Tanya Khovanova's Math Blog


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My Ancestry

I always wanted to be a person of the world. I wanted my genes to be a mixture of everything. I was glad that I had a great-grandfather from Poland and a great-great-great-grandmother from France. I was also thrilled when my mom told me that her Asian students think she is one of theirs. So I decided to send my DNA to 23andMe and really see what I have.

To my surprise, my world is not as mixed as I expected: I am 99.5% European. My Asian part is minuscule: 0.2%, out of which only 0.1% is assigned as Yakut. My African part is also 0.2%.

My European part is a mixture of mostly eastern and northern European. I am 2.8% Ashkenazi.

My Ancestry

In addition to my genetic profile, 23andMe sent me the list of a thousand of my distant relatives. They also sent me a report about the most common last names among my relatives. The list starts with Cohen and continues with Levine, Levin, Goldberg, and Rubin.

You might be surprised by this list of Jewish names when I am only 2.8% Jewish. But the list is based on people who decided to send their DNA to 23andME and provided their last names. All my Russian relatives remained in Russia. Russia has its own company, I-gene, that provides a similar service, and the two databases are not shared.

Only my distant relatives who moved to the US and who are curious about their ancestry and who are willing to share their last names will appear on this list. So maybe this list is not surprising.


No Averages

Here is an old Olympiad problem:

Prove that you can choose 2k numbers from the set {1, 2, 3, …, 3k−1} in such a way that the chosen set contains no averages of any two of its elements.

Lazy Jokes

* * *

—Describe yourself in three words.
—Lazy.

* * *

Internet forum:
—Tell me about yourself.
—I am lazy and I like to eat.
—Tell me some more.
—I am tired of typing. I'll go grab a snack.

* * *

—Why do you want to divorce your wife?
—She nags too much. For the last half six month, she's been bugging me everyday to throw away the Christmas tree.

* * *

Yesterday I realized that I'm not the laziest person in the world. I saw my neighbor walking the dog on a leash through his window.

* * *

The list of symptoms of laziness:
1)


A Logic Quiz

This is a variation on an old quiz. Can you answer the last question?

—An airplane carries 500 bricks. One of the bricks falls out. How many bricks are left in the airplane?
—This is easy: 499!
—Correct. Next question. How do you put a giraffe into a refrigerator?
—Open the refrigerator, put in the giraffe, and close the refrigerator door.
—Good, next. How do you put an elephant into a refrigerator?
—Open the refrigerator, take out the giraffe, put in the elephant and close the door.
—Correct. The Lion King is hosting his birthday party. All the animals come to congratulate him—except one. Why?
—The elephant couldn't come because it is in the refrigerator.
—Fantastic, next. A man needs to cross a river inhabited by crocodiles and he doesn't have a boat. What should he do?
—He can just swim: all the crocodiles are attending Lion King's birthday party.
—Amazing! The last question: The man swims across the river, and dies. What happened?


ApSimon's Mints

Hugh ApSimon described the following coin puzzle in his book Mathematical Byways in Ayling, Beeling and Ceiling.

New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material: fake coins weigh the same independently of the mint. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings so that you can deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?

I will follow ApSimon's notation. Suppose Pr and Qr is the number of coins from the mint r used in the first and the second weighing correspondingly. That is, we are minimizing Σmax(Pr,Qr). (All my summations are over all the mints. I skip the summation limits because it is difficult to write math in html.) Let us denote by W the weight of the genuine coin and by W(1 + ε) the weight of the fake coin. We do not know ε, except that it is not zero.

Let dr be either 0 or 1, depending on what material the r-th mint uses. Thus, the coin from the r-th mint weighs W(1 + drε). We know the results of these two weighings and the weight of the genuine coin. Therefore, we can calculate the following two values: a = ΣPrdrε and b = ΣQrdrε.

It is clear that we need to request at least one coin from each mint and use it in at least one weighing: Pr + Qr > 0. If both sums a and b are zero, then all the mints are producing genuine coins. Neither of the two values gives us much information as we do not know ε. We can get rid of ε by dividing a by b.

There are 2n − 1 combinations of possible answers: these are subsets of the set of mints producing fake coins given that there is at least one. Thus we need to select numbers Pr and Qr, so that a/b produces 2n − 1 possible answers for different sets of values of dr.

Let us consider cases in which the total number of mints is small. If there is one mint we can take one coin and we won't even need a second weighing. For two mints we need one coin from each mint for a total of 2. For three mints, one coin from each mint is not enough. I leave this statement as an exercise. It is possible to test three mints with four coins: one each from the first and second mints and two from the third mint. The coins from each mint for the first and second weighings are (0,1,2) and (1,1,0) respectively.

To prove that this works we need to calculate (d2 + 2d3)/(d1 + d2) for seven different combinations of dr. I leave this as an exercise.

This puzzle seems to be very difficult. We only know the answer if the number of mints is not more than seven. The corresponding sequence A007673 in the OEIS is: 1, 2, 4, 8, 15, 38, 74. It is possible to give bounds for this sequence, but they are so far apart. The lower bound is n. And the ApSimon's book offers a construction for two weighings were Pr = r! and Qr = 1.

You can try to find a better construction, or you can try calculating more terms of the sequence. You can also read more about this problem in my short paper Attacking ApSimon's Mints.

I do not want to leave the readers with the puzzle that might end up being intractable. So I suggest the following easy puzzle. Solve the ApSimon's Mints problem assuming that the weight of the fake coin is known.


Masturbating With an Accent

I once took an accent reduction course, to modify my Russian accent in English. In the first class the teacher explained that the biggest reason people have strong accents is that they stop learning and trying to improve their speech as soon as they can be understood. I promised myself to never stop learning and to continue working on my accent reduction forever.

Once I was giving a lecture on probability and statistics at the IAP mathematical series. My last slide was about the research on the correlation between masturbation male habits and prostate cancer. Their interpretation of the data had been wrong and a very good example of what not to do.

So I looked directly into the eyes of the course coordinator, who was observing my lecture, and without realizing what I was saying, asked, "Do we have time for masturbation?"

Everyone started laughing and I had to present my slide in order to explain myself.

The news of my double entendre spread. Soon after that I was asked to give a lecture at the Family Weekend at MIT. I wonder if that is why the lecture coordinator asked me not to discuss masturbation as small children might be present.

Luckily that was the only fallout from my blooper. Anyway, I decided to stop working on my accent. When people understand that English is not my first language they forgive more readily my slips of tongue.


Computer Security Jokes

* * *

—Honey, have you blocked our computer?
—Yes.
—What's the password?
—Our wedding date.
—%?#!!

* * *

—What's the pin on our card?
—We're on a public chat, honey. Why don't I sms it?
—But I forgot my phone. Please tell me, cupcake!
—Okay. By digit: the second digit of our apartment number, the fourth digit of your phone number, the month of my birthday, and the number of our children.
—Got it. How clever! 8342, right?

* * *

—Where is the report?
—We are stuck. The tech people took our monitor with passwords.
—What!?!
—Our monitor got broken so the techs took it for repair. Our passwords were written on the stand.


Hat Puzzle: Create a Distribution

Here is a setup that works for the several puzzles that follow it:

The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat—from his 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 together on a strategy.

I wrote about puzzles with this setup before in my essay The Wizards' Hats. My first request had been to maximize the number of wizards who are guaranteed to survive. It is easy to show that you cannot guarantee more than 50 survivors. Indeed, each wizard will be right with probability 0.5. That means whatever the strategy, the expected number of wizards guessing correctly is 50. My second request had been to maximize the probability that all of them will survive. Again, the counting argument shows that this probability can't be more than 0.5.

Now here are some additional puzzles, including the first two mentioned above, based on the same setup. Suggest a strategy—or prove that it doesn't exist—in which:

  1. 50 wizards will be guaranteed to survive.
  2. 100 wizards will survive with probability 0.5.
  3. 100 wizards will survive with probability 0.25 and 50 wizards will survive with probability 0.5.
  4. 75 wizards will survive with probability 1/2, and 25 wizards survive with probability 1/2.
  5. 75 wizards will survive with probability 2/3.
  6. The wizards will survive according to a given distribution. For which distributions is it possible?

As I mentioned, I already wrote about the first two questions. Below are the solutions to those questions. If you haven't seen my post and want to think about it, now is a good time to stop reading.

To guarantee the survival of 50 wizards, designate 50 wizards who will assume that the total number of red hats is odd, and the rest of the wizards will assume that the total number of red hats is even. The total number of red hats is either even or odd, so one of the groups is guaranteed to survive.

To make sure that all of them survive together with probability 0.5, they all need to assume that the total number of red hats is even.


Linear Algebra on a Mission Impossible

I love making math questions out of the movies. Here is a Mission Impossible III question.

Tom Cruise is cute. He plays Ethan Hunt in Mission Impossible movies. In Mission Impossible III he needs to steal the Rabbit's Foot from a secure skyscraper in Shanghai. He arrives in Shanghai and studies the skyscraper looking out his window. He decides to break in through the roof. And the way to get to the roof is to use a rope and swing across from another, even taller, skyscraper. 1:21 minutes into the movie, Ethan Hunt calculates the length of the rope he will need by using the projection of a skyline on his window, as seen on the first picture.

MI3 Skyline

Explain why the projection is not enough to calculate the length of the rope. What other data does he need for that? Ethan Hunt does request extra data. But he makes one mistake. He uses his pencil as a compass to draw the end of the rope curve, as seen on the second picture. Explain what his mistake is.

MI3 Rope

Nim Automaton

I mentor three PRIMES projects. One of them, with Joshua Xiong from Acton-Boxborough Regional High School, is devoted to impartial combinatorial games. We recently found a connection between these games and cellular automata. But first I need to remind you of the rules of Nim.

In the game of Nim there are several piles with counters. Two players take turns choosing a pile and removing several counters from it. A player loses when he or she who doesn't have a turn. Nim is the most famous impartial combinatorial game and its strategy is well known. To win, you need to finish your move in a so called P-position. Nim P-positions are easy to calculate: Bitwise XOR the number of counters in all the piles, and if the result is zero then it's a P-position.

The total number of counters in a P-position is even. So we calculated the sequence a(n): the number of P-positions in the game of Nim with three piles with the total number of counters equal to 2n. As soon as we got the sequence we plugged it into the OEIS, and voilà it was there: The sequence A130665 described the growth of the three branches of the Ulam-Warburton cellular automaton.

U-W automaton

The first picture shows the automaton after 6 generations. The automaton consists of cells that never die and it grows like this: start with a square on a square grid. In the next generation the squares that share exactly one side with the living squares are born. At the end remove the Southern branch.

Everything fell into place. We immediately realized that the language of the automata gives us the right words to describe what we know about the game of Nim.

Now we want to describe the automaton related to any impartial combinatorial game. Again, the cells never die and the initial cells correspond to terminal P-positions. People who write programs for calculating P-positions will find a notion of the next generation very natural. Indeed, the program usually starts with the terminal P-positions: they are generation 0. Then we can proceed by induction. Suppose we have found P-positions up to generation i. Denote the positions that are one move away from generation i and earlier as Ni. Then the P-positions that do not belong to generation i and earlier and from which all moves belong to Ni are the P-positions from generation i + 1.

This description explains the generations, but it doesn't explain who is the parent of a particular P-position. The parent-child relations are depicted as edges on the cellular automaton graph. The parent of position P1 from generation i + 1 is a P-position P2 in generation i that can be reached from P1 in the game.

The parent-child relationship in the game of Nim is especially easy to explain. A P-position P1 is a parent of a P-position P2 if P1 differs from P2 in exactly two piles and it has one fewer counter in each of these piles. For example, a P-position (1,3,5,7) has six parents, one of them is (1,3,4,6). In the game with thee piles a P-position always has exactly one parent.

A position in the game of Nim with three piles is naturally depicted as a triple of numbers, that is as a point in 3D. The picture below shows the Nim automaton in 3D at generation 6.

Nim Automaton

Our paper, Nim Fractals, about sequences enumerating P-positions and describing the automaton connection in more detail is posted at the arXiv:1405.5942. We give a different, but equivalent definition of a parent-child relationship there. A P-position P1 is a parent of P2 if there exists an optimal game such that P1 is achieved from P2 in exactly two moves in a game which takes the longest number of possible moves.


Kolmogorov Student Olympiad in Probability

There are too many Olympiads. Now there is even a special undergraduate Olympiad in probability, called Kolmogorov Student Olympiad in Probability. It is run by the Department of Probability Theory of Moscow State University. I just discovered this tiny Olympiad, though it has been around for 13 years.

A small portion of the problems are accessible for high school students. These are the problems that I liked. I edited them slightly for clarity.

Second Olympiad. Eight boys and seven girls went to movies and sat in the same row of 15 seats. Assuming that all the 15! permutations of their seating arrangements are equally probable, compute the expected number of pairs of neighbors of different genders. (For example, the seating BBBBBBBGBGGGGGG has three pairs.)

Third Olympiad. One hundred passengers bought assigned tickets for a 100-passenger railroad car. The first 99 passengers to enter the car get seated randomly so that all the 100! possible permutations of their seating arrangements are equally probable. However, the last passenger decides to take his reserved seat. So he arrives at his seat and if it is taken he asks the passenger in his seat to move elsewhere. That passenger does the same thing: she arrives at her own seat and if it is taken, she asks the person to move, and so on. Find the expected number of moved passengers.

Third Olympiad. There are two 6-sided dice with numbers 1 through 6 on their faces. Is it possible to "load" the dice so that when the two dice are thrown the sum of the numbers on the dice are distributed uniformly on the set {2,...,12}? By loading the dice we mean assigning probabilities to each side of the dice. You do not have to "load" both dice the same way.

Sixth Olympiad. There are M green and N red apples in a basket. We take apples out randomly one by one until all the apples left in the basket are red. What is the probability that at the moment we stop the basket is empty?

Seventh Olympiad. Prove that there exists a square matrix A of order 11 such that all its elements are equal to 1 or −1, and det A > 4000.

Twelfth Olympiad. In a segment [0,1] n points are chosen randomly. For every point one of the two directions (left or right) is chosen randomly and independently. At the same moment in time all n points start moving in the chosen direction with speed 1. The collisions of all points are elastic. That means, after two points bump into each other, they start moving in the opposite directions with the same speed of 1. When a point reaches an end of the segment it sticks to it and stops moving. Find the expected time when the last point sticks to the end of the segment.

Thirteenth Olympiad. Students who are trying to solve a problem are seated on one side of an infinite table. The probability that a student can solve the problem independently is 1/2. In addition, each student will be able to peek into the work of his or her right and left neighbor with a probability of 1/4 for each. All these events are independent. Assume that if student X gets a solution by solving or copying, then the students who had been able to peek into the work of student X will also get the solution. Find the probability that student Vasya gets the solution.


IQ Migration

The Russian website problems.ru has a big collection of math problems. I use it a lot in my work as a math Olympiad coach. Recently I was giving a statistics lesson. While there was only one statistics problem on the website, it was a good one.

Assume that every person in every country was tested for IQ. A country's IQ rating is the average IQ of the population. We also assume that for the duration of this puzzle no one is born and no one dies.

Math Kangaroo's Logic Puzzle

My AMSA students loved the following puzzle from the 2003 Math Kangaroo contest for grades 7-8:

The children A, B, C and D made the following assertions. How many of the children were telling the truth?
A) 0   B) 1   C) 2   D) 3   E) Impossible to determine

Express 6

I was given this puzzle at the last Gathering for Gardner.

Use arithmetic operations to express 6 using three identical digits. For each digit from 0 to 9 find at least one way to express 6.

For example, I can express 6 using three twos in many ways: 2 + 2 + 2, or 2 · 2 + 2, or 22 + 2. But the problem doesn't ask for many ways. One way is enough, but you need to do it for every digit. So nine more cases to go: 0, 1, 3, 4, 5, 6, 7, 8, and 9.


The Virtue of Laziness

My son, Alexey Radul, is a programmer. He taught me the importance of laziness in programming.

One of his rules:

Not to write the same line of code in the same program twice.

If you need the same line of code in the same program, that means you should either use a loop or outsource the line to a function. This style of coding saves time; it makes programs shorter and more elegant. Such programs are easier to debug and understand.

I remember how I copied and pasted lines of code before he taught me this rule. Then I needed to change parameters and missed some of the lines during changing. Debugging was such a headache.

Mathematicians are way lazier than programmers. Consider the system of two equations: x+2y=3 and 4x+5y=6. There are no repeating lines here. Only letters x and y appear twice. Mathematicians invented the whole subject of linear algebra and matrices so that they would not need to rewrite variables.

Mathematicians are driven by laziness. Once ancient mathematicians first solved a quadratic equation, they didn't want to do it again. So they invented a formula that solves all quadratic equations once and for all.

I try to keep up with tradition. I try to make my theorems as general as possible. When I write my papers, I try to make them short and simple. When I think about mathematics I try to get to the stage where the situation is so clear I can think about it without paper and pencil. I often discover new theorems while I am in bed, about to fall asleep. Sometimes I wake up with a good idea. So I do my job while I sleep.

I love my profession. I get paid for being lazy.


Beer Jokes and Hat Puzzles

This is one of my favorite jokes:

Three logicians walk into a bar. The waitress asks, "Do you all want beer?"
The first logician answers, "I do not know."
The second logician answers, "I do not know."
The third logician answers, "Yes."

This joke reminds me of hat puzzles. In the joke each logician knows whether or not s/he wants a beer, but doesn't know what the others want to drink. In hat puzzles logicians know the colors of the hats on others’ heads, but not the color of their own hats.

This is a hat puzzle which has the same answers as in the beer joke. Three logicians walk into a bar. They know that the hats were placed on their heads from the set of hats below. The total number of available red hats was three, and the total number of available blue hats was two.

Red Hat Red Hat Red Hat Red Hat Red Hat
Three logicians walk into a bar. The waitress asks, "Do you know the color of your own hat?'"
The first logician answers, "I do not know."
The second logician answers, "I do not know."
The third logician answers, "Yes."

The puzzle is, what is the color of the third logician's hat?

This process of converting jokes to puzzles reminds me of the Langland's Program, which tries to unite different parts of mathematics. I would like to unite jokes and puzzles. So here I announce my own program:

Tanya's Program: Find a way to convert jokes into puzzles and puzzles into jokes.


How Well Do You Know Your Dice?

Each time I see John Conway he teaches me something new. At the Gathering for Gardner he decided to quiz me on how well I know a regular six-sided die. I said with some pride that the opposite sides sum up to 7. He said, "This is the first level of knowledge." So much for my pride. I immediately realized that the next level would be to know how all the numbers are located relative to each other. I vaguely remembered that in the corner where 1, 2, and 3 meet, the numbers 1, 2, and 3 are arranged in counter-clockwise order.

Here's how John taught me to remember every corner. There are two types of corners. In the first type numbers form an arithmetic progression. John calls such numbers counters. He chose that name so that it would be easy to remember that counters are arranged in counter-clockwise order. The other numbers he calls chaos: their increasing sequence goes clockwise.

Once I grasped that, I relaxed thinking that now I know dice. "What about the third level?" he asked. "What third level?" "Now that you know which number goes on which side, you need to know how the dots are arranged." Luckily, there are only three sides on which the dots are not placed with rotational symmetry: 2, 3, and 6. And they all meet in a corner, which John calls the home corner. The rule is that the diagonals formed by the dots on the sides with 2, 3, and 6, meet in the home corner. You might argue that 6 doesn't have a diagonal. But if you look at 6, you can always connect the dots to form the letters N or Z, depending on the orientation of the die. When you lay the letter N on its side, it becomes the letter Z. Thus they define the same diagonal. This diagonal has to meet the diagonals from 2 and 3 in the corner.

When I came home from the conference I picked up a die and checked that the rules work. There are 8 corners. It is enough to remember one corner of numbers to recover the other numbers by using the opposite sum rule. But it is nice to have a simple rule that allows us to bypass the calculation. Four of the corners have numbers in arithmetic progression: 1:2:3, 1:3:5, 2:4:6, and 4:5:6. They are counters and they are arranged counter-clockwise. The other four corners are: 1:2:4, 1:4:5, 2:3:6, and 3:5:6, and they are arranged clockwise.

I wanted to provide a picture of a die for this post and went online to see if I could grab one. Many of the graphic images of dice, as opposed to photographs, were arranged incorrectly. Clearly these visual artists did not study dice with John Conway.

Then I decided to check my own collection of dice. Most of them are correct. The ones that are incorrect look less professional. Here is the picture. The ones on the right are correct.

Dice

My New Yellow Road

I started my Yellow Road a year ago on February 9, 2013, when my weight was 245.2 pounds. My system worked for eight months. I lost 25 pounds. Then I went to two parties in a row and gained four pounds. According to my plan, I was supposed to eat only apples after lunch. It was too difficult to stick to that, and I got off-target. My target weight continued decreasing daily, as per my plan, while I got stuck. The growing difference between my real weight and my target weight was very discouraging, so I lost my momentum.

I decided to reset the target weight and restart the plan. I changed my plan slightly to incorporate the lessons I had learned about myself.

On February 9, 2014, I started my New Yellow Road. I weighed 223.2 pounds. So I reset my target weight to be 223.2 on February 9. Each day my target weight goes down by 0.1 pounds. I weigh myself each morning. If I am within one pound of my target weight, I am in the Yellow zone and I will eat only fruits and vegetables after 5:00 pm. If I am more than one pound over my target weight, I am in the Red Zone and will eat only apples after 5:00 pm. If I am more than one pound below my target weight, I can eat anything.


Salary Negotiations

I want to tell you the story of my "successful" salary negotiation. The year was 2003 and I had a temporary visiting position at Princeton University. I wanted to move to Boston and my friend showed my resume to Alphatech. The interview went well and they offered me a position as Lead Analyst with a salary of $110.000. This was particularly good news considering that the tech job market was very weak in 2003.

At that time, I was working very hard on building my self-esteem. I read lots of books and was in therapy for two years. I decided to practice what I had learned to try to negotiate a bigger paycheck. While speaking to the HR guy on the phone, I was standing up, as I was taught, and projecting my voice firmly with my chest opened up. I could hardly believe it when I heard myself ask for a $10,000 increase.

Despite my wonderful posture, the human resource person refused. However, he remembered that they had forgotten to give me a moving bonus. He asked me about my living conditions. I told him that I lived in a four-bedroom house. I didn't elaborate: it was a tiny four-bedroom house made out of a garage. I made a counter offer: forget the moving bonus, but give me my salary increase as I asked. He agreed.

For the first year, there was no real difference, because the salary increase was equal to the moving bonus. But I was planning to stay with the company for a long time, so by the second year, my clever negotiation would start to pay off. My negotiations were a success. But were they?

Things change. The boss who hired me and appreciated me stopped being my boss. The company was bought by BAE Systems who were not interested in research. To my surprise, I started getting non-glowing performance reviews. Luckily, by that time I had made a lot of friends at work, and one of them not only knew what was going on, but was willing to tell me. My salary was higher than that of other employees in the same position. Salary increases were tied to performance. They wanted to minimize my increases to bring my salary into the range of others at my level. So to justify it, they needed a negative performance review. After one negative review it is difficult to change the trend. A negative review stays on the record and affects the future reputation.

In a long run I am not sure that my salary negotiations were a success.


The Three Light Bulbs Puzzle

There is a famous puzzle about three light bulbs, that is sometimes given at interviews.

Suppose that you are standing in a hallway next to three light switches that are all off. There is another room down the hall, where there are three incandescent light bulbs—each light bulb is operated by one of the switches in the hallway. You can't see the light bulbs from the hallway. How would you figure out which switch operates which light bulb, if you can only go to the room with the light bulbs one and only one time?

This puzzle worked much better in the past when we only had incandescent light bulbs and so didn't need to specify the type of bulbs. Unfortunately, the standard solution only works with incandescent bulbs and the word "incandescent" nowadays needs to be stated. But the use of "incandescent" is a big hint. Indeed, incandescent light bulbs generate heat when they are on, so the standard solution is to turn on the first light switch, to keep the second switch off, and to turn the third switch on for five minutes before turning it off. In the room, the light bulb corresponding to the first switch will be lit, and out of the two unlit bulbs, the one corresponding to the third switch will be warm.

It's a cute solution, but there could be so many other approaches:

I invite my readers to invent other methods to solve this problem. Be creative. After all, if I were to interview you for a job, I would be more impressed by a new solution than the one that is all over the Internet.


Liars and Their Motivation

You arrive at an archipelago of many islands. On each island there are two villages. In one village truth-tellers live, and they always tell the truth. In the other village liars live, and they always lie. The islanders all know each other.

On the first island you stumbled upon three islanders and you ask each of them your question:

How many truth-tellers are there among you?

Here are their answers:

A: One.
B: A is wrong.
C: A and B are from the same village.

Can you determine who is a truth-teller and who is a liar?

This island is called a classic island, where all behave as if they were in a standard logic puzzle. It is a perfectly nice puzzle but B and C didn't answer the question: B ratted on A, and C went on a tangent. When I was younger, I never cared about the motivations of A, B, or C. Their answers are enough to solve the puzzle. But now that I am older, I keep wondering why they would choose these particular answers over other answers. So I invented other islands to impose rules on how the villagers are allowed to answer questions.

Now you travel to the next island that is called a straightforward island, where everyone answers your question exactly. You are in the same situation, and ask the same question, with the following result:

A: One.
B: One.
C: Ten.

Can you determine who is a truth-teller and who is a liar?

Once again we wonder about their motivation. This time C told an obvious lie, an answer that is impossible. Why on earth did he say 10? Isn't the goal of lying to deceive and confuse people? There is nothing confusing in the answer "ten."

Now you come to the third island, which is a straightforward inconspicuous island. To answer your question, a liar wouldn't tell you an obvious lie. For this particular situation, the liar has to choose one of the four answers that are theoretically possible: zero, one, two, or three. You are again in the same situation of asking three people how many truth-tellers are among them, and these are the answers:

A: Two.
B: Zero.
C: One.

Can you determine who is a truth-teller and who is a liar?

When you think about it, a truth-teller cannot answer zero to this question. So although zero is a theoretically possible answer, we can deduce that the person who said it is a liar. If liars are trying to confuse a stranger, and they're smart, they shouldn't answer "zero."

The next island is a straightforward inconspicuous smart island. The liars on this island are smart enough not to answer zero. You are in the same situation again and ask the same question with the following outcome:

A: Two.
B: Two.
C: One.

Can you determine who is a truth-teller and who is a liar? You shouldn't be able to. There are three possibilities. There are two truth-tellers (A and B), one truth-teller (C), or zero truth-tellers.

Let us assign probabilities to liars' answers. Assume that liars pick their answers randomly from the subset of wrong answers out of the set: one, two, three. If two of these answers are incorrect, they pick a wrong answer with probability one half. If all three of the answers are incorrect, they pick one of them with probability one-third. Suppose the people you meet are picked at random. Suppose that the probability that a random person is a truth-teller is 1/2. Given the answers above, what is more probable: that there are two truth-tellers, one truth-teller, or zero truth-tellers?


Reverse Bechdel Test

A movie passes the Bechdel Test if these three statements about it are true:

Surely there should be a movie where two women talk about the Bechdel test. But I digress.

The Bechdel test website rates famous movies. Currently they have rated 4,683 movies and 56% pass the test. More than half of the movies pass the test. There is hope. Right? Actually they have a separate list of the top 250 famous movies. Only 70 movies, or 28%, from this list pass the test.

My son Alexey suggested the obvious reverse Bechdel test, which is more striking than the Bechdel test. A movie doesn't pass the test if it

I can't think of any movie like that. Can you?


More Math Jokes

* * *

A Roman walks into a bar, holds up two fingers, and says, "Five beers, please."

* * *

To understand what a recursion is, you must first understand recursion.

* * *

A guy is complaining to his mathematician friend:
— I have a problem. I have difficulty waking up in the morning.
— Logically, counting sheep backwards should help.

* * *

— Can I ask you a question?
— You can, but you have already just done that.
— Darn, what about two questions?
— You can, but that was your second question.

* * *

The Internet ethics committee worked hard to generate a list of words that should never be used on the Internet. The problem is, now they can't post it.

* * *

Quantum entanglement of a pair of socks: As soon as one is designated as the left, the other instantly becomes the right.


Mathy Problems from the 2014 MIT Mystery Hunt

The last MIT Mystery Hunt was well-organized. It went smoothly—unlike the hunt that my team designed the year before. Sigh. As I do every year, here is the list of 2014 puzzles related to math.

There were also several puzzles requiring decoding or having a CS flavor.

I want to mention one non-mathematical puzzle.


Ambiguities in Logic

You visit an island of three towns: Trueton, Lieberg and Alterborough. Folks living in Trueton always tell the truth. Those who live in Lieberg, always lie. People from Alterborough alternate strictly between truth and lie. You meet an islander who says:

Two plus two is five. Also, three plus three is six.

Can you determine which town he is from?

It should be easy. He made two statements: the first one is false, the second is true. So he must be from Alterborough.

But what about "also"? How should we interpret this transition? There are many ways to interpret this "also." On one hand it could mean: In addition to the previous statement I am making another statement. On the other hand it could mean: The previous pause shouldn't be considered as the end of the statement; the whole thing should be interpreted as one statement. Besides this person was speaking not writing. Are we sure that the first period was not meant to be a comma or a semi-colon? If we assume that the quote is one statement, then the speaker might be either a liar or an alternator.

Here is a puzzle for you from the same island:

One night a call came into 911: "Fire, help!" The operator couldn't ID the phone number, so he asked, "Where are you calling from?" "Lieberg." Assuming no one had overnight guests from another town, is there an emergency? If so, where should help be sent? And was it a fire?

Now find the ambiguities.


Prepare for the Hunt

The MIT Mystery Hunt starts on Friday. My old team—Manic Sages— fell apart after last years' hunt. My new team—Death and Mayhem—started sending us daily practice puzzle to prepare for the hunt.

Today's puzzle was written by Paul Hlebowitsh. As usual, the answer is a word or a phrase.

Puzzle 3: Humans

"After a long day taking care of the animals, it's good to unwind by letting the taps flow."

Tap 1: Sierra Nevada Pale Ale
Tap 2: Lagunitas Pale Ale
Tap 3: 21st Amendment "Brew Free! or Die IPA"
Tap 4: Wachusett Blueberry
Tap 5: Harpoon UFO

Alice, Bob, Carol, Danny, Erica, Fred, and Gregario, the seven children of Noah, went to a local bar before the flood to get drinks. Accounts of night vary, and no one remembers exactly what happened, but some facts have become clear:

  1. Everyone had two drinks, in some order.
  2. Danny liked his first drink so much he had it again. Everyone else had drinks from different breweries.
  3. Erica was the only person to drink an IPA.
  4. Four people had the Wachusett Blueberry as their first drink.
  5. One person had the Sierra Nevada Pale Ale as their first drink.
  6. Alice only drank beers with headquarters in California, in order to spite Bob and Danny.
  7. Two Harpoon UFOs were ordered, as well as two Sierra Nevada Pale Ales. 8
  8. Alice's second drink was the same as Fred's first drink.
  9. Bob and Danny hate California and refuse to drink any beer from a company that's headquartered there.
  10. Alice and Erica had the same first drink.
  11. Fred had a Harpoon UFO.

Who ordered which drinks and in what order?


A Bump on My Yellow Road

I stopped losing weight. My Yellow Road plan stopped working. If you recall I draw a line on the weight/time plane which I call my target weight. If I'm more than one pound above my target weight, then I'm in the red zone and must restrict my evening food to apples. The hour at which the evening starts depends on how many pounds I'm above my target weight.

And now, back to my bump.

First I stopped following the plan exactly. I realized that I didn't need to restrict myself to eating only apples in the evening when I am in the red zone. I can use salad or anything light.

One day I found myself in the red zone weighing two pounds over my target weight. I was invited to dinner that evening. I decided to accept the invitation and skip the plan for one day. At the party the food was so good I couldn't resist it. The next day I was four pounds over my target weight.

My Yellow Road plan requires me in this situation to eat only apples from 2:00pm onward, but I knew that applying this restriction after 6:00pm worked for me. I decided not to torture myself and started to restrict my food only from 6:00pm. The weight didn't go down. Even though I went to bed very hungry for two weeks, it didn't work.

After these two weeks, I started to feel hungry all the time and even began dreaming about food. As a result, my food intake increased. Now I am seven pounds over my target weight. I've reached a plateau. For the last two months I've been stuck at 220 pounds.

There is some good news: I now have a partner on this journey. After I started my program, I received an email from Natalia Grinberg from Germany. She offered to join forces. We send each other weekly updates on our progress and cheer each other along. Natalia's path wasn't smooth from the start, so she tried to supplement her diet with Almased, which is very popular in Europe. Because Natalia likes it, I looked into it. While I am afraid of pills and chemical ingredients, Almased seems to be okay. It contains soy, yogurt, honey, and vitamins. I bought one can. It is expensive and tastes awful. I'll experiment with cinnamon or pepper and see if that helps. Will Almased help me get over the Bump?


Intellectual Jokes

I collect geeky jokes. I think I've heard most of them. So I was surprised to stumble upon a website with many new ones: 50 People On 'The Most Intellectual Joke I Know'. These are some of them:

* * *

Q: What does the "B" in Benoit B. Mandelbrot stand for?
A: Benoit B. Mandelbrot.

* * *

Entropy isn't what it used to be.

* * *

There are two types of people in the world: those who can extrapolate from incomplete data sets

* * *

This sentence contains exactly threee erors.

* * *

There's a band called 1023MB. They haven't had any gigs yet.

* * *

A logician's wife is having a baby. The doctor immediately hands the newborn to the dad. His wife asks impatiently, "So, is it a boy or a girl"? The logician replies, "Yes."

* * *

The barman says, "We don't serve time travelers in here."
A time traveler walks into a bar.

* * *

The first rule of Tautology club, is the first rule of Tautology club.

* * *

A woman walks in on her husband, a string theorist, in bed with another woman. He shouts, "I can explain everything!"

* * *

What do you get when you cross a joke with a rhetorical question?


arXiv's Police

I used to love arXiv. I've long thought that it was one of the best things that happened to mathematics. arXiv makes mathematical research available for free and without delay. Moreover, it is highly respected among mathematicians. For example, Grigori Perelman never submitted his proof of the Poincaré conjecture to any journal: he just posted it on arXiv.

When I came back to mathematics, all my math friends explained to me that I should submit my paper to arXiv on the same day that I submit it to a journal. As my trust in arXiv grew, I started submitting to arXiv first, waiting one week for comments, and then submitting to a journal.

Now it seems that arXiv might not love its contributors as much as they used to. arXiv moderators seem to be getting harsher and harsher. Here is my story.

In June of 2013, I submitted my paper "A Line of Sages" to arXiv. This paper is about a new hat puzzle that appeared at the Tournaments of the Towns in March 2013. The puzzle was available online at the Tournaments of the Towns webpage in Russian. After some thought I decided that it is better to cite the Tournament itself inside the body of the paper, rather than to have a proper reference. Online references in general are not stable, and this particular one was in Russian. Very soon this competition will be translated into English and the puzzle will appear in all standard math competition archives.

arXiv rejected my paper. A moderator complained that I didn't have a bibliography. So I created a bibliography with the link to the puzzle. My paper was rejected again saying that the link wasn't stable. Duh. That's the reason why I didn't put it there from the start. I Goggled the puzzle and I still didn't find any other links.

I argued with my moderator that the standards for papers in recreational mathematics are different from the standards for purely research papers. Short recreational notes do not require two pages of history and background, nor a long list of references. A recreational paper doesn't need to have theorems and lemmas.

Meanwhile, the moderator complained that the paper was "not sufficiently motivated to be interesting to the readership."

I got tired of exchanging emails with this moderator and submitted my paper to The Mathematical Intelligencer, where it was immediately accepted. So I dropped my submission to arXiv.

Now my paper is not available for free on arXiv. But anyone can freely buy it from Springer for $39.95.


Industry vs Academia

I started my life wanting to be a mathematician. At some point I had to quit academia in order to feed my children. And so I went to work in industry for ten years. Now that my children have grown, I am trying to get back to academia. So I am the right person to compare the experience of working in the two sectors. Just remember:

Money. The pay is much better in industry. About twice as high as academia.

Time. I almost never had to work overtime while working in industry. That might not be true for programmers and testers. As a designer, I worked at the beginning of the project stage. Programmers and testers are closer to deadlines, so they have more pressure on them. The industrial job was more practical than conceptual, so I didn't think about it at home. My evenings and weekends were free, so I could relax with my children. In academia I work 24/7. There are 20 mathematical papers that I have started and want to finish. This is a never-ending effort because I need those papers to find my next job. Plus, I want to be a creative teacher, so I spend a lot of time preparing for classes. I do not have time to breath.

Respect. When I was working in industry, some of my co-workers would tell me that I was the smartest person they ever met. In any case, I always felt that my intelligence and my skills were greatly appreciated. In academia, I am surrounded by first-class mathematicians who rarely express respect and mostly to those who supersede them in their own fields.

Social Life. Mathematics is a lonely endeavor. Everyone is engrossed in their own thoughts. There is no urge to chat at the coffee machine. In industry we were working in teams. I knew everyone in my group. I was closer to my co-workers when I worked in industry.

Freedom. In both industry and academia there are bosses who tell you what to do. But while building my university career, a big part of my life is devoted to writing papers. It is not a formal part of my job, but it is a part of the academic life style. And in my papers I have my freedom.

Motivation. In academia, one must be self-motivated.

Rejection. The output of an academic job is published papers. Most journals have high rejection rates. For me, it's not a big problem because from time to time I get fantastic reviews and I usually have multiple papers awaiting review. I have enough self-confidence that if my paper is rejected, I don't blink. I revise it and send it to a different journal. But this is a huge problem for my high school students who submit their first paper and get rejected. It is very discouraging.

Perfectionism. In industry I was working on deadlines. The goal was to deliver by the deadline a project that more or less worked. Time was more important than quality. My inner perfectionist suffered. When I write papers, I decide myself when they are ready for publication.

Impact. When I was working at Telcordia I felt that I was doing something useful. For example, we were building a local number portability feature, the mechanism allowing people to take their phone numbers with them when they moved. I wish Verizon had bought our product. Just a couple of months ago I had to change my phone number when I moved five blocks from Belmont to Watertown. Bad Verizon. But I digress. When I was working at Alphatech/BAE Systems, I was designing proofs of concepts for future combat systems. I oppose war and the implementation was sub-standard. I felt I was wasting my time. Now that I am teaching and writing papers, I feel that I am building a better world. My goal is to help people structure their minds and make better decisions.

Fame. All the documents I wrote in industry were secret. The world would never know about them. Plus, industry owns the copyright and takes all the credit. There is no trace of what I have done; there is no way to show off. People in academia are much more visible and famous.

Happiness. I am much happier now. I do what I love.


More Geeky Jokes

* * *

I just received a call from the delivery man working for the store where I ordered a new GPS device. He got lost and asked directions to my house.

* * *

I have two problems: I can't count.

* * *

Do you want to double your cash?
Hold your money in front of a mirror.

* * *

- Doctor, I think I have paranoia.
- Why is that?
- Yesterday I left my computer for ten minutes to go to the bathroom, and when I came back all my browser windows were filled with bathroom tissue ads.

* * *

- Why is your disc drive so noisy?
- It is reading a disc.
- Aloud?

* * *

If only DEAD people understand hexadecimal, how many people understand hexadecimal?


An Irresistible Cannonball

I gave the following puzzle from Raymond Smullyan's book What is the Name of this Book? to my AMSA students.

What happens if an irresistible cannonball hits an immovable post?

This puzzle is known as the Irresistible Force Paradox. The standard answer is that the given conditions are contradictory and the two objects cannot exist at the same time.

My AMSA student gave me a much cuter answer: The post falls in love with the cannonball as it is so irresistible.


My Life: An Update

It has been a while since I wrote my last essay and my readers have started to worry. Sorry for being out of touch, but let me tell you what is going on in my life.

In September I received an offer from MIT that changes my status there. In exchange for a slight increase in pay, I am now conducting recitations (supplemental seminars) in linear algebra In addition to my previous responsibilities.

My readers will know that just a slight increase in money in exchange for significant demands on my time would not appeal to me. But this offer comes with perks. First, my position at MIT changes from an affiliate to a lecturer, which looks so much better on my CV. Second, it includes benefits, the most important of which is medical insurance.

I lived without insurance for three years. On the bright side, lack of insurance made me conscious of my health. I developed many healthy habits. I read a lot about the treatments for colds and other minor problems that I had. On the other hand, it is a bit scary to be without insurance.

Many people are surprised to hear that I didn't have any insurance: Doesn't Massachusetts require medical insurance for everyone?

The Commonwealth levies a fine on those who do not have insurance. But I was in this middle bracket in which my income was too high for a subsidized plan, and too low to be fined. You see, the fine is dependent on one's income and is pro-rated. So I didn't have to pay it at all.

I got my insurance from MIT in October, but ironically my doctor's waiting list is so long, that my first check-up will not be until January.

Anyway, I sort of have four jobs now. I am coaching students for math competitions at the AMSA charter school six hours a week. I am the head mentor at the RSI summer program where I supervise the math research projects of a dozen high school students. I do the same thing for the PRIMES program, in which I have the additional responsibility of mentoring my own students. And as I mentioned, I am also teaching two recitation groups in linear algebra at MIT.

Teaching linear algebra turned out to be more difficult than I expected. I love linear algebra, but I had to learn the parts of it that are related to applications and engineering. Plus, I didn't know linear algebra in English. And my personality as a perfectionist didn't help because to teach linear algebra up to my standards would have taken more time than I really had.

This semester I barely had time to breathe, and I certainly couldn't concentrate on essay pieces. Now that this semester is almost over, my as yet unwritten essays are popping up in my head. It's nice to be back.


Lost a Digit in Kilos

Hooray! My weight is down to two digits in kilograms: below 100. This is a big deal for me. I reached the desired number of digits. In pounds it means I weigh less than 220 and I've lost 25 pounds.

My friends have started to notice. The chubbier ones ask me to tell them about my Yellow Road. And I don't actually know what to reply, because the Yellow Road is not a solution. I took many steps before I approached the Yellow Road. The Yellow Road is, I hope, the end of the road.

The idea of the Yellow Road is simple. If I weigh more than I want, I decrease food. All my skinny friends have always lived that way. The problem is that the rest of us do not know how exactly to reduce food intake and then how to sustain that reduction.

So today, I would like to explain to my friends and my readers what I really think helped me to lose weight.

1. I got desperate. I was ready to do whatever it takes. I was prepared not to ever eat again. If I had to extract my calories from the air, I was prepared to do that. I was ready to be hungry and restrain myself for the rest of my life. In short, I was totally motivated.

2. I fought my sugar addiction. I used to crave sugar. I used to think that sugar helps my brain. But once I looked into it, I realized that I might be wrong. I decided to experiment and cut off my carbohydrates intake significantly. That was the most painful thing I had to do. But after a week of withdrawal symptoms, I felt better and stopped craving foods and sugars as much.

3. I wrote my weight down everyday. Having numbers staring me in the face reminded me what I ate the day before. This moment of reflection allowed me to understand what causes the increase or decrease in my weight. Now I know that some foods provoke my appetite: carbohydrates, dairy, mayonnaise. I eat them in small portions, but I do not start my day with them. It's better to have an increased appetite for a couple of hours in the evening, than for the whole day.

4. I had already changed some bad habits. I tried to build new healthy habits before I started my Yellow Road. These alone didn't help me lose weight, but I think they contribute to my weight loss. I still do the following:

I feel that an internal switch was turned off. I don't feel that hungry anymore. My son thinks that I'm in my hibernating state. I wonder if he is right and I will awake one day as hungry as ever.


Alexey's Conversations

My son, Alexey Radul, is gainfully unemployed. While looking for a new job he wrote several essays about his programming ideas. I am a proud and happy mother. While I can't understand his code, I understand his cutting-edge essays. Below are links to the four essays he has posted so far. He is also a superb writer. You do not need to take my word for it. Each link is accompanied by the beginning of the essay.


Four More Papers

I submitted four papers to the arXiv this Spring. Since then I wrote four more papers:


Gelfand's Centennial

This is my toast at the Gelfand's Centennial Conference:

I moved to the US twenty years ago, right after I got my Ph.D in mathematics under the supervision of Israel Gelfand. My first conversation with an American mathematician went like this:

The guy asks me, "What do you do?"

I say, "Mathematics."

"No, I mean what is your field?"

I can't understand what he wants, and repeat "Mathematics."

He says, "No, no, I mean, I do differential geometry. What do you do?"

I do not know how to answer him. My teacher, Israel Gelfand, never mentioned that mathematicians divide mathematics into pieces. So I had to repeat, "My field is mathematics."

I got asked this question many times and I couldn't figure out how to give a satisfactory answer, so I quit academia. Well, I quit it not because of the question, but for many other reasons... But answering the question became so much easier when I worked for industry.

A guy asks me, "What do you do?"

I say, "Battle management."

He says, "What?"

I say, "Battle management. I manage battles, in case there is a war." And this is it, he doesn't ask any more questions … ever.

I always knew that industry was not the right place for me. Five years ago, when my children grew up, I realized that it was time to take some risks. So I resigned from my job, and came back to mathematics. But now I know how to answer the question. When someone asks me, What is your field in mathematics? I say, … brag, "I am a student of Israel Gelfand, I just do mathematics."

I would like to drink to the Unity of Mathematics.


My Exercise Plan

Now that my weight loss is under way, I want to build an exercise program.

I already exercise. I am a member of the MIT ballroom dance team and I go to the gym, where I use the machines, swim, and take gentle yoga and Zumba classes.

Doesn't sound bad, right? But the reality is that I went to Zumba class a total of three times last year. I skipped half the dance classes I signed up for because I was too tired to attend. I had sincerely planned to go to them, but they're held in the evenings, when I'm already drooping.

The yoga situation is the worst. I'm scheduled to go to yoga twice a week, but just as it is time to leave for yoga, I get very hungry and cannot resist having a snack. Then I remember that they do not recommend practicing yoga after eating a meal, and voilà—I have found my excuse. I am all set up to exercise five hours a week, but in reality most weeks I do not exercise at all.

How can I motivate myself? I know that with food, if I have gained weight one day, I eat less the next. So if I've missed my goal of exercising one week, do I add those hours to the next week? That's unlikely to actually help, since the problem is that I'm not meeting my exercise goal, period.

Many people suggested that I punish myself. For example, if I do not meet my goal, I should donate money to a cause I do not support. This feels wrong. If I fail, I'll feel doubly guilty. I'll go broke paying for psychotherapy.

I can try to reward myself. But what should the reward be? Should I reward myself with a piece of tiramisu? Since I am trying to persuade myself that sugar is bad, I shouldn't create a situation that makes sugar desirable. So rewarding with food won't work. Should I buy myself something? If I really want it, I will buy it anyway.

I've been thinking about a plan for a long time. Finally I realized that I should find other people to reward me. I do have a lot of friends, and I came up with an idea of how they could help.

The next time I saw my friend Hillary I asked her if she wanted to sponsor my new exercise plan. She said, "I'm in," without even hearing the plan. Hillary is a true friend. This is what she blindly signed up for.

I decided to push myself to exercise five hours a week. Because of weather and health fluctuations, I pledged to spread 20 hours of exercise over four weeks. I will sent Hillary weekly reports of what I do. This is in itself a huge motivating factor. After the four weeks, we will go to lunch together, which is a great reward for me to look forward to. If I succeed with my plan, she pays for lunch. If I fail, I pay.

Once I saw how enthusiastic Hillary was, I lined up four other friends for the next four-week periods. I hope that after several months of exercise, I will learn to enjoy it. Or at least, I will start feeling the benefits and that itself will be a motivating factor.

Hillary liked my plan so much that she designed a similar exercise plan for herself. Now I am looking forward to two lunches with Hillary.


Smart Brake Lights

I was driving on Mass Pike, when the cars in front of me stopped abruptly. I hit the brakes and was lucky to escape the situation without a scratch.

Actually, it wasn't just luck. First of all, I always keep a safe distance from the other cars. Second, if I see the brake lights of the car in front of me, I automatically remove my foot from the gas pedal and hold it over the brake pedal until I know what the situation is.

On a highway, if the car in front of me has its brake lights on, usually that means that the driver is adjusting their speed a little bit. So, most of the time I don't have to do anything. Seeing that the car in front of me has its brake lights on is not a good predictor of what will happen next. Only after I see that the distance between me and the car in front of me is decreasing rapidly, do I know to hit my brakes. That means that brake lights alone are not enough information. Differentiating between insignificant speed adjustments and serious braking requires time and can cost lives.

I have a suggestion. Why not create smart brake lights. The car's computer system can recognize the difference in the strength with which the brakes are hit and the lights themselves can reflect that. They can be brighter or a different color or pulsing, depending on the strength of the pressure.

The drivers behind will notice these things before they will notice the decrease in the distance. This idea could save lives.


Parallel Weighings Solution

I recently posted the following coin weighing puzzle invented by Konstantin Knop:

We have N indistinguishable coins. One of them is fake and it is not known whether it is heavier or lighter than all the genuine coins, which weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?

The author's solution in Russian is available at his blog. Also, two of my readers, David Reynolds and devjoe, solved it correctly.

Here I want to explain the solution for any number of required weighings.

It is easy to see that for n weighings the information theoretical bound is 5n. Indeed, each weighing divides coins into five groups: four pans and the leftover pile. To distinguish between coins, there can't be two coins in the same pile at every weighing.

Suppose we know the faking potential of every coin, that is, each coin is assigned a value: potentially light or potentially heavy. If a potentially light coin is ever determined to be fake, then it must be lighter than a real coin. The same story holds for potentially heavy coins. How many coins with known potential can we process in n weighings?

If all the coins are potentially light then we can find the fake coin out of 5n coins in n weighings. What if there is a mixture of coins? Can we expect the same answer? How much more complicated could it be? Suppose we have five coins: two of them are potentially light and three are potentially heavy. Then on the first scale we compare one potentially light coin with the other such coin. On the other scale we compare one potentially heavy coin against another potentially heavy coin. The fake coin can be determined in one weighing.

The discussion above shows that there is a hope that any mixture of coins with different potential can be resolved. After each weighing, we want the number of coins that are not determined to be real to be reduced by a factor of 5. If one of the weighings on one scale is unbalanced, the potentially light coins on the lighter pan, plus the potentially heavy coins on the heavier pan would contain the fake coin. We do not want this number to be bigger than one-fifth of the total number of coins we are processing. So we divide coins in pairs with the same potential, and from each pair we put the coins on different pans of the same scale. So in one weighing we can divide the group into five equal groups. If there is an odd number of coins with the same potential, then the extra coin doesn't go on the scales.

The only thing that we is left to check is what happens if the number of coins is small. Namely, we need to check what happens when the number of potentially light coins is odd and the number of potentially heavy coins is odd, and the total number of coins is not more than five. In this case the algorithm requires us to put aside the extra coin in each group, but the put-aside pile can't have more than one coin.

After checking small cases, we see that we can't resolve the problem in one weighing when there are 2 coins of different potential, or when the 4 coins are distributed as 1 and 3.

On the other hand, if we have extra coins that are known to be real, then the above cases can be resolved. Hence, any number of coins with known potential greater than four can be resolved in ⌈log5n⌉ weighings.

Now let's go back to the original problem in which we do not know the coins' potential at the start. After a weighing, if both scales balance, then all the coins on the scale are real and the fake coin is in the leftover pile and we do not know its potential. If a scale doesn't balance then the fake coin is in one of its two pans: the lighter pan has coins that are potentially light and the heavier pan has coins that are potentially heavy.

Let's add an additional assumption to the original problem. Suppose we have an unlimited supply of coins that we know to be real. Let u(n) be the maximum number of coins we can process in n weighings if we do not know their potential.

What would be the first weighing? Both scales might be balanced, meaning that the fake coin is in the leftover pile of coins with unknown potential. So we have to leave out not more than u(n−1) coins. On the other hand, exactly one scale might be unbalanced. In this case, all the coins on this scale will get their potential known. The number of these coins can't be more than 5n-1. But this is an odd number, so we can use one extra real coin to make this number even, in order to put the same number of coins in each pan on this scale.

So u(n) = 2 · 5n-1 + u(n−1), and u(1) = 3. This gives the answer of (5n+1)/2. Now we need to go back and remember that we got this bound using an additional assumption that we have an unlimited supply of real coins. Looking closer, we do not need our additional supply of real coins to be unlimited; we just need not more than two real coins. The good news is that we will have these extra real coins after the first weighing. The bad news is that for the first weighing we do not have extra real coins at all. So in the first weighing we should put unknown coins against unknown coins, not more than 5n-1 on each scale, and as the number on each scale must be even, the best we can do is put 5n-1−1 coins on each scale.

Thus the answer is (5n−3)/2 for n more than 1.

We can generalize this problem to any number of scales used in parallel. Suppose the number of scales is k. Suppose the number of weighings is more than 1, then the following problems can be solved in n weighings:

The methods I described can be used to answer another common question in the same setting: Find the fake coin and say whether it is heavier or lighter. Let us denote by U(n) the number of coins that can be resolved in n weighings when there is an unlimited supply of extra real coins. Then the recurrence for U(n) is the same as the recurrence for u(n): U(n) = 2·5n-1 + U(n−1). The only difference is in the initial conditions: U(1) = k. This means that U(n) = ((2k+1)n−1)/2. If we don't have extra real coins then the answer is: U(n) = ((2k+1)n−1)/2 − k.

When we don't need to say whether the fake coin is heavier or lighter, we can add one extra coin to the mix: the coin that doesn't participate in any weighing and is fake if the scales always balance.


Ode to My Digital Scale

I valued my previous analog scale: I brought it from Europe and it showed my weight in kilograms.

I don't remember why, a year ago, I decided to buy a digital scale: after all, my old scale is still functioning. But my new digital scale became an important instrument in my weight loss program.

Now I understand how my old scale and my lazy nature played tricks on my mind. For example, let's say I decide that as soon as I reach one hundred kilograms (220 pounds), I would do something about it. Some time passes, I step on the scale, and it shows 100 kilograms — and maybe more. Is it more or not quite? If I lean to the right, it looks lower. The scale is not precise. So, I step back and adjust the scale at zero a little bit, to make sure it is not more than zero, but it actually could be slightly less. So I have reached my limit, but the scale's imprecision allows me to pretend that there is a chance I am not over 100 kilograms, and thus do not have to do anything. More time passes, the scale shows 103 kilograms. I realize that I have been deceiving myself, but then it is too late for a fast fix that would lower my weight below 100. So I push the limit, and decide to wait until 105 kilograms.

My digital scale erased any ambiguity. The scale, however, has its own doubts. When I stand on it, the display flips between two numbers for a while. One of the numbers means no dinner tonight. But at the end the scale calls it: it spits out the final number with a beep. I had already decided that the scale is the boss. The flipping is irrelevant; the decision is irrevocable. If it means no dinner, so be it.

I weigh 228.6

My new scale prevents inaction. It also allowed me to design my new plan: my Yellow Road. In this plan, my target weight decreases by 0.1 pounds per day. My behavior depends on my real weight with respect to my target weight. A small difference changes my behavior. So precision is essential.

As you can see in the picture the plan continues to work. I've lost 16 pounds since the start of the plan. Actually, I've lost 18 pounds, if I weigh myself without the camera.


Discussing a Problem from the Moscow Olympiad

I recently posted the following problem from the Moscow Olympiad:

There were n people at a meeting. It appears that any two people at the meeting shared exactly two common acquaintances.

Here is the proof for the first bullet. Choose a person X. Take a pair of X's acquaintances. These two acquaintances have to share two acquaintances between themselves one of whom is X. In other words, we have described a function from all pairs of X's acquaintances to people who are not X. On the other hand, for every person who is not X, s/he and X share a pair of acquaintances. Hence, there is a bijection between people other than X and all pairs of X's acquaintances. If the number of X's acquaintances is a, and the total number of people is t, then we have shown that (a choose 2) = t−1. As this is true for any X, we see that everyone has the same number of acquaintances. Moreover, this situation can happen only if t−1 is a triangular number.

But wait. There is more work that needs to be done. The smallest triangular number is 1. That means that t might be 2. If there are two people at the meeting, then the condition holds: they have 0 common acquaintances. The next triangular number is 3. So we need to see what would happen if there are four people. In this case, if everyone knows each other, it works. This is why the second bullet asks us to find an example of the situation with more than four people, because four people is too easy.

Let's look at larger triangular numbers. The situation described in the problem might also happen when there are:

The official Olympiad solution suggests the following example for 16 people total. Suppose we put 16 people in a square formation so that everyone knows people in the same row and column. I leave it to the reader to check that every two people share exactly two acquaintances.

Let me prove that there is no solution for a total of seven people. If there were a solution, then each person would have to know four people. My first claim is that the acquaintance graph can't contain a four-clique. Suppose there is a four-clique. Then each person in the clique has to have another acquaintance outside of the clique to make it up to four. In addition, this extra acquaintance can't be shared with anyone in the clique, because the clique contains all the acquaintances that they share. This means we need to have at least four more people.

Next, suppose two people a and b know each other and share an acquaintance c. Any two people in this group of three has to have another shared acquaintance, who is not shared with the third person. That is, there should be another person who is the acquaintance of a and b, a different person who is an acquaintance of a and c, and a third person who is acquainted with b and c. These three extra people are all the acquaintances of a, b, and c. Which means the last person who is not acquainted with a, b, or c, has less than for four acquaintances.

Let's look at a more difficult problem that I offered at the same posting:

There were n people at a meeting. It appears that any k people at the meeting shared exactly k common acquaintances.

As in the previous solution, we see that a, the number of acquaintances of a person and t, the total number of people, satisfy the following equation: (a choose k) = (t−1 choose k−1).

For example, if k = 3, the equation becomes (a choose 3) = (t−1 choose 2). This is a question of finding numbers that are both tetrahedral and triangular. They are known and their sequence, A027568, is finite: 0, 1, 10, 120, 1540, 7140. The corresponding number of acquaintances is 3, 5, 10, 22, 36 and the total number of people is 3, 6, 17, 57, 121. The first trivial example involves 3 people who do not know each other. The next example is also simple: it has 6 people and everyone knows everyone else.

What about non-trivial examples? If there are 17 people in the group, then each person has to know 10 people. Does the acquaintance graph exist so that every group of three people share 3 acquaintances?

We see that the problem consists of two different parts. First, we have to solve the equation that equates two binomial coefficients. And second, we need to build the acquaintance graph. Both questions are difficult. We see that for k = 2 we have an infinite number of solutions to the equation with binomial coefficients. For k = 3, that number is finite. What happens with other k? If there are 2k people and they all know each other, then this works. But are there other non-trivial solutions? I am grateful to Henry Cohn for directing me to the works of Singmaster who studied non-trivial repetitions of numbers in Pascal's triangle. In particular, Singmaster showed that the equation (n+1 choose k+1) = (n choose k+2) has infinitely many solutions given by n = F2i+2F2i+3−1 and k = F2iF2i+2−1.

This sequence generates the following non-trivial examples (15 choose 5) = (14 choose 6), (104 choose 39) = (103 choose 40), and so on. That means it might be possible that there is a group of 16 people so that every 6 people share 6 acquaintances. In this situation every person must know everyone else except for one other person. That leads us to the structure of the acquaintance graph: it is a complement to the perfect matching graph. I leave it to my readers to check that the corresponding acquaintance graph doesn't exist. Are there examples of two binomial coefficients that equal each other and that lead to an acquaintance graph that can be built?

Now that I've tackled the solution to this Olympiad problem, I see that I generated more questions than I answered.


Next Tanya Khovanova

Many years ago at Gelfand's seminar in Moscow, USSR, someone pointed out a young girl and told me: "This is Natalia Grinberg. In her year in the math Olympiads, she was the best in the country. She is the next you."

We were never introduced to each other and our paths never crossed until very recently.

Several years ago I became interested in the fate of the girls of the IMO (International Math Olympiad). So, I remembered Natalia and started looking for her. If she was the best in the USSR in her year, she would have been a gold medalist at the IMO. But I couldn't find her in the records! The only Grinberg I found was Darij Grinberg from Germany who went to the IMO three times (2004, 2005, and 2006) and won two silver medals and one gold.

That was clearly not Natalia. I started doubting my memory and forgot about the whole story. Later I met Darij at MIT and someone told me that he was Natalia's son.

I was really excited when I received an email from Natalia commenting on one of my blog posts. We immediately connected, and I asked her about past events.

Natalia participated in the All-Soviet Math Olympiads three times. In 1979 as an 8th grader she won a silver medal, and in 1980 and 1981 she won gold. That indeed was by far the best result in her year. So she was invited to join the IMO team.

That year the IMO was being held in the USA, which made Soviet authorities very nervous. At the very last moment four members of the team did not get permission to travel abroad. Natalia was one of them. The picture below, which Natalia sent to me, was taken during the Soviet training camp before the Olympiad. These four students were not allowed to travel to the IMO: Natalia Grinberg, Taras Malanyuk, Misha Epiktetov, and Lenya Lapshin.

1981 IMO training camp

Because of the authorities' paranoia, the Soviet team wasn't full-sized. The team originally contained eight people, but as they rejected four, only six traveled to the USA, including two alternates.

I have written before how at that time the only way for a Jewish student to get to study mathematics at Moscow State University was to get to the IMO. I wrote a story about my friend Sasha Reznikov who trained himself to get to the IMO, but because of some official machinations, still was not accepted at MSU.

Natalia's story surprised me in another way. She didn't get to the IMO, but she was accepted at MSU. It appears that she was accepted at MSU as a member of the IMO team, because that decision was made before her travel documents were rejected.

Natalia became a rare exception to the rule that the only way for a Jewish person to attend MSU was to participate in the IMO. It was a crack in the system. They had to block visas at the last moment, so that people wouldn't have time to make a fuss and do something about it. Natalia slipped through the crack and got to study at the best university in the Soviet Union.

Unfortunately, the world lost another gold IMO girl. Three Soviet team members won gold medals that year. Natalia, being better then all of them, would have also won the gold medal.


Missing Coin

I recently published the following coin puzzle:

There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

My readers, David Reynolds and ext_1973756, wrote to me that I am missing a coin of 4 grams. Indeed, the same puzzle with five coins—1, 2, 3, 4, and 5—is a more natural and a better puzzle.

David Reynolds also suggested to go all the way up to 9 coins:

There are nine silver coins marked 1, 2, 3, 4, 5, 6, 7, 8, and 9. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

It is impossible to resolve this situation with more than nine coins as two weighings provide nine different answers to differentiate coins. But indeed it is possible to solve this problem for nine coins. It is even possible to suggest a non-adaptive algorithm, that is to describe the weighings before knowing the results.

To find such a strategy we need to satisfy two conditions. First, we have to weigh groups of coins of the same supposed weight, otherwise we do not get any useful information. Second, there shouldn't be any two coins together (in or out of the pan) in both weighings, because it would then be impossible to differentiate between them.

Here is one possible solution of the problem:

David Reynolds also suggested a problem in which we do not know whether the fake coin is heavier or lighter:

There are four silver coins marked 1, 2, 3, and 4. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

Again, four coins is the best we can do when in addition to find it, we also want to determine if it is heavier or lighter. Indeed, if there were five coins we would have needed to cover ten different answers, which is too many for two weighings.

Here is the solution for four coins:

The two weighings are 1+3=4, and 1+2=3. If the first weighing balances, then the fake coin is 2 and the second weighing shows if it is heavier or lighter than it should be. Similarly, if the second weighing balances, then the fake coin is four and we can see whether it is heavier or lighter than it should be. If the left pan is lighter/heavier for both weighings, then the fake coin is 1 and is lighter/heavier. But if one pan is heavier on the first of two weighings and the other pan is heavier on the second weighing, then the fake coin is 3. In both cases it is easy to determine whether the fake coin is heavier or lighter.

Now David is missing a coin. If we just want to find the fake coin without determining whether it is heavier of lighter, we can do it with five coins:

There are five silver coins marked 1, 2, 3, 4, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

We can use the same solution as the previous (four coins) problem. If the scale balances both times, then the fake coin is 5. However, in this case we will not know whether the coin is heavier or lighter.

We can't extend this problem to beyond five coins. Suppose we have six coins. We can't use more than three coins in the first weighing. This is because if the scale unbalances, we can't resolve more than three coins in one remaining weighing. Suppose the first weighing balances; then we have at least three leftover coins we know nothing about and one of them is fake. These three coins should be separated for the next weighing. That means one of the coins needs to be on the left pan and one on the right pan. We can add real coins any way we need. But if the second weighing unbalances we do not know if the fake coin is on the left and lighter or on the right and heavier.


Fraternal Birth Order and Fecundity

Two interesting research results about male homosexuality are intertwined. The first one shows that the probability of homosexuality in a man increases with the number of older brothers. That is, if a boy is the third son in a family, the probability of him being a homosexual is greater than the probability of a first son in a family being homosexual. The second research result shows that the probability of homosexuality increases with the number of children the mother has. So if a woman is fertile and has many children, the probability that each of her sons is a homosexual is greater than the probability that an only child is a homosexual.

Many people conclude from the first result that a woman undergoes hormonal or other changes while being pregnant with boys that influence the probability of future boys being homosexual. Looking at the second result, researchers conclude that homosexuality has a genetic component. Moreover, that component is tied up with the mother's fecundity. The same genes are responsible for both the mother having many children and for her sons being homosexual. This assumption explains why homosexuality is not dying out in the evolution process.

In one of my previous essays I showed that the first results influences the second result. If each next son is homosexual with higher probability, then the more children a mother has the more probable it is that her sons are homosexuals. That means that the second result is a mathematical consequence of the first result. Therefore, the conclusion that the second result implies a genetic component might be wrong. The correlation between homosexuality and fecundity could be the consequence of hormonal changes.

Now let's look at this from the opposite direction. I will show that the first result is the mathematical consequence of the second result: namely, if fertile women are more probable to give birth to homosexuals, then the probability that the second sons are is higher than the probability that the first sons are gay.

For simplicity let's only consider mothers with one or two boys. Suppose the probability of a son of a one-son mother to be a homosexual is p1. Suppose the probability of a son of a two-sons mother to be a homosexual is p2. The data shows that p2 is greater than p1. What is the consequence? Suppose the number of mothers with one son is m1 and the number of mothers with two sons is m2. Then in the whole population the probability of a boy who is the first son to be gay is (p1m1+p2m2)/(m1+m2) and the probability of a boy who is the second son to be gay is p2. It is easy to see that the first probability is smaller than the second one.

Let me create an extreme hypothetical example. Suppose mothers of one son always have straight sons, and mothers of two sons always have gay sons. Now consider a random boy in this hypothetical setting. If he's the second son, he is always gay, while if he is the first son he is not always gay.

We can conclude that if the probability of having homosexual sons depends on fecundity, then the higher numbered children would be gay with higher probability than the first-born. This means that if the genetics argument is true and being a homosexual depends on the mother's fecundity gene, then it would follow mathematically that the probability of homosexuality increases with birth order. The conclusion that homosexuality depends on hormonal changes might not be valid.

So what is first, chicken or egg? Is homosexuality caused by fecundity, while birth order correlation is just the consequence? Or vice versa? Is homosexuality caused by the birth order, while correlation with fecundity is just the consequence?

What do we do when the research results are so interdependent? To untangle them we need to look at the data more carefully. And that is easy to do.

To show that homosexuality depends on the order of birth independently of the mother's fertility, we need to take all the families with two boys (or the same number of boys) and show that in such families the second child is more probable to be homosexual than the first child.

To show the dependence on fertility, without the influence of the birth order, we need to take all first-born sons and show that they are more probable to be homosexuals if their mothers have more children.

It would be really interesting to look at this data.


My New Favorite Hat Puzzle

My new favorite hat puzzle was invented by Konstantin Knop and Alexander Shapovalov. It appeared (in a different wording) in March 2013 at the Tournament of the Towns:

A sultan decides to give 100 of his sages a test. The sages will stand in line, one behind the other, so that the last person in the line sees everyone else. The sultan has 101 hats, each of a different color, and the sages know all the colors. The sultan puts all but one of the hats on the sages. The sages can only see the colors of the hats on people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. They are not allowed to repeat a color that was already announced. Each person who guesses his color wrong will get his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

I loved it so much that I wrote a paper about it. You can find the solution there.


Samples from My AMSA Homework

I particularly like these two problems that I gave my AMSA students for homework:

Athos, Porthos, and Aramis were rewarded with six coins: three gold and three silver. Each got two coins. Athos doesn't know what kind of coins the others got, but he knows his own coins. Ask him one question to which he can answer "Yes," "No," or "I do not know," so that you will be able to figure out his coins.
There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

Parallel Weighings

We've all been hearing about parallel computing, and now it has turned up in a coin-weighing puzzle invented by Konstantin Knop.

"We have N indistinguishable coins. One of them is fake and it is not known whether it is heavier or lighter, but all genuine coins weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts a minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?"
This puzzle reminds me of another coin-weighing problem, where in a similar situation you need to find a fake coin by using one scale with four pans. The answer in this variation would be 55 = 3125. We can divide coins in five groups with the same number of coins and put four groups on the scale. If one of the groups is different (heavier or lighter), then this group contains the fake coin. Otherwise, the leftover group contains the fake coin. This way each weighing reduces the pile with the fake coin by a factor of five.

One scale with four pans gives you more information than two scales with two pans used in parallel. We can conclude that Knop's puzzle should require at least the same number of weighings as the four-pan puzzle for the same number of coins. So we can expect the answer to Knop's puzzle will not be bigger than 3125. But what will it be?


Was I Dead?

Once when I was working at Telcordia, I received a phone call from my doctor's office. Here is how it went:

— Are you Tanya Khovanova?
— Yes.
— You should come here immediately and redo your blood test ASAP.
— What's going on?
— Your blood count shows that you are dead.
— If I'm dead, then what's the hurry?

Given that I wasn't dead, the conclusion was that there had been a mistake in the test. If there had been a mistake, the probability that something was wrong after the test was the same as it was before the test. There was no hurry.


I've Lost 10 Pounds

I started my Yellow Road plan on February 9 when I was 245.2 pounds.

I decided that my first target weight would be my actual weight on February 9: 245.2. Every day this target weight goes down by 0.1 pounds. I weigh myself every morning and compare my actual weight to my target weight. My actions depend on the difference.

My Yellow Zone is plus or minus one pound of my target weight. My Green Zone means I am doing even better: my weight is less than my target weight minus one pound. My Red Zone means that I am not doing so great: my weight is more than my target weight plus a pound.

If I am within the Yellow corridor, I continue building my healthy habits as I have been doing. If I am in the Green Zone, I can afford to digress from healthy habits and indulge myself a bit. If I am in the Red Zone, I have to reduce my evening meals to apples only, which I do not particularly like. The Red Zone has different shades: if I am one pound over my target weight I have to start my apple restrictions after 8:00 pm. If I am two pounds over, then after 6:00 pm, and so on.

Today I am ten pounds lighter. In the process, I have made these discoveries:

Writing down my weight daily is very important. When I look at yesterday's number and today's number, I start thinking about what caused the increase or decrease. Now I have more clarity about which foods are better for me.

I have a better picture of how much I should eat. One day I held a party and I didn't eat much. In fact, I had only one small desert. I didn't feel full and went to bed feeling proud of myself. The next morning I weighed myself and was surprised to find I had gained three pounds. The amount of food I should be eating is much smaller than I expected. I think it may be three times less than what I was used to eating. My plan might not be aggressive enough. Currently when I am in my lightest Red Zone, I have to eat just apples after 8:00 pm. I discovered that I can still gain weight with this regime.

A half-empty stomach is not such a bad feeling. I was so afraid of starting a plan where I might feel hungry. Now I discovered that there are several hours between my first signal that I should eat and real hunger. My first sense that I should eat something might not be actual hunger at all. I do experience a light feeling in my stomach, but now I am starting to learn to enjoy it.

The system works for me. That's the bottom line. For the first time in my life, I found a way to lose weight. All my friends ask me about this system. I explain that this Yellow Road plan is not a panacea. My plan is based on many other things that I did before. If it continues working, I promise to discuss it further: to analyze what exactly works and why.

My next step is to adjust my plan in light of my discoveries. From now on, I'll eat only apples after 8:00 pm—not as an exception, but as a rule.


A Problem from the Moscow Olympiad

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.

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.

Happy Nobel Prize Winners

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.


Four Papers in Three Weeks

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:


Integers and Sequences Solution

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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My Yellow Road to Healthy Weight

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.


Skyscrapers

Tanya Khovanova and Joel Brewster Lewis

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.

Skyscraper Puzzle

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:

Skyscraper Formula 1

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=2b=3b=4b=5b=6b=7
n=21
n=311
n=4231
n=561161
n=6245035101
n=712027422585151

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:

Skyscraper Formula 2

Let's have a table for f7(a,b), of which we already calculated the first row:

b=1b=2b=3b=4b=5b=6b=7
a=1012027422585151
a=21205486753407560
a=32746755101501500
a=422534015020000
a=58575150000
a=615600000
a=71000000

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.


Vampires versus Mathematicians

I just compared two searches on Google Trends:

Vampires versus Mathematicians

Integers and Sequences

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.

*****

*****

*****

*****

*****

*****

*****


Weighing Coins during the Mystery Hunt

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

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.


February Jokes

* * *

Grigori Perelman's theorem: There is no offer you can't refuse.

* * *

A conversation between two Russians:
— Run to the store and fetch a couple bottles of vodka.
— How much is a couple?
— Seven.

* * *

— 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.

* * *

I saw our system administrator's shopping list. The first line was tomatoes.zip for ketchup.


Solving In the Details

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

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.

Ben's Locations

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.

Pesto with sarongs
Pesto under a white sarong

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.


Open Secrets Revealed

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.


Turnary Reasoning

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.

Chess Position from Turnary Reasoning

The Most Difficult Hunt Puzzle: 50/50

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.


In the Details

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:


Something in Common

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:


Portals

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.


Random Walk

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.


Open Secrets

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.


Apologies

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.


Children's Riddle

The father of my son has four children. My son is my only child. How many children do we have in total together?


Affirmations

"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?


Halving Lines

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.

Path

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.


I am on the Air

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?


Helpmate

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.

Helpmate

Me and Chess

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.


Challenge Problems

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?

Nerdy and Flirtatious Jokes

* * *

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'.


2012!

In what base does 2012! have more trailing zeros: base 15 or 16?

Explain why the result shouldn't be too surprising.


Why I Eat

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.


Three out of Three

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

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.


Number Gossip is Back

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.


A Visit to Smullyan

Raymond Smullyan 2012

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.


A Measure of Central Symmetry

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.

Five Discs

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.

Triangle's Symmetricity

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.


Great Ideas that Haven't Worked. Yet.

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.