example, to calculate RATS(732) we need to reverse 732 getting 237, then add 732 and 237 together getting 969, then sort the digits. Thus, RATS(732) = 699.
Let's look at the RATS sequence starting with one: 1, 2, 4, 8, 16, 77, 145, 668, 1345, 6677, 13444, 55778, 133345, 666677, 1333444, 5567777, 12333445, 66666677, 133333444, 556667777, 1233334444, 5566667777, 12333334444, 55666667777, 123333334444, 556666667777, …. We can prove that this sequence is infinite, because numbers fall into a repetitive pattern with an increasing number of digits. This is the first such example in this discussion. John Conway calls the destiny of 1 "the creeper". Conway conjectured that RATS destinies are either the creeper or a cycle.
New destinies do not appear too often in this sequence. That is why the sequence of new destinies might be of interest: 1, 3, 9, 29, 69, 2079, 3999, 6999, 10677, 20169, …. This sequence is A161590 in the Online Encyclopedia of Integer Sequences and it needs more terms. The length of the corresponding periods starting from the second term are: 8, 2, 18, 2, 2, 2, 14, which is the sequence A161593.
Destinies are kinda fun.
Decades ago there was a study in Russia that claimed that a woman worked four more hours a day than a man on average. Men and women were equal in Russia and all had the same 40-hours-a-week jobs. Women were not, by and large, housewives, for they worked full-time.
So where did the additional four hours come from? They were devoted to house chores. In Russia, women did everything at home — at a time when life in Russia was much more difficult. For example, my family didn't have a washer, or a dryer or a dish-washing machine. Plus, everything was in deficit, so to buy milk or a sweater, women had to stand in lines, sometimes for hours.
My mother was very bitter because her husband, my father, never helped her. So I always hoped that when I got married, my husband would take on some of the house chores.
When I married Andrey, he was somewhat helpful — better than the average Russian husband. Then, when I was at grad school, we had a baby named Alexey. Andrey convinced me that I had to take over all the child care because only women could get academic maternity leave. It seemed logical and I agreed.
In a year, when the leave was over, I felt that Andrey should take over some of these duties. He refused. He insisted that since I already had published a paper when I was an undergrad, and since he still didn't have his research results for his PhD, that he had to stay focused on his work. I wasn't strong enough to resist.
We signed up for government child care — private care didn't exist — but we were on the waiting list for a couple of years. Almost no one in Russia — certainly not graduate students — could afford a private babysitter. I couldn't really work on my PhD research because between caring for the house and the baby, I never had big chunks of time. The best I could do was to start preparing for my qualifying exams.
Allow me to digress from my main story for a moment to mention my gray notebook. This notebook was our baby diary. Initially I recorded important baby data — like the first time Alexey smiled. But later, as soon as Alexey turned one year old, he became very eloquent; and this notebook became my son's quote book.
One day Andrey and I went out and my mom babysat Alexey, who was two years old. When we returned, my mother recited the following quote from Alexey:
When will Daddy be back from the university and Mommy from the store?
I don't really remember the long hours in stores or the cooking and cleaning. I remember the quote.
My son Alexey taught me to play "The Settlers of Catan
Lesson 1. Trading is beneficial for both traders.
When you agree to exchange your two rocks for one grain, one grain is more valuable to you than two rocks. The opposite is true for your trading partner.
Presumably, the same principle works for the economy. If I buy a sweater at T.J.Maxx for $20, I need the sweater more than $20. And if the store sells this sweater for $20, they are hoping to make some profit, that is, that the sweater cost them less than $20. Supposedly, shopping transactions are profitable for both parties.
Lesson 2. Trading is bad for non-trading players.
This is the consequence of the fact that in "Settlers of Catan" there is only one winner. If something is good for someone, it is bad for everyone else. In real life you do not have to lose if someone wins. With each shopping transaction everyone gains. This is the reason why shopping must be good for the economy.
Lesson 3. Powerful players can persuade other players to trade against their best interests.
Shortly after I moved to the US, I became very aware of my own smell. My smell didn't change with my move from Russia, nor did my sense of smell change. I was just bombarded with deodorant advertisements, and due to the vulnerability of my self-perception, in one year I bought more deodorants than in all my previous 30 years. I have a friend who has an exceptional sense of smell. He told me that people often use much more deodorant and perfume than they need.
Lesson 4. You pay a lot for storage.
In Settlers, if you have more than seven cards and the dice rolls seven, you need to discard half of your hand. So if you have six cards and someone offers you three grains for one sheep, consider the storage price before jumping into this bargain deal.
Once I bought so much discounted toilet paper that it lasted me for months and months. When it was time to move to a different apartment, I had to pay for the largest truck available to fit all my junk.
Lesson 5. It is important to understand the goal of the person you are trading with.
A profitable deal becomes a big mistake when, as a result of the trade, your trading partner builds a settlement right in the spot where you were planning to build.
Similarly, if your doctor prescribes you a medication, it would behoove you to know whether he will reap any profit from it himself.
Lesson 6. If a player is the only receiver of rock in the game he dictates the price.
This is like a monopoly. I needed my last laptop more than the $1,000 I paid for it. But this price included pre-installed Windows, which I didn't want and which I immediately deleted. I was forced to pay extra for Windows because of Microsoft's monopoly.
So, is shopping good for the economy?
What about that skirt I bought and never used and eventually threw away? I wasted $20 on it. But the store didn't gain that $20; they only gained their profit margin, which could have been $5. That means that together we wasted $15.
I do not throw away every piece of clothing I buy, but it is true that we buy more things than we need.
I think that going shopping to help our country get out of an economic crisis is a ridiculous idea. If you are shopping for other reasons than necessity, you do not help anyone and as a group we lose.
My son Alexey wins almost every game of "Settlers of Catan" he plays. So does my friend Mark Shiffer. The main reason is that they both know how to use trading effectively. To me that indicates that there are probably other people out there who know how to effectively sell deodorants, pills, clothing and other junk to us. I suspect that I lose in every shopping transaction, as I am an unskilled trader. If most folks are like me, could it be that shopping is actually bad for the economy?
Many years ago I conducted an experiment. I asked several sets of friends who had written joint math papers what they thought their individual contributions were. I asked them separately, of course. As the result of this experiment I formulated the conjecture:
The total of what joint authors estimate their contributions to be is always more than 100%.
Here is an actual example of answers I received from the two authors of a joint paper.
Author 1: My contribution is 80%. I suggested a breakthrough idea that made this paper possible. He just typed everything.
Author 2: My contribution is 80%. I did all the work. She just suggested a good idea.
You can see how the answers are synchronized. It is clear that both are telling the truth. People just tend to over-value their own input.
In other cases each author thinks that she or he generated the main idea. It doesn't mean that one of them is lying. Very often they are absolutely sincere. Take this example of Alice and Bob, who are working on a paper together. Alice suggests that they might have better progress on their theorem if they consider graphs with symmetries first. Bob is engrossed in his thoughts and doesn't register Alice's suggestion. Next day, he comes up with an idea to add a group action on graphs. He sincerely believes that this was his own idea. It would be hard to know whether this had been provoked by Alice's suggestion, or had come to Bob independently. Alice assumes that they are working on her idea.
When you acknowledge other people's contribution, keep in mind that their perception might be different from yours. If you do not want to hurt other people's feelings, you might consider inflating your gratitude.
The conjecture doesn't apply to single-author papers. First of all, mathematicians never claim their contribution is 110% as non-mathematicians do. In many cases, especially when there are acknowledgements in the paper, it would be illogical to claim 100% contribution. Most mathematicians are logical, so if they are gracious enough to acknowledge the help of others, they are unlikely to claim 100%.
I would be curious to continue the experiment and either prove or disprove my conjecture. I'd appreciate your help. If you want to be part of this experiment, you can provide the following numbers in your comments: your average contribution to your own papers; and also your weighted average contribution to your joint papers.
I always thought that the famous equation
102 + 112 + 122 = 132 + 142
is sort of a miracle, a random fluke. I enjoyed this cute equation, but never really thought about it seriously. Recently, when my son Sergei came home from MOP, he told me that this equation is not a fluke; and I started thinking.
Suppose we want to find five consecutive integers such that the sum of the squares of the first three is equal to the sum of the squares of the last two. Let us denote the middle number by n, which gives us the equation:
(n–2)2 + (n–1)2 + n2 = (n+1)2 + (n+2)2.
After simplification we get a quadratic equation: n2 – 12n = 0, which has two roots, 0 and 12. Plugging n = 0 into the equation above gives us (–2)2 + (–1)2 + 02 = 12 + 22, which doesn't look like a miracle at all, but rather like a trivial identity. If we replace n with 12, we get the original miracle equation.
If you looked at how the simplifications were done, you might realize that this would work not only with five integers, but with any odd number of consecutive integers. Suppose we want to find 2k+1 consecutive integers, such that the sum of the squares of the first k+1 is equal to the sum of the squares of the last k. Let us denote the middle number by n. Then finding those integers is equivalent to solving the equation: n2 = 2k(k+1)n. This provides us with two solutions: the trivial solution 0, and the non-trivial solution n = 2k(k+1).
So our miracle equation becomes a part of the series. The preceding equation is the well-known Pythagorean triple: 32 + 42 = 52. The next equation is 212 + 222 + 232 +242 = 252 + 262 + 272. The middle numbers in the series are triangular numbers multiplied by four.
Actually, do you know that 102 + 112 + 122 = 132 + 142 = 365, the number of days in a year? Perhaps there are miracles or random flukes after all.
My son, Sergei Bernstein, got accepted to MIT through early action. Because the financial costs of studying at MIT worried me, I insisted that Sergei also apply to Princeton and Harvard, as I had heard they give generous financial packages. In the end, Sergei was rejected by Princeton and wait-listed and finally rejected by Harvard. Though many people have been rejected by Princeton and Harvard, not too many of them have won places on US teams for two different international competitions — one in mathematics and the other in linguistics. To be fair, Sergei was accepted by these teams after Princeton had already rejected him. Nonetheless, Sergei has an impressive mathematical resume:
I am trying to analyze why he was rejected and here are my thoughts.
Many people told me of surprising decisions by Ivy League schools this year. The surprises were in both directions: students admitted to Ivy League colleges who didn't feel they had much of a chance and students not admitted that had every right to expect a positive outcome. I should mention that I personally know some very deserving kids who were admitted.
I wonder if there has been a change in the financial demographics of the students Harvard and Princeton have accepted this year. If so, this will be reflected in the data very soon. We will be able to see if the average SAT scores of students go down relative to the population and previous years.
I do not know why Sergei wasn't accepted; perhaps I'm missing something significant. But if it was because of our finances, it would be ironic: Sergei wasn't admitted to Princeton and Harvard for the same reason he applied there.
I recently got a new job — to coordinate math students at RSI (Research Science Institute). RSI provides a one-month research experience based at MIT for high school juniors. The program is highly competitive and kids from all over the world apply for it.
Before the program started, I asked around among mathematicians for advice on how to do a great job with these talented kids. I was surprised by the conflicting opinions on the value of the program. I thought you'd be interested in hearing those opinions, although I confess that I do not remember who said what, or anyone's exact words. I will just repeat the gist of it.
Former participants:
Potential participants:
Grad students, former and potential mentors:
Professors on the program in general:
I asked some math professors to suggest problems for these students:
The 2009 RSI has just begun. We have awesome students, great mentors and quite interesting problems to solve. I am positive we'll prove the negativists wrong.
Let me remind you of a very interesting problem from my posting Oleg Kryzhanovsky's Problems.
You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same, except for their labels. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?
I do not want you to find the weight of each coin; I just want you to say yes if the labels are correct, or no if they are not.
I have given this problem to a lot of people, and not one of them solved it. Some of my students mistakenly thought that they succeeded. For example, they would start by putting the coins labeled 1 and 2 on the left cup of the scale and 3 on the right cup. If these coins balanced, the students assumed that the coins on the left weighed 1 and 2 grams and that the coin on the right weighed 3 grams. But they'd get the same result if they had 1 and 4 on the left, for example, and 5 on the right. I am surprised that no one has solved it yet, as I thought that this problem could be offered to middle-schoolers, since it does not actually require advanced mathematical skills.
If you want to try to solve this problem, pause here, as later in this essay I will be providing a number of hints on how to do it. The problem is fun to solve, so continue reading only if you are sure you're ready to miss out on the pleasure of solving it.
I propose the following sequence a(n). Suppose we have a set of n coins of different weights weighing exactly an integer number of grams from 1 to n. The coins are labeled from 1 to n. The sequence a(n) is the minimum number of weighings we need on a balance scale to confirm that the labels are correct. The original Oleg Kryzhanovsky's problem asks to prove that a(6) = 2. It is easy to see that a(1) = 0, a(2) = 1, a(3) = 2. You will enjoy proving that a(4) = 2 and a(5) = 2.
In general, we can prove that a(n) ≤ n-1. For any k < n, the k-th weighing compares coins labeled k and k+1. If we get the expected result every time, then we can confirm that the weights are increasing according to the labels.
On the other hand, we can prove that a(n) ≥ log3(n). Indeed, suppose we conducted several weighings and confirmed that the labels are correct. To every coin we can assign a sequence of three letters L, R, N, corresponding to where the coin was placed during each weighing — left cup, right cup or no cup. If two coins are assigned the same letters for every weighing, then we can't confirm that the labels on these two coins are accurate. Indeed, if we switch the labels on these two coins, the results of all the weighings will be the same.
My son, Alexey Radul, sent me the proof that a(10) = a(11) = 3. As 3 is the lower bound, we just need to describe the weighings that will work.
Here is the procedure for 10 coins. For the first weighing we put coins labeled 1, 2, 3, and 4 on one side of the scale and the coin labeled 10 on the other. After this weighing, we can divide the coins into three groups (1,2,3,4), (5,6,7,8,9) and (10). We know to which group each coin belongs, but we do not know which coin in the group is which. The second weighing is 1, 5 and 10 on the left, and 8 and 9 on the right. The left side should weigh less than the right side. The only possibility for the left side to weigh less is when the smallest weighing coins from the first and the second group and 10 are on the left, and the two largest weighing coins from the first two groups are on the right. After the second weighing we can divide all coins into groups we know they belong to: (1), (2,3,4), (5), (6,7), (8,9) and (10). The last weighing contains the lowest weighing coin from each non-single-coin group on the left and the largest weighing coin on the right, plus, in order to balance them, the coins whose weights we know. The last weighing is 2+6+8+5 = 4+7+9+1.
Here is Alexey's solution, without explanation, for 11 coins: 1+2+3+4 < 11; 1+2+5+11 = 9+10; 6+9+1+3 = 8 +4+2+5.
Let me denote the n-th triangular number as Tn. Then a(Tn) ≤ a(n) + Tn - n - 1. Proof. The first weighing is 1+2+3 ... +n = Tn. After that we can divide coins into groups, where we know that the labels stay within the group: (1,2,...,n), (n+1,n+2,...,Tn-1), (Tn). We can check the first group in a(n) weighings, the second group in Tn - n - 2 weighings, and we already used one. QED.
Similarly, a(Tn+1) ≤ a(n) + Tn - n.
For non-triangular numbers there are sometimes weighings that divide coins into three groups such that the labels can only be permuted within the same group. For example, with 13 coins, the first weighing could be 1+2+3+4+5+6+7+8 = 11+12+13. After that weighing we can divide all coins into three groups (1,2,3,4,5,6,7,8), (9,10), (11,12,13).
In all the examples so far, each weighing divided all the coins into groups. But this is not necessary. For example, here is Alexey's solution for 9 coins. The first weighing is 1+2+3+4+5 < 7+9. When we have five coins on the left weighing less than two coins on the right, we have several different possibilities of which coins are where. Other than the case above, we can have 1+2+3+4+6 < 8+9 or 1+2+3+4+5 < 8+9. But let's look at the next weighing that Alexey suggests: 1+2+4+7 = 6+8. Or, three coins from the previous weighing's left cup, plus one coin from the previous weighing's right cup equals the sum of the two coins that were left over. This can only be true if the coins in the first weighing were indeed 1+2+3+4+5 on the left and 7+9 on the right. After those two weighings everything divides into groups (1,2,4), (3,5), (6,8), (7) and (9). The last weighing 1+7+9 = 4+5+8, resolves the rest.
To check 7 or 8 coins in three weighings is simpler than the cases for 9, 10, and 11 coins, so I leave it as an exercise. As of today I do not know if it is possible to check 7, 8 or 9 coins in two weighings. Consider this a starred exercise.
I invite you to play with this amusing sequence and calculate some bounds. Also, let me know if you can prove or disprove that this sequence is non-decreasing.
Langton's ant travels on the infinite square grid, colored black and white. At each time step the ant moves one cell forward. The ant's direction changes according to the color of the cell he moves onto. The ant turns 90 degrees left if the cell is white, and 90 degrees right if the cell is black. After that, the cell he is on changes its color to the opposite color.
There is a symmetry of time and space for this ant. If at any point of the ant's travel, someone interferes and reverses the ant's direction in between the cells, the ant and the grid will traverse the steps and stages back to the starting point.
Let's give this ant a life. I mean, let's place him inside the Game of Life invented by John H. Conway. In addition to the Langton's ant's rules, I want the cells to change colors according to the rules of the Game of Life.
Let me remind you of the rules of Conway's Game of Life. We call black cells live cells and white cells dead cells. Black is life and white is death. The cell has eight neighbors — horizontal, vertical, diagonal. At each time step:
So, our ant will be traveling in this dying and reproducing population and correcting nature's mistakes. He revives dead cells and kills live cells.
There is an ambiguity in this ant's life description. The life can happen at two different moments. In the first ant's world, the ant jumps from one cell to the next, and while he is in the air, the cells have time to copulate, give birth and die. Upon landing, the ant changes direction and uses his magic wand to change the life status of its landing cell. In the second ant's world, the ant moves to the destination cell, changes its own direction and the status of the cell and then takes a smoke. All the fun, sex and death happen while he is enjoying his cigarette.
The ant's life has symmetry in a way that is similar to the symmetry of the ant without life. If we reverse the ant's direction back and also switch his life-style from the first to the second or vice versa, then the ant and the grid will go backwards in their states.
The parameters for the Langton's ant were chosen to make the ant's behavior interesting. The parameters of the Game of Life were chosen to make the Game of Life's behavior interesting. To make the ant's life fascinating, we might want to modify the ant's behavior or the Life's rules. The synergy of the ant and the Life might be intriguing only if the ant changes its behavior and the Life changes its rules.
Let's experiment and discover how we need to change the rules in order to make the ant's life interesting.
John H. Conway is teaching me his doomsday algorithm to calculate the day of the week for any day. The first lesson was devoted to 2009. "The 2009's Doomsday is Saturday" is a magic phrase I need to remember.
The doomsday of a particular year is the day of the week on which the last day of February falls. February 28 of 2009 is Saturday, thus 2009's doomsday is Saturday. For leap years it is the day of the week of February 29. We can combine the rules for leap years and non-leap years into one common rule: that the doomsday of a particular year is the day of the week of March 0.
f you know the day of the week of one of the days in 2009, you can theoretically calculate the day of the week of any other day that year. To save yourself time, you can learn by heart all the days of the year that fall on doomsday. That is actually what Conway does, and that is why he is so fast with calculations. The beauty of the algorithm is that the days of the doomsday are almost the same each year. They are the same for all months other than January and February; and in January and February you need to make a small adjustment for a leap year. That gives me hope that after I learn how to calculate days in 2009 I can easily move to any year.
To get us going we do not need to remember all the doomsday days in 2009. It is enough to remember one day for each month. We already know one for February, which works for March too. As there are 28 days in February, January 31 happens on a doomsday. Or January 32 for leap years.
Now we need to choose days for other months that are on doomsday and at the same time are easy to remember. Here is a nice set: 4/4, 6/6, 8/8. 10/10. For even months the days that are the same as the month will work. The reason it works so nicely is that two consecutive months starting with an even-numbered month, excluding February and December, have the sum of days equaling 61. Hence, those two months plus two days are 63, which is divisible by 7.
Remembering one of the doomsdays for every other month might be enough to significantly simplify calculations. But if you want a day for every month, there are additional doomsday days to remember on odd numbered months: 5/9, 9/5, 7/11 and 11/7. These days can be memorized as a mnemonic "9-5 job at 7-11," or, if you prefer, "I do not want to have a 9-5 job at 7-11."
If you throw in March 7, then the rule will fit into a poem John recited to me:
The last of Feb., or of Jan. will do
(Except that in leap years it's Jan. 32).
Then for even months use the month's own day,
And for odd ones add 4, or take it away*.
*According to length or simply remember,
you only subtract for September or November.
Let's see how I calculate the day of the week for my friend's birthday, July 29. The 11th of July falls on the doomsday, hence July 25 must be a doomsday. So we can see that my friend will celebrate on Wednesday this year.
You might ask why I described this trivial example in such detail. The reason is that you might be tempted to subtract 11 from 29, getting 18 and saying that you need to add four days to Saturday. In the method I described the calculation is equivalent, but as a bonus you calculate another day for the doomsday and consequently, you are getting closer to John Conway who remembers all doomsdays.
My homework is the same as your homework: practice calculating the days of the week for 2009.
Visitors to the math department of Princeton University used to stop by John Conway's office. Even if it were closed, they could peek through the window in the door to see the many beautiful, symmetric figures hanging in his office.
The figures, which John Conway had made, were there for 20 years. Just recently John received a letter informing him that his office had been inspected by the State Fire Marshall and that "those things hanging from your ceiling are against the State's fire code and must be taken down." The math department was worried about a possible fine.
So John threw away the "things." I wanted to cry as I watched these huge garbage bags being taken away. I rescued several figures, but that was all that I could fit into my car. For 20 years no one complained, but now the bureaucracy has beat out beauty and mathematics.
This picture is the last view of the "hazardous" office.
On one of my visits to Princeton, I stopped by the math department and, as usual, asked John H. Conway what he was up to. He told me that he was turning numbers inside out. He explained that to perform this procedure on a number you need to reverse every prime factor, multiply the reversed factors back and reverse the result. For example, for 34, which is the product of 2 and 17, we need to reverse 2 and 17 (turning inside), changing them into 2 and 71, multiply back, getting 142, and reversing again (turning outside), leading to the resulting number 241.
He started with a number, turned it inside out, then turned the result inside out, and so on, thus getting an infinite sequence for any number. Every sequence he had calculated up to this point ended with a cycle.
Before I had interrupted him, he was calculating the sequence starting with 78 and it was growing. I suggested that Mathematica could do this calculation faster than John could do in his head. Although that was very rude considering his reputation for speed, John agreed, and we moved to a computer. The computer confirmed that the sequence starting with 78 was growing wildly.
While playing around with this, I became very interested in numbers that are fixed under this turning inside-out operation. First, prime numbers do not change — you just reverse them twice. Second, palindromes with palindromic primes do not change, as every reversal encounters a palindrome to apply itself to. I started to wonder if there are palindromes that are fixed under the turning inside-out procedure, but are not products of palindromic primes.
Here is where John had his revenge. He told me that he would be able to find such a number faster than I could write a program to find it. And he won! He found such a number while I was still trying to debug my program. The number he found was 1226221.
Here is how he beat me. If you have two not-too-big primes that consist of zeroes and ones and that are reversals of each other, their product will be a palindrome. And John is really fast in checking primes for primality. See his lesson in my essay Remember Your Primes.
The next day, when I stumbled on John again, he was doing something else. I asked him about the numbers and he told me that he was no longer interested. Initially he had hoped that every sequence would end in a cycle. The turning inside-out operation doesn't produce much growth in a number. On top of that, prime numbers are stable. That means that if the turning inside-out operation was a random operation with a similar growth pattern, there would have been a very high probability of every sequence eventually hitting a prime. But the operation is not random, as it doesn't change remainders modulo 9. In particular, sequences that start with a composite number divisible by 3 would never hit a prime. Our experiment with 78 discouraged him by showing no hope for a cycle.
I asked him, "Why not do it in binary?" He answered that he had sinned enough playing with a base 10 sequence.
A year later when I next visited Princeton and saw John again, I asked him if he had published or done something with the operation. He had not. He agreed to submit the sequence to the online database, but only if we came up with a name he liked. And we did. We now call this operation TITO (turning inside, turning outside). Please welcome TITO.
When my son Sergei made it to the International Linguistics Olympiad I got very excited. After I calmed down I realized that training for this competition is not easy because it is very difficult to find linguistics puzzles in English. This in turn is because these Olympiads started in the USSR many years ago and were adopted here only recently. So I started translating problems from Russian and designing them myself for my son and his team. For this particular problem I had an ulterior motive. I wanted to remind my son and his team of rare words in English with Greek origins. Here is the problem:
We use many words that have Greek origins, for example: amoral, asymmetric, barometer, chronology, demagogue, dermatology, gynecologist, horoscope, mania, mystic, orthodox, philosophy, photography, polygon, psychology, telegram and telephone. In this puzzle, I assume that you know the meanings of these words. Also, since I am a generous person, I will give you definitions from Answers.com of some additional words derived from Greek. If you do not know these words, you should learn them, as I picked words for this list that gave me at least one million Google results.
In the list below, I picked very rare English words with Greek origins. You can derive the meanings of these words without looking in a dictionary, just by using your knowledge of the Greek words above.
Here are some other words. You do not have enough information in this text to derive their definitions, but you might be able to use your erudition to guess the meaning.
Robert Calderbank and Ingrid Daubechies jointly taught a course called "The Theory of Games" at Princeton University in the spring. When I heard about it I envied the students of Princeton — what a team to learn from!
Here is a glimpse of this course — a problem on Evolutionarily Stable Strategy from their midterm exam with a poem written by Ingrid:
On an island far far away, with wonderful beaches
Lived a star-bellied people of Seuss-imagin'd Sneetches.
Others liked it there too — they loved the beachy smell,
From their boats they would yell "Can we live here as well?"
But it wasn't to be — steadfast was the "No" to the Snootches:
For their name could and would rhyme only with booches …
Until with some Lorxes they came!
These now also enter'd the game;
A momentous change this wrought
As they found, after deep thought.
Can YOU tell me now
How oh yes, how?
In what groupings or factions
Or gaggles and fractions
They all settled down?
Sneetches and Snootches only:
| Sneetches | Snootches | |
| Sneetches | 4 | 3 |
| Snootches | 3 | 2 |
Sneetches and Snootches and Lorxes:
| Sneetches | Snootches | Lorxes | |
| Sneetches | 4 | 3 | 8 |
| Snootches | 3 | 2 | 16 |
| Lorxes | 8 | 16 | -60 |
Find all the ESSes, in both cases.
My guest blogger is a friend and a wordsmith Sue Kelman:
* * *
Rizzo (Ratso to his friends) was my cage mate. We had a nifty pad at Children's Hospital — all we could eat with no scrounging, clean beds, quiet surroundings, and plenty of activity to keep us occupied.
Rizzo's favorite was the maze. Each week he bragged to us about how fast he made it through to the cheese. Larry over in Row-D was always the slowest. All the guys used to razz him about it. No matter how hard he tried, Larry took the wrong turn every time. I think his mother spent some time out at a psych hospital, so maybe they messed with her brain and that affected Larry. Who knows? I suspect that know-it-all visiting researcher from MIT knows what happened to Larry but he's probably keeping it under his hat until he publishes his results in JAMA. Putz!
Okay, so one day, Rizzo just came back from one of those tests where they make us hit a little button when the red and green lights go on. Personally this is my favorite gig because of course there's no running around, but Rizzo likes to throw a monkey wrench into the research data. So every now and then, even when he knows how we should respond, he does just the opposite. I told you, he's one smart rodent.
Rizzo's pretty famous, too. Oh he's not as famous as that talking grey parrot that used to be over at Harvard, but he's been around. For a while he was a top gun — the big performer for a group of genetics guys. He's had his DNA tested more times than Mike Tyson.
Then they lent him out to Hematology where, I swear, the guy's already had 15 blood transfusions. No wonder he's healthy as a horse.
Me, I'm just your average lab rat. I know the drill: wake up, eat a few pellets, perform, eat some more pellets, doze off, and wake up to do it all over again. Not a bad life if you can stay away from those vivisection weirdos. They're like Dr. Mengele all over again.
I'd tell you more but Rizzo's gonna tell us about the time he got out of his cage and made it almost all the way to the Starbucks wagon before they caught him. Great story and he's a real raconteur. None of that Stuart Little crap. We fall over laughing every time we hear it. Gotta go.
Due to the popularity of my previous posting of linguistics puzzles, I've translated some more puzzles from the online book Problems from Linguistics Olympiads 1965-1975. I've kept the same problem number as in the book; and I've used the Unicode encoding for special characters.
Problem 180. Three Tajik sentences in Russian transliteration with their translations are below:
Your task is to assign a meaning to each out of four used Tajik words.
Problem 185. For every sequence of words given below, explain whether it can be used in a grammatically correct English sentence. If it is possible show an example. In the usage there shouldn't be any extra signs between the given words.
Problem 241. In a group of relatives each person is denoted by a lower-case letter and relations by upper-case letters. The relations can be summarized in a table below:
| a | b | c | d | e | f | g | |
| a | — | A | A | B | D | E | E |
| b | A | — | A | E | D | E | E |
| c | F | F | — | G | H | I | I |
| d | H | J | J | — | K | L | L |
| e | B | B | B | N | — | N | N |
| f | O | O | D | L | Q | — | A |
| g | J | J | H | L | K | F | — |
The table should be read as following: if the intersection of the row x and the column y has symbol Z, then x is Z with respect to y. It is known that e is a man.
You task is to find out the meaning of every capital letter in the table (each letter can be represented as one English word).
There is interesting data to show that MIT takes math students more seriously than Harvard and Princeton. By Michael Sipser's suggestion I looked at the Putnam Competition results. Out of the top 74 scorers of 2007, 21 were from MIT, 9 from Harvard and 7 from Princeton. Keep in mind that the total freshman enrollment at MIT is much lower than at Harvard or Princeton. This story repeated itself in 2008: out of top 79 scorers 23 were from MIT, 11 from Harvard and 11 from Princeton.
Ironically, MIT's team didn't win Putnam in those years. MIT's team won the third place after Harvard and Princeton. If you look at the results more closely, you will notice that had MIT arranged teams differently, MIT would have won.
It appears that MIT put their three top scorers from the previous year on their lead team. MIT shouldn't assume that those three continue to be their strongest competitors. Instead they should probably test their students right before the Putnam competition, because if you look at MIT's top individual performers, had they been on a team together, they would have won.
Maybe MIT should rethink its algorithm for creating teams, or maybe we should just wait. As it is obvious that MIT is more serious about math, all top math students may want to go to MIT in coming years. If this happens, the mathematics field will be absolutely dominated by MIT.
Of course you can. Can you do it in Russian? You do not need to know Russian to do it; you just need to solve my puzzle. Below are some numerals written in Russian. You have enough information to write any number from 1 to 99 inclusive in Russian.
If you are too lazy to write all the Russian numerals I requested, try the most difficult ones: 16, 17, 19, 67 and 76.
If you know Russian, then I have a back-up puzzle for you. Do the same thing for French:
And again, if you are lazy, you can concentrate on translating 15, 18, 19, 41, 51, 56, 65, 78 and 99 into French.
I invite my readers to create similar puzzles in all languages.
I recently read an article titled "Think having children will make you happy?" that discusses studies correlating happiness and having children. Some studies show that parents and non-parents have the same level of happiness. But other studies show that non-parents are happier. So, do children make us less happy?
There are two major reasons that kids might make people less happy in a long run. First, children require a lot of resources; they put a strain on our budget, time and careers. As my friend Sue Katz puts it: parental unhappiness could stem from poverty, illness, fighting the educational institutions, feeling stuck in a violent relationship because of the kids — a million things, depending on class and options.
Second, children might not live up to our expectations. Parents often dream that their children will have wonderful careers, be supportive of their parents later in life and most importantly be good people. But in reality, children choose their own careers, not necessarily a path approved by their parents. Plus they might live at a distance or the relationship might be strained. They might even develop completely different values from their parents.
The article claims that on average kids will bring more problems than joy to our lives. Do not rush to cancel unprotected sex with your spouse tonight yet.
My friend Peggy Boning suggested that the study should have separately checked parents who wanted children and parents who didn't. It could be that parents who didn't want children are less happy than parents who wanted them. Which means that if you do not want children, make sure you have protected sex. If you do want children, you might be happier with children than without.
Anyone who has studied statistics knows that correlation doesn't mean causality. An individual who wants to have children might be happier as a result, and at the same time the statistics data may well be true. I'd like to find arguments that can make peace between these two suppositions.
Feel free to add your own ideas.
In Psilvania no one knows English, except for one retired professor Mary Bobs. That is why every year the organizers of the linguistics Olympiad in Psilvania beg Mary to design a puzzle in English. Kids in Psilvania know other languages — which gives individuals an advantage if the puzzle is in those languages. An English puzzle would create a level playing field.
Here is the puzzle that Mary proposed. I'm omitting the Psilvanian text, because the characters do not match anything in Unicode tables.
Professor Bobs provided the following sentences in English, accompanied by their translations into Psilvanian. She called these sentences Raw Materials:
The first task that she required was to translate the following sentences into Psilvanian:
Professor Mary Bobs had quit smoking that very week and she couldn't concentrate. It seems that she may have given more information than is necessary. Is it possible to remove any of the Raw Materials (one or more translated sentences) and keep the puzzle solvable? If so, what is the largest number of Raw Materials you can eliminate? Explain.
Her second task was to translate some sentences from Psilvanian into English, and the answers she hoped the students would calculate were:
For each of the three English sentences above, decide whether the participants of the Olympiad will be capable of getting this particular answer. If for any of these three sentences you suspect that they will not be able to arrive at the correct answer, explain why.
Just out of curiosity I googled two phrases: "male mathematician" and "female mathematician". The results for these phrases on May 3, 2009 were:
Why do you think that female mathematicians are more popular than male mathematicians? I think it is because when people hear the word mathematician, by default they picture a man, so the phrase "male mathematician" is perceived as pleonastic.
I decided to look at some of the 824 sites talking about "male mathematicians". Many web-pages containing the words "male mathematician" are actually pages about female mathematicians, where there is a need to mention a mathematician of the opposite sex. Many other sites are dating pages where a mathematician looks for a partner, and it is wise to start with a description of the sex of the seeker.
Speaking of dating, did I ever mention that I am a female mathematician who was married three times, and all of my spouses were male mathematicians?
One of the best inventions of recent years is toilet seat covers. So whenever I visit a bathroom without seat covers, I curse and make my own out of toilet paper. This is not very effective, as toilet paper doesn't stay nicely in place and it takes a lot of time, when time might be of the essence. Besides it wastes a lot of tissue.
When I come into a toilet and see a lot of long pieces of clean toilet paper lying around, I know that someone else tried to create a seat cover before me.
So here is my idea. Why not make individual packages with folded toilet seat covers, like kleenex packets? When I couldn't find toilet seat covers in my local pharmacy, I started wondering how to patent my idea and dreaming about a lot of money. But before I let myself get too excited, I checked the Internet. My new toilet-seat-covers-to-go were already available. So I bought some. Now, every time I flush them, I watch my potential patent going down the toilet.
I would like to discuss the solution to one of the linguistics puzzles I posted a while ago. Here is problem number 211 from the online book Problems from Linguistics Olympiads 1965-1975:
You are given words in Swahili: mtu, mbuzi, jito, mgeni, jitu and kibuzi. Their translations in a different order are: giant, little goat, guest, goat, person and large river. Make the correspondence.
First, lets say that a giant is a large man. The Swahili translation of "giant" may have elements of Swahili words for a "man" and a "large river". Next we notice that each of these Swahili words naturally divides into two parts. We can put them in a table such that the first part is the same for every row and the second part is the same for every column.
| m-tu | m-buzi | m-geni | |
| ji-tu | ji-to | ||
| ki-buzi |
When I gave this problem to my students, they loved the idea that the word "giant" is comprised of the two words "large" and "man", so they assumed that in Swahili a "guest" would also have a two-part translation, such as a "man who visits." In the list of words we have three different types of "man": man, giant and guest. Once they noticed that "m" appears three times, they concluded that "m" must mean a man. Therefore, the object must be the first part of a Swahili word, while the second part contains its description.
Next, they noted that the first part "ji" appears twice. They decided that "ji" must be a goat and thus "ki" must be a river. All of this gives us sufficient information to derive the translations: "mgeni" a guest, "kibuzi" a large river, "mbuzi" a giant, "mtu" a man, "jitu" a goat and "jito" a little goat.
My students were very proud of themselves, but I was dissatisfied with this solution. Here are the problems I've identified:
I would suggest a different approach. Let's say that the puzzle is about sizes, and we have three objects (man, guest, goat) of normal size, two large objects (giant and large river), and one small object (little goat). That means "m" must mean normal, and the size description is in front. If the first part is the size, then "ki" is small, "ji" large. From here "mgeni" is a guest, "kibuzi" a little goat, "mbuzi" a goat, "mtu" a man, "jitu" a giant, and "jito" is a large river.
I love this puzzle because it teaches us to continue pondering, even after everything seems to fit. If you stumble upon the first solution you need to go back and think some more. Only after you discover the second solution does it become clear that the second one is right.
My son, Alexey Radul, married Rebecca Frankel recently. They had a nerdy wedding. You can judge just how nerdy by this poem written and presented by Gremio (Gregory Marton), the best man:
In this summation, may there be no subtraction;
May you multiply blissfully, and find no division;
May the roots of the power of your love run deep;
May the logs of your joys be exponentially steep;
May you derive greatest pleasure from integrating your lives,
And well past your primes, retain composite rhymes.
This group gathered here, on that lush green field:
May we help you build proofs on the vows you just sealed.
I have a tiny book written by Vladimir Arnold Problems for Kids from 5 to 15. A free online version of this book is available in Russian. The book contains 79 problems, and problem Number 6 criticizes American math education. Here is the translation:
(From an American standardized test) A hypotenuse of a right triangle is 10 inches, and the altitude having the hypotenuse as its base is 6 inches. Find the area of the triangle.
American students solved this problem successfully for 10 years, by providing the "correct" answer: 30 inches squared. However, when Russian students from Moscow tried to solve it, none of them "succeeded". Why?
Arnold has inflated expectations for kids. The book presents the problems according to the increasing order of difficulty, and this suggests that he expects kids under 10 to solve Number 6.
Arnold claimed that every student from Moscow would notice what is wrong with this problem. I can forgive his exaggeration, because I've met such kids. Anyways, I doubt that Arnold ever stumbled upon an average Russian student.
My own fundamental interest is in the state of American math education, so I decided to check his claim concerning American students. I asked my students to calculate the area of the triangle in the above puzzle.
Here are the results of my experiment. Most of them said that the answer is 30. Some of them said that it is 24. In case you're wondering where the 24 is coming from, I can explain. They decided that a right triangle with hypotenuse 10 must have two other legs equal to 8 and 6.
Some of the students got confused, not because they realized that there was a trick, but because they thought the way to calculate the area of the right triangle is to take half the product of its legs. As lengths of legs were not given, they didn't know what to do.
There was one student. Yes, there was one student, who decided that he could calculate the legs of the triangle from the given information and kept wondering why he was getting a negative number under the square root.
You decide for yourself whether there is hope for American math education. Or, if you are a teacher, try running the same experiment yourself. I hope that one day I will hear from you that one of your students, upon reading the problem, immediately said that such a triangle can't exist because the altitude of the right triangle with the hypotenuse as the base can never be bigger than half of the hypotenuse.
A long ago my son Sergei went to the Streamline School Olympiad. Some of the problems were really nice and I asked the organizer, Oleg Kryzhanovsky, where he took the problems from. It seems that he himself supplied all the problems, many of which are his original creations. He told me that he can invent a math Olympiad problem on demand for any level of difficulty on any math topic. No wonder that he is the author of almost all math problems at the Ukraine Olympiad.
The following is a sample of his problems from the Streamline Olympiad. For my own convenience I have chosen problems without figures and equations. Note: I edited some of them.
1998 (8th - 9th grade). Find three numbers such that each of them is a square of the difference of the two others.
1999 (9th - 10th grade). The positive integers 30, 72, and N have a property that the product of any two of them is divisible by the third. What is the smallest possible value of N?
1999 (9th - 10th grade). You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?
1999 (11th - 12th grade). In how many ways can the numbers 1, 2, 3, 4, 5, 6 be ordered such that no two consecutive terms have a sum that is divisible by 2 or 3.
2000 (6th - 7th grade). Let A be the least integer such that the sum of all its digits is equal to 2000. Find the left-most digit of A.
2000 (8th grade). You have six bags with coins that look the same. Each bag has an infinite number of coins and all coins in the same bag weigh the same amount. Coins in different bags weigh 1, 2, 3, 4, 5 and 6 grams exactly. There is a label (1, 2, 3, 4, 5, 6) attached to each bag that is supposed to correspond to the weight of the coins in that bag. You have only a balance scale. What is the least number of times do you need to weigh coins in order to confirm that the labels are correct?
I've been translating a lot of linguistics puzzles lately. Now it is my turn to create a new linguistics puzzle. Here are some English phrases with their Russian translations:
Your task is to translate into Russian the following sentences:
Bonus question. Have you noticed any signs that I am getting tired of linguistics?
I stumbled on a Russian linguistics competition called The Russian Little Bear. Most of the puzzles are Russian-specific; but some of them can be translated. I concentrated on puzzles for grades six through nine and used Unicode for uncoding strange characters.
Problem 1. Here are some Latin words with their English translations:
Pick the line of words from A to E that best translates these phrases into Latin: You are loved, I ask, He invites.
Problem 2. The first astronauts from India (I), Hungary (H), France (F) and Germany (G) were Bertalan Farkas (1), Sigmund Jähn (2), Rakesh Sharma (3) and Jean-Loup Chrétien (4). Match the astronauts to the countries:
Problem 3. You do not need to know Russian to solve this puzzle. It is enough to know the modern Russian alphabet: А, Б, В, Г, Д, Е, Ё, Ж, З, И, Й, К, Л, М, Н, О, П, Р, С, Т, У, Ф, Х, Ц, Ч, Ш, Щ, Ъ, Ы, Ь, Э, Ю. Before XVIII century, numbers in Russian were denoted by letters, for example: ТЛЕ — 335, РМД — 144, ФЛВ — 532.
How was 225 written in old Russian?
(A) ВВФ; (B) ВВЕ; (C) СКЕ; (D) СКФ; (E) ВНФ.
Problem 4. Here are several Turkish words and phrases with their English translations:
Pick the line of words from A to E that best translates these phrases into Turkish: thirty isles, men?
Courtney Gibbons gave me her permission to add her webcomics to my collection of Funny Math Pictures.
Abstruse Goose gave me his permission to add his webcomics to my collection of Funny Math Pictures.
Until the introduction of the Abel prize, the Fields medal was the most prestigious prize in mathematics. The medal has been awarded 48 times and all of the recipients have been men. Can we conclude that women are inferior to men when it comes to very advanced mathematics? I do not think so.
The Fields medal was designed for men; it is very female-unfriendly. It is the prize for outstanding achievement made by people under age 40. Most people start their research after graduate school, meaning that people have 10-15 years to reach this outstanding achievement. If a woman wants to have children and devote some time to them, she needs to do it before she is 40. That puts her at a big disadvantage for winning the medal.
Recently the Abel prize for mathematics was introduced. This is the math equivalent of a Nobel prize and nine people have received the prize, all of them male. The Wolf prize is another famous award: 48 people have received it so far and they too have all been male.
On the grand scale of things, women have only recently had the option of having a career in mathematics. Not so long ago it was considered quite exceptional for a woman to work in mathematics. The number of female mathematicians is increasing, but as this is a new trend, they are younger people. At the same time, Abel prizes and Wolf prizes are given to highly accomplished and not-so-young people. That means the increase in the percentage of women PhDs in mathematics might affect the percentage of females getting the prize, but with a delay of several dozen years.
There are other data covering extreme math ability. I refer to the International Math Olympiad. The ability that is needed to succeed in the IMO is very different from the ability required to succeed in math research. But still they are quite similar. The IMO data is more interesting in the sense that the girls who participate are usually not yet distracted by motherhood. So in some sense, the IMO data better represents potential in women's math ability than medals and prizes.
Each important math medal or prize is given to one person a year on average. So the IMO champion would be the equivalent of the Fields medal or the Wolf prize winner. While no girl was the clear best in any particular year, there were several years when girls tied for the best IMO score with several other kids. For example:
In one of those years, a girl might have been the best, but because the problems were too easy, she didn't have a chance to prove it. Evgenia Malinnikova was an outstanding contender who twice had a perfect score. In 1990, she was one out of four people, and she was younger than two of them, as evidenced by the fact they they were not present in 1991. Only one other person — Vincent Lafforgue — got a perfect score in 1990 and 1991. We can safely conclude that Evgenia was one of two best people in 1990, because she was not yet a high school senior.
This might be a good place to boast about my own ranking as IMO Number Two, but frankly, older rankings are not as good as modern ones. Fewer countries were participating 30 years ago, and China, currently the best team, was not yet competing.
Girls came so close to winning the IMO that there is no doubt in my mind that very soon we will see a girl champion. The Fields medal is likely to take more time.
Due to the popularity of my previous posting of linguistics puzzles, I've translated some more puzzles from the online book Problems from Linguistics Olympiads 1965-1975. All of them are from the phonetics section and I've kept the same problem number as in the book. I've used the Unicode encoding for special characters.
Problem 20. In the table below there are numerals from some Polynesian languages. Note that I couldn't find the proper English translation for one of the languages, so I used transliteration from Russian. The language sounds like "Nukuhiva" in Russian.
| Languages | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Hawaiian | kahi | lua | ha | lima | ono | hiku | walu | ***** | ||
| Māori | tahi | rua | toru | wha | ono | whitu | waru | iwa | ***** | |
| Nukuhiva | tahi | to'u | ha | ono | va'u | ***** | ||||
| Rarotongan | ta'i | 'a | rima | ono | 'itu | varu | iva | ŋa'uru | ||
| Samoan | tasi | lua | lima | ono | fitu | iva | ŋafulu |
Your task is to find the words that should be in the empty cells. Note that wh, ', and ŋ denote special consonants.
Problem 21. Below you will find words in several relative languages. You can group these words into pairs or triples of words with the same origin and the same or a similar meaning.
āk, dagr, bōk, leib, fōtr, waʐʐar, buoh, dæʒ, plōgr, hām, wæter, hleifr, pfluog, eih, heimr, fuoʐ, plōʒ.
Task 1. Divide the words into groups so that the first group has words from the same language, the second group has words from another language and so on.
Task 2. (optional) List your suggestions about the meanings of the words and about the identity of the languages.
Problem 22. These words from the Aliutor language are followed by their translations. The stresses are marked by an apostrophe in front of the stressed vowel.
Your task is to put the stresses in the following words: sawat 'lasso', pantawwi 'fur boots', nəktəqin 'solid', ɣətɣan 'late autumn', nəminəm 'bouillon', nirvəqin 'sharp', pujɣən 'spear', tilmətil 'eagle', wiruwir 'red fish', wintatək 'to help', nəmalqin 'good', jaqjaq 'seagull', jatək 'to come', tavitətkən 'I will work', pintətkən 'he attacks (someone)', tajəsqəŋki 'in the evening'.
Note that the vowel ə is similar to many unstressed syllables in English words, such as the second syllable in the words "taken" and "pencil". This vowel is shorter than other vowels in the Aliutor language.
Last time Sue refinanced her mortgage was six years ago. She received a 15-year fixed loan with 5.5% interest. Her monthly payment is $880, and Sue currently owes $38,000.
Sue is considering refinancing. She has been offered a 5-year fixed loan with 4.25% interest. You can check an online mortgage calculator and see that on a loan of $38,000, her monthly payments will be $700. The closing costs are $1,400. Should Sue refinance?
Seems like a no-brainer. The closing costs will be recovered in less than a year, and then the new mortgage payments will be pleasantly smaller than the old ones. In addition, the new mortgage will last five years instead of the nine years left on the old mortgage.
What is wrong with this solution? What fact about Sue's old mortgage did I wickedly neglect to mention? You need to figure that out before you decide whether Sue should really refinance.
So many people liked the puzzles I posted in Subtraction Problems, Russian Style, that I decided to present a similar collection of multiplication and division puzzles. These two sets of puzzles have one thing in common: kids who go for speed over thinking make mistakes.
Humans have 10 fingers on their hands. How many fingers are there on 10 hands?
This one is from my friend Yulia Elkhimova:
Three horses were galloping at 27 miles per hour. What was the speed of one horse?
Here is a similar invention of mine:
Ten kids from Belmont High School went on a tour of Italy. During the tour they visited 20 museums. How many museums did each kid go to?
Another classic:
How many people are there in two pairs of twins, twice?
Can you add more puzzles to this collection?
Today I have my first invited guest blogger, John Berman. John is a 2006, 2007, 2008 and 2009 USAMO qualifier. He was also selected to be on the US team at the Romanian Masters in Mathematics competition. Also, he placed 6th at the North American Computational Linguistics Olympiad. Here is his piece:
The analysis is based on the list of 2009 USAMO qualifiers.
There is a rule that if nobody naturally qualifies for the USAMO from a state, then the highest scoring individual will qualify. Unfortunately, this means that we must remove those states with only one USAMO qualifier. We have 33 states remaining. If we sort these strictly by number of USAMO qualifiers, then we find the following result.
States with at least 4 USAMO qualifiers (24 total) voted for Obama, with the following exceptions: Georgia, Texas, South Carolina, and Missouri. In addition, of the two states with 3 USAMO qualifiers, one voted for Obama and one for McCain. The remaining states with 2 qualifiers (5 total) voted Republican.
Now this is not really unexpected. States with very large populations tend to be democratic and also produce more USAMO qualifiers. The most notable exceptions are Georgia and Texas, both of which were indeed exceptions (major outliers, in fact) above. This prompts the following consideration.
States with at least 8 USAMO qualifiers per 10 million residents (25 total) voted for Obama, with the following exceptions: Florida, Wisconsin, South Carolina, Missouri, and Georgia. Of these, all but Georgia fall within 50% of the target 8 USAMO qualifiers per 10 million residents. Georgia has 18 qualifiers per 10 million residents. Note also that the entire USA has 16 qualifiers per 10 million residents.
Furthermore, if USAMO qualifiers had been used instead of population for determining electoral votes, Obama would have won with 86% of the vote rather than 68%. In general, if the Democrat can secure all those states with at least 1 qualifier per million residents (plus DC), he will win with 303 votes. He can even lose the three red states in that category (Georgia, Missouri, and South Carolina) for exactly 269.
USAMO qualifiers per 10 million residents (for states with more than one qualifier) are:
The states with only one USAMO qualifier are WY, VT, ND, AK, SD, DE, MT, RI, HI, ID, NE, WV, NV, AR, MS, OK, and AZ. The only blue one of these which falls below 8 qualifiers per 10 million is Nevada (we would expect it to have at least 2 qualifiers to fit the expected pattern). Otherwise, it is at least possible that each state fits the pattern of 8 qualifiers per 10 million residents if and only if it votes Democratic.
More than a year ago, when I had my employment benefits with BAE Systems, I called my benefits center with a general question. The customer service representative refused to answer until I gave her my password. I didn't have a password, so she told me that they would mail my new password to me.
But I needed an answer, so I tried the website, only to be informed that my new password is in the mail and I should wait for its arrival.
In a week, a letter with a password arrived and I called the benefits center again. I happily told them my new password and opened my mouth to ask my question. However, they didn't accept my password. Obviously, they had changed my password twice, first when I called and then again when I tried their website. Since only ten minutes passed between these two events, both passwords should have arrived on the same day. But that didn't happen. So my valid password was still in the mail.
In the second it took me to recover from this news, the customer representative told me that they would be sending me a new password and hung up before I could tell her not to.
A new password arrived the next day. I knew that they had already reset that password, and that I'd have to wait a week for the third password to arrive.
I was tempted to call them again and try to create an infinite password resetting loop, but I actually needed to ask my question. So I threw away my freshly arrived, but no-longer-valid password and waited for a week for the next one.
I was lucky to figure it out so quickly, for otherwise my problem could have spiraled out forever. As a professional specifications writer, here are my suggestions to all benefits centers that have that kind of software on what they should do:
I had to wait two weeks to ask a simple question. Now I am writing and complaining about it in the hopes that someone who can fix the problem will read this. Maybe it would have been more productive to write a program that clicks on the "I forgot my password" button every second. This would have daily generated thousands of letters with new passwords to me. Maybe then this problem would have drawn attention sooner.
My son Sergei brought back the Flip-Flop game from Canada/USA Mathcamp, and now I teach it to my students. This game trains students in the multiplication table for seven and eight. These are the most difficult digits in multiplication. This game is appropriate for small kids who just learned the multiplication table, but it is also fun for older kids and adults.
This is a turn-based game. In its primitive simplification kids stand in a circle and count in turn. But it is more interesting than that. Here's what to say and do on your turn, and how the game determines who is next.
First I need to tell you what to say. On your turn, say the next number by default. However, there are exceptions when you have to say something else. And this something else consists of flips and/or flops.
So what are flips? Flip is related to seven. If a number is divisible by seven or has a digit seven, instead of saying this number, we have to say "flip" with multiplicities. For example, instead of 17 we say "flip" because it contains one digit seven. Instead of 14 we say "flip", because ii is divisible by seven once. Instead of 7 we say "flip-flip", as it is both divisible by seven and has a digit seven. Instead of 49, we say "flip-flip" as 49 is divisible by the square of seven. Instead of 77 we say "flip-flip-flip" as it has two digits seven and is divisible by seven once.
Flop relates to eight the same way as flip relates to seven. Thus, instead of 16 we say "flop" as it is divisible by eight; instead of 18 we say "flop" as it contains the digit eight; and for 48 we say "flop-flop&
