This blog has a modern version at http://blog.tanyakhovanova.com/, where you can leave comments. This page contains the copies of blog entries.
If you enjoy this page you can support it by shopping at amazon through this link. Thank you.
Do you remember how to divide three apples among four people? Make apple sauce, of course. In the following two puzzle you are not allowed to cut apples. Here is an old riddle:
There are four apples in a basket. How can you divide them among four people, so that one apple remains in the basket?
Here is a variation from Konstantin Knop's blog:
There are four apples in a basket. How can you divide them among three people, so that no one has more than the others and one apple remains in the basket?
Imagine a magician who comes on stage and performs the following magic trick:
He asks someone in the audience to think of a two-digit number, then subtract the sum of its digits. He waves his wand and guesses that the result is divisible by nine. Ta-Da!
This is not magic. This is a theorem. To make it magical we need to disguise the theorem.
First, there are many ways to hide the fact that we subtract the sum of the digits. For example, we can ask to subtract the digits one by one, while chatting in between. It is better to start with subtracting the first digit. Indeed, if we start with subtracting the second digit, the audience might notice that the result is divisible by 10 and start suspecting that some math is involved here. You can be more elaborate in how you achieve the subtraction of the sum of digits. For example, subtract twice the first digit, then the second, then add back the original number divided by 10.
Second, we need to disguise that the result is divisible by 9. A nice way to do this is implemented in the online version of this trick. The website matches the resulting number to a gift that is described on the page in pale letters. Paleness of letters is important as it is difficult to see that the same gift reappears in a pattern. In my work with students I use the picture on the left. At the end I tell them, "Ta-Da! the resulting number is blue." Here is the full sized version of the same picture that you can download.
My students are too smart. They see through me and guess what is going on. Then I ask them the real question, "Why do I have some cells with question marks and other symbols?" To give you a hint, I can tell you that the symbols are there for the same reason some blue numbers are not divisible by 9.
We got this problem from Rados Radoicic:
A 7 by 7 board is covered with 38 dominoes such that each covers exactly 2 squares of the board. Prove that it is possible to remove one domino so that the remaining 37 still cover the board.
Let us call a domino covering of an n by n board saturated if the removal of any domino leaves an uncovered square. Let d(n) be the number of dominoes in the largest saturated covering of an n by n board. Rados' problem asks us to prove that d(7) < 38.
Let's begin with smaller boards. First we prove that d(2) = 2. Suppose that 3 dominoes are placed on a 2 × 2 board. Let us rotate the board so that at least two of the dominoes are horizontal. If they coincide, then we can remove one of them. If not, they completely cover the board and we can remove the third one. Similarly, you can check all the cases and show that d(3) = 6.
Now consider a saturated domino covering of an n × n board. We can view the dominoes as vertices of a graph, joining two if they share a cell of the board. No domino can share both cells with other dominoes, or we could remove it. Hence, each domino contains at most one shared cell. This means that all the dominoes in a connected component of the graph must overlap on a single shared cell. Hence, the only possible connected components must have the following shapes:
The largest shape in the picture is the X-pentomino. We can describe the other shapes as fragments of an X-pentomino, where the center and at least one more cell is intact. We call these shapes fragments.
A saturated covering by D dominoes corresponds to a decomposition of the n × n board into F fragments. Note that a fragment with k cells is made from k − 1 dominoes. Summing over the dominoes gives: D = n2 − F. Thus, in order to make D as large as possible, we should make F as small as possible. Let f(n) be the minimal number of fragments that are required to cover an n by n board without overlap. Then d(n) = n2 − f(n).
Consider the line graph of the n by n board. The vertices of the line graph correspond to cells in the original board and the edges connect vertices corresponding to neighboring cells. Notice that in the line graph our fragments become all star graphs formed by spokes coming out from a single central node. Thus a decomposition of a rectangular board into fragments corresponds to a covering of its line graph by star graphs. Consider an independent set in the line graph. The smallest independent set has the same number of elements as the smallest number of stars that can cover the graph. This number is called a domination number.
Now let's present a theorem connecting domino coverings with X-pentomino coverings.
Theorem. f(n) equals the smallest number of X-pentominoes that can cover an n by n board allowing overlaps and tiles that poke outside, which is the same as the domination number of the corresponding line graph.
The proof of this theorem and the solution to the original puzzle is available in our paper: "Saturated Domino Coverings." The paper also contains other theorems and discussions of other boards, not to mention a lot of pictures.
The practical applications of star graph coverings are well-known and widely discussed. We predict a similar future for saturated domino coverings and its practical applications, two examples of which follow:
First, imagine a party host arranging a plate of cookies. The cookies must cover the whole plate, but to prevent the kids sneaking a bite before the party, the cookies need to be placed so that removal of just one cookie is bound to expose a chink of plate. This means the cookies must form a saturated covering of the plate. Of course the generous host will want to use a maximal saturated covering.
For the second application, beam yourself to an art museum to consider the guards. Each guard sits on a chair in a doorway, from where it is possible to watch a pair of adjacent rooms. All rooms have to be observed. It would be a mistake to have a redundant guard, that is, one who can be removed without compromising any room. Such a guard might feel demotivated and then who knows what might happen. This means that a placement of guards must be a saturated domino covering of the museum. To keep the guards' Union happy, we need to use a maximal saturated covering.
We would welcome your own ideas for applications of saturated coverings.
Several years ago my son Sergei started a new movement: Sergeism. Followers of this philosophy seek to maximize Sergei's happiness. Since Sergei's happiness involves everyone being happy, becoming happy is a consequential goal of his followers.
Let me explain why this might be a perfect religion for many people, not the least myself. My parents didn't teach me to love myself. They taught me to sacrifice myself and put other peoples' interests ahead of my own. After reading tons of books and spending hours in therapy, I've learned to love myself — well, somewhat. But the truth is, I still feel guilty when I pamper myself. Sergeism eliminates this guilt. I can freely invest in my happiness as a committed member of this movement.
I became a Sergeist when I lost all hope of losing weight. I realized that my own health wasn't a strong enough motivation. But I'm always glad to skip a cookie in tribute to Sergeism. If, like me, you put others ahead of yourself and never find the time to exercise or the will to refuse deserts, join me. Become a Sergeist and lose weight for Sergei.
I recently posted an essay Binary Bulls without Cows with the following puzzle:
The test Victor is taking consists of n "true" or "false" questions. In the beginning, Victor doesn't know any answers, but he is allowed to take the same test several times. After completing the test each time, Victor gets his score — that is, the number of his correct answers. Victor uses the opportunity to re-try the test to figure out all the correct answers. We denote by a(n) the smallest numbers of times Victor needs to take the test to guarantee that he can figure out all the answers. Prove that a(30) ≤ 24, and a(8) ≤ 6.
There are two different types of strategies Victor can use to succeed. First, after each attempt he can use each score as feedback to prepare his answers for the next test. Such strategies are called adaptive. The other type of strategy is one that is called non-adaptive, and it is one in which he prepares answers for all the tests in advance, not knowing the intermediate scores.
Without loss of generality we can assume that in the first test, Victor answers "true" for all the questions. I will call this the base test.
I would like to describe my proof that a(30) ≤ 24. The inequality implies that on average five questions are resolved in four tries. Suppose we have already proven that a(5) = 4. From this, let us map out the 24 tests that guarantee that Victor will figure out the 30 correct answers.
As I mentioned earlier, the first test is the base test and Victor answers every question "true." For the second test, he changes the first five answers to "false," thus figuring out how many "true" answers are among the first five questions. This is equivalent to having a base test for the first five questions. We can resolve the first five questions in three more tests and proceed to the next group of five questions. We do not need the base test for the last five questions, because we can figure out the number of "true" answers among the last five from knowing the total score and knowing the answers for the previous groups of five. Thus we showed that a(mn) ≤ m a(n). In particular, a(5) = 4 implies a(30) ≤ 24.
Now I need to prove that a(5) = 4. I started with a leap of faith. I assumed that there is a non-adaptive strategy, that is, that Victor can arrange all four tests in advance. The first test is TTTTT, where I use T for "true" and F for "false." Suppose for the next test I change one of the answers, say the first one. If after that I can figure out the remaining four answers in two tries, then that would mean that a(4) = 3. This would imply that a(28) ≤ 21 and, therefore, a(30) ≤ 23. If this were the case, the problem wouldn't have asked me to prove that a(30) ≤ 24. By this meta reasoning I can conclude that a(4) ≠ 3, which is easy to check anyway. From this I deduced that all the other tests should differ from the base test in more than one answer. Changing one of the answers is equivalent to changing four answers, and changing two answers is equivalent to changing three answers. Hence, we can assume that all the other tests contain exactly two "false" answers. Without loss of generality, the second test is FFTTT.
Suppose for the third test, I choose both of my "false" answers from among the last three questions, for example, TTFFT. This third test gives us the exactly the same information as the test TTTTF, but I already explained that having only one "false" answer is a bad idea. Therefore, my next tests should overlap with my previous non-base tests by exactly one "false" answer. The third test, we can conclude, will be FTFTT. Also, there shouldn't be any group of questions that Victor answers the same for every test. Indeed, if one of the answers in the group is "false" and another is "true," Victor will not figure out which one is which. This uniquely identifies the last test as FTTFT.
So, if the four tests work they should be like this: TTTTT, FFTTT, FTFTT, FTTFT. Let me prove that these four tests indeed allow Victor to figure out all the answers. Summing up the results of the last three tests modulo 2, Victor will get the parity of the number of correct answers for the first four questions. As he knows the total number of correct answers, he can deduce the correct answer for the last question. After that he will know the number of correct answers for the first four questions and for every pair of them. I will leave it to my readers to finish the proof.
Knop and Mednikov in their paper proved the following lemma:
If there is a non-adaptive way to figure out a test with n questions by k tries, then there is a non-adaptive way to figure out a test with 2n + k − 1 questions by 2k tries.
Their proof goes like this. Let's divide all questions into three non-overlapping groups A, B, and C that contain n, n, and k − 1 questions correspondingly. By our assumptions there is a non-adaptive way to figure out the answers for A or B using k tries. Let us denote subsets from A that we change to "false" for k − 1 non-base tests as A1, …, Ak-1. Similarly, we denote subsets from B as B1, …, Bk-1.
Our first test is the base test that consists of all "true" answers. For the second test we change the answers to A establishing how many "true" answers are in A. In addition we have k − 1 questions of type Sum: we switch answers to questions in Ai ∪ Bi ∪ Ci; and type Diff: we switch answers to (A ∖ Ai) ∪ Bi. The parity of the sum of "false" answers in A − Ai + Bi and Ai + Bi + Ci is the same as in A plus Ci. But we know A's score from the second test. Hence we can derive Ci. After that we have two equations with two unknowns and can derive the scores of Ai and Bi. From knowing the number of "true" answers in A and C, we can derive the same for B. Knowing A and Ai gives all the answers in A. Similarly for B. QED.
This lemma is powerful enough to answer the original puzzle. Indeed, a(2) = 2 implies a(5) ≤ 4, and a(3) = 3 implies a(8) ≤ 6.
The following variation of a Bulls and Cows problem was given at the Fall 2008 Tournament of the Towns:
A test consists of 30 true or false questions. After the test (answering all 30 questions), Victor gets his score: the number of correct answers. Victor doesn't know any answer, but is allowed to take the same test several times. Can Victor work out a strategy that guarantees that he can figure out all the answers after the 29th attempt? after the 24th attempt?
Let's assume that we have a more general problem. There are n questions, and a(n) is the smallest number of times we need to take the test to guarantee that we can figure out the answers. First we can try all combinations of answers. This way we are guaranteed to know all the answers after 2n attempts. The next idea is to start with a baseline test, for example, to say that all the answers are true. Then we change answers one by one to see if the score goes up or down. After changing n − 1 answers we will know the answers to the first n − 1 questions. Plus we know the total number of true answers, so we know the answers to all the questions. We just showed that a(n) ≤ n.
This is not enough to answer the warm-up question in the problem. We need something more subtle.
Let's talk about the second part of the problem. As we know, 24 = 4 ⋅ 6. So to solve the second part, on average, we need to find five correct answers per four tests. Is it true that a(5) ≤ 4? If so, can we use it to show that a(30) ≤ 24?
The following three cases are the most fun to prove: a(5) = 4, a(8) ≤ 6, and a(30) ≤ 24. Try it!
By the way, K. Knop and L. Mednikov wrote a paper (available in Russian) where they proved that a(n) is not more than the smallest number k such that the total number of ones in the binary expansion of numbers from 1 to k is at least n − 1. Which means they proved that a(30) ≤ 16.
On the left is a puzzle from the 2000 Qualifying Test for USA and Canada teams to compete in the world puzzle championship. A set of all 21 dominoes has been placed in a 7 by 6 rectangular tray. The layout is shown with the pips replaced by numbers and domino edges removed. Draw the edges of the dominoes into the diagram to show how they are positioned.
We would like to discuss the mathematical theory behind this puzzle using a toy example below. Only three dominoes: 1-1, 1-2, 2-2 are positioned on the board and the goal is to reconstruct the positioning:
Let's connect adjacent numbers with segments to show potential dominoes and color the segments according to which domino they represent. The 1-1 edge is colored green, the 1-2 — blue, and the 2-2 — red. Now our puzzle has become a graph, where the numbers are vertices, the segments are edges, and the edges are colored. In this new setting, the goal of the puzzle is to find edges of three different colors so that they do not share vertices.
The next picture represents the line graph of the previous graph. Now the colors of the vertices correspond to different potential dominoes. Vertices are connected if the corresponding dominoes share a cell. In the new setting finding dominoes that do not share a cell is equivalent to finding an independent set. The fact that we need to use all possible dominoes means that we want the most colorful independent set.
* * *
— If a black cat crosses in front of you and then crosses back, what does it mean? Is your bad luck doubled or canceled?
— Is this a scalar or a vector cat?
— Huh?
— A scalar cat doubles and a vector cat cancels.
* * *
Unbuttered bread, unable to cause the usual harm, tries to fall on the dirtiest spot.
* * *
Chance is a design carefully planned by someone else.
* * *
Wikipedia: I know everything.
Google: I can find anything.
Facebook: I know everyone.
Internet: You are nothing without me.
Electricity: Shut up, jerks.
* * *
Yesterday I bought pills to increase my IQ. Couldn't open the jar.
* * *
Today I opened my desktop's case and finally understood whither my trash is emptied.
I just discovered a Russian Internet Linguistics Olympiad. Even though most linguistics problems are not translatable, this time we are lucky. My favorite problem from this Olympiad is related to chemical elements — their names in Russian have the same logical structure as in English. Keep in mind, the problem doesn't assume any knowledge of chemistry. Here is the problem:
The formulae for chemical elements and their names are given below in mixed order:
C3H8, C4H6, C3H4, C4H8, C7H14, C2H2;
Heptene, Butine, Propane, Butene, Ethine, Propine.
- Match the formulae with their names. Explain your solution.
- Write the names of the elements with the following formulae: C2H4, C2H6, C7H12.
- Write the formulae for the following elements: Propene, Butane.
I have two admirers, Alex and Mike. Alex lives next to my home and Mike lives next to my MIT office. I have a lousy memory, so I invented the following system to guarantee that I see both of my friends and also manage to come to my office from time to time. I have a sign hanging on the inside of my home door that says Office on one side and Alex on the other. When I approach the door, I can see right away where I went last time. So I flip the sign and that tells me where next to go. I have a similar sign inside my office door that tells me to go either to home or to Mike. Every evening I spend with one of my admirers discussing puzzles or having coffee. Late at night I come home to sleep in my own bed. Now let's see what happens if today my home sign shows Office and the office sign shows Mike:
After three days the signs return to their original positions, meaning that the situation is periodic and I will repeat this three-day pattern forever.
Let's get back to reality. I am neither memory-challenged nor addicted to coffee. I invented Alex and Mike to illustrate a rotor-router network. In general my home is called a source: the place where I wake up and start the day. There can only be one source in the network. My admirers are called targets and I can have an infinite number of them. The network needs to be constructed in such a way that I always end up with a friend by the end of the day. There could be many other places that I can visit, other than my office: for example, the library, the gym, opera and so on. These places are other vertices of a network that could be very elaborate. Any place where I go, there is a sign that describes a pattern of where I go from there. The sign is called a rotor.
The patterns at every rotor might be more complicated than a simple sign. Those patterns are called rotor types. My sign is called 12 rotor type as it switches between the first and the second directions at every non-friend place I visit.
The sequence of admirers that I visit is called a hitting sequence and it can be proved that the sequence is eventually periodic. Surprisingly, the stronger result is also true: the hitting sequence is purely periodic.
The simple 12 rotor is universal. That means that given a set of friends and a fancy periodic schedule that designates the order I want to visit them in, I can create a network of my activities where every place has a sign of this type 12 and where I will end up visiting my friends according to my pre-determined periodic schedule.
It is possible to see that not every rotor type is universal. For example, palindromic rotor types generate only palindromic hitting sequences, thus they are not universal. The smallest such example, is rotor type 121. Also, block-repetitive rotor types, like 1122, generate block-repetitive hitting sequences.
It is a difficult and an interesting question to describe universal rotor types. My PRIMES student Xiaoyu He was given a project, suggested by James Propp, to prove or disprove the universality of the 11122 rotor type. This was the smallest rotor type the universality of which was not known. Xiaoyu He proved that 11122 is universal and discovered many other universal rotor types. His calculations support the conjecture that only palindromic or block-repetitive types are not universal. You can find these results and many more in his paper: On the Classification of Universal Rotor-Routers.
My co-author Konstantin Knop wrote a charming book, Weighings and Algorithms: from Puzzles to Problems. The book contains more than one hundred problems. Here are a couple of my favorites that I translated for you:
There is one gold medal, three silver medals and five bronze medals. It is known that one of the medals is fake and weighs less than the corresponding genuine one. Real medals made of the same metal weigh the same and from different metals do not. How can you use a balance scale to find the fake medal in two weighings?
There are 15 coins, out of which not more than seven are fake. All genuine coins weigh the same. Fake coins might not weigh the same, but they differ in weight from genuine coins. Can you find one genuine coin using a balance scale 14 times? Can you do it using fewer weighings?
You might get the impression that the latter problem depends on two parameters. Think about it: It is necessary that the majority of the coins are genuine in order to be able to solve the problem. In fact, the number of weighings depends on just one parameter: the total number of coins. Denote a(n) the optimal number of weighings needed to find a genuine coin out of n coins, where more than half of the coins are genuine. Can you calculate this sequence?
Hint. I can prove that a(n) ≤ A011371(n-1); that is, the optimal number of weighings doesn't exceed n − 1 − (number of ones in the binary expansion of n−1).
We all heard this paradoxical statement:
This statement is false.
Or a variation:
True or False: The correct answer to this question is 'False'.
Recently we received a link to the following puzzle, which is similar to the statement above, but has a cute probabilistic twist:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 60%
- 25%
There are four answers, so you can choose a given answer with probability 25%. But oops, this answer appears twice. Is the correct answer 50%? No, it is not, because there is only one answer 50%. You can see that none of the answers are correct, hence, the answer to the question—the chance to be correct—is 0. Now is the time to introduce our new puzzle:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 0%
- 25%
This fractal was designed by Ross Hilbert and is named "Weathered Steel Weave." You can find many other beautiful pictures in his fractal gallery.
The fractal is based on iterations of the following fractal formula znew = cos(c zold), where the Julia Constant c is equal to −0.364444444444444+0.995555555555556i. To produce the image, you need to start with a complex value of z and iterate it many times using the formula above. The color is chosen based on how close the iteration results are to the border of the unit circle.
I found a new Russian Olympiad for high schools related to universities. I translated my favorite problems from last year's final round. These are the math problems:
8th grade. In a certain family everyone likes their coffee with milk. At breakfast everyone had a full cup of coffee. Given that Alex consumed a quarter of all consumed milk and one sixth of all coffee, how many people are there in the family?
8th grade. How many negative roots does the equation x4 − 5x3 − 4x2 − 7x + 4 = 0 have?
10th grade. Find a real-valued function f(x) that satisfies the following inequalities for any real x and y: f(x) ≤ x and f(x+y) ≤ f(x) + f(y).
I liked the physics problems even more:
8th grade. Winnie-the-Pooh weighs 1 kg. He hangs in the air with density 1.2kg/m3 next to a bee hive. He is holding a rope connected to a balloon. Estimate the smallest possible diameter of the balloon, assuming that this happens on Earth.
10th grade. Two containers shaped like vertical cylinders are connected by a pipe underneath them. Their heights are the same and they are on the same level. The cross-sectional area of the right container is twice bigger than the left's. The containers are partially filled with water of room temperature. Someone put ice into both containers: three times more ice into the right one than into the left one. After that, the containers are closed hermetically. How will the water level will change after the ice melts completely:
- The levels will not change.
- The level on the left will be higher than on the right.
- The level on the left will be lower than on the right.
- The answer depends on the initial volume of water in the containers.
Once I talked to my friend Michael Plotkin about IQ tests, which we both do not like. Michael suggested that I run an experiment and send a standard IQ question for children to my highly-educated friends. So I sent a mass email asking:
What's common between an apple and an orange?
I believe that the expected answer is that both are fruits.
Less than half of my friends would have passed the IQ test. They gave four types of answer. The largest group chose the expected answer.
The second group related the answer to language. For example, apples and oranges both start with a vowel and they both have the letters A and E in common.
The third group connected the answer to what was on their minds at the time:
And the last group were people who just tried to impress me:
If my friends with high IQs have given so many different answers, I would expect children to do the same. The variety of answers is so big that no particular one should define IQ. By the way, my own well-educated kids' answers are quoted above — and they didn't go with the standard answer. I'm glad they never had IQ tests as children: I'm sure they would never have passed.
I have an idea for a start-up medical insurance company for Massachusetts. My insurance will have an infinite deductible. That means you pay your own bills. The cost of insurance can be very low, say $100 a year, as I do not need to do anything other than to send you a letter confirming that you have medical insurance. People who otherwise will be fined up to $900 for being uninsured will run in droves to buy my insurance.
I have an even better idea. For an extra fee, I will negotiate with doctors so that you will pay the same amount as medical insurance companies pay to them, which is often three times less than you would pay on your own.
Who am I kidding? I am not a business person, I can't build a company. But I am looking to buy the insurance I just described.
I am just wondering:
What is the largest integer consisting of distinct digits such that, in its English pronunciation, all the words start with the same letter?
I continue to wonder:
What is the largest integer consisting of the same digit such that, in its English pronunciation, all the words start with distinct letters?
When you name your child there are many considerations to take into account. For example, you should always check that your kids' initials don't embarrass them. For example, if the Goldsteins want to name their son Paz, because it means golden in Biblical Hebrew, the middle name shouldn't be Isaak, or anything starting with I.
Contemporary culture adds another consideration: how easy would it be to find your child on the Internet? I personally find it extremely convenient to have a rare name, because my fans can find my webpage and blog just by googling me. Parents need to decide whether they want their children to be on the first page of the search engine or hidden very far away when someone googles them.
When I named my son Sergei, I knew that there was another mathematician named Sergei Bernstein. But I didn't think about the Internet. As a result, I confused the world: is my son more than a hundred years old or did Sergei Natanovich Bernstein compete at Putnam?
I decided to see the film The Oxford Murders
At the core of the movie are sequences of numbers and symbols. When the characters started a discussion about how to continue a sequence, I immediately tensed up. Why? Because when people ask what the next element in the sequence is, I get ready to confront them, by explaining that there are many ways to continue a sequence. For example, the sequence — 1, 2, 4 — could be powers of two, or could be Tribonacci numbers, or any of 10,000 sequences that the Online Encyclopedia of Integer Sequences spills out if you plug in 1, 2, 4. That is, if we do not count the infinity of sequences that are not in the Encyclopedia.
To my surprise and relief, the logic Professor, one of the main characters in the movie, explained that there is no unique way to continue a sequence. From that moment on, I relaxed and fell in love with the movie.
The movie is a detective story with a lot of twists and turns. The crimes are related to symbols. The first two symbols are in the picture below. Can you guess the next symbol?
I cannot. There is an irony in the film at this point, because the Professor and the student need to guess the sequence in order to solve the crimes. But the Professor has already explained that there is no unique way to continue. So illogical for a movie about logic.
And what's worse, the sequence of symbols they finally discover doesn't make sense. I guess I fell in love with this movie too quickly.
* * *
— I've noticed that fools are always sure of themselves, while clever people are doubtful.
— No doubt.
* * *
— What happened to your girlfriend, that really cute math student?
— She's no longer my girlfriend. I caught her cheating on me.
— I don't believe that she cheated on you!
— Well, a couple of nights ago I called her, and she told me that she was in bed wrestling with three unknowns.
* * *
A programmer calls the library:
— Can I talk to Kate please?
— She's in the archive.
— Can you unzip her?
* * *
To protect the population from airplane disasters, Congress has ratified an addendum to the law of gravity.
* * * (invented by David Bernstein)
Energy conservation: it's not just a good idea; it's the law.
* * *
— Your computer is such a mess.
— It got a nasty virus.
— And it poured coffee on your keyboard?
* * *
After little Tom learned to count, his father had to start dividing dumplings evenly.
* * *
In spite of the crisis, inflation, and erratic fluctuations of the market, Russian mathematicians promised the president to keep number Pi between 3 and 4 until at least the end of the year.
* * *
A logician rides an elevator. The door opens and someone asks:
— Are you going up or down?
— Yes.
My webpage and my blog generate a lot of emails. I love receiving most of the emails, but if I reply to them, I won't have time to work on my blog. My favorite type of message is one that is full of compliments, with a note that the writer doesn't expect a reply.
I am grateful to people who send me things I requested, like pictures of Russian plates, or some interesting number properties. I apologize that it takes me so long to reply.
The emails that I don't enjoy reading contain amazing elementary proofs of Fermat's last theorem, or any other theorem on the Millennium list, for that matter. I also do not like when my readers ask me for help with their homework.
Like most people, I'm already dealing with spammers who want to enlarge the body parts I do not have or to slim the ones I do have. However, if you do need to send me millions of dollars that I won in your lottery, there is no reason to waste time on email exchanges: you can process them through my "donate" button.
You are welcome to contact me, but ….
I already gave an example of the kinds of problems that were given to Jewish people at the oral entrance exam to the math department of Moscow State University. In fact, I have a whole page with a collection of such problems, called Jewish problems or Coffins. That page was one of the first pages I created when I started my website more than ten years ago.
When my son Alexey was in high school, I asked him to help me type these problems into a file and to recover their solutions from my more than laconic notes, and solve the problems that I didn't have notes for. He did the job, but the file was lying dormant on my computer. Recently I resurrected the file and we prepared some of the solutions for a publication.
The problems that were given during these exams were very different in flavor: some were intentionally ambiguous questions, some were just plain hard, some had impossible premises. In our joint paper "Jewish Problems" we presented problems with a special flavor. These are problems that have a short and "simple" solution, that is nonetheless very difficult to find. This way the math department of MSU was better protected from appeals and complaints.
Try the following problem from our paper:
Find all real functions of real variable F(x) such that for any x and y the following inequality holds: F(x) − F(y) ≤ (x − y)2.
I will give a talk on the subject for UMA at MIT on October 18, at 5pm.
What's "plagiarism"? It's when you take someone else's work and claim it's your own. It's basically STEALING.
Ideas improve. The meaning of words participates in the improvement. Plagiarism is necessary. Progress implies it. It embraces an author's phrase, makes use of his expressions, erases a false idea, and replaces it with the right idea.
Perhaps the Russians have done the right thing, after all, in abolishing copyright. It is well known that conscious and unconscious appropriation, borrowing, adapting, plagiarizing, and plain stealing are variously, and always have been, part and parcel of the process of artistic creation. The attempt to make sense out of copyright reaches its limit in folk song. For here is the illustration par excellence of the law of Plagiarism. The folk song is, by definition and, as far as we can tell, by reality, entirely a product of plagiarism.
If you copy from one author, it's plagiarism. If you copy from two, it's research.
In my essays The Oral Exam and A Math Exam's Hidden Agenda, I gave some examples of math problems that were used during the entrance exams to Moscow State University. The problems were designed to prevent Jewish and other "undesirable" students from studying at the University. My readers might have supposed that an occasional bright student could, by solving all the problems, get in. Here is the story of my dear friend Mikhail (Misha) Lyubich; it shows that being extremely bright was not enough.
Misha passed the first three exams and was facing his last exam: oral physics. He answered all the questions. None of his answers were accepted: all of them were declared wrong. Misha insisted that he was right and requested that the examiners explain themselves. Every time their reply was the same:
This is not a consultation, it's an exam.
Misha failed the exam. The solution to the last problem was a simple picture: a document that seemed to be impossible to deny, so Misha decided that he had grounds for an appeal. The person in charge denied the appeal. When Misha requested an explanation, can you guess the answer?
This is not a consultation, it's an appeal.
Misha ended up studying at Kharkov State University. Now he is a professor at Stony Brook and the director of the Institute for Mathematical Sciences at Stony Brook.
You know that the negation of a true statement is a false statement, and the negation of a false statement is a true statement. You also know that you can negate a sentence by preceding it with "It is not true that …."
Now look at the following statement and its negation, invented by David Bernstein. Which one is true?
How about this pair?
My son Alexey taught me to always plug unused power strips into themselves, so that we can call them "The Rings of Power." These are my Borromean Rings of Power:
Let's call a projection of a body L onto a hyperplane a shadow. Here is a mathematical way to hide behind. An object K can hide behind an object L if in any direction the shadow of K can be moved by a translation to be inside the corresponding shadow of L. If K can hide inside L, then obviously K can hide behind L. Dan Klain drew my interest to the following questions. Is the converse true? If K can hide behind L can it hide inside L? If not, then if K can hide behind L, does it follow that the volume of K is smaller than that of L?
We can answer both questions for 2D bodies by using objects with constant width. Objects with constant width are ones that have the same segment as their shadow in every direction. The two most famous examples are a circle and a Reuleaux triangle:
Let's consider a circle and a Reuleaux triangle of the same width. They can hide behind each other. Barbier's Theorem states that all objects of the same constant width have the same perimeter. We all know that given a fixed perimeter, the circle has the largest area. Thus, the circle can hide behind the Reuleaux triangle which has smaller area and, consequently, the circle can't hide inside the Reuleaux triangle. By the way, the Reuleaux triangle has the smallest area of all the objects with the given constant width.
To digress. You might have heard the most famous Microsoft interview question: Why are manhole covers round? Presumably because round manhole covers can't fall into slightly smaller round holes. The same property is true for manhole covers of any shape of constant width. On the picture below (Flickr original) you can see Reuleaux-triangle-shaped covers.
Let's move the dimensions up. Dan's questions become both more difficult and more interesting, because the shadows are not as simple as segments any more.
Before continuing, I need to introduce the concept of "Minkowski sums." Suppose we have two convex bodies in space. Let's designate the origin. Then a body can be represented as a set of vectors from the origin to the points in the body. The Minkowski sum of two bodies are all possible sums of two vectors corresponding to the first body and the second body.
Another way to picture the Minkowski sum is like this: Choose a point in the second body. Then move the second body around by translations so that the chosen point covers the first body. Then the area swept by the second body is the Minkowski sum of both of them.
Suppose we have two convex bodies K and L. Their Minkowski interpolation is the body tK + (1-t)L, where 0 ≤ t ≤ 1 is a scaling coefficient. The picture below made by Christina Chen illustrates the Minkowski interpolation of a triangle and an inverted triangle.
If two bodies can hide behind L, then their Minkowski interpolation can hide behind L for any value of parameter t. In particular if K can hide behind L, then the Minkowski interpolation tK + (1-t)L can hide behind L, for any t.
In my paper co-authored with Christina Chen and Daniel Klain "Volume bounds for shadow covering", we found the following connection between hiding inside and volumes. If L is a simplex, and K can hide behind it, but can't hide inside L, then there exists t such that the Minkowski interpolation tK + (1-t)L has a larger volume than the volume of L.
In the paper we conjecture that the largest volume ratio V(K)/V(L) for a body K that can hide behind another body L is achieved if L is a simplex and K is a Minkowski interpolation of L and an inverted simplex. The 3D object that can hide behind a tetrahedron and has 16% more volume than the tetrahedron was found by Christina Chen. See her picture below.
The main result of the paper is a universal constant bound: if K can hide behind L, then V(K) ≤ 2.942 V(L), independent of the dimension of the ambient space.
Question 1. Holodeck. After a long and difficult assignment on an uninhabited planet, Commander Riker went to Holodeck III to unwind. While there he ate three cheeseburgers generated by the holodeck program. Is Commander Riker hungry after he ends the program?
Question 2. Relativity. We know that speed in space is relative, there is no absolute speed. What does Captain Picard mean when he orders a "full stop"?
Question 3. The Replicator. Captain Picard approached a replicator and requested: "Tea, Earl Grey. Hot." The replicator immediately created a glass with hot Earl Grey tea. How much energy would the Enterprise have saved in seven years if they used a dish-washing machine, rather than creating glasses from atoms each time and dissolving them afterwards?
Question 4. Contractions. Commander Data hasn't mastered contractions in English speech. In what year do you think the first program was written to convert formal English into English with contractions?
Question 5. Data. Commander Data is fully functional and absolutely superior to a vibrator. Given that there are more than a thousand people on board the Enterprise, estimate how many times a year on average Data will receive sexual requests.
The next two questions are related to particular episodes.
Question 6. "Up The Long Ladder". Mariposans reproduce by cloning. Why do all the identical sets of clones appear to be the same age? Does it mean that upon the reproduction the clone is the age of the host? If so, they all should be 300 years old.
Mariposans steal sample DNA from Commander Riker and Dr. Pulaski. If Riker and Pulaski didn't destroy their maturing clones what age would those clones be? Would they know how much two plus two is when they awaken? If clones awaken as adults, what is their life span?
Question 7. "Force of Nature". Serova sacrifices herself to save her world from the effects of warp drive, but in doing so, she herself creates the rift that will destroy her world. Explain the logic.
* * *
Logic: if an empty yogurt container is in the sink, a spoon is in the garbage can.
* * *
Logically, a wireless mouse should be called a hamster.
* * *
— I started a new life today.
— You quit smoking and drinking?
— No, I changed my email and Facebook accounts.
* * *
— The reviewer has rejected your paper submitted to our math journal because it doesn't contain any theorems or fomulae or even numbers.
— Wait a minute. Your reviewer is mistaken. There are page numbers on every page.
* * *
A kyboard for sal: only on ky dosn't work.
* * *
My computer always beats me in chess. In revenge, I always beat it in a boxing match.
* * *
— Were your parents married when you were born?
— 50%.
— 50%?
— Yes, my father was married and my mother was not.
* * *
Two programmers are talking:
— I can't turn on my oven.
— What's the error message?
Have you ever been punished for being too good at spider solitaire? I mean, have you ever been stuck because you collected too many suits? Many versions of the game don't allow you to deal from the deck if you have empty columns, nor do they allow you to get back a completed suit. If the number of cards left on the table in the middle of the game is less than ten — the number of columns — you are stuck. I always wondered what the probability is of being stuck. This probability is difficult to calculate because it depends on your strategy. So I invented a boring version of spider solitaire for the sake of creating a math problem. Here it goes:
You start with two full decks of 104 cards. Initially you take 54 cards. At each turn you take all full suits out of your hand. If you have less than ten cards left in your hand, you are stuck. If not, take ten more cards from the leftover deck and continue. What is the probability that you can be stuck during this game?
Let us simplify the game even more by playing the easy level of the boring spider solitaire in which you have only spades. So you have a total of eight full suits of spades. I leave it to my readers to calculate the total probability of being stuck. Here I would like to estimate the easiest case: the probability of being stuck before the last deal.
There are ten cards left in the deck. For you to be stuck, they all should have a different value. The total number of ways to choose ten cards is 104 choose 10. To calculate the number of ways in which these ten cards have different values we need to choose these ten values in 13 choose 10 ways, then multiply by the number of ways each card of a given value can be taken from the deck: 810. The probability is about 0.0117655.
I will leave it to my readers to calculate the probability of being stuck before the last deal at the medium level: when you play two suits, hearts and spades.
No, I will not tell you how many times I played spider solitaire.
Is there a way to put a sequence of numbers to music? The system that comes immediately to mind is to match a number to a particular pitch. The difference between any two neighboring integers is the same, so it is logical to assume that the same tone interval should correspond to the same difference in integers. After we decide which tone interval corresponds to the difference of 1, we need to find our starting point. That is, we need to choose the pitch that corresponds to the number 1. After that, all numbers can be automatically matched to pitches.
After we know the pitches for our numbers, to make it into music we need to decide on the time interval between the notes. The music should be uniquely defined by the sequence, hence the only logical way would be to have a fixed time interval between two consecutive notes.
We see that there are several parameters here: the starting point, the pitch difference corresponding to 1, and the time interval between notes. The Online Encyclopedia of Integer Sequences offers the conversion to music for any sequence. It gives you freedom to set the parameters yourself. The sequences do not sound melodic because mathematical sequences will not necessarily follow rules that comply with a nice melody. Moreover, there are no interesting rhythms because the time interval between the notes is always the same.
One day I received an email from a stranger named Michael Blake. He sent me a link to his video on YouTube called "What Pi Sounds Like." He converted the digits of Pi to music. My stomach hurt while I was listening to his music. My stomach hurts now while I am writing this. He just numbered white keys on the piano from 1 to 9 starting from C. Then he played the digits of Pi. Clearly, Michael is not a mathematician, as he does not seem to know what to do with 0. Luckily for him the first 32 digits of Pi do not contain zero, so Michael played the first several digits over and over. My stomach hurts because he lost the basic math property of digits: the difference between the neighboring digits is the same. In his interpretation the digits that differ by one can have a tone interval of minor or major second in a random order corresponding to his random starting point.
I am not writing this to trash Michael. He is a free man in a free country and can do whatever he wants with the digits of Pi. Oops, I am sorry, he can't do whatever he wants. Michael's video was removed from YouTube due to an odd copyright infringement claim by Lars Erickson, who wrote a symphony using the digits of Pi.
Luckily for my readers Michael's video appears in some other places, for example at the New Scientist channel. As Michael didn't follow the symmetry of numbers and instead replaced the math rules with some music rules, his interpretation of Pi is one of the most melodic I've heard. The more randomly and non-mathematically you interpret digits, the more freedom you have to make a nice piece of music. I will say, however, that Michael's video is nicely done, and I am glad that musicians are promoting Pi.
Other musicians do other strange things. For example, Steven Rochen composed a violin solo based on the digits of Pi. Unlike Michael, he used the same tone interval for progressing from one number to the next, like a mathematician would do. He started with A representing 1 and each subsequent number corresponded to an increase of half a tone. That is, A# is 2 and so on. Like Michael Blake he didn't know what to do with 0 and used it for rest. In addition, when he encountered 10, 11, and 12 as part of the decimal expansion he didn't use them as two digits, but combined them, and used them for F#, G, G# respectively. To him this was the way to cover all possible notes within one octave, but for me, it unfortunately caused another twinge in my stomach.
In August I visited my son Alexey Radul, who currently works at the Hamilton Institute in Maynooth, Ireland. One of the greatest Irish attractions, Broom Bridge, is located there. It's a bridge over the railroad that connects Maynooth and Dublin. One day in 1843, while walking over the bridge, Sir William Rowan Hamilton had a revelation. He understood how the formulae for quaternions should be written. He scratched them into a stone of the bridge. Now the bridge has a plaque commemorating this event. The plaque contains his formulae. I don't remember ever seeing a plaque with math, so naturally I rushed off to make my pilgrimage to Broom Bridge.
Quaternions have very pronounced sentimental value for me, since my first research was related to them. Let's consider a simple graph. We can construct an algebra associated with this graph in the following way. For each vertex we have a generator of the algebra. In addition we have some relations. Each generator squared is equal to −1. If two vertices are connected the corresponding generators anti-commute, and they commute otherwise. The simplest non-commutative algebra associated with a graph corresponds to a graph with two vertices and one edge. If we call the generators i and j, then the we get the relations: i2 = j2 = −1, and ij = −ji. I we denote ij as k, the algebra as a vector space has dimension 4 and a basis: 1, i, j, k. These are exactly the quaternions. In my undergraduate research I studied such algebras related to Dynkin diagrams. Thirty years later I came back to them in my paper Clifford Algebras and Graphs. But I digress.
I was walking on the bridge hoping that like Hamilton I would come up with a new formula. Instead, I was looking around wondering why the Broombridge Station didn't have a ticket office. I already had my ticket, but I was curious how other people would get theirs. I asked a girl standing on the platform where to buy tickets. She said that there is no way to buy tickets there, so she sometimes rides without a ticket. The fine for not having tickets is very high in Ireland, so I expressed my surprised. She told me that she just says that she is from the town of Broombridge if she is asked to present her ticket.
Being a Russian I started scheming: obviously people can save money by buying tickets to Broombridge and continuing without a ticket wherever they need to go. If the tickets are checked, they can claim that they are traveling from Broombridge. Clearly Ireland hasn't been blessed with very many Russians visitors.
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and is offered a bet. She may pay $550 in which case she will get $1000 if the coin was tails. If the coin was tails, she is put back to sleep with her memory erased, and awakened on Tuesday and given the same bet again. She knows the protocol. Should she take the bet?
As we discussed in our first essay about Sleeping Beauty, she should take the bet. Indeed, if the coin was heads her loss is $550. But if the coin was tails her gain is $900.
To tell you the truth, when Beauty is offered the bet, she dreams: "It would be nice to know the day of the week. If it were Tuesday, then the coin must have been tails and I would gladly take the winning bet."
In our next variation of the riddle her dream comes true.
Every time she is awakened she is offered to buy the knowledge of the day of the week. How much should she be willing to pay to know the day of the week?
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning and instead of the expected questions she is offered a bet. She may pay $600 in which case she will get $1000 if the coin was tails. Should she take the bet?
We know that tripling the triangular number 1 yields the triangular number 3. The figure shows how we can use this fact to conclude that tripling the triangular number 15 yields the triangular number 45.
Using this new fact, can you modify the figure to find even larger examples of tripling triangles?
This post is inspired by the following problem:
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning: What is her probability that the coin was heads?
Some people argue: asleep or awake, the probability of a fair coin being heads is one half, so her probability should be one half.
Other people, including us, argue that those people didn't study conditional probability. On the information of the setup to the problem and the information of having awakened, the three situations "Coin was heads and it is Monday", "Coin was tails and it is Monday", and "Coin was tails and it is Tuesday" are symmetric and therefore equiprobable; thus the probability that the coin was tails is, on this information, two thirds.
So who is right? We are, of course. A good way to visualize probability judgements is to turn them into bets. Suppose each time Beauty wakes up she is offered the following bet: She pays $600 and gets $1000 if the coin was tails. Should she take it? If her probability of the coin being tails were one half, then obviously not; if her probability of the coin being tails were two thirds, obviously yes. So which is it? Consider the situation from her perspective as of Sunday. She can either always take this bet or always refuse it. If she always refuses, she gets nothing. If she always accepts: If the coin turns up heads, she will be asked the question once and will lose $600. If the coin turns up tails, she will be asked the question twice and will gain $800. So on average she will win, so she should take the bet. By this thought experiment, her probability of tails is clearly not one half.
To make matters more interesting, let's try another bet. Suppose she is given the above bet just once, in advance, on Sunday. She pays $600, and she gets paid $1000 on Wednesday if the coin was tails. This has nothing to do with sleeping and awakening. If she takes the bet she loses $600 with probability one half and gains $400 otherwise. So she shouldn't take the bet. Her probability on Sunday that the coin will come up heads is, of course, one half. The point is that just as these two bets are different bets, the sets of information Beauty has on Sunday vs at awakening are different, and lead to different conclusions. On Sunday she knows that the next time she wakes up it will be Monday, but when she then wakes up, she doesn't know that it's Monday.
Parting thought: The phenomenon of predictably losing information leads to the phenomenon of predictably changing one's assessments. Suppose for some reason she decided to take that unprofitable bet on Sunday. When she wakes up during the experiment, should she feel happy or sad? From her perspective during the experiment, the odds of gaining $400 vs losing $600 are two to one, so she should be happy. Given that she knows on Sunday how she will (with complete certainty!) feel about this bet on Monday, should she take it, even given her Sunday self's assessment that it's a bad bet?
If you buy one Mega Millions ticket, your probability of hitting the jackpot is one in 175,000,000. For all practical purposes it is zero. When I give my talk on lotteries, there is always someone in the audience who would argue that "but someone is winning and so can I." The fact that someone is winning depends on the number of people buying tickets. It is difficult to visualize the large number of people buying tickets and the miniscule odds of winning. For example, the probability of you dying from an impact with a meteorite is larger than the odds of winning the jackpot.
I receive a lot of emails from strangers asking me to advertise their websites on my blog. I always check out their websites and I often find them either unrelated to math or boring. That is why I was pleasantly surprised when I was asked to write about a useful website: Understanding Big Numbers. In each post Liam Gray takes a big number and puts it into some perspective. For example, he estimates Mark Zuckerberg's Hourly Wage by dividing Mark's estimated wealth in 2011 by the number of hours Mark might have worked on Facebook. Facebook has existed for 7 years and, assuming 10 hours of work a day every day, we get 25,000 work hours. That is more than half a million dollars an hour.
Imagine someone calls Mark Zuckerberg and asks to talk to him for a minute. Mark wouldn't be out of line to request nine thousand dollars for that. Lucky am I, that I do not need to talk to Mark Zuckerberg.
Here is a game that John Conway popularizes. It is called "Finchley Central," which is a station of the London Underground. The game goes as follows. Alice and Bob take turns naming London Underground stations, in any order. The first person to say "Finchley Central" wins.
Alice, who starts, can just name the station. But then Bob will give her a look. It is not fun to win a game on the first turn. To avoid appearing rude, Alice will not start with "Finchley Central." It would be impolite of Bob to take advantage of Alice's generosity, so he also won't say "Finchley Central." The game might continue like this for a while.
The game has a hidden agenda: winning it after 10 turns will supply many more bragging rights than winning it right away would. We can make this hidden agenda explicit by assigning a value to the honor of continuing the game. For example, suppose every time Alice (or Bob) says a station, she puts one dollar into the pile. The person who says "Finchley Central" first takes all the money from the pile. The implicit goal of the game becomes explicit: you want to say "Finchley Central" right before your opponent says it.
By the way, Finchley Central is not actually a particularly central station — it is the station between Finchley East and Finchley West, serving the relatively small place called Finchley; and is not even under ground. It has the distinction of being one of the oldest still-standing pieces of London Underground physical plant, because plans to rebuild it were interrupted on account of World War II and never resumed. It also has the distinction of having served the home of the guy (an employee of the Underground system) who had the brilliant idea that since the Underground was, indeed, mostly under ground, the right way to map it was topologically, rather than geographically.
Here is another way to model the game. Alice writes an odd number on a piece of paper, and Bob writes an even number. When they compare, the person who wrote a smaller number wins that number of dollars. This version loses the psychological aspect. When you take turns, it is to your advantage to read the non-verbal signs of your opponent to see when s/he is getting ready to drop the bomb.
People play this game in real life. Here are Alice and Bob looking at the last piece of a mouth-watering Tiramisu:
At this point Alice wins with some extra brownie points for being polite.
We can model the honor points differently. We can say you will be the most proud of the game if you name the station write before you opponent is about to do so. Then the model is: everyone writes down their next move; if your move is Finchley Central when your opponent's next move was going to be Finchley Central, then you win.
Here we suggest another game that we call "Reverse Finchley Central." Alice and Bob name London Underground stations in turns and the person who names "Finchley Central" first loses. This game can continue until all the stations are exhausted, if the players are forbidden to repeat them, or it can continue indefinitely otherwise. But this is quite tiresome. The hidden agenda would be to not waste too much time. Clearly the person who values time less will win.
But let us model this game. We want to fix the value of winning. Let us set aside ten dollars for the winner. On their turn, each player puts one dollar into the pile, and as soon as one of the players says "Finchley Central," the other one wins and takes the ten dollars. The pile goes to charity. Alternatively, Alice and Bob can each write a number. The person with the larger number wins the prize, while both have to pay the smaller number to charity.
We play this game with our parents. They nag us to do the dishes. We resist. Then they give up and do the dishes themselves. They lose, but we all pay with our nerves for nagging or being nagged at. Later our parents get their revenge when we have children of our own.
At the 2011 IMO, Lisa Sauermann received yet another gold medal. Now she tops the Hall of Fame of the IMO with four gold medals and one silver medal.
In addition, in 2011 she achieved the absolute best individual result and was the only person with a perfect score. In previous years, there were several girls who tied for first place, but she is the first girl ever to have an absolute rank of 1.
I told you so. In my 2009 essay Is There Hope for a Female Fields Medalist?, I predicted that a girl will soon become an absolute champion of the IMO.
In that essay I draw a parallel between the absolute champion of IMO and a Fields medalist. Indeed, we get one of each per year. Lisa Sauermann is the best math problem solver in her year. Will she grow up to receive a Fields medal? I am not so sure: the medal is still unfriendly to women. Lisa Sauermann is the best math problem solver ever. Will she grow up to be the best mathematician of our century? I wonder.
My e-friend and coauthor, Konstantin Knop, designed the following problem for the 2011 All-Russia Olympiad:
Some cells of a 100 by 100 board have one chip placed on them. We call a cell pretty if it has an even number of neighboring cells with chips. Neighbors are the cells that share a side. Is it possible for exactly one cell to be pretty?
The problem is not easy. Only one person at the Olympiad received full credit for it.
From time to time my female colleagues share stories with me of great unfairness or horrible sexual harassment in the world of mathematics. I can't reciprocate — certainly not on that level.
I do not have any horror stories to tell. Generally I am treated with great respect, at least to my face. In fact, some men have told me that I am the smartest person they ever met.
The stories I want to share are not about harassment. No single incident is a big deal. But when these things happened time after time after time, I realized: this is gender bias.
First story. A guy told me, "Your proof is unbelievably amazing."
What can I say? It is just a compliment. Though I am not sure why the word "unbelievable" was included. Is it difficult to believe that I can produce an amazing proof? I encounter surprise too often to my taste.
Second story. Another guy tells me after I explain a solution to a math problem, "I didn't realize it was so simple."
Actually it wasn't simple. When I explained the solution, it may have seemed simple, but that was because I was able to explain it to him with such clarity. People tend to downgrade their opinion of the problem, rather than upgrade their opinion of my ability. It actually affects my reputation as a mathematician.
Third story. Another guy said to me (and I quote!), "I am so dumb. I tried for a week to write the program that computes these numbers and you did it in one hour. I feel so dumb. I didn't expect myself to be so dumb. Why am I so dumb?"
After the fourth "dumb", I started wondering what it was all about. Many guys try to compete with me. And they hate losing to a woman. It creates a strong motivation for them to discard my brilliance and to explain away my speed, even if they have to claim temporary dumbness.
Fourth story. Someone asked me, "What is the source of the solutions and math ideas in your blog? Can you refer me to the literature?"
I do invest extra effort in citing the sources of the math puzzles I discuss. Everything else — the solutions, the ideas, new definitions, new sequences — I invent myself. I have even started inventing math puzzles. This is my blog. I thought of it myself, I wrote it myself. Has anyone ever asked Terence Tao where he takes the solutions for his blog from?
Unfortunately, this attitude damages my career. When people think that my ideas come from someone else, they do not cite me.
But all these stories however minor happen all the time, not only to me but to all my female colleagues. Gender bias is real. Next time someone tells me how unbelievably amazing my proof is, I will explode.
Recently I stumbled on a cute xkcd comic with the hidden message:
Wikipedia trivia: if you take any article, click on the first link in the article text not in parentheses or italics, and then repeat, you will eventually end up at "Philosophy".
Naturally, I started to experiment. The first thing I tried was mathematics. Here is the path: Mathematics — Quantity — Property — Modern philosophy — Philosophy.
Then I tried physics, which led me to mathematics: Physics — Natural science — Science — Knowledge — Fact — Information — Sequence — Mathematics.
Then I tried Pierre de Fermat, who for some strange reason led to physics first: Pierre de Fermat — French — France — Unitary state — Sovereign state — State — Social sciences — List of academic disciplines — Academia — Community — Living — Life — Objects — Physics.
The natural question is: what about philosophy? Yes, philosophy goes in a cycle: Philosophy — Reason — Rationality — philosophy.
The original comic talks about spark plugs. So I tried that and arrived at physics: Spark plug — Cylinder head — Internal combustion engine — Engine — Machine — Machine (mechanical) — Mechanical system — Power — Physics.
Then I tried to get far away from philosophy and attempted sex, unsuccessfully: Sex — Biology — Natural science. Then I tried dance: Dance — Art — Sense — Physiology — Science.
It is interesting to see how many steps it takes to get to philosophy. Here is the table for the words I tried:
| Word | # Steps |
|---|---|
| Mathematics | 4 |
| Physics | 11 |
| Pierre de Fermat | 24 |
| Spark plug | 19 |
| Sex | 12 |
| Dance | 13 |
Mathematics wins. It thoroughly beats all the other words I tried. For now. Fans of sex might be disappointed by these results and tomorrow they might change the wiki essay about sex to start as:
Modern philosophy considers sex …
Not personally. Someone hacked into my website.
I would like to thank my readers Qiaochu Yuan, Mark Rudkin, "ano" and Paul who alerted me to the problem. Viewers who were using the Google Chrome browser and who tried to visit my website got this message: "This site contains content from howmanyoffers.com, a site known to distribute malware."
It took me some time to figure out what was going on. It appears that on June 19 someone from 89-76-135-50.dynamic.chello.pl hacked into my hosting account and added a script to all my html files and to my blog header. It seems that the script was dormant and wasn't yet doing bad things.
As soon as I grasped what was going on, I replaced all the affected files.
I have had my website for many years without changing my hosting password. Unfortunately, passwords, not dissimilar to humans, have this annoying tendency to become weaker with age. I wasn't paying attention to the declining strength of my password and so I was punished.
Now I have fixed the website and my new password is: qwP35q2054uWiedfj052!@#$%.
Just kidding.
I recently had the following chat with a particular calculator:
It seems odd to me that putting a few more e's down the bottom should result in it thinking there were the same number of extra 10s at the bottom. In fact, I've never seen a calculator answer in this form at all. I'm especially intrigued that the final power of ten seems to be the same in all three cases, so it can't even just be estimating. Do you have any thoughts on what screwy counting could be behind these particular answers?
May the Mass times the Acceleration be with you!
John Conway taught me how to tell time at night. But first I need to explain the notions of the "time in the sky" and the "time in the year."
The clock in the sky. Look at Polaris and treat it as the center of a clock. The up direction corresponds to 12:00. Now we need to find a hand. If you find Polaris the way I do, first you locate the Big Dipper. Then you draw a line through the two stars that are furthest away from the Big Dipper's handle. The line passes through Polaris and is your "hour" hand. Now you can read the time in the sky.
The hand of the clock in the sky makes a full rotation in approximately 24 hours. So if you stare at the sky for a long time, you can calculate the time you spent staring. Keep in mind that the hand in the sky clock is twice as slow as the hour hand, and it turns counter-clockwise. So to figure out how long you're looking into the sky, take the sky-time when you start staring, subtract the sky-time when you stop staring and multiply the result by 2.
To calculate the absolute time, we need to adjust for the day in the year.
The clock in the year. A year has twelve months and a clock has twelve hours. How convenient. You can treat each month as one hour. In addition as a month has about 30 days and an hour has exactly 60 minutes, we should count a day as two minutes. Thus, January 25 is 1:50.
Fact: on March 7th at midnight the clock in the sky shows 12:00. March 7th corresponds to 3:15. So to calculate the solar time you need to add up the time in the sky and the time in the year and multiply it by 2. Then subtracting the result from 6:30, which is twice 3:15, you get the solar time.
You are almost ready. You might need to adjust for daylight savings time or for peculiarities of your time zone.
This time formula is not very precise. But if you are looking into the sky and you do not have your watch or cell phone with you, you probably do not need to know the time precisely.
In my life as a female mathematician I have quite often encountered a mathematician's wife who, despite not knowing me, already hated me. It was clear that it had nothing to do with me personally, so being clueless and naive, I assumed that most men were cheaters and that their wives were extremely insecure and jealous.
Then one day one of the wives decided to be frank about her feelings. It wasn't about cheating, she told me. It was that she felt distant from her husband. He lived in a world of mathematics from which she was excluded. I on the other hand shared this world with him.
It was very sad. It meant that I incurred their jealousy, not because of my sins, but because I am a female mathematician.
Let me tell you another story that helped me realize how all-encompassing this world of mathematics can be for some people. Once I had a very close friend who we will call Jack. I do not want to name him as he is a famous mathematician. Jack told me that the strongest emotions he feels are related to mathematics. He can only feel close to someone if he can share a mathematical discussion with them.
Now I understand the wives better. Husbands like Jack invest so much more in their math world and their colleagues than they do in their home life, that it is not surprising the wives are jealous. Because women mathematicians are scarce, when I appear in their husbands' world, it adds another layer of worry.
Another thing that Jack told me is that he gets such a euphoric feeling when he discovers a new math idea that it is better than any orgasm. Of course, this statement made me question the quality of Jack's orgasms, but in any case, for some mathematicians math is an aphrodisiac.
If math is an aphrodisiac, then tattooing a formula on the lover's body may well enhance the orgasm. I just remembered the movie by Ed Frenkel. But I digress.
If math is an aphrodisiac, then I understand jealous wives even better. Without sex I can give their husbands pleasure they can't.
* * *
I am taking my dog to tweet. He'll check other dog's posts at every pole and will leave his comments.
* * *
Not many people know that 1000 chameleons is a chabillion.
* * *
The Internet paradox: it connects people who are far apart, and disconnects those who are close.
* * *
We bought a cell phone for our TV set. We attached it to the remote control, so that we can call our TV when the remote is lost.
* * *
Mary's mom failed arithmetic. Actually, that is why Mary was born.
* * *
Your call is very important to us. Please, hold. And in the meantime, to protect your health, our customer care team encourages you to drink a glass of water at least every two hours.
* * *
Who is your favorite computer game character?
The stick from Tetris.
* * *
Our new boss invited everyone to bring their keyboards to his office. He kept the employees who had worn letters and laid off the ones with worn arrows.
* * *
My son will be a hacker. He started his career before he was born: he found a flaw in the condom.
I recently stumbled upon some notes (in Russian) of a public lecture given by Vladimir Arnold in 2006. In this lecture Arnold defines a notion of complexity for finite binary strings.
Consider a set of binary strings of length n. Let us first define the Ducci map acting on this set. The result of this operator acting on a string a1a2…an is a string of length n such that its i-th character is |ai − a(i+1)| for i < n, and the n-th character is |an − a1|. We can view this as a difference operator in the field F2, and we consider strings wrapped around. Or we can say that strings are periodic and infinite in both directions.
Let's consider as an example the action of the Ducci map on strings of length 6. Since the Ducci map respects cyclic permutation as well as reflection, I will only check strings up to cyclic permutation and reflection. If I denote the Ducci map as D, then the Ducci operator is determined by its action on the following 13 strings, which represent all 64 strings up to cyclic permutation and reflection: D(000000) = 000000, D(000001) = 000011, D(000011) = 000101, D(000101) = 001111, D(000111) = 001001, D(001001) = 011011, D(001011) = 011101, D(001111) = 010001, D(010101) = 111111, D(010111) = 111101, D(011011) = 101101, D(011111) = 100001, D(111111) = 000000.
Now suppose we take a string and apply the Ducci map several times. Because of the pigeonhole principle, this procedure is eventually periodic. On strings of length 6, there are 4 cycles. One cycle of length 1 consists of the string 000000. One cycle of length 3 consists of the strings 011011, 101101 and 110110. Finally, there are two cycles of length 6: the first one is 000101, 001111, 010001, 110011, 010100, 111100, and the second one is shifted by one character.
We can represent the strings as vertices and the Ducci map as a collection of directed edges between vertices. All 64 vertices corresponding to strings of length 6 generate a graph with 4 connected components, each of which contains a unique cycle.
The Ducci map is similar to a differential operator. Hence, sequences that end up at the point 000000 are similar to polynomials. Arnold decided that polynomials should have lower complexity than other functions. I do not completely agree with that decision; I don't have a good explanation for it. In any case, he proposes the following notion of complexity for such strings.
Strings that end up at cycles of longer length should be considered more complex than strings that end up at cycles with shorter length. Within the connected component, the strings that are further away from the cycle should have greater complexity. Thus the string 000000 has the lowest complexity, followed by the string 111111, as D(111111) = 000000. Next in increasing complexity are the strings 010101 and 101010. At this point the strings that represent polynomials are exhausted and the next more complex strings would be the three strings that form a cycle of length three: 011011, 101101 and 110110. If we assign 000000 a complexity of 1, then we can assign a number representing complexity to any other string. For example, the string 111111 would have complexity 2, and strings 010101 and 101010 would have complexity 3.
I am not completely satisfied with Arnold's notion of complexity. First, as I mentioned before, I think that some high-degree polynomials are so much uglier than other functions that there is no reason to consider them having lower complexity. Second, I want to give a definition of complexity for periodic strings. There is a slight difference between periodic strings and finite strings that are wrapped around. Indeed, the string 110 of length 3 and the string 110110 of length 6 correspond to the same periodic string, but as finite strings it might make sense to think of string 110110 as more complex than string 110. As I want to define complexity for periodic strings, I want the complexity of the periodic strings corresponding to 110 and 110110 to be the same. So this is my definition of complexity for periodic strings: let's call the complexity of the string the number of edges we need to traverse in the Ducci graph until we get to a string we saw before. For example, let us start with string 011010. Arrows represent the Ducci map: 011010 → 101110 → 110011 → 010100 → 111100 → 000101 → 001111 → 010001 → 110011. We saw 110011 before, so the number of edges, and thus the complexity, is 8.
The table below describes the complexity of the binary strings of length 6. The first column shows one string in a class up to a rotation or reflections. The second column shows the number of strings in a class. The next column provides the Ducci map of the given string, followed by the length of the cycle. The last two columns show Arnold's complexity and my complexity.
| String s | # of Strings | D(s) | Length of the end cycle | Arnold's complexity | My complexity |
|---|---|---|---|---|---|
| 000000 | 1 | 000000 | 1 | 1 | 1 |
| 000001 | 6 | 000011 | 6 | 9 | 8 |
| 000011 | 6 | 000101 | 6 | 8 | 7 |
| 000101 | 6 | 001111 | 6 | 7 | 6 |
| 000111 | 6 | 001001 | 3 | 6 | 5 |
| 001001 | 3 | 011011 | 3 | 5 | 4 |
| 001011 | 12 | 011101 | 6 | 9 | 8 |
| 001111 | 6 | 010001 | 6 | 7 | 6 |
| 010101 | 2 | 111111 | 1 | 3 | 3 |
| 010111 | 6 | 111001 | 6 | 8 | 7 |
| 011011 | 3 | 101101 | 3 | 4 | 3 |
| 011111 | 6 | 100001 | 6 | 9 | 8 |
| 111111 | 1 | 000000 | 1 | 2 | 2 |
As you can see, for examples of length six my complexity doesn't differ much from Arnold's complexity, but for longer strings the difference will be more significant. Also, I am pleased to see that the sequence 011010, the one that I called The Random Sequence in one of my previous essays, has the highest complexity.
I know that my definition of complexity is only for periodic sequences. For example, the binary expansion of pi will have a very high complexity, though it can be represented by one Greek letter. But for periodic strings it always gives a number that can be used as a measure of complexity.
I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here's his version:
The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?
When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:
A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.
* * * A Generic Limerick (submitted by Michael Chepovetsky)
There once was an X from place B,
Who satisfied predicate P,
The X did thing A,
In a specified way,
Resulting in circumstance C.
* * *
I just learned that 4,416,237 people got married in the US in 2010. Not to nitpick, but shouldn't it be an even number?
* * *
We are happy to announce that 100% of Russian citizens are computer-savvy and use the Internet on a regular basis (according to a recent Internet survey).
* * *
Two math teachers had a fight. It seems they couldn't divide something.
* * *
Do you know that if you start counting seconds, once you reach 31,556,926 you discover that you have wasted a whole year?
* * *
What I need after a visit to the hairdresser is a "Save" button.
* * *
— Hello! Is this a fax machine?
— Yes.
* * *
— I am not fat at all! My girlfriend tells me that I have a perfect figure.
— Your girlfriend is a mathematician. For her a perfect figure is a sphere.
* * *
A: Hi, how are you?
B: +
A: Will you come to classes today?
B: -
A: You will be kicked out!
B: =
A: Are you using your calculator to chat?
I wonder what the largest number is that can be represented with one character. Probably 9. How about two characters? Is it 99? What about three or four?
I guess I should define a character. Let's have two separate cases. In the first one you can only use keyboard characters. In the second one you can use any Unicode characters.
I'm awaiting your answers to this.
The last time I talked to John H. Conway, he taught me to walk up the stairs. It's not that I didn't know how to do that, but he reminded me that a nerd's goal in climbing the steps is to establish the number of steps at the end of the flight. Since it is boring to just count the stairs, we're lucky to have John's fun system.
His invention is simple. Your steps should be in a cycle: short, long, long. Long in this case means a double step. Thus, you will cover five stairs in one short-long-long cycle. In addition, you should always start the first cycle on the same foot. Suppose you start on the left foot, then after two cycles you are back on the left foot, having covered ten stairs. While you are walking the stairs in this way, it is clear where you are in the cycle. By the end of the staircase, you will know the number of stairs modulo ten. Usually there are not a lot of stairs in a staircase, so you can easily estimate the total if you know the last digit of that number.
I guess I am not a true nerd. I have lived in my apartment for eight years and have never bothered to count the number of steps. That is, until now. Having climbed my staircase using John's method, I now know that the ominous total is 13. Oh dear.
The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year's Olympiad:
Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?
Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, "My age is equal to the sum of the digits of my birth year." "Mine too," replied Paul. By how many years is Paul younger than Peter?
Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?
Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?
Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.
Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.
A mathematician is someone who pauses when asked "How much is two and two?"
Indeed, the answer might be:
I already wrote about the research of my friend Olga Amosova who studied the sickle-cell anemia mutation. She and her colleagues needed to store short fragments of hemoglobin genes for their experiments. All the fragments were identical. They noticed that with time the fragments always broke down in the same place. It was a mystery. When good scientists stumble on a mystery, they start digging.
They found that one of the nucleotides rips off the DNA fragment at the site of the Sickle-cell mutation. That place on the DNA becomes fragile and later breaks down. These sites need to be repaired. The repair is very error-prone and often leads to a mutation.
When DNA strands are left unattended, they want to pair up. There are four types of nucleotides: A, C, G and T. So mathematically the fragment of DNA is a string in the alphabet A, C, G, T. These nucleotides are matched to each other. When two DNA strands pair up, A on one strand always matches T and C matches G. So it is logical that if there are two complementary DNA pieces on the same fragment, they will find each other and pair up. They form a hydrogen bond. For example, a piece AACGT matches perfectly another piece TTGCA. Suppose a substring of DNA consists of a piece AACGT and somewhere later the reverse of the match: ACGTT. Such a string is called an inverted repeat. The DNA fragment I mentioned contains a string AACGT****ACGTT. Two pieces AACGT and ACGTT are complementary and not too far from each other in space. So it is easy for them to find each other and to bond to form a so-called stem-loop or a hairpin structure. The site of Sickle-cell mutation falls into the loop.
Olga and her colleagues discovered that for some particular loops the orientation in space becomes awkward and one of the nucleotides rips off. Such a rip off is called depurination. In further investigation, Olga found examples of when depurination happens. The first sequence of the pair that will bond later has to have at least five nucleotides and has to end in T. Correspondingly the second part in the pair has to begin with A. In the middle there needs to be four nucleotides GTGG. The first G flies away. Enzymes rush like a first aid squad to repair it and introduce mistakes that lead to mutation and diseases like cancer.
DNA was thought to be simply a passive information storage system, not capable of any action. Now we see that DNA is capable of action. DNA can damage itself. Damage provokes a mutation. For all practical purposes it is self-mutilation. Olga and her colleagues scanned the human genome for other sequences that are capable of self-mutilation. They found that such sequences are overwhelmingly present. They are present in much higher numbers than would be expected statistically. The pieces that are capable of damaging themselves occur 40 times more often than would occur if the nucleotides were distributed randomly. They are especially overrepresented in genes linked to cancer.
Self-damaging shouldn't happen in normal situations. It can be provoked by the environment, for example, the chemistry of the cell. That means, that our cancers are not only in our genes but also in our life-style. There was, for example, a suggestion in a recent NY Times article, Is Sugar Toxic?, that too much sugar in a diet might provoke cancer. If the rate of mutation depends on the environment, we can influence it and prolong our lives.
It is not clear why the ability to self-mutilate survives in the evolutionary process. It is quite possible that if something very bad happens to our planet, we need our genes to be able to mutate very fast in order to adjust to the environment so that humans can survive.
Though I never tried to donate my sperm to a sperm bank, because of my inability to produce it, I know that sperm banks look for people who have ancestors who lived for a very long time. Such sperm is in bigger demand as everyone wants their children to live longer. I wonder if this tendency is a mistake. Global warming is upon us. People with longevity genes might not be flexible enough for their children to survive the changing of the Earth.
As you might have guessed from the title, this essay is about domino tilings.
Suppose a subset of a square grid has area N, and the number of possible domino tilings is T. Let's imagine that each cell is contributing a factor of x tilings to the total independently of the others. Then we get that xN = T. This mental exercise suggests a definition: we call the nth root of T the degree of freedom per square for a given region.
Let's consider a 1 by 2k rectangle. There is exactly one way to tile it with dominoes. So the degree of freedom per square of such a rectangle is 1. Now consider a 2 by k rectangle. It has the same area as before, and we know that there should be more than one tiling. Hence, we expect the degree of freedom to be larger than the one in the previous example. The number of tilings of a 2 by k rectangle is Fk-2, where Fk is kth Fibonacci number. So the degree of freedom for large k will be approximately the square root of the golden ratio, which is about 1.272.
You might expect that squares should give larger degrees of freedom than rectangles of the same area. The degree of freedom for a large square is about 1.3385. You can find more information in the beautiful paper Tilings by Federico Ardila and Richard P. Stanley.
Let's move from rectangles to Aztec diamonds. They are almost like squares but the side of the diamond is aligned with diagonals of the dominoes rather than with their sides. See the sample diamonds in the picture above, which Richard Stanley kindly sent to me for this essay.
It is easier to calculate the degree of freedom for Aztec diamonds than for regular squares. The degree is the fourth root of 2, or 1.1892…. In the picture below created by James Propp's tiling group you can see a random tiling of a large Aztec diamond.
Look at its colors: horizontal dominoes are yellow and blue; vertical ones are red and aquamarine. You might wonder what rule decides which of the horizontal dominoes are yellow and which are blue. I will not tell you the rule; I will just hint that it is simple.
Back to freedom. As you can see from the picture, freedom is highly non-uniform and depends on where you live. Freedom is concentrated inside a circle called the arctic circle, perhaps because the areas outside it are frozen for lack of freedom.
Now I would like to expand the notion of freedom to give each cell its own freedom. For a large Aztec diamond, I will approximate freedom with a function that is one outside the arctic circle and is uniform inside. The Aztec diamond AZ(n) consists of 2n(n+1) squares, shaped like a square with side-length n√2. So the area of the circle is πn2/2. Hence we can calculate the freedom inside the circle as the πth root of 2, which is about 1.247. This number is still much less than the degree of freedom of a cell in a large square.
Jorge Tierno sent me a link to the following puzzle:
There is a certain country where everybody wants to have a son. Therefore, each couple keeps having children until they have a boy, then they stop. What fraction of the children are female?
If we assume that a boy is born with probability 1/2 and children do not die, then every birth will produce a boy with the same probability as a girl, so girls will comprise half of all children.
Now, I wonder why everyone would want a boy? Y-chromosomes are much shorter than X-chromosomes. If a man wants to pass his genes to the next generation, a daughter should be preferable as she keeps more genes from the father. I am a mother of two boys, so my granddaughters will have my X-chromosome while my grandsons will have my ex-husband's Y-chromosome, so to keep my genes in the pool I should be more interested in granddaughters.
But I digress. I started writing this essay because in the original puzzle link the answer was different from mine. Here is how the other argument goes:
Half of all families have zero girls, a quarter have 1/2 girls, 1/8 have 2/3 girls, and so on. If we sum this up the expected ratio of girls to boys is (1/2)0 + (1/4)(1/2) + (1/8)(2/3) + (1/16)(3/4) + ... which adds to 1 − ln 2, which is about 30%.
What's wrong with this solution?
I found this cute problem in the Russian book Sharygin Geometry Olympiad by Zaslavsky, Protasov and Sharygin.
Find numbers p and q that satisfy the equation: x2 + px + q = 0.
The book asks you to find a mistake in the following solution:
By Viète's formulae we get a system of equations p + q = − p, pq = q. Solving the system we get two solutions: p = q = 0 and p = 1, q = −2.
What is wrong with this solution?
I wrote how the written entrance exam was used to keep Jewish students from studying at Moscow State University, but the real brutality happened at the oral exam. Undesirable students were given very difficult problems. Here is a sample "Jewish" problem:
Solve the following equation for real y:
Here is how my compatriots who studied algebra in Soviet high schools would have approached this problem. First, cube it and get a 9th degree equation. Then, try to use the Rational Root Theorem and find that y = 1 is a root. Factoring out y − 1 gives an 8th degree equation too messy to deal with.
The most advanced students would have checked if the polynomial in question had multiple roots by GCDing it with its derivative, but in vain.
We didn't study any other methods. So the students given that problem would have failed it and the exam.
Unfortunately, this problem is impossible to appeal, because it has an elementary solution that any applicant could have understood. It goes like this:
Let us introduce a new variable: x = (y3 + 1)/2. Now we need to solve a system of equations:
This system has a symmetry which we can exploit. The graphs of the functions x = (y3 + 1)/2 and y = (x3 + 1)/2 are reflections of each other across the line x = y. As both functions are increasing, the solution to the system of equations should lie on the line x = y. Hence, we need to solve the cubic y = (y3 + 1)/2, one of whose roots we already know.
Now I offer you another problem without telling you the solution:
Four points on a plane used to belong to four different sides of a square. Reconstruct the square by compass and straightedge.
Recently I asked my readers to look at the 1976 written math exam that was given to applicants wishing to study at the math department of Moscow State University. Now it's time to reveal the hidden agenda. My readers noticed that problems 1, 2, and 3 were relatively simple, problem 4 was very hard, and problem 5 was extremely hard. It seems unfair and strange that problems of such different difficulty were worth the same. It is also suspicious that the difficult problems had no opportunity for partial credit. As a result of these characteristics of the exam, almost every applicant would get 3 points, the lowest passing score. The same situation persisted for many years in a row. Why would the best place to study math in Soviet Russia not differentiate the math abilities of its applicants?
In those years the math department of Moscow State University was infamous for its antisemitism and its efforts to exclude all Jewish students from the University. The strange structure of the exam accomplished three objectives toward that goal.
1. Protect the fast track. There was a fast track for students with a gold medal from their high school who got 5 points on the written exam. The structure of the exam guaranteed that very few students could solve all 5 problems. If by chance a Jewish student solved all 5 problems, it was not much work to find some minor stylistic mistake and not count the solution.
2. Avoid raising suspicion at the next exams. The second math entrance exam was oral. At such an exam different students would talk one-on-one with professors and would have to answer different questions. It was much easier to arrange difficult questions for undesirable students and fail all the Jewish students during the oral exam than during the written exam. But if many students with perfect scores on the written exam had failed the oral exam, it might have raised a lot of questions.
3. Protect appeals. Despite these gigantic efforts, there were cases when Jewish students with a failing score of 2 points were able to appeal and earn the minimum passing score of 3. If undesirable students managed to appeal all the exams, they would only get a half-passing grade at the end and would not be accepted because the department was allowed to choose from the many students that the exams guaranteed would have half-passing scores.
I have only heard about one faculty member who tried to publicly fight the written exam system. It was Vladimir Arnold, and I will tell the story some other time.
The following problem appeared at the Gillis Math Olympiad organized by the Weizmann Institute:
A foreign government consists of 12 ministers. Each minister has 5 friends and 6 enemies amongst the ministers. Each committee needs 3 ministers. A committee is considered legitimate if all of its members are friends or all of its members are enemies. How many legitimate committees can be formed?
Surprisingly, this problem implies that the answer doesn't depend on how exactly enemies and friends are distributed. This meta thought lets us calculate the answer by choosing an example. Imagine that the government is divided into two factions of six people. Within a faction people are friends, but members of two different factions dislike each other. Legitimate committees can only be formed by choosing all three members from the same faction. The answer is 40.
We would like to show that actually the answer to the problem doesn't depend on the particular configuration of friendships and enmities. For this, we will count illegitimate committees. Every illegitimate committee has exactly two people that have one enemy and one friend in the committee. Let's count all the committees from the point of view of these "mixed" people. Each person participates in exactly 5*6 committees as a mixed person. Multiply by 12 (the number of people), divide by 2 (each committee is counted twice) and you get the total 180. This gives an answer of 40 for the number of legitimate committees without using a particular example.
What interests us is the fact that the number of illegitimate, as well as legitimate, committees is completely defined by the degree distribution of friends. For any set of people and who are either friends or enemies with each other, the number of illegitimate committees can be calculated from the degree distribution of friends in the same way as we did above.
Any graph can be thought of as representing friendships of people, where edges connect friends. This cute puzzle tells us that the sum of the number of 3-cliques and 3-anti-cliques depends only on the degree distribution of the graph.
As a non mathematical comment, the above rule for legitimate committees is not a bad idea. In such a committee there is no reason for two people to gang up on the third one. Besides, if at some point in time all pairs of friends switch to enemies and vice versa, the committees will still be legitimate.
In 1976 I was about to become a student in the math department at Moscow State University. As an IMO team member I was accepted without entrance exams, but all of my other classmates had to take the exams. There were four exams: written math, oral math, physics, and an essay.
The written math exam was the first, and here are the problems. I want my non-Russian readers to see if they notice anything peculiar about this exam. Can you explain what is peculiar, and what might be the hidden agenda?
Problem 1. Solve the equation
Problem 2. Solve the inequality
Problem 3. Consider a right triangle ABC with right angle C. Angle B is 30° and leg CA is equal to 1. Let D be the midpoint of the hypotenuse AB, so that CD is a median. Choose F on the segment BC so that the angle between the hypotenuse and the line DF is 15°. Find the area of CDF. Calculate its numeric value with 0.001 precision.
Problem 4. Three balls, two of which are the same size, are tangent to the plane P, as well as to each other. In addition, the base of a circular cone lies on the plane P, and its axis is perpendicular to the plane. All three balls touch the cone from the outside. Find the angle between a generatrix of the cone and the plane P, given that the triangle formed by the points of tangency of the balls and the plane has one angle equal to 150°.
Problem 5. Let r < s < t be real numbers. If you set y equal to any of the numbers r, s or t in the equation x2 − (9 − y)x + y2 − 9y + 15 = 0, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that −1 < r < 1.
Let me describe some background to this exam. Applicants who solve fewer than two problems fail the exam and are immediately rejected. People who solve two or three problems are given 3 points. Four problems earn 4 points, and five problems earn 5 points.
If you still do not see the hidden agenda, here is another clue. People who get 5 points on the first exam and, in addition, have a gold medal from their high school (that means all As) are admitted right after the first exam. For the others, if they do not fail any of the exams, points are summed up with their GPAs to compute their scores. The so-called half-passing score is then calculated. Scores strictly higher than the half-passing score qualify applicants for admission. However, there are too many applicants for the available openings with at least the half-passing score. As a result only some people with exactly the half-passing score are accepted, at the discretion of the department.
Now my readers have enough information to figure out the hidden agenda behind that particular exam.
We worked for several years with RSI where we supervised summer math research projects by high school students. Now, we've started an additional program at MIT's math department called PRIMES, where local high school students do math research during the academic year. In this essay we would like to discuss what makes a good math research project for a high school student.
A doable project. The project should not be believed to be extremely difficult to yield at least results. It is very discouraging for an aspiring mathematician not to produce anything during their first project.
An accessible beginning. The student should be able to start doing something original soon after the start of the project. After all, they don't come to us for coursework, but for research.
Flexibility. It is extremely important to offer them a project that is adjustable; it should go in many directions with many different potential kinds of results. Since we do not know the strength of incoming students in advance, it is useful to have in mind both easier and harder versions of the project.
Motivation. It is important for the project to be well motivated, which means related to other things that have been studied and known to be interesting, to research of other people, etc. Students get more excited when they see that other people are excited about their results.
A computer component. This is not a must for a good project. But modern mathematics involves a lot of computation and young students are better at it than many older professors. Such a project gives young students the opportunity to tackle something more senior people are interested in but might not have enough computer skills to solve. In addition, through computer experiments students get exposed to abstract notions (groups, rings, Lie algebras, representations, etc.) in a more "hands-on" way than when taking standard courses, and as a result are less scared of them.
A learning component. It is always good when a project exposes students to more advanced notions.
The student should like their project. This is very difficult to accomplish when projects are chosen in advance before we meet the students. However, we try to match them to great projects by using the descriptions they give of their interests on their applications. It goes without saying that mentors should like their project too.
Having stated the desired properties of a good project, let us move on to giving examples: bad projects and good projects. We start with a bad one:
Prove that the largest power of 2 that doesn't contain 0 is 286.
The project satisfies only one requirement: it contains a computer component. Otherwise, it doesn't have an accessible beginning. It is not very flexible: if the student succeeds, the long-standing conjecture will be proven; if s/he doesn't, there is not much value in intermediate results. The question is not very interesting. The only motivation is that it has been open for a long time. Also, there is not much to learn. Though, almost any theoretical question can be made flexible. We can start with the question above and change its direction to make it more promising and enticing.
Another bad example is a project where the research happens after the programs are written. This is bad because it is difficult to estimate the programming abilities of incoming students. It doesn't have an accessible beginning and there is no flexibility until the programming part is finished. If the student can't finish the programming quickly, s/he will not have time to look at the results and produce conjectures. For example, almost any project in studying social networks may fall into this category:
Study an acquaintance graph for some epic movies or fiction, for example Star Wars or The Lord of the Rings. In this graph people are vertices and two people are connected by an edge if they know each other. The project is to compare properties of such graphs to known properties of other social networks.
Though the networks in movies are much smaller than other networks that people study, the amount of programming might be substantial. This project can be a good project for a person with a flexible time frame or a person who is sure in advance that there will be enough time for him/her to look at the data.
Now on to an example of a good project. Lynnelle Ye and her mentor, Tirasan Khandhawit, chose to analyze the game of Chomp on graphs during RSI 2009.
Given a graph, on each turn a player can remove an edge or a vertex together with all adjacent edges. The player who doesn't have a move loses. This game was previously solved for complete graphs and forest graphs, so the project was to analyze the game for other types of graphs.
It is clear how to analyze the game for any particular new graph. So that could be a starting point providing an accessible beginning. After that the next step could be to analyze other interesting sets of graphs. The flexibility is guaranteed by the fact that there are many sets of graphs that can be used. In addition, the project entails learning some graph theory and game theory. And the project has a computational component.
Lynnelle Ye successfully implemented this project and provided a complete analysis of complete n-partite graphs for arbitrary n and all bipartite graphs. She also gave partial results for odd-cycle pseudotrees. The paper is available at the arxiv. Not surprisingly, Lynelle got fourth place in the Intel Science Talent Search and second place in the Siemens Competition.
Suppose that we choose all families with two children, such that one of them is a son named Luigi. Given that the probability of a boy to be named Luigi is p, what is the probability that the other child is a son?
Here is a potential "solution." Luigi is a younger brother's name in one of the most popular video games: Super Mario Bros. Probably the parents loved the game and decided to name their first son Mario and the second Luigi. Hence, if one of the children is named Luigi, then he must be a younger son. The second child is certainly an older son named Mario. So, the answer is 1.
The solution above is not mathematical, but it reflects the fact that children's names are highly correlated with each other.
Let's try some mathematical models that describe how the parents might name their children and see what happens. It is common to assume that the names of siblings are chosen independently. In this case the first son (as well as the second son) will be named Luigi with probability p. Therefore, the answer to the puzzle above is (2-p)/(4-p).
The problem with this model is that there is a noticeable probability that the family has two sons, both named Luigi.
As parents usually want to give different names to their children, many researchers suggest the following naming model to avoid naming two children in the same family with the same name. A potential family picks a child's name at random from a distribution list. Children are named independently of each other. Families in which two children are named the same are crossed out from the list of families.
There is a problem with this approach. When we cross out families we may disturb the balance in the family gender distributions. If we assume that boys' and girls' names are different then we will only cross out families with children of the same gender. Thus, the ratio of different-gender families to same-gender families will stop being 1/1. Moreover, it could happen that the number of boy-boy families will differ from the number of girl-girl families.
There are several ways to adjust the model. Suppose there is a probability distribution of names that is used for the first son. If another son is born, the name of the first son is crossed out from the distribution and following that we proportionately adjust the probabilities of all other names for this family. In this model the probability of naming the first son by some name and the second son by the same name changes. For example, the most popular name's probability decreases with consecutive sons, while the least popular name's probability increases.
I like this model, because I think that it reflects real life.
Here is another model, suggested by my son Alexey. Parents give names to their children independently of each other from a given distribution list. If they give the same name to both children the family is crossed-out and replaced with another family with children of the same genders. The advantage of this model is that the first child and the second child are named independently from each other with the same probability distribution. The disadvantage is that the probability distribution of names in the resulting set of families will be different from the probability distribution of names in the original preference list.
I would like my readers to comment on the models and how they change the answer to the original problem.
I am reading the book Eat to Live by Joel Fuhrman. It contains a formula that as a math formula doesn't make any sense. But as an idea, it felt like a revelation. Here it is:
HEALTH = NUTRIENTS/CALORIES
The idea is to choose foods that contain more nutrients per calorie. The formula doesn't make sense for many reasons. Taken to its logical conclusion, the best foods would be vitamins and tea. The formula doesn't provide bounds: it just emphasizes that your calories should be nutritious. However, too few calories — nutritious or not — and you will die. And too many calories — even super nutritious — are still too many calories. In addition the formula doesn't explain how to balance different types of nutrients.
Let's see why it was a revelation. I often crave bananas. I assumed that I need bananas for some reason and my body tells me that. Suppose I really need potassium. As a result I eat a banana, which contains 800 milligrams of potassium and adds 200 calories as a bonus. If I ate spinach instead, I would get the same amount of potassium at a price of only 35 calories.
The book suggests that if I start eating foods that are high in nutrients, I will satisfy my need for particular nutrients, and my cravings will subside. As a result I will not want to eat that much. If I start my day eating spinach, that might eliminate my banana desire.
I've been following an intuitive eating diet. I am trying to listen to my body hoping that my body will tell me what is better for it. It seems that my body sends me signals that are not precise enough. It's not that my body isn't communicating with me, but it is telling me "potassium" and all I hear is "bananas." What I need to do is use my brain to help me decipher what my body really, really wants to tell me.
As Dr. Fuhrman puts it, we are a nation of overfed and malnourished people. But Fuhrman's weight loss plan is too complicated and time-consuming for me, so I designed my own plan based on his ideas:
I will start every meal with vegetables, as they are the most nutritious. I hope that vegetables will provide the nutrients I need. That in turn will make me less hungry by the next meal, at which time I'll take in fewer calories. I will report to my readers whether or not my plan works. I'm off to shop for spinach. Will I ever love it as much as bananas?
The Cookie Monster is a peculiar creature that appeared in The Inquisitive Problem Solver (Vaderlind, Guy & Larson, MAA, P34). Presented with a set of cookie jars, the Cookie Monster will try to empty all the jars with the least number of moves, where a move is to select any subset of the jars and eat the same number of cookies from each jar in the subset.
Even an untalented Cookie Monster would be able to empty n jars in n moves: to fulfill this strategy the Monster can devour all the cookies of one jar at a time. If the Monster is lucky and some jars have the same number of cookies, the Monster can apply the same eating process to all these identical jars. For example, if all the jars have the same number of cookies, the Monster can gulp down all of them in one swoop.
Now, let us limit our discussion to only cases of n non-empty jars that contain distinct numbers of cookies. If indeed all the numbers are distinct, can the Monster finish eating faster than in n moves?
The answer depends on the actual number of cookies in each jar. For example, if the number of cookies in jars are different powers of 2, then even the most talented Monsters can't finish faster than in n steps. Indeed, suppose the largest jar contains 2N cookies. That would be more than the total number of cookies in all the other jars together. Therefore, any strategy has to include a step in which the Monster only takes cookies from the largest jar. The Monster will not jeopardize the strategy if it takes all the cookies from the largest jar in the first move. Applying the induction process, we see that we need at least n steps.
On the other hand, sometimes the Monster can finish the jars faster. If 2k−1 jars contain respectively 1, 2, 3, …, 2k−1 cookies, the Cookie Monster can empty them all in k steps. Here is how. For its first move, the Monster eats 2k-1 cookies from each of the jars containing 2k-1 cookies or more. What remains are 2k-1−1 pairs of identical non-empty jars containing respectively 1, 2, 3, …, 2k-1−1 cookies. The Monster can then continue eating cookies in a similar fashion, finishing in k steps. For instance, for k=3 the sequences of non-empty jars are: 1,2,3,4,5,6,7 → 1,1,2,2,3,3 → 1,1,1,1 → all empty.
Now we would like to prove a theorem that shows that the example above is the lowest limit of moves even for the most gifted Cookie Monsters.
Theorem. If n non-empty jars contain distinct numbers of cookies, the Cookie Monster will need at least ⌈log2(n+1)⌉ steps to empty them all.
Proof. Suppose that n jars contain distinct numbers of cookies and let f(n) be the number of distinct non-empty jars after the first move of the Cookie Monster. We claim that n ≤ 2f(n)+1. Indeed, after the first move, there will be at least n − 1 non-empty jars, but there cannot be three identical non-empty jars. That means, the number of jars plus 1 can't decrease faster than twice each time.
Now here is something our readers can play with. Suppose a sequence of numbers represents the number of cookies in the jars. Which sequences are interesting, that is, which can provide interesting solutions for the Cookie Monster problem?
I read an interesting article on the paradoxes involved in allocating seats for the Congress. The problem arises because of two rules: one congressperson has one vote, and the number of congresspeople per state should be proportional to the population of said state.
These two rules contradict each other, because it is unrealistic to expect to be able to equally divide the populations of different states. Therefore, two different congresspeople from two different states may represent different sizes of population.
Let me explain how seats are divided by using as an example a country with three states: New Nevada (NN), Massecticut (MC) and Califivenia (C5). Suppose the total number of congresspeople is ten. Also suppose the population distribution is such that the states should have the following number of congresspeople: NN — 3.33, MC — 3.34 and C5 — 3.33. As you know states generally do not send a third of a congressperson, so the situation is resolved using the Hamilton method. First, each state gets an integer portion of the seats. In my example, each state gets three seats. Next, if there are seats left they are allocated to states with the largest remainders. In my example, the remainders are 0.33, 0.34 and 0.33. As Massecticut has the largest reminder it gets the last seat.
This is not fair, because now each NN seat represents a larger population portion than each MC seat. Not only is this not fair, but it can also create some strange situations. Suppose there have been population changes for the next redistricting: NN — 3.0, MC — 3.4 and C5 — 3.6. In this case, NN and MC each get 3 seats, while C5 gets the extra seat for a total of 4. Even though MC tried very hard and succeeded in raising their portion of the population, they still lost a seat.
Is there any fair way to allocate seats? George Szpiro in his article suggests adding fractional congresspersons to the House of Representatives. So one state might have three representatives, but one of those has only a quarter of a vote. Thus, the state's voting power becomes 2 1/4.
We can take this idea further. We can use the Hamilton method to decide the number of representatives per state, but give each congressperson a fractional voting power, so the voting power of each state exactly matches the population. This way we lose one of the rules that each congressperson has the same vote. But representation will be exact. In my first example, NN got three seats, when they really needed 3.33. So each congressperson from New Nevada will have 1.11 votes. On the other hand MC got four seats, when they needed 3.34. So each MC representative gets 0.835 votes.
Continuing with this idea, we do not need congresspeople from the same state to have the same power. We can give proportional voting power to a congressperson depending on the population in his/her district.
Or we can go all the way with this idea and lose the districts altogether, so that every congressperson's voting power will be exactly proportionate to the number of citizens who voted for him/her. This way the voting power will reflect the popularity — rather than the size of the district — of each congressperson.
Lionel Levine invented a new hat puzzle.
The sultan decides to torture his hundred wise men again. He has an unlimited supply of red and blue hats. Tomorrow he will pile an infinite, randomly-colored sequence of hats on each wise man's head. Each wise man will be able to see the colors of everyone else's hats, but will not be able to see the colors of his own hats. The wise men are not allowed to pass any information to each other.
At the sultan's signal each has to write a natural number. The sultan will then check the color of the hat that corresponds to that number in the pile of hats. For example, if the wise man writes down "four," the sultan will check the color of the fourth hat in that man's pile. If any of the numbers correspond to a red hat, all the wise men will have their heads chopped off along with their hats. The numbers must correspond to blue hats. What should be their strategy to maximize their chance of survival?
Suppose each wise man writes "one." The first hat in each pile is blue with a probability of one-half. Hence, they will survive as a group with a probability of 1 over 2100. Wise men are so wise that they can do much better than that. Can you figure it out?
Inspired by Lionel, I decided to suggest the following variation:
This time the sultan puts two hats randomly on each wise man's head. Each wise man will see the colors of other people's hats, but not the colors of his own. The men are not allowed to pass any info to each other. At the sultan's signal each has to write the number of blue hats on his head. If they are all correct, all of them survive. If at least one of them is wrong, all of them die. What should be their strategy to maximize their chance of survival?
Suppose there is only one wise man. It is clear that he should write that he has exactly one blue hat. He survives with the probability of one-half. Suppose now that there are two wise men. Each of them can write "one." With this strategy, they will survive with a probability of 1/4. Can they do better than that? What can you suggest if, instead of two, there is any number of wise men?
* * *
Today I saw an ad — "A printer for sale" — handwritten. Hmm.
* * *
What do you call a motherboard on your spouse's computer?
The motherboard-in-law.
In a puzzle book by Mari Berrondo (in Russian), I found the following logic problem:
Alfred, Bertran and Charles are asked about their profession. One of them always lies; another one always tells the truth; and the third one [who I will refer to as a "half-liar"] sometimes lies and sometime tells the truth. Here are their answers:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The solution in the book is the following. As Alfred is right about what Bertran would say, Alfred can't be a liar. If Alfred is a half-liar then the other two people would give the opposite statements, since one will be a truth-teller and the other a liar. But they both say that Alfred is a piano-tuner, therefore Alfred must be a truth-teller. Hence, Alfred's statement about everyone's profession must be the truth. Now we know that Charles is an insurance agent. As Charles confirms that, thus telling the truth in this instance, we recognize that he must be a half-liar. The answer to the problem is that the half-liar is an insurance agent.
But I have a problem with this problem. You see, a liar can say many things. He can say that he is a conductor, a mathematician, a beekeeper or whatever. So there is no way of knowing what a person who decides to lie can say. Let's just analyze the statement by Alfred: "Concerning Bertran, if you ask him, he will tell you that he is a painter."
If Alfred tells the truth about what Bertran would say, he needs to know for sure that Bertran will say that he is a painter. Hence, Bertran must be a truth-teller and a painter. If Alfred lies, he needs to be sure that Bertran won't say that he is a painter. So Bertran must be either a truth-teller and not a painter, or a liar and a painter. Bertran can't be a half-liar, because a half-liar can say that he is a painter as well as he can say something else, no matter what his real profession.
There is one interesting aspect of this that many people overlook. There are different types of people who are half-liars. In some books half-liars are introduced as people who, before making a statement, flip a coin to decide whether to lie or to tell the truth. Such a person needs to know in advance exactly what other people are saying, in order to construct a statement about what those people might say that corresponds to the coin flip. On the other hand, other types of half-liars exist. One half-liar can say something and then see later whether it is true. If Alfred is a half-liar who doesn't care in advance about the truth of his statement, he can say that Bertran will claim that he is a painter.
I leave it to my readers to finish my analysis and see that the problem doesn't have a solution. To end my essay on a positive note, I decided to slightly change the problem, so that there is no contradiction. In the same setting:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is not a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The data from annual surveys carried out by the American Mathematical Society shows the same picture year after year: the percentage of females in different categories decreases as the category level rises. For example, here is the data for 2006:
| Category | Percentage of Women |
|---|---|
| Graduating Math Majors | 41 |
| PhDs Granted | 32 |
| Fresh PhD hires in academic jobs | 27 |
| Full-time Faculty | 27 |
| Full-time tenured or tenure-track faculty | 12 |
The high percentage of female math majors means that a lot of women do like mathematics. Why aren't women becoming professors of mathematics? In the picture to the left, little Sanya fearlessly took her first integral. I hope, even as an adult, she will never be afraid of integrals.
I am one of the organizers of the Women and Mathematics Program at the Institute for Advanced Study at Princeton In 2009 we had a special seminar devoted to discussing this issue. Here is the report of our discussion based on the notes that Rajaa Al Talli took during the meeting.
Many of us felt, for the following three reasons, that the data doesn't represent the full picture.
First, the different stages correspond to women of different ages; thus, the number of tenured faculty should be compared, not to the number of current math majors, but rather to women who majored in math many years ago. The percentage of female PhDs in mathematics has been increasing steadily for the past several years. As a result, we expect an eventual increase in the number of full-time female faculty.
Second, international women mathematicians might be having a great impact on the numbers. Let's examine a hypothetical situation. If many female professors come to the US after completing their studies in other countries, it would be logical to assume that they would raise the numbers. But since the numbers are falling, we might be losing more females than we think. Or, it could be the opposite: international graduate students complete a PhD in mathematics in the USA and then go back to their own countries. In this case we would be losing fewer females to professorships than the numbers seem to suggest. Unfortunately, we can't really say which case is true as we do not know the data on international students and professors.
Third, many women who major in mathematics also have second majors. For example, the women who have a second major in education probably plan to become teachers instead of pursuing an academic career. It would be interesting to find the data comparing women who never meant to have careers in science with those women who left because they were discouraged. If we are losing women from the sciences because they decide not to pursue scientific careers, then at least that is their choice.
It is also worth studying why so few women are interested in careers in mathematics in the first place. Changing our culture or applying peer pressure in a different direction might change the ambitions of a lot of people.
We discussed why the data in the table doesn't represent the full picture. On the other hand, there are many reasons why women who can do mathematics and want to do mathematics might be discouraged from pursuing an academic career:
Our group proposed many solutions to help retain women in mathematics:
At the end of our meeting, everyone accepted Ingrid Daubechies' proposal that we do the following:
Each woman in mathematics should take as her responsibility the improvement of the mathematical environment in which she works. If every woman helps change what's going on in her university or the school where she teaches, that will help solve the problem on the larger scale.
The book How to Drive Your Man Wild in Bed
The list of bad features also includes professions to avoid. Can you guess the first profession on the list? OK, I think you should be able to meta-guess given the fact that I am writing about it. Indeed, the list on page 64 starts:
Avoid, on the whole, mathematicians…
I am an expert on NOT avoiding mathematicians: in fact, I've married three of them and dated x number of them. That isn't necessarily because I like mathematicians so much; I just do not meet anyone else.
When I was a student I had a theory that mathematicians are different from physicists. My theory was based on two conferences on mathematical physics I attended in a row. The first one was targeted for mathematicians and the second for physicists. The first one was very quiet, and the second one was all boozing and partying. So I decided that mathematicians are introverts and physicists are extroverts. I was sure then that my second husband chose a wrong field, because he liked booze and parties.
By now, years later, I've met many more mathematicians, and I have to tell you that they are varied. It is impossible and unfair to describe mathematicians as a type. One mathematician even became the star of an erotic movie. I write this essay for girls who are interested in dating mathematicians. I am not talking about math majors here, I am talking about mathematicians who do serious research. Do I have a word of advice?
I do have several words of caution. While they don't apply to all mathematicians, it's worth keeping them in mind.
First, there are many mathematicians who, like my first husband, are very devoted to mathematics. I admire that devotion, but it means that they plan to do mathematics on Saturday nights and prefer to spend vacation at their desks. If they can only fit in one music concert per year, it is not enough for me. Of course, this applies to anyone who is obsessed by his work.
Second, there are mathematicians who believe that they are very smart. Smarter than many other people. They expand their credibility in math to other fields. They start going into biology, politics and relationships with the charisma of an expert, when in fact they do not have a clue what they are talking about.
Third, there are mathematicians who enjoy their math world so much that they do not see much else around them. The jokes are made about this type of mathematician:
What is the difference between an extroverted mathematician and an introverted one? The extroverted one looks at your shoes, rather than at his own shoes.
Yes, I have met a lot of mathematicians like that. Do you think that their wives complain that their husbands do not notice their new haircuts? No. Such triviality is not worth mentioning. Their wives complain that their husbands didn't notice that the furniture was repossessed or that their old cat died and was replaced by a dog. My third husband was like that. At some point in my marriage I discovered that he didn't know the color of my eyes. He didn't know the color of his eyes either. He wasn't color-blind: he was just indifferent. I asked him as a personal favor to learn the color of my eyes by heart and he did. My friend Irene even suggested creating a support group for the wives of such mathematicians.
While you need to watch out for those traits, there are also things I like about mathematicians. Many mathematicians are indeed very smart. That means it is interesting to talk to them. Also, I like when people are driven by something, for it shows a capacity for passion.
Mathematicians are often open and direct. Many mathematicians, like me, have trouble making false statements. I stopped playing —Mafia— because of that. I prefer people who say what they think and do not hold back.
There is a certain innocence among some mathematicians, and that reminds me of the words of the Mozart character in Pushkin's poetic drama, Mozart and Salieri: —And genius and villainy are two things incompatible, aren't they?— I feel this relates to mathematicians as well. Many mathematicians are so busy understanding mathematics, they are not interested in plotting and playing games.
Would I ever date a mathematician again? Yes, I would.
33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?
The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.
Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.
First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.
As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.
Let's look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.
Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.
We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.
Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.
If we continue this, we'll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.
We get two interesting new sequences out of this. The sequence of the fastest rider's speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.
The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.
I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.
One day we may all face the necessity of hiring a lawyer. If the case is tricky the lawyer must be smart and inventive. I am collecting puzzles to give to a potential lawyer during an interview. The following puzzle is one of them. It was given at the second Euler Olympiad in Russia:
At a local Toyota dealership, you are allowed to exchange brand new cars. You can exchange three Camrys for one Prius and one Avalon, and three Priuses for two Camrys and one Avalon. Assuming an unlimited supply of cars at the dealership, can collector Vasya exchange 700 Camrys for 400 Avalons?
The beauty of this puzzle is that the answer I may find acceptable from a mathematician is not the same as I want from my future lawyer.
Have I intrigued you? Get to work and send me the solutions.
I am sitting in front of my computer and scheming, or, more precisely, scamming. I am inventing scams as a way of raising awareness of how probability theory can be used for deception.
My first scam is my lottery project. Suppose I create and run a private lottery. I will award minor payments to some participants, while promising a grand prize of one hundred million dollars. However, there will be a very small probability that anyone will win the big payout. My plan is to live lavishly on my proceeds, hoping no one ever wins the big ticket.
The beauty of this scheme is that nobody will complain until someone scores the top prize. After all, everyone has been receiving what I promised, and no one realizes my fraud. If nobody wins the big award until I retire, I will have built my life style on deception without having been caught.
Suppose someone wins the hundred million dollars. Oops. I am in big trouble. On the other hand, maybe I can avoid jail time. I could tell the winner that the money is gone and if s/he complains to the police, I will declare bankruptcy and we will all lose. Alternatively, I can suggest a settlement in exchange for silence. For example, we could share future proceeds. Probability theory will help me run this lottery with only a small chance of being exposed.
But even a small chance of failure will cause me too much stress, so I have come up with an idea for another scam. I will write some complicated mathematical formulas with which to persuade everyone that global warming will necessarily produce earthquakes in Boston in the near future. Then I'll open an insurance company and insure everyone against earthquakes. As I really do not expect earthquakes in my lifetime, I can spend the money. I'll just need to keep everyone scared about earthquakes. This time I can be sure that I won't be caught as no one will have a reason to complain. The only danger is that someone will check my formulas and prove that I used mathematics to lie.
Perhaps I need a scam that covers up the lie better. Instead of inventing an impossible catastrophe, I need to insure against a real but rare event. Think Katrina. I collect the money and put aside money for payouts and pocket the rest. But I actually tweak my formulas and put aside less than I should, boosting my bank account. I will be wealthy for many years, until this event happens. I might die rich but if this catastrophe happens while I'm still alive, I'll declare bankruptcy.
Though I was lying to everyone, I might be able to avoid jail time. I might be able to prove that it was an honest mistake. Mathematical models include some subjective parameters; besides, everyone believes that nature is unpredictable. Who would ever know that I rigged my formulas in my favor? I can claim that the theory ended up being more optimistic than reality is. Who could punish me for optimism?
Maybe I can be accused of lying if someone proves that I knew that the optimistic model doesn't quite match the reality. But it is very difficult for the courts to punish a person for a math mistake.
When I started writing this essay, I wanted to write about the financial crisis of 2008. I ended up inventing scams. In a way, I did write about the financial crisis. My scams are simplified versions of what banks and hedge funds did to us. Will we ever see someone punished?
Most movies related to mathematics irritate me because of simplifications. I especially do not like when a movie pretends to be intelligent and then dumbs it down. I recently watched the Spanish movie Fermat's Room
The movie opens with people receiving invitations to attend a meeting for geniuses. To qualify for the meeting they need to solve a puzzle. Within ten days, they must guess the order underlying the following sequence: 5, 4, 2, 9, 8, 6, 7, 3, 1. Right away, at the start of the movie, I was already annoyed because of the simplicity of the question. You do not have to be a genius to figure out the order, not to mention how easy it would be to plug this sequence into the Online Encyclopedia of Integer Sequences to find the order in five minutes.
The participants were asked to hide their real names, which felt very strange to me. All famous puzzle solvers compete in puzzle championships and mystery hunts and consequently know each other.
The meeting presumably targets the brightest minds and promises to provide "the greatest enigma." During the meeting they are given seven puzzles to solve. All of them are from children's books and the so-called "greatest enigma" could easily be solved by kids. Though I have to admit that these were among the cutest puzzles I know. For example:
There are three boxes: one with mint sweets, the second with aniseed sweets, and the last with a mixture of the two. The boxes are labeled, but all the labels are wrong. What is the minimum number of sweets you need to taste to correctly re-label all the boxes?
Another of the film's puzzles includes a light bulb in a room and three switches outside, where you have to correctly find the switch that corresponds to the bulb, but you can only enter the room once. In another puzzle you need to get out of prison by deciding which of two doors leads to freedom. You are allowed to ask exactly one question to one of the two guards, one of whom is a truth-teller and the other is a liar.
The other four puzzles are similar to these three I have just described. To mathematicians they are not the greatest enigmas. They are nice material for a children's math club. For non-mathematicians, they may be fascinating. Certainly it's a good thing that such tasteful puzzles are being promoted to a large audience. But they just look ridiculous as "the greatest enigmas."
So what is it about this film that I so enjoyed?
The intensity of the movie comes from the fact that the people are trapped in a room that starts shrinking when they take more than one minute to solve a puzzle.
I well remember another shrinking room from Star Wars: A New Hope. When Princess Leia leads her rescuers to a room, it turns out to be a garbage compactor. The bad guys activate the compactor and two opposite walls start moving in. In contrast, Fermat's room is shrinking in a much more sophisticated way: all four walls are closing in. Each of the walls in the rectangular room is being pressured by an industrial-strength press. The walls in the corners do not crumble, but rather one wall glides along another. I was more puzzled by this shrinking room than I was by the math puzzles. I recommend that you try to figure out how this can be done before seeing the movie or its poster.
However, the best puzzle in the movie is the plot itself. Though I knew all the individual puzzles, what happened in between grabbed me and I couldn't wait to see what would happen next. I saw the movie twice. After the first time, I decided to write this review, so I needed to check it again. I enjoyed it the second time even better than the first time. The second time, I saw how nicely the plot twists were built.
Maybe I shouldn't complain about the simplicity and the familiarity of the puzzles. If they were serious new puzzles I would have started solving them instead of enjoying the movie. The film's weakness might be its strength.
I collect hats puzzles. A puzzle about hats that I hadn't heard before appeared on the Konstantin Knop's blog (in Russian):
The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat — for both of which he has an inexhaustible supply — on every wizard's head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan's signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide on a strategy to maximize the number of survivors. Suggest a strategy for them.
I'll start the discussion with a rather simple idea: Each wizard writes down a color randomly. In this case the expected number of survivors is 50. Actually, if each wizard writes "red", then the expected number of survivors is 50, too. Can you find a better strategy, with a greater expected number of survivors or prove that such a strategy doesn't exist?
As a bonus question, can you suggest a strategy that guarantees 50 survivors?
Now that you've solved that issue, here's my own variation of the problem.
The wizards are all very good friends with each other. They decide that executions are very sad events and they do not wish to witness their friends' deaths. They would rather die themselves. They realize that they will only be happy if all of them survive together. Suggest a strategy that maximizes the probability of them being happy, that is, the probability that all of them will survive.
The following two problems appeared together in Martin Gardner's Scientific American column in 1959.
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Many people, including me and Martin Gardner, wrote a lot about Mr. Smith. In his original column Martin Gardner argued that the answer to the first problem is 1/3. Later he wrote a column titled "Probability and Ambiguity," where he corrected himself about Mr. Smith.
… the answer depends on the procedure by which the information "at least one is a boy" is obtained.
This time I would like to ignore Mr. Smith, as I wrote a whole paper about him that is now under consideration for publication at the College Mathematics Journal. I would rather get back to Mr. Jones.
Mr. Jones failed to stir a controversy from the start and was forgotten. Olivier Leguay asked me about Mr. Jones in a private email, reminding me that the answer to the problem about his children also depends on the procedure.
One of the reasons Mr. Jones was forgotten is that for many natural procedures the answer is 1/2. For example, the following procedures will produce an answer of 1/2:
There are many other procedures that lead to the answer 1/2. However, there are many procedures that lead to other answers.
Suppose I know Mr. Jones, and also know that he has two children. I meet Mr. Jones at a mall, and he tells me that he is buying a gift for his older daughter. Most probably I would assume that the other child is a daughter, too. In my experience, people who have a son and a daughter would say that they are buying a gift for "my daughter." Only people with two daughters would bother to specify that they are buying a gift for "my older daughter."
In some sense I didn't forget about Mr. Jones. I wrote about him implicitly in my essay Two Coins Puzzle. His name was Carl and he had two coins instead of two children.
My blog is getting more famous. Now I don't need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov's Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:
Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?
Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:
Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?
Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.
86 is conjectured to be the largest power of 2 not containing a zero. This simply stated conjecture has proven itself to be proof-resistant. Let us see why.
What is the probability that the nth power of two will not have any zeroes? The first and the last digits are non-zeroes; suppose that other digits become zeroes randomly and independently of each other. This supposition allows us to estimate the probability of 2n not having zeroes as (9/10)k-2, where k is the number of digits of 2n. The number of digits can be estimated as n log102. Thus, the probability is about cxn, where c = (10/9)2 ≈ 1.2 and x = (9/10)log102 ≈ 0.97. The expected number of powers of 2 without zeroes starting from the power N is cxN/(1-x) ≈ 40 ⋅ 0.97N.
Let us look at A007377, the sequence of numbers such that their powers of 2 do not contain zeros: 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86. Our estimates predicts 32 members of this sequence starting from 6. In fact, the sequence has 30 conjectured members. Similarly, our estimate predicts 2.5 members starting from 86. It is easy to check that the sequence doesn't contain any more numbers below 200 and our estimate predicts 0.07 members after 200. As we continue checking larger numbers and see that they do not belong to the sequence, the probability that the sequence contains more elements vanishes. With time we check more numbers and become more convinced that the conjecture is true. Currently, it has been checked up to the power 4.6 ⋅ 107. The probability of finding something after that is about 1.764342396 ⋅10-633620.
Let us try to approach the conjecture from another angle. Let us check the last K digits of powers of two. As the number of possibilities is finite, these last digits eventually will start cycling. If we can show that all the elements inside the period contain zeroes, then we need to check the finite number of powers of two until this period starts. If we can find such K, we can prove the conjecture.
Let us look at the last two digits of powers of two. The sequence starts as: 01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04. As we would anticipate, it starts cycling. The cycle length is 20, and 90% of numbers in the cycle don't have zeroes.
Now let's continue to the last three digits. The period length is 100, and 19 of them either start with zero or contain zero. The percentage of elements in the cycle that do not contain zero is 81%.
The cycle length for the last n digits is known. It is 4 ⋅ 5n-1. In particular the cycle length grows by 5 every time. The number of zero-free elements in these cycles form a sequence A181610: 4, 18, 81, 364, 1638, 7371, 33170. If we continue with our supposition that the digits are random, and study the new digits that appear when we move from the cycle of the last n digits to the next cycle of the last n+1 digits, we can expect that 9/10 of those digits will be non-zero. Indeed, if we check the ratio of how many numbers do not contain zero in the next cycle compared to the previous cycle, we get: 4.5, 4.5, 4.49383, 4.5, 4.5, 4.50007. All of these numbers are quite close to our estimation of 4.5. If this trend continues the portion of the numbers in the cycle that don't have zeroes tends to zero; however, the total of such numbers grows exponentially. We can even estimate that the expected growth is 4 ⋅ 4.5n-1. From this estimation, we can derive the conjecture:
Conjecture. For any number N, there exists a power of two such that its last N digits are zero-free.
Indeed, the last N digits of powers of two cycle, and there are an increasing number of members inside that cycle that do not contain zeroes. The corresponding powers of two don't have zeroes among N rightmost digits.
So, how do we combine the two results? First, the expected probability of finding the power of two larger than 86 that doesn't contain zero is minuscule. And second, we most certainly can find a power of two that has as many zeroless digits at the end as we want.
To combine the two results, let us look at the sequences A031140 and A031141. We can deduce from them that for the power 103233492954 the first zero from the right occupies the 250th spot. The total number of digits of that power is 31076377936. So 250 is a tiny portion of the digits.
As time goes by we grow more and more convinced that 86 is the largest power of two without zeroes, but it is not at all clear how we can prove the conjecture or whether it can be proven at all.
My son, Sergei, suggested that I claim that I have a proof of this conjecture, but do not have enough space in the margin to fit my proof in. The probability that I will ever be shamed and disproven is lower than the probability of me winning a billion dollars in the lottery. Though then, if I do win the big bucks, I will still care about being shamed and disproven.
Janet Mertz encouraged me to find IMO girls and compare their careers to that of their teammates. I had always wanted to learn more about the legendary Lida Goncharova — who in 1962 was the first girl to win an IMO gold medal. So I located her, and after an interview, wrote about her. Only 14 years later, in 1976, did the next girl get a gold medal. That was me. I was ranked overall second and had 39 points out of 40.
As I did in the article about Lida, I would like to compare my math career to that of my teammates.
I got my PhD in 1988 and moved to the US in 1990. My postdoc at MIT in 1993 was followed by a postdoc at Bar-Ilan University. In 1996 I got a non-paying visiting position at Princeton University. In 1998 I gave up academia and moved to industry, accepting an offer from Bellcore. There were many reasons for that change: family, financial, geographical, medical and so on.
On the practical level, I had had two children and raising them was my first priority. But there was also a psychological element to this change: my low self-esteem. I believed that I wasn't good enough and wouldn't stand a chance of finding a job in academia. Looking back, I have no regrets about putting my kids first, but I do regret that I wasn't confident enough in my abilities to persist.
I continued working in industry until I resigned in January 2008, due to my feeling that I wasn't doing what I was meant to do: mathematics. Besides, my children were grown, giving me the freedom to leave a job I did not like and return to the work I love. Now I am a struggling freelance mathematician affiliated with MIT. Although my math blog is quite popular and I have been publishing research papers, I am not sure that I will ever be able to find an academic job because of my non-traditional curriculum vitae.
The year 1976 was very successful for the Soviet team. Out of nine gold medals our team took four. My result was the best for our team with 39 points followed by Sergey Finashin and Alexander Goncharov with 37 points and by Nikita Netsvetaev with 34 points.
Alexander Goncharov became a full professor at Brown University in 1999 and now is a full professor at Yale University. His research is in Arithmetic Algebraic Geometry, Teichmuller Theory and Integral Geometry. He has received multiple awards including the 1992 European Math Society prize. Sergey Finashin is very active in the fields of Low Dimensional Topology and Topology of Real Algebraic Varieties. He became a full professor at Middle East Technical University in Ankara, Turkey in 1998. Nikita Netsvetaev is an expert in Differential Topology. He is a professor at Saint Petersburg State University and the Head of the High Geometry Department.
Comparing my story to that of Lida, I already see a pattern emerging. Now I'm curious to hear the stories of other gold-winning women. I believe that the next gold girl, in 1984, was Karin Gröger from the German Democratic Republic. I haven't yet managed to find her, so can my readers help?
I wrote a series of essays about AMC competitions:
This essay is next in the series. Although it is not strictly about AMC, it should be useful during any test when you need to check your answers. There are several important rules which are helpful.
Rule 0. Checking is important. If wrong answers are punished, then correcting a mistake brings more points than solving a new problem. In addition, problems that were solved are often easier than problems yet to be solved, so finding a mistake might be faster than solving a new problem.
Rule 1. Your checking methods must be fast. The tests are generally timed. This means that in order to check your answers, you need to sacrifice your work on the next problem.
Rule 2. Customize how you check according to your strengths and weaknesses. For example, if you tend to jump to conclusions about what the question is going to be, and as a result answer your anticipated question instead of the one that is actually on the test, then when you are checking you should start reading the problem from the question. Or, if you usually make mistakes in geometry problems, you should allocate more time to geometry problems when you are checking. If you never make mistakes in arithmetic problems then you do not need to check those.
Rule 3. Mark problems that might need checking. If you do not have enough time to check all the problems, check only those you are not sure about.
Rule 4. Do not repeat your solution when you check. While solving the problem your brain often creates a pathway from start to finish. If on this pathway your brain decided to believe that two plus two is five, very often during checking, your brain will make the same mistake again. Because of that it is crucial to use other methods for checking than repeating your reasoning. In case you can't find a way to check your answers using a different method and have to repeat your reasoning, you should repeat it in a different order.
This rule is so important, that I am providing some methods to change your brain pathway when you are checking your answers.
Plug in. Plugging in the answer you found is much faster than finding it. Use this method whenever possible. It is perfect for problems like this one below from 2004 AMC10-A:
What is the value of x if |x – 1| = |x – 2|?
Plug in an intermediate result. Sometimes you can't plug in the answer, but you can plug in an intermediate result. In the following problem from 2004 AMC10-B you can plug in the number of nickels and dimes:
Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?
Calculate something else related to your answer. For example a negation. Here is a problem from 2004 AMC10-B:
How many two-digit positive integers have at least one 7 as a digit?
If you calculated the answer directly, to check it you may want to calculate the number of two-digit positive integers that do not contain 7.
Create an example. Sometimes you solve a problem by reasoning, but to check it you might create a particular example. Here is a problem from 2001 AMC10:
Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N?
If we denote the tens digit by a and the units digit by b, then N = 10a + b, P(N) = a*b, and S(N) = a + b. We get an equation a(b+1) = 10a, from which the answer is 9. To check the answer we do not need to repeat the reasoning. It is enough to check that 19 is the sum of the product of its digits plus the digits.
Here is another problem from 2001 AMC10:
Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?
The list of choices is: 6, 14, 21, 28, 42. Your solution might go like this: the product of three consecutive numbers is divisible by 6. Hence, n is divisible by 42. So, the answer must be 28. To check you might consider a product of three consecutive numbers: 5*6*7=210 and see that it is not divisible by 4, hence it is not divisible by 28.
Rule 5. Embrace the partial check. It is very important to check your answers fast. Sometimes you can gain speed if you do not check the problem completely, but check it partially. For example, you can check that your answer is one of the two correct answers. There are many methods for partial checking.
Try an example. Sometimes an example doesn't guarantee that your choice is correct, but it increases your confidence in your answer. Here is another problem from 2001 AMC10:
The sum of two numbers is S. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
The choices are: 2S + 3, 3S + 2, 3S + 6, 2S + 6, 2S + 12. You can reason that increasing each summand by 3, increases the sum by 6. After that doubling each summand increases the resulting sum twice, so the answer is 2S + 12. To check the answer you can use an example. Usually an example doesn't guarantee the confirmation of your answer, but it might help you eliminate some of the wrong answers. For example, if you choose zero and zero as your initial two numbers, then S = 0, and your transformation brings the result to 12, which confirms your answer 2S + 12. In this particular case, a very easy specific example excluded all the wrong answers.
Divisibility. Sometimes it is faster to calculate the remainder of the answer by some number.
For example, look at the following problem from 2003 AMC10:
What is the units digit of 132003?
The choices are 1, 3, 7, 8, 9. We can immediately say that the answer must be an odd number.
Approximation check. One important example of a partial check is an approximation check. By estimating an approximate answer you might exclude most of the wrong answers. Consider this problem from 2001 AMC12:
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
The divisibilities by 3, 4 or 5 shouldn't correlate with each other. Approximately one third of those number are multiples of 3 and one quarter are multiples of 4. Let's say that one twelfth are multiples of both 3 and 4. Hence, we estimate the portion of numbers that are multiples of 3 or 4 as 1/3 + 1/4 – 1/12 = 1/2. We have about 1,000 such numbers. The number of numbers that are, in addition, not divisible by 5, are less than that. So out of the given choice of (A) 768, (B) 801, (C) 934, (D) 1067, (E) 1167, we can immediately confirm that the answer is among the first three.
The methods above can be useful even if you do not have multiple choices. But if you do…
Rule 6. Use given choices as extra information. In the previous examples you saw how to use a partial check to exclude some of the choices. Here is a specific example from 2006 AMC10-A of how to exclude choices:
What non-zero real value for x satisfies (7x)14 = (14x)7?
The choices are: 1/7, 2/7, 1, 7, 14. If you solved the problem directly, to check it you can reason why other choices do not work. In this particular case it can be done very fast. 1/7 doesn't work because the left part of the equation becomes 1 when the right is clearly not. 1 and 7 do not work because the left part is odd and the right is even; 14 doesn't work because the left is clearly bigger than the right.
Rule 7. Use meta considerations. If you get into the mind of the designers you can better anticipate when you should check more thoroughly. Consider this problem from 2006 AMC10-A:
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
The most common mistake would be to assume that 12:59 supplies the largest sum, which is 17. But look at the choices: 17, 19, 21, 22, 23. When the designers are asking to find the largest number with some property, they assume that some students will make a mistake and chose a smaller number over a larger one. That means the designers would include this potential mistake among the choices. So the answer is extremely unlikely to be the smallest number on the list of choices. Thus, if you think the answer is 17, understanding how these problems are constructed should alert you to thoroughly check your answer. Indeed, the correct answer is 23 which corresponds to 9:59. Not surprisingly, it is the largest on the list of choices.
AMC 10/12 is coming on February 8 and HMMT on February 12. Happy checking.
I've been celebrating my 29th birthday for many years. Once, when I was actually 45 and wanted to have a big party, I invited everyone to the 5th anniversary of my 29th birthday.
Last week my son turned 29 and I realized that it is time to drop this beautiful, prime, evil, deficient, lazy-caterer number, that in addition is the largest power of two to have all different digits. No more celebrating 29.
For my next age, I picked 42. Not because it is the smallest abundant odious number, but rather because it is the answer to life, the universe and everything.
Thank you everyone who congratulated me on my birthday two days ago. For your information, from now on I am 42.
As I did for 2010 and for previous years, here are math-related puzzles from the MIT Mystery Hunt 2011.
Two more puzzles deserve a special mention for their nerdiness. My teammates loved them.
Tired of the same old sudoku? Here's an opportunity to try many variations of it. Thomas Snyder and Wei-Hwa Huang wrote a book called Mutant Sudoku
The book contains about 180 fun puzzles. Look at the variety:
Wei-Hwa Huang kindly sent me this sample Thermo Sudoku puzzle from the book to use on my blog. The grey areas represent thermometers. Every particular thermometer has to have numbers in increasing order (not necessarily consecutive) starting from the bulb.
The second book by the same authors Sudoku Masterpieces: Elegant Challenges for Sudoku Lovers
I bought both books and immediately started scribbling in the first one. My bad handwriting would seem so out of place in the beautiful second book that I have not even started working in it yet. Maybe I will give it as a gift to someone with better penmanship.
Two days ago I threw at my readers the following problem:
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
The purpose of throwing this problem was to discuss the nature of the implicit assumptions that we are asked to make when solving math problems, and the implicit assumptions we teach our children to make when we teach them to solve math problems. This is especially important for problems like this, that are phrased in terms of a situation in the real world. The real world is too complex to model all of; the great power of mathematics is that sufficiently idealized situations are predictable. But which idealizations are appropriate? How does one choose? How does one teach youngsters what to choose?
Before I get to the actual discussion, however, I want to re-throw this problem at my readers, in an effort to highlight what originally jumped out at me as being wrong with it.
Neglecting the effects of altitude, differential wind, acceleration, relativity, measurement error, finite size and non-superimposability of the planes, and the Earth's deviations from perfect sphericity,
- Find how much time it takes them to become 2000 miles apart, assuming that the planes are starting from Boston and the distance is measured as
- a straight line in 3-space.
- the shortest surface distance.
- How far from the closest pole may the starting point be located, so that the answer to the problem is "never"? Solve separately for
- the 3D distance.
- the shortest surface distance.
- What portion of the Earth's surface do the "never"-locations of the previous question occupy?
- under the 3D distance?
- under the shortest surface distance?
Hint: The easiest question is 2b.
My Jewish ex-father-in-law, Naum Bernstein, is 96 years old and is full of life. He has a joke for every situation. In the last decade he wrote several volumes of memoirs in Russian. One of the books was a collection of his favorite jokes and his explanations of them. I decided to retell some of the jokes from his selection.
An arithmetic teacher calls the student Rabinovich to the blackboard. "It is known that from 1 kilogram of sour cream you can make 200 grams of butter. Imagine, Rabinovich, that your father bought 2 kilograms of sour cream. How much butter can he make?"
"Five hundred grams," Rabinovich replies.
The teacher frowns, "Rabinovich, you do not know arithmetic!"
Rabinovich answers, "Sir, you do not know my father."
An astronomy teacher explains that in the future the Earth will lose its heat energy, continents will collide, and solar radiation will increase. In six billion years life will be extinct. A student looking really scared raises his hand and asks, "In how many years will life become extinct?"
"In about six billion years," the teacher repeats.
"Whew," says the student, "you got me so scared. I thought you said six million."
Two professors are chatting while watching a soccer game. The first one says, "They say that soccer players have their brains in their legs. So their heads are really empty."
"Not quite," the second professor replies. "The player on the right passed my exam yesterday."
The first professor expresses interest, so the second one elaborates. "As a rule, I ask two questions. If the student gives a correct answer to one of them, he passes."
"So, what did you ask that guy?"
"My first question was 'What color are red blood cells?' He answered 'Yellow.' That was an incorrect answer. The second question was 'How is sulfuric acid produced?' To this he replied, 'I do not know,' which was absolutely true, so he passed."
A Russian literature teacher asks a pupil, "Who wrote Eugene Onegin?" The pupil gets scared that he is being blamed for something and replies, "No, not me! I swear I didn't write it!" Everyone laughs. The teacher decides that the pupil disrupted the class on purpose and asks for his father to come by.
The father arrives and after the teacher explains what happened, the father says, "Maybe he is not guilty; maybe he really didn't write it. I doubt that he is capable of writing anything."
The teacher is stunned and later tells the whole story in the teachers' lounge to her colleagues. An astronomy teacher comes home and retells the story to her husband who works for the KGB. The husband comments, "Do not worry, we are on it. Three people already confessed to writing it."
A hardcore anti-Semite was dying. As he got weaker he made a last request. He wanted to convert to Judaism. Everyone was extremely surprised, but decided not to interfere. After the conversion, his wife summoned the courage to ask him what was going on. "Do you think you were mistaken, hating Jews all your life?"
"No," he replied happily, "But now with my death, the world will get rid of one more Jew."
An old Jew comes to a Rabbi and asks if he can shave his beard off, because his children think that he is old-fashioned. The Rabbi tells him that by Jewish law he is not allowed to shave. The old man turns to go home when he realizes that the Rabbi himself doesn't have a beard. He stops and asks, "Dear Rabbi, you just forbade me to shave my beard, but how come you are clean-shaven yourself?"
The Rabbi replies, "I didn't ask anyone's permission."
When Rabinovich came to a bureaucrat with a request, the bureaucrat replied, "Come back tomorrow." Rabinovich returned the next day and received the same reply. Rabinovich was very persistent and returned day after day.
Finally, the bureaucrat lost his patience and attacked Rabinovich, "This is outrageous! Don't you understand simple language? I keep telling you to come tomorrow and you keep coming today."
The communist committee of a supermarket in the USSR received a lot of complaints about the rudeness of their salespeople. The committee decided to improve the quality of service and provided special training in which salespeople were taught politeness. The training emphasized what to do in case a particular item was unavailable. The salespeople were supposed to politely explain that the item is temporarily unavailable and to offer a substitute.
The next thing one of their salespeople said to a customer was "I am very sorry, we are temporarily out of toilet paper. May I offer some sandpaper?"
There is panic in an apartment on the 13th floor. The wife recognizes the sound of her husband's approach, even though he was supposed to be on a business trip. The lover asks, "What should I do, honey?"
"What do people do in such cases? Jump out the window!"
"But we are on the 13th floor!"
"This is no time for superstition!"
A young man had smelly feet, plus he always forgot to change his socks. His girlfriend got tired of it and asked him to promise that he would always change his socks before coming to see her.
Next visit the young man smelled as bad as ever. Outraged, the girl said, "But you promised to change your socks!"
The young man answered, "I did as I promised."
"I don't believe you, you smell awful."
"I was sure you wouldn't believe me. Good thing I brought my dirty socks with me as proof."
At the 50th anniversary of a very happy couple, someone asked the husband for their secret. He said that right before the wedding they agreed that the husband would decide all the crucial and very important things, and the wife would be responsible for all minor decisions. "For example," he continued, "yesterday I decided that the US should withdraw their troops from Iraq, and my wife decided where to buy our vacation house."
Two long-time girlfriends meet after several years without being in touch. "How are your children?" asks one of them.
The other replies, "My daughter is fine, she married a nice young man, who is providing for her. He also helps her with chores and even brings her coffee in bed every morning."
"What about your son?"
"It's a disaster. I don't know what to do. He married a really lazy woman. Even though she's not working, she wants him to help her with the chores. Can you imagine that? She even dared to ask him to bring her coffee in bed every morning."
Two friends are walking along a street very late at night. Robbers attack them with guns, demanding their wallets. One of the friends asks the robbers, "Can you give me 30 seconds?" The robbers agree. He takes out $100 from his wallet and gives it to his friend, "Remember I owed you $100? I am paying back my debt in front of witnesses."
Life is a struggle. Before lunch with hunger, after lunch with sleepiness.
A mother says to her son, "Please, close the window — it's cold outside."
The son replies, "Do you think it will get warmer outside if I close the window?"
How are pessimists and optimists different from normal people?
A pessimist uses both a belt and suspenders, an optimist uses neither.
In a cemetery there is a beautiful monument with a picture of a bald, wrinkled old man. He is smiling, showing his perfect white teeth. His epitaph says:
Here lies Mr. X, who lived more than 100 years, lost his hair, became all wrinkled, but kept his perfect teeth. That is because he always used our company's toothpaste.
A nearby monument has a picture of an old toothless woman with beautiful, voluminous hair. The inscription explains which brand of shampoo she used.
Many other tombstones with ads are scattered throughout the cemetery. But in the middle there is a huge mausoleum with an inscription reading:
No one is buried here and no one ever will be, because his or her parents used condoms made by our company.
A Russian man marries an American woman. After a while he writes a letter home.
My wife must be very dirty. She showers every day.
Rabinovich was asked why he didn't attend the last committee meeting. He replied, "If I knew it was the last, I would certainly have come."
I stumbled upon the following problem in Mathematics Teacher v.73 (September 1980):
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
Ooh, boy!
Question for my readers: explain my reaction.
I have a leased Toyota Corolla, and I am happily enrolled in AutoCheck payments with Toyota's Financial Services. So I do not even look at my bills. Once I opened my bill and noticed that the requested payment was twice as high as I expected. I looked closer and the bill had a car tax included in it. I looked even closer and read that:
Your Current Payment Due will be automatically withdrawn from your checking or savings account on the above Payment Due Date or the next banking day.
I decided that everything was taken care of and continued my relaxed life. After several months I checked my bill again, and the car tax was still there. After more careful study of my bill I discovered that Toyota's "Current Payment Due" doesn't include my car tax. Obviously they assume that their definition of "Current Payment Due" is crystal clear to everyone.
I got worried about this delayed car tax payment and went online to pay it. I tried to make this payment, but Toyota's website rejected it. The website informed me that because I am enrolled in AutoCheck, I am not allowed to make separate online payments. I couldn't believe it: to do so, I would have to de-enroll first!
So I just wrote a check.
In one day my feelings for my Toyota Corolla were turned around. If their financial system is designed so stupidly, what can we say about their car designs? Suddenly the sound of my brakes and the squeak in my steering wheel worry me.
A year ago I posted a chessboard puzzle. Recently I stumbled on a September 2008 issue of "Math Horizons" where it was presented as a magic trick.
When the magician leaves the room, the trickees lay out eight coins in a row deciding which side is turned up according to their whim. They also think of a number between 1 and 8 inclusive. The magician's assistant then flips exactly one of the coins, before inviting the magician back in. The magician looks at the coins and guesses the number that the trickees thought of.
The magician's strategy can be derived from the solution to the chessboard puzzle. The assistant numbers the coins from zero to seven from left to right. Then s/he flips the coin so that the parity addition (XORing) of all the numbers corresponding to heads is the number that the magician needs to guess. For this trick to work, the number of coins needs to be a power of 2.
Andrey Zelevinsky posted (in Russian) a cool variation of this trick with two decks of cards.
The magician has two identical card decks and he is out of the room for now. A random person from the audience thinks of a card. Next, the audience chooses several cards from the first deck. Then the assistant adds one card from the second deck to the set of chosen cards, lays them on a table, and then invites the magician back. The magician looks at the cards on the table and guesses the card that was thought of.
Unlike in the coin trick above, the number of cards in the deck doesn't need to be a power of 2. This flexibility is due to the fact that the magician has two decks of cards, as opposed to one set of coins. Having the second deck is equivalent to the assistant in the coin trick being allowed to flip one or ZERO coins.
Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.
An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?
The other one is from 1995.
There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles' orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?
Perfidy is to parity as odious is to odd and evil is to even. As a reminder, odious numbers are numbers with an odd number of ones in their binary expansions. From here you can extrapolate what the evil numbers are and the fact that the perfidy of an integer is the parity of the number of ones in its binary expansion. We live in a terrible world: all numbers are perfidious.
So why are we writing about the perfidy of negative numbers? One would expect it to be a natural extension of the perfidy of positive numbers, but it turns out that the naive way of defining it doesn't work at all. Is there hope? Could negative numbers be innocent of evil and free of odiousness? Is zero an impenetrable bulwark against perfidy? Not quite, but something interesting does happen to evil as it tries to cross zero. Read on.
To define perfidy for negative numbers, let us study how perfidy behaves for positive numbers. It is easiest to think about the perfidies of power-of-two-sized chunks of non-negative integers at a time. Let us denote by Tn the string of perfidies of the integers from 0 to 2n−1, with evil being zero and odious being 1. So T0 = 0, T1 = 01, T2 = 0110, T3 = 01101001, …. The recurrence relation governing the Tn is Tn+1 = TnTn, where T is the bitwise negation of the string T, and juxtaposition is concatenation. The limit of this as n tends to infinity is the (infinite) sequence of perfidies of non-negative integers. This sequence is called the Thue-Morse sequence: 0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,….
So defining the perfidy of negative numbers is equivalent to extending the Thue-Morse sequence to the left. If we are to define "the" perfidy of negative numbers, that definition should preserve most of the properties of the Thue-Morse sequence after extension.
So, let's see. We asked around, and most people said that the binary expansion of a negative integer should be the binary expansion of its absolute value, but with a minus sign. Defining perfidy as parity of number of ones in this binary expansion corresponds to the following extended Thue-Morse sequence in which we mark values corresponding to negative indices with bold font: … 0, 1, 1, 0, 1, 1, 0, ….
One of the major properties of the Thue-Morse sequence is its fractal property: if you replace every zero of the Thue-Morse sequence by 0,1 and every one by 1,0, you will get the Thue-Morse sequence back. Clearly, our new extended sequence doesn't have this property.
Another set of properties for the Thue-Morse sequence, called avoidance properties, is a long list of patterns that the sequence avoids. For example, the Thue-Morse sequence doesn't contain any overlapping squares — patterns axaxa, where a is a character and x is a word. But you can see above, our first extension contains it. So this definition is wrong, not just once but twice (and two wrongs only make a right under very unusual circumstances). Perfidy is stymied by the cross-over from zero to minus one. Are negative numbers protected from the ravages of evil? (and odiousness?)
Unfortunately, there are many people, for example John Conway, who inadvertently extend the reach of perfidy by arguing that the binary expansion of a negative integer should be different. Indulge in a flight of fancy and imagine the binary expansion that consists of infinitely many ones to the left: …1111. What happens when you add 1 to it? The carry gets pushed infinitely far away, and you get …000000 — zero. So it is quite reasonable to let …1111 be the binary expansion of −1. Similarly, the string …1110 represents −2, …1101 represents −3, etc. Continuing this we see that the binary expansion of a negative integer −n is the bitwise negation of the binary expansion of n − 1 (including the leading zeros). This is called the Two's complement representation.
Why is two's complement a reasonable representation? Suppose you were trying to invent a binary notation for negative numbers, but you wanted to pursue uniformity by not using a minus sign. The problem is that the standard definition of the binary representation allows you to represent only positive numbers. But you can solve this problem with modular arithmetic: modulo any fixed N, every negative number is equivalent to some positive number (by adding enough multiples of N), so you can just represent it by representing that positive number. If you choose N to be a power of two, modding out by it is just truncation of the binary representation. If you let those powers of two tend to infinity, you get the two's complement representation described above.
Aside: When you are building a computer, uniformity is money, because special cases cost special transistors. The two's complement idea lets one build arithmetic units that just operate on positive numbers with some number of bits (effectively doing arithmetic modulo 2k), and leave the question of negativeness to the choice of representatives of those equivalence classes.
If we take two's complement as the binary expansion of negative numbers, how will we define the perfidy? Is the number of ones in the infinite string …1111 corresponding to −1 even or odd?
We can't answer that question, but we know for every binary expansion of negative numbers the parity of the number of zeroes. Thus we can divide all negative integers in two classes with different perfidy. We just do not know which one is which.
Let us consider two cases. In the first case we consider a negative number odious if the number of zeroes in its binary expansion is odd. The corresponding extended Thue-Morse sequence is: … 0, 1, 1, 0, 0, 1, 1, 0, …. The negative half is the reflection of the classical Thue-Morse sequence. In the second case we consider a negative number odious if the number of zeroes in its binary expansion is even. The corresponding extended Thue-Morse sequence is: … 1, 0, 0, 1, 0, 1, 1, 0, …. The negative half is the bitwise negation of the reflection of the classical Thue-Morse sequence.
Can we say that one of the sequences is better than the other? Both of them respect the fractal property of the classical Thue-Morse sequence. Let us look at the avoidance properties. The avoidance properties are symmetric with respect to switching zeroes with ones and with respect to changing the direction of the sequence. Hence, the negation, the reflection, and the reflection of the negation of the Thue-Morse sequence will continue to respect these properties.
Thus, we only need to check the avoidance properties of the finite subsequences that span both negative and non-negative indices. We claim that for both definitions of perfidy, any finite middle subsequence of the extended Thue-Morse sequence occurs as a subsequence in the classical Thue-Morse sequence. So any avoidance properties that are true for the Thue-Morse sequence will also be true for both extensions.
Indeed, it is easy to show that the strings T2n defined above are palindromes. So for the first definition of perfidy the string in the middle will be a substring of T2nT2n for some large n, and for the second definition a substring of T2nT2n. But the classical Thue-Morse sequence contains the subsequence T2nT2nT2nT2nT2nT2nT2nT2n. So whichever way we extend the Thue-Morse sequence to the left any finite middle part will always be a repetition of a piece in the classical Thue-Morse sequence. Thus, all the avoidance properties will hold.
We see that there are two logical ways to define perfidy for negative integers. There are two clear groups of numbers with the same perfidy, but which is called evil and which odious is interchangeable. So evil doesn't stop at zero after all, but at least it gets an identity crisis.
* * *
A note posted on the door of the tech-support department:
"Theory — you know everything, but nothing works. Practice — everything works, but nobody knows why. In our department we merge theory with practice: nothing works and nobody knows why.
* * *
A plus is two minuses at each others' throats.
I was teaching my students PIE, the Principle of Inclusion and Exclusion. This was the last lesson of 2010 and it seemed natural to have a party and bring some pie. It appears that the school has a new rule. If I want to bring any food to class, I need to submit a request that includes all food ingredients. The administrators send it to the parents asking them to sign a permission slip and then, if I receive all the slips back in time, I can bring pie to school. We had to study PIE without pie.
Our most important task as parents and teachers is to teach kids to make their own decisions. They are in high school; they know by now about their own allergies and diets; they should be able to avoid foods that might do them harm. I understand why schools create such rules, but we are treating the students like small children. We can't protect them forever; they need to learn to protect themselves.
Next semester, we will study the mathematics of fair division. I will have to teach them how to cut a cake without a cake.
I once wrote a story about a mistake that my medical insurance CIGNA made. They had a typo in the year of the end date of my insurance coverage in their system. As a result of this error, they mistakenly thought they had paid my doctors after my insurance had expired and tried to get their money back. While I was trying to correct all this mess, an interesting thing happened.
To help me explain, check out the following portion of my bill. (If it looks a bit funny, it's because I cut out some details including the doctor's name).
On the bill you can see that I had a mammography for which I was charged $493.00, but CIGNA paid only $295.80. The remaining $197.20 was removed from the bill as an adjustment, as frequently happens because of certain agreements between doctors and insurance companies. A year later when CIGNA made their mistake, they requested that the payment be returned. You can see on the bill that once the payment was reversed, my doctors reversed the adjustment too.
When CIGNA fixed the typo, they repaid the doctors, but the adjustment stayed on the bill, which the doctors then wanted me to pay. And that was only one of many such bills. It took me a year of phone calls to get the adjustments taken off, but this is not what I am writing about today.
If not for this mistake, I would have never seen these bills and the revealing information on the different amounts doctors charge to different parties, and how much they really expect to receive. As you can see my doctors wanted 67% more for my mammogram than they later agreed to.
The difference in numbers for my blood test was even more impressive. I was charged $173.00, and the insurance company paid $30.28 — almost six times less.
If I ever need a doctor and I don't have insurance, I will take these bills with me to support my request for a discount. I do not mind if you use this article for the same purpose.
I met Ed (Edik) Frenkel 20 years ago at Harvard when he was a brilliant math student of my now ex-husband, and a handsome young man. Now, at 42, he is a math professor at Berkeley and he is even hotter. He made a bizarre move for a mathematician: he produced and starred in an erotic short movie, Rites of Love and Math. If he wants to be known as the sexiest male mathematician alive, he just might get the title.
The movie created a controversy when Mathematical Sciences Research Institute (MSRI) withdrew its sponsorship for the first screening after a lot of objections based on the trailer. My interest was piqued by a painting that dominated the visual of the trailer's erotica scene. The black and white amateur painting is of the integral sign with Russian letters stylized as math symbols that spell the word "Truth". In addition, the name of the woman in the movie, Mariko, means "truth" in Japanese. Though it felt pretentious, I was hoping that the movie would be symbolic. When I heard that the actors do not talk in the movie, my expectations of symbolism grew. I love movies that are open to interpretation. So I bought the movie, watched it and wrote the following review. Before getting to the review itself I would like to thank Ed Frenkel for sending me the photos and giving me permission to use them in my frank assessment of his work.
Here is the plot:
A Mathematician, hoping to serve humanity, discovers a formula of Love. Bad guys find an evil way to use the formula to destroy humanity and are hunting for the Mathematician, who is hiding in his lover Mariko's home. The Mathematician fears for his own life. Although it would make sense to destroy all the papers with the formula, the Mathematician loves his formula even more than his lover and himself. He wants to preserve the formula and tattoos it on her body with her consent.
There is much about the film that I like, including the slow pace and the visuals, with their minimalistic background and palette of black, white and red. The camera work is superb.
I welcomed the idea of a Love formula, because mathematics is ready to broaden the scope of its models, including venturing into love. Of course, some mathematical models of relationships already exist.
As I mentioned, I was looking forward to the movie, hoping that it would encourage the imagination of viewers in their interpretations. To my disappointment, every scene in the movie is preceded by text that describes the plot, removing any flexibility of interpretation. Besides that, the emotions portrayed didn't quite match the written plot, in no small part because Ed Frenkel is not a good actor.
The idea of preserving a formula by tattooing it on someone is beyond strange. He could have used a safe-deposit box. Or put the formula in an envelope and given it to the lover to keep, or just encrypted it, etc. With narcissistic lack of consciousness, the Mathematician seems unaware of the implications of his action of imprinting this dangerous secret on Mariko. She can never go swimming, or go to the gym, or be intimate with anyone else. Moreover, if the bad guys discover that Mariko is the Mathematician's lover, her life will be in grave danger. Not to mention that tattooing is painful.
Something that could have been interesting and watchable in a historic movie, in this contemporary movie seems pointlessly cruel, dehumanizing and senseless.
I know for sure that Ed Frenkel is not stupid, so what are his reasons for constructing the plot in this way? Before investigating his reasons, I have a mathematical complaint about the movie. Every mathematician and teacher knows that when asserting a formula you need to indicate its interpretation: what its symbols refer to in the real world. For example, suppose I tell you my own great Formula of Love: Cn = (2n)!/(n+1)!n!. You may recognize Cn as the Catalan numbers, but what does this have to do with Love? To give the formula meaning I need to tell you that Cn is the number of ways you can seat n loving couples at a round table with 2n chairs, so that each couple can join hands (assuming the arms are long enough to reach across the table) without any two pairs of arms crossing. Assigning an interpretation makes the Catalan numbers part of the world's growing body of romantic research.
Writing a formula without mentioning what the variables mean fails to preserve it for the future. Ed Frenkel knows that. Wait a minute. The formula in the movie is actually not the Formula of Love, but a real formula from Ed's paper on instantons. It's right there, formula 5.7 on page 74. Every variable is explained in the paper. Ah-ha! So his movie isn't actually about art, but rather about Ed's formula. Indeed, there is no real Formula of Love. In such situations in other movies, they have simply shown fragments of a formula. However, in Rites of Love and Math, Frenkel's formula — which has nothing to do with Love — is shot in full view, zooming in slowly.
The movie is a commercial. Ed is using our fascination with sex to popularize his formula, and using his formula and his scientific standing to advertise his body.
I was so disappointed that the default interpretation of the movie was imposed on me by those pre-scene texts, that I decided to watch the movie for a second time, trying to ignore the text, hoping to find some new meaning.
If you decide to see the movie, you'll probably come up with your own interpretation of the plot. I actually came up with several. I had a funny one and an allegorical one, but the most interesting task for me was to try create an interpretation matching the emotions portrayed:
Mariko knows that something is wrong in her sex life with the Mathematician. But she still loves him and writes him a love letter. The Mathematician comes to Mariko's place. He is distant and cold. They cuddle. He explains to her that sex doesn't bring him pleasure anymore and that moreover, he can't even perform. He tells her that the only thing that brings him joy is mathematics and suggests that his sexual dysfunction and lack of pleasure will be fixed if they tattoo his favorite formula on her body. She agrees, but first they decide to give sex a last try. They try real hard. But he can't relax and he doesn't enjoy it, so she agrees to the tattoo. He does get excited during the tattooing process itself, but once he finishes his whole formula, he is no longer turned on. Mariko's suffering has been in vain.
Janet Mertz wrote several papers about the gender gap in mathematics. One of her research ideas was to find girls who went to the International Math Olympiad (IMO) and compare their fate to that of their teammates with a similar score. She asked me to find Soviet and Russian IMO girls. All my life I had heard about Lida Goncharova, the first girl on a Soviet team, and the first girl in the world who took a gold medal, but I had never dared to reach out to her. A little push by Janet Mertz was enough for me to find Lida's phone number in Moscow and call her.
Lida got interested in mathematics when she was five years old. Luckily, many of her relatives were mathematicians and she started bugging them for math puzzles.
Her involvement with math was interrupted by the death of her parents — her mother when she was seven, and her father when she was nine. She ended up living with her sister, but felt very lonely.
After several years of personal turmoil, she renewed her pursuit of mathematics. Lida started discussing math with her mother's first husband. She joined a math circle which was run at Moscow State University. When she was 13 she went to a summer camp and found a mentor there to study trigonometry. Eventually she ended up at School Number 425, one of the first schools in Moscow that opened for children gifted in math.
At the end of high school she went to the IMO as part of the Soviet team and won a gold medal there. After that she enrolled in the most prestigious Soviet institute for the study of math — Moscow State University (MSU).
Half of her high school classmates went to MSU, including her high school sweetheart Alexander Geronimus. Lida married Alexander when she was a sophomore and they had their first son in her fourth year of undergraduate school.
Meanwhile, she wasn't doing as well in her studies as she had hoped. Lida was very fast to pick up math ideas during conversations, but she had difficulty reading books. As ideas were becoming more complicated and involved, this became a problem. She started feeling that she was falling behind her friends. When her friends gathered together to discuss mathematics she couldn't understand everything. She wanted to ask questions, but was too shy. Plus, she didn't want to impose on them. She made a decision to be silent. As a result she started ignoring the conversations of others and became discouraged as she fell behind.
She had her second child at the beginning of graduate school, where she studied under the supervision of Dmitry Fuchs. Lida was already losing her self-esteem and so she chose a self-contained problem that didn't require a lot of outside knowledge. The solution involved some combinatorial methods, but Lida didn't quite understand the big picture and the problem's goal.
I contacted Dmitry Fuchs and asked him about Lida's thesis. He told me that Lida's main result is extremely important and widely cited. It is called Goncharova's theorem.
Meanwhile, her husband finished his PhD in math and secured a great job in an academic institution. They had started as peers, but her work was interrupted by having their children. Lida finished her PhD a couple of years after her husband and got a very boring job as an algorithm designer. She even wrote some papers at the job, but she was not much interested. She continued her attempts to do mathematics and continued asking everyone for problems, but it didn't go anywhere. Her friends were not very interested in her calculations and after the birth of her third child she began to lose hope in her research.
When Lida and her husband entered graduate school they became religious. Ten years later, Alexander decided to pursue the Russian Orthodox religion as a career and got a parish 600 km from Moscow. They didn't want to move their children away from Moscow, with its educational and cultural opportunities. So they started living in two places with long commutes. This didn't help her math either.
Eight years after the third son, the fourth son was born. Although Lida sporadically continued her calculations, she still didn't talk about them to anyone.
When the older children went to high school, Lida enjoyed solving their math problems tremendously. In 1990 perestroika started and Lida lost her job. She got an offer to create a private school and teach there. By this time she had had two more children, a son and a daughter. Lida continued working for the private school until her six children grew out of it. Lida enjoyed teaching and inventing methods to teach mathematics. The school ended in 2004. But she continues working with kids sharing with them her joy of mathematics.
Lida believes that she has had an extremely lucky life in many ways. The only exception was her unsuccessful math career. She can't live without math, and will continue working with kids, solving fun problems and doing her private research.
When I first called her and said I wanted to talk about her and math, she told me: —There is nothing to talk about. I stopped doing math after my PhD. Almost.— That —almost— kept me asking questions.
Janet Mertz was considering a serious research project comparing the fates of IMO medal girls with the fates of their teammates, to see whether gender plays a role later. However, due to the language and cultural differences and the fact that most of the girls changed their last name, it was difficult to locate them. So Mertz put this research project on hold.
She had asked me to find and contact the Russian women and I was so fascinated with Lida's story that I decided to write it up in this article. And because the research is on hold, I decided to include the fates of Lida's teammates.
Lida Goncharova got her gold medal in the 1962 IMO with 42 points and was ranked third. The teammate with the closest score was Joseph Bernstein with another gold medal and 46 points. I don't even have to check Wikipedia to tell you about Joseph, as I was once married to him. He used to be a professor at Harvard University and is now a professor at Tel-Aviv University. He is a member of the Israel Academy of Sciences and Humanities and the United States National Academy of Sciences. He achieved a lot and is greatly respected by his peers.
Joseph Bernstein might not be the best person to compare Lida to as he had a perfect score. Some might argue that a perfect score indicates that he might have done better if the problems had been more difficult.
The two Soviet teammates whose scores were the closest to that of Lida, but below her, were Alexey Potepun with 37 points and Grigory Margulis with 36 points.
Alexei Potepun got a PhD in mathematics and is now a professor at Saint-Petersburg University. He has published eleven papers.
Next to Alexey Potepun is Grigory Margulis, who is a professor at Yale and was awarded the Fields Medal and the Wolf Prize. He is a member of the U.S. National Academy of Sciences.
You might notice that the two people who moved to the US are much more famous than those who stayed in Russia. You might say that moving to the US is a better predictor of success than gender. Sure, living in a free country helps, but Margulis got his Fields medal while he was in the USSR. And Bernstein invented his famous D-Modules while in Russia also.
My conversation with Lida was personally inspiring. I loved the tone of her voice when she talked about mathematics. There were many elements that prevented her from having the mathematical career she might have had: the untimely death of her parents, her shyness, raising six children, many years of long commutes. When we look at the achievements of her closest teammates, we can't help but wonder what kind of mathematics we lost.
This conversation was very encouraging for me. I felt there were similarities between Lida and myself in more ways than I expected. What we share most of all is a love for mathematics. I could hear that in her voice.
I gave my students a problem from the 2002 AMC 10-A:
Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, …, 10}. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is: (A) 2/5, (B) 9/20, (C) 1/2, (D) 11/20, (E) 24/25.
Here is a solution that some of my students suggested:
On average Tina gets 6. The probability that Sergio gets more than 6 is 2/5.
This is a flawed solution with the right answer. Time and again I meet a problem at a competition where incorrect reasoning produces the right answer and is much faster, putting students who understand the problem at a disadvantage. This is a design flaw. The designers of multiple-choice problems should anticipate mistaken solutions such as the one above. A good designer would create a problem such that a mistaken solution leads to a wrong answer — one which has been included in the list of choices. Thus, a wrong solution would be punished rather than rewarded.
Readers: here are three challenges. First, to ponder what is the right solution. Second, to change parameters slightly so that the solution above doesn't work. And lastly, the most interesting challenge is to explain why the solution above yielded the correct result.
This is a version of the standard charades game that my son, Sergei Bernstein, invented.
Unlike in regular charades, the person who acts out the phrase doesn't know what the phrase is and has to guess it. The viewers on the other hand, know the phrase but they are not allowed to talk.
So the actor is blindfolded and the viewers are not just watching; they are actively moving the actor and his/her body parts around to communicate the phrase. For example, if the actor is on the right track, since the viewers can't say, "Yes, good!", they might communicate it by nodding the actor's head.
Sounds like fun, especially for people who enjoy touching and being touched.
Once again I am one of the organizers of the Women and Math Program at the Institute for Advanced Study in Princeton, May 16-27, 2011. It will be devoted to an exciting modern subject: Sparsity and Computation.
In case you are wondering about the meaning of the picture on the program's poster (which I reproduce below), let us explain.
The left image is the original picture of Fuld Hall, the main building on the IAS campus. The middle image is a corrupted version, in which you barely see anything. The right image is a striking example of how much of the image can be reconstructed from the corrupted image using clever algorithms.
Female undergraduates, graduates and postdocs are welcome to apply to the program. You will learn exactly how the corrupted image was recovered and much more. The application deadline is February 20, 2011.
Eugene Brevdo generated the pictures for our poster and agreed to write a piece for my blog explaining how it works. I am glad that he draws parallels to food, as the IAS cafeteria is one of the best around.
The three images you are looking at are composed of pixels. Each pixel is represented by three integers corresponding to red, green, and blue. The values of each integer range between 0 and 255.
The image of Fuld hall has been corrupted: some pixels have been replaced with all 0s, and are therefore black; this means the pixel was not "observed". In this corrupted version, 85% of the pixel values were not observed. Other pixels have been modified to various degrees by stationary Gaussian noise (i.e. independent random noise). For the 15% observed pixel values, the PSNR is 6.5 db. As you can see, this is a badly corrupted image!
The really interesting image is the one on the right. It is a "denoised" and "inpainted" version of the center image. That means the pixels that were missing were filled in and the observed pixel integer values were re-estimated. The algorithm that performed this task, with the longwinded name "Nonparametric Bayesian Dictionary Learning," had no prior knowledge about what "images should look like". In that sense, it's similar to popular wavelet-based denoising techniques: it does not need a prior database of images to correct a new one. It "learns" what parts of the image should look like from the original image, and fills them in.
Here's a rough sketch of how it works. The idea is to use a new technique in probability theory — the idea that a a patch, e.g. a contiguous subset of pixels, of an image is composed of a sparse set of basic texture atoms (from the "Dictionary"). Unfortunately for us, the number of atoms and the atoms themselves are unknowns and need to be estimated (the "Nonparametric Learning" part). In a way, the main idea here is very similar to Wavelet-based estimation, because while Wavelets form a fixed dictionary, a patch from most natural images is composed of only a few Wavelet atoms; and Wavelet denoising is based on this idea.
We make two assumptions that allow us to simplify and solve this problem, which is unwieldy-sounding and vague when the texture atoms have to be estimated. First, there may be many atoms, but a single patch is a combination of only a sparse subset of them. Second, because each atom appears in part in many patches, even if we observe some noisily, once we know which atoms appear in which patches, we can invert and average together all of the patches associated with an atom to estimate it.
To explain and programmatically implement the full algorithm that solves this problem, probability theorists like to explain things in terms of going to a buffet. Here's a very rough idea. There's a buffet with a (possibly infinite) number of dishes. Each dish represents a texture atom. An image patch will come up to the buffet and, starting from the first dish, begins to flip a biased coin. If the coin lands on heads, the patch takes a random amount of food from the dish in front of it (the atom-patch weight), and then walks to the next dish. If the coin lands on tails, the patch skips that dish and instead just walks to the next. There it flips its coin with a different bias and repeats the process. The coins are biased so the patch only eats a few dishes (there are so many!). When all is said and done, however, the patch has eaten a random amount from a few dishes. Rephrased: the image patch is made from a weighted linear combination of basic atoms.
At the end of the day, all the patches eat their own home-cooked dessert that didn't come from the buffet (noise), and some pass out from eating too much (missing pixels).
If we know how much of each dish (texture atom) each of the patches ate and the biases of the coins, we can estimate the dishes themselves — because we can see the noisy patches. Vice versa, if we know what the dishes (textures) are, and what the patches look like, we can estimate the biases of the coins and how much of a dish each patch ate.
At first we take completely random guesses about what the dishes look like and what the coins are, as well as how much each patch ate. But soon we start alternating guesses between what the dishes are, the coin biases, and the amounts that each patch ate. And each time we only update our estimate of one of these unknowns, on the assumption that our previous estimates for the others is the truth. This is called Gibbs sampling. By iterating our estimates, we can build up a pretty good estimate of all of the unknowns: the texture atoms, coin biases, and the atom-patch weights.
The image on the right is our best final guess, after iterating this game, as to what the patches look like after eating their dishes, but before eating dessert and/or passing out.
I've heard many fun problems in which blindfolded parachutists are dropped somewhere and they need to meet up once they're on the ground. They can't shout or purposefully leave traces behind. They will recognize each other as soon as they bump into each other. Their goal is to get to the same assembly point. They can design their strategy in advance.
Here is the first problem in a series that gets increasingly difficult:
Two parachutists are dropped at different locations on a straight line at the same time. Both have an excellent sense of direction and a good geographical memory, so both know where they are at any moment with respect to their starting point on the line. What's their strategy?
The strategy is that the first person stands still and the second one goes forward and back repeatedly, increasing the distance of each leg until they collide.
In the next variation, both are required to execute the same program, that is, if one stands still, then both stand still. To compensate for this increased difficulty, they are allowed to leave their parachutes anywhere. And both of them will recognize the other's parachute if they bump into it.
In the third variation, the set-up is similar to the previous problem, but they are not allowed to change the direction of their movement. To their advantage, they know which way East is.
I recently heard a 2-D version from my son Sergei in which the parachutists are ghosts. That means that when they bump into each other they go through each other without even recognizing the fact that they met:
Several blindfolded men are sleeping at different locations on a plane. Each wakes up, not necessarily at the same time. At the moment of waking up, each of them receives the locations of all the others in relation to himself at that moment. They are not allowed to interact, nor will they receive any further information as time passes. They need to get together in one place. How can they do that, if they are allowed to decide on their strategy in advance?
They do not know where North is. So they can't go to the person at the most Northern point. Also they do not know how locations correspond to people, so they can't all go to where, say, Peter is. Let us consider the case of two men. Suppose they decide to go to the middle of the segment of two locations they receive when they awake. But they get different locations because they wake up at different times. Suppose the first person wakes up and goes to the middle. The fact that he walks while the other is sleeping, means that he changes the middle. So when the second person wakes up, his calculated middle is different from the one calculated by the first person. Consequently, they will never manage to meet. Hence, the solution should be different.
Actually Sergei gave me a more difficult problem:
Not only do they need to meet, but they need to stay together for a predefined finite time period.
Here is as bonus problem.
If there are three parachutists, it is possible to end up in a meeting place and stay there indefinitely. For four people it is often possible too.
Darth Maul killed Qui-Gon Jinn. Obi-Wan Kenobi killed Darth Maul. Palpatine killed Mace Windu. Darth Vader killed Obi-Wan Kenobi and Palpatine. I am mentally drawing the kill graph of Star Wars, where people are vertices and kills are edges. The graph is not very interesting. In movies where no one gets resurrected, the kill graph is a forest.
I'm interested in studying social networks in the movies and how they differ from social networks in real life. As we saw, the kill graph is not very exciting mathematically.
Now let's try the acquaintance graph, where edges mark two people who know each other. Unfortunately, in the movies there are often many nameless people and we learn very little about their acquaintances. On the other hand, all the "nameful" people usually know each other, thus their acquaintance graph is a complete graph. The richest acquaintance graphs would be for epic movies like Star Wars, in which the events span two generations and many planets. As a result, there are characters who never meet each other. For example, Leia, Luke and Han from the original trilogy never meet people who died in the prequel, such as Anakin's mother and Count Dooku.
But I think that the most intriguing type of filmic social network is the fight graph, where edges represent characters who fight each other. Usually such graphs are bipartite, reflecting the division between bad guys and good guys. When an epic film is more complex and has traitors, the fight graph is no longer bipartite. Consider Darth Vader who fought and killed a lot of good guys including Obi-Wan Kenobi as well as many bad guys including Count Dooku and the Emperor.
I would like to immortalize Darth Vader in mathematics. He did restore the balance to the Force. If there is a graph which is not bipartite and can become bipartite by removing one highly connected node, I would like to name such a node Darth Vader.
Dear Nita Palmira,
I do not recall your name and I'm not sure where you got my email address from, but I really appreciate you contacting me. I am excited by your Two-Procedures-For-The-Price-Of-One offer. I am really looking forward to my enlarged penis and my DDD breasts.
Meanwhile, I can give you a unique group discount on IQ tests. I can test the IQ of all your company employees for the price of one test. Moreover, you do not need to waste even a minute. Actually, no one even needs to answer any questions. You can send me your $500 check to the address below and I will promptly send you the IQ report, the accuracy of which I can guarantee.
Sincerely yours.
I am coaching my AMSA students for math competitions. Recently, I gave them the following problem from the 1964 MAML:
The difference of the squares of two odd numbers is always divisible by:
A) 3, B) 5, C) 6, D) 7, E) 8?
The fastest way to solve this problem is to check an example. If we choose 1 and 3 as two odd numbers, we see that the difference of their squares is 8, so the answer must be E. Unfortunately this solution doesn't provide any useful insight; it is just a trivial calculation.
If we remove the choices, the problem immediately becomes more interesting. We can again plug in numbers 1 and 3 to see that the answer must be a factor of 8. But to really solve the problem, we need to do some reasoning. Suppose 2k + 1 is an odd integer. Its square can be written in the form 4n(n+1) + 1, from which you can see that every odd square has remainder 1 when divided by 8. A solution like this is a more profitable investment of your time. You understand what is going on. You master a method for solving many problems of this type. As a bonus, if students remember the conclusion, they can solve the competition problem above instantaneously.
This is why when I am teaching I often remove multiple choices from problems. To solve them, rough estimates and plugging numbers are not enough. To solve the problems the students really need to understand them. Frankly, some of the problems remain boring even if we remove the multiple choices, like this one from the 2009 AMC 10.
One can holds 12 ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?
It's a shame that many math competitions do not reward deep analysis and big-picture understanding. They emphasize speed and accuracy. In such cases, plugging in numbers and rough estimates are useful skills, as I pointed out in my essay Solving Problems with Choices.
In addition, smart guessing can boost the score, but I already wrote about that, too, in How to Boost Your Guessing Accuracy During Tests and To Guess or Not to Guess?, as well as Metasolving AMC 8.
As the AMC 10 fast approaches, I am bracing myself for the necessity to include multiple choices once again, thereby training my students in mindless arithmetic.
Many people ask me when is a good time to teach kids math. In my experience, it can never be too early. You just need to keep some order. Multiplication should be taught after addition, and negative numbers after subtraction. Kids should remember multiplication by heart at the age of seven. They can understand negative numbers as early as four.
In the picture I am explaining Platonic solids to four-month-old Eli, the son of my friends. His homework is to chew on a dodecahedron.
Suppose Romeo is encouraged by love and attention. If Juliet likes him, his feelings for Juliet grow and flourish. If she doesn't like him, he loses his interest in her.
Juliet, on the other hand, is the opposite. If Romeo doesn't like her, she needs to win him over and her attraction for him grows. If he likes her, she feels that her task is accomplished and she loses her interest in him. Juliet likes the challenge more than the relationship.
Steven Strogatz used differential equations to model the dynamics of the relationship between Romeo and Juliet. This is a new and fascinating area of applied mathematical research; you can read more about the roller-coaster relationship between Romeo and Juliet in Steven Strogatz's Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering
Mathematicians like symmetry: in math literature they switch the roles between Romeo and Juliet randomly. So in some papers they give Romeo the role of preferring a challenge over love and in some papers they give that role to Juliet.
When I teach this subject of love, Alexander Pushkin's famous quote always pops into my mind. The quote comes from the first lines of Chapter Four of Eugene Onegin, and in Russian it is:
Чем меньше женщину мы любим,
Тем легче нравимся мы ей…
I didn't like the English translations that I found, so I asked my son Alexey to provide a more literal translation:
The less we love a woman, the more she likes us in return…
I blame Pushkin for my tendency to always pick Juliet as the character who thrives on the challenge, even though men are often assumed to be the chasers. I'd like to ask my readers to comment on these roles: Do you think both genders play these roles equally? If not, then who is more prone to be into the chase?
Let's return to mathematical models. In the original model, the reactions of Romeo and Juliet are a linear function of feelings towards them. I would like to suggest two other roles, in which people react to the absolute value of feelings towards them. They do not care if it is love or hate: they care about intensity.
First, there is the person, like my friend Connie, who feeds on the emotions of other people. She's turned on by guys who love her as well as by guys who hate her. If they're indifferent, she's turned off.
Second, there is the opposite type, like my colleagues George, Joseph, David and many others. They hate emotion and prefer not to be involved. They lose all interest in people who feel strongly about them and they like people who are distant. I know the name for this role: it's a mathematician!
My friend Olga Amosova worked as a molecular biologist at Princeton University. Last time I visited her, we talked about her research.
She told me that she and her group designed a repair for a DNA mutation that is highly localized. "What's the point," I asked her, "of repairing DNA mutation in one cell?"
I was amazed to learn that not only is there a practical use to her research, but that there is something urgent that I myself must do.
There are many diseases that are caused by localized (so called "point") mutations. The most famous one is Sickle-cell disease. In Sickle-cell disease, defective hemoglobin causes erythrocytes to adopt a sickle shape that makes it difficult to pass through blood vessels. It is a very painful and debilitating disease. However, it turns out that the results of the research of Olga and her group could make the lives of people with such mutations much easier.
Stem cells have two amazing abilities. They grow fast and they can be turned into any type of cells in the human body. If the mutation is repaired in just one stem-cell, it can be selected and turned into a blood progenitor cell. These progenitor cells produce erythrocytes that actually transport oxygen. If these repaired cells are added to the patient's blood, they would produce good hemoglobin for half a year. This would improve the patient's quality of life tremendously.
So what do the rest of us learn from Olga's research? That we must save all left-over stem-cells that are produced in childbirth, like the umbilical cord and the placenta. It's not only Sickle-cell, but many other diseases that could benefit from using stem-cells. Research is moving so fast that these frozen stem-cells might become relevant in surprising ways — not only for the child, but also for relatives of the child — like you yourself!
So what's the urgent thing I must do? My son recently got married, so I must finish this post and send it to my son in case they get pregnant.
Mikhail Zhvanetsky is the most prolific and famous Russian humorist. Here are my own translations of some of his best lines.
I love the TV series of Angel
Where do vampires take their energy from? Usually oxygen is the fuel for the muscles of living organisms, but vampires do not breathe. Vampires are not living organisms, and yet they have to get their energy from somewhere.
When you kill a vampire, it turns to dust. If organisms are 60% water, then a 200-pound vampire should generate 80 pounds of dust. So why, in the series, do you get just a little puff of dust whenever someone plunges a stake into a vampire? Plus 120 pounds of water apparently evaporates instantly during staking. Can someone who is less lazy than me please calculate the energy needed to evaporate 120 pounds of water in one second? Because my first reaction is that you would need an explosion, not just one stab with Buffy's stake.
All these unscientific elements do not actually bother me that much. What does bother me are inconsistencies in logic. For example, at the end of Season One of Buffy, Angel refuses to give Buffy CPR, claiming that as a vampire he can't breathe. But then how can Spike and other vampires smoke? If they can smoke that means they are capable of inhaling and exhaling. Not to mention that these vampires talk: wouldn't they need an airflow through their throats to produce sounds?
It would make more sense for the show to state that vampires do not need to breathe, but are nonetheless capable of inhaling and exhaling. So Angel should have given Buffy CPR. It would have created a great plot twist: Angel saves Buffy at the end of Season One, only for her to send him to the hell dimension at the end of Season Two.
Back to breathing. I remember a scene in "Bring On the Night" in which Spike was tortured by Turok-Han holding his head in water. But if Spike can't breathe, why is this torture?
Another thing that bothers me in the series is not related to what happens but to what doesn't happen. For example, vampires do not have reflections. So I don't understand why every vampire-aware person didn't install a mirror on the front door of their house to check for reflections before inviting anyone in.
Also, it looks like producers do not care about backwards compatibility. Later in the series we get to know that vampires are cold. Watch the first season of Buffy with that knowledge. In the very first episode, Darla is holding hands with her victim, but he doesn't notice that she is cold. Later Buffy kisses Angel, before she knows that he is a vampire, and she doesn't notice that he's cold either. Unfortunately, the series also isn't forward compatible. In the second season of Angel in the episode "Disharmony", when we already know that vampires are cold, Harmony is trying to reconnect with Cordelia. They hug and touch each other. Such an experienced demon fighter as Cordelia should have noticed that Harmony is cold and, therefore, dead.
Finally, let's look at Spike in the last season of Angel. Spike is non-corporeal for a part of the season; we see him going through walls and standing in the middle of a desk. Yet, one time we see him sitting on a couch talking to Angel. In addition, he can take the stairs. He can go through the elevator wall to ride in an elevator instead of falling down through its floor. And what about floors? Why isn't he falling through floors? Some friends of mine said that we can assume that floors are made from stronger materials. But, if there is a material that can prevent Spike from penetrating it, they ought to use this material to make a weapon for him.
I've never been involved in making a show, but these producers clearly need help. Perhaps they should hire a mathematician like me with an eye for detail to prevent so many goofs.
The jokes are a rough translation from a Russian collection, except the last one I invented myself.
* * *
— Moishe, do you know how many cuckolds are there in Odessa not counting you?
— What? What do you mean by saying "not counting you"?
— Sorry. Okay then, how many counting you?
* * *
At a very prestigious Russian nursery school a teacher talks to a four-year-old applicant.
"Mike, can you count for me?"
Mike counts very fast and with a lot of enthusiasm, "Fifty-nine, fifty-eight, fifty-seven…"
"Super," says the teacher, "But how did you learn to count backwards?"
Mike replies proudly, "I can heat my own lunch — in the microwave."
* * *
The curl of the curl equals the gradient of the divergence minus the Laplacian. Why do I remember this shit that I never need, but can't remember where I put my keys yesterday?
* * *
In a bike store:
Customer: "Can you show me your finest helmet? I've already spent $200,000 on my head, so I don't want to take any risks."
Clerk, sympathetically: "You had a head trauma?"
Customer: "No, I went to college."
* * *
A topologist walks into a cafe:
— Can I have a doughnut of coffee, please.
In my old essay I presented the following coin problem.
We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weigh the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale. Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.
Here is my solution to this problem. Let us start with small values of N. For one coin you can't do anything. For two coins there isn't much you can do either. I will leave it to the readers to solve this for three coins, while I move on to four coins.
Let us compare two coins against the other two. The weighing has to unbalance. Then put aside the two coins from the right pan and compare one coin from the left pan with the other coin from the left pan. If they balance, then the right pan in the first weighing contained the fake coin. If they are unbalanced then the left pan in the first weighing contained the fake coin. Knowing where the fake coin was in the first weighing gives us the answer.
It is often very useful to go through the easy cases. For this problem we can scale the solution for three and four coins to get a solution for any number of coins that is divisible by three and four by just grouping coins accordingly. Thus we have solutions for 3k and 4k coins.
For any number of coins we can try to merge the solutions above. Divide all coins into three piles of size a, a and b, where a ≤ b ≤ 2a. In the first weighing compare the first two piles. If they balance, then the fake coin must be among the b remaining coins. Now pick any b coins from both pans in the first weighing and compare them to the remaining b coins. If the first weighing is unbalanced, then the remaining coins have to be real. For the second weighing we can pick a coins from the remaining pile and compare them to one of the pans in the first weighing.
The solution I just described doesn't cover the case of N = 5. I leave it to my readers to explain why and to solve the problem for N = 5.
Among ten given coins, some may be real and some may be fake. All real coins weigh the same. All fake coins weigh the same, but have a different weight than real coins. Can you prove or disprove that all ten coins weigh the same in three weighings on a balance scale?
When I first received this puzzle from Ken Fan I thought that he mistyped the number of coins. The solution for eight coins was so easy and natural that I thought that it should be eight — not ten. It appears that I was not the only one who thought so. I heard about a published paper with the conjecture that the best you can do is to prove uniformity for 2n coins in n weighings.
I will leave it to the readers to find a solution for eight coins, as well as for any number of coins less than eight. I'll use my time here to explain the solution for ten coins that my son Sergei Bernstein suggested.
First, in every weighing we need to put the same number of coins in both pans. If the pans are unbalanced, the coins are not uniform; that is, some of them are real and some of them are fake. For this discussion, I will assume that all the weighings are balanced. Let's number all coins from one to ten.
Consider two sets. The first set contains only the first coin and the second set contains the second and the third coins. Suppose the number of fake coins in the first set is a and a could be zero or one. The number of fake coins in the second set is b where b is zero, one or two. In the first weighing compare the first three coins against coins numbered 4, 5, and 6. As they balance the set of coins 4, 5, and 6 has to have exactly a + b fake coins.
In the second weighing compare the remaining four coins 7, 8, 9, and 10 against coins 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among the coins 7, 8, 9, and 10 is 2a + b.
For the last weighing we compare coins 1, 7, 8, 9, and 10 against 2, 3, 4, 5, and 6. The balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. This in turn means that either a = b = 0 and all the coins are real, or that a = 1, and b = 2 and all the coins are fake.
Now that you've solved the problem for eight and less coins and that I've just described a solution for ten coins, can we solve this problem for nine coins? Here is my solution for nine coins. This solution includes ideas of how to use a solution you already know to build a solution for a smaller number of coins.
Take the solution for ten coins and find two coins that are never on the same pan. For example coins 2 and 10. Now everywhere where we need 10, use 2. If we need both of them on different pans, then do not use them at all. The solution becomes:
The first weighing is the same as before with the same conclusion. The set containing the coin 1 has a fake coins, the set containing the coins 2 and 3 has b fake coins and the set containing coins 4, 5, and 6 has to have exactly a + b fake coins.
In the second weighing compare the four coins 7, 8, 9, and 2 against 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among 7, 8, 9, and 2 is 2a + b.
For the last weighing we compare coins 1, 7, 8, and 9 against 3, 4, 5, and 6. If we virtually add the coin number 2 to both pans, the balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. Which in turn means, similar to above, that either all the coins are real or all of them are fake.
It is known (see Kozlov and Vu, Coins and Cones) that you can solve the same problem for 30 coins in four weighings. I've never seen an elementary solution. Can you provide one?
I am starting yet another part-time job as the Head Mentor at PRIMES, a new MIT research program for high schoolers. I am very excited about this program, for it will be valuable not only to kids who want to become researchers, but also to kids who just want to see what research is like. Kids who want to learn to think in a new way will also find it highly useful.
PRIMES is in many ways similar to RSI, which it augments and complements. There are also a lot of differences. Keep in mind that I am only comparing PRIMES to the math part of RSI, with which I was working as a coordinator for two years. I do not know how RSI handles other sciences.
Different time scale. RSI lasts six weeks; PRIMES will take about a year. I already wrote about some peoples' skepticism towards RSI in my piece called "Fast Food Research?." PRIMES creates a more natural pace for research.
Choices. Because of the time schedule at RSI, students get their project as soon as they start. Students who realize by the end of the second week that they do not like their project are at a disadvantage: if they do not change their project, they're stuck with something that does not inspire them or is too difficult, and if they do change their project, they won't have enough time to do a great job. At PRIMES students will have time to talk to the mentors before starting their project, so that they can participate in choosing their project. Depending on how it goes later, they'll have time to try several different directions. I believe that the best research comes from the heart: students who have the time and opportunity to shape their choices will be more invested in their project.
Application process. At RSI, The Center for Excellence in Education reviews the applications. Even though they usually do a superb job at sending us great students, I believe it would be an advantage if mentors were able to influence the review process, for they might find even better matches to their projects. At PRIMES, the mentors will have this opportunity to review the applications.
Geography. RSI accepts students from all over the US and from some other countries. PRIMES can only accept local students — those who live close enough to visit MIT once a week for four months. Because of this restriction, PRIMES is recruiting from a smaller pool of students than RSI. But for local students it means that it will be easier to get accepted to PRIMES than to RSI.
Coaching. At RSI, students get a lot of coaching. I think that every student is in close contact with four adults. Two of them are from the math department — mentor and coordinator (that's me!) — and two tutors from CEE. PRIMES will have less coaching. A student will have a mentor and me, the head mentor. In addition, mentors might arrange for students to talk to the professors who originated their projects.
Immersion. RSI students are physically present. They are housed at MIT with the expectation that they completely devote their time to their research. Students at PRIMES will be integrating their research into the rest of their lives and their commitments. That will require good organizational skills and a lot of self-discipline. RSI students have discipline imposed on them by their situation — which may be an advantage to them.
Olympiads. While they are at RSI, students can't go to IMO or other summer activities. This is why many strong Olympiad students choose not to go to RSI, or they turn down an RSI acceptance if in the meantime they have gotten on to an Olympic team. At PRIMES you can do both. It is possible to go to an Olympiad, in addition to writing a paper.
Grade. RSI students have to be juniors. There are no grade limitations for PRIMES. Thus, it is possible to go to PRIMES in one's senior year. In this case, it may be too late to use PRIMES on college applications, but it is perfectly fine for the sake of research itself. Or it might be possible to go to PRIMES as a sophomore, and then apply for RSI the next year. This will strengthen the student's application for RSI.
RSI is well-established and has proven itself. PRIMES is new and hopefully will offer young mathematicians additional opportunities to try research.
I think that the American system of education creates a lot of pressure for teachers to drill their students for standardized tests and multiple choice questions. This blocks creative thinking. Every program like PRIMES is very good for unleashing students' creativity and contributing to the development of the future thinkers of American society.
Silvio Micali taught me cryptography. To explain one-way functions, he gave the following example of encryption. Alice and Bob procure the same edition of the white pages book for a particular town, say Cambridge. For each letter Alice wants to encrypt, she finds a person in the book whose last name starts with this letter and uses his/her phone number as the encryption of that letter.
To decrypt the message Bob has to read through the whole book to find all the numbers. The decryption will take a lot more time than the encryption. If the book increases in size the time it takes Alice to do the encryption almost doesn't increase, but the decryption process becomes more and more draining.
This example is very good for teaching one-way functions to non-mathematicians. Unfortunately, the technology changes and the example that Micali taught me fifteen years ago isn't so cute anymore. Indeed you can do a reverse look-up online of every phone number in the white pages.
I still use this example, with an assumption that there is no reverse look-up. I recently taught it to my AMSA students. And one of my 8th graders said, "If I were Bob, I would just call all the phone numbers and ask their last names."
In the fifteen years since I've been using this example, this idea never occurred to me. I am very shy so it would never enter my mind to call a stranger and ask for their last name. My student made me realize that my own personality affected my mathematical inventiveness.
Since modern technology is murdering my 15-year-old example, I would like to ask my readers to suggest other simple examples of one-way functions or ways to resurrect the white pages example.
I received a message at the beginning of October: "This month has 5 Fridays, 5 Saturdays and 5 Sundays; this only happens every 823 years."
Wait a minute. The Gregorian calendar cycles every 400 years. Where is the figure of 823 coming from?
Wait another minute. Within a century the calendar repeats itself every 28 years. So we are guaranteed that October 2038 will be the same as October 2010.
Wait one more minute. To have a month with five Fridays, Saturdays and Sundays, we need a month that has 31 days and starts on a Friday. There are seven months a year with 31 days, so on average we would expect to have such a month once a year.
If you study the calendar you can see that the seven long months start on six different days. This means that two of the months start on the same day and one of the days is skipped altogether. We see this in both leap years and non-leap years.
Ironically, 2010 is the year with two long months starting on Friday — October and January. Despite the claims of the email about this only happening every 823 years, in fact the same phenomenon occurred twice this year. The next time this will happen is in July 2011.
For those people who get all excited when a month has five Fridays, five Saturdays and five Sundays, I have good news for you. The month following each of these months has to start on Monday. And unless it is a February of a non-leap year, it will have five Mondays.
I asked my son Alexey Radul what exactly he is doing for his postdoc at the Hamilton Institute in Ireland. Here is his reply:
The short, jargon-loaded version: We are building an optimizing compiler for a programming language with first-class automatic differentiation, and exploring mathematical foundations, connections, applications, etc.
Interpretation of jargon:
Automatic differentiation is a technique for turning a program that computes a function into a program that computes that function together with its derivative; with a constant factor overhead. This is better than the usual symbolic differentiation that, say, Mathematica does because there is no intermediate-expression bulge. For example, if your function is a large product
Product f1(x) f2(x) ... fn(x),
the symbolic derivative has size n2
Sum (Product f1'(x) f2(x) ... fn(x)) (Product f1(x) f2'(x) ... fn(x)) ... (Product f1(x) f2(x) ... fn'(x))
automatic differentiation avoids that cost. Automatic (as opposed to symbolic) differentiation also extends to conditionals, data structures, higher-order functions, and all the other wonderful things that distinguish a computer program from a mathematical expression.
First-class means that the differentiation operations are normal citizens of the programming language. This is not the case with commonly used automatic differentiation systems, which are all preprocessors that rewrite C or Fortran source code. In particular, we want to be able to differentiate any function written in the language, even if it is a derivative of something, or contains a derivative of something, etc. The automatic differentiation technique works but becomes more complicated in the presence of higher order, multivariate, or nested derivatives.
We are building an optimizing compiler because the techniques necessary to get good performance and correct results with completely general automatic differentiation are exactly the techniques used to produce aggressive optimizing compilers for functional languages, so we might as well go all the way.
It appears that the AD trick (or at least half of it) is just an implementation of synthetic differential geometry in the computer. This leads one to hope that a good mathematical foundation can be found that dictates the behavior of the system in all the interesting special cases; there is lots of math to be thought about in the vicinity of this stuff.
Applications are also plentiful. Any time you want to optimize anything with respect to real parameters, gradients help. Any time you are dealing with curves, slopes help. Computer graphics, computer vision, physics simulations, economic and financial models, probabilities — there's so much stuff to apply a high quality such system to that we don't know where to begin.
I usually give a lot of lectures and I never used to announce them in my blog. This time I will give a very accessible lecture at the MIT "Women in Mathematics" series. It will be on Wednesday October 6th at 5:30-6:30 PM in room 2-135. If you are in Boston, feel free to join. Here is the abstract.
I will discuss several coin-weighing puzzles and related research. Here are two examples of such puzzles:
1. Among 10 given coins, some may be real and some may be fake. All real coins weigh the same. All fake coins weigh the same, but have a different weight than real coins. Can you prove or disprove that all ten coins weigh the same in three weighings on a balance scale?
2. Among 100 given coins, four are fake. All real coins weigh the same. All fake coins weigh the same, but they are lighter than real coins. Can you find at least one real coin in two weighings on a balance scale?
You are not expected to come to my talk with the solutions to the above puzzles, but you are expected to know how to find the only fake coin among many real coins in the minimum number of weighings.
In 2009 I was working at MIT coordinating math research for Research Science Institute for high school students. One of our students Jacob Hurwitz got a project on decycling graphs.
"Decycling" means removing vertices of a graph, so that the resulting graph doesn't have cycles. The decycling number of a graph is the smallest number of vertices you need to remove.
Decycling is equivalent to finding induced forests in a graph. The set of vertices of the largest induced forest is a complement to the smallest set of vertices you need to remove for decycling.
Among other things, Jacob found induced trees and forests of the highest densities on graphs of all semi-regular tessellations. On the pictures he provided for this essay, you can see an example of a tessellation, a corresponding densest forest, and a corresponding densest tree. The density of the forest and the tree is 2/3, meaning that 1/3 of the vertices are removed.
To motivate RSI students I tried to come up with practical uses for their projects. When I was talking to Jacob about decycling, the only thing I could think of was terrorists. When terrorists create their cells, they need to limit connections among themselves, in order to limit the damage to everyone else in the cell if one of them gets busted.
That means the graph of connections of a terrorist cell is a tree. Suppose there is a group of people that we suspect, and we know the graph of their contacts, then the decycling number of the graph is the number of people that are guaranteed to be innocent.
Have you noticed how Facebook and LinkedIn are reasonably good at suggesting people you might know? The algorithm they use to analyze the data is fairly effective in revealing potential connections. Recently, someone was able to download all of the Facebook data, which means that any government agency ought to be able to do the same thing. They could analyze such data to discover implicit connections. As a byproduct of looking for terrorists, they would also discover all of our grudges.
Oh dear. What are they going to think when they find out I'm not connected to my exes?
In the name of privacy, I have changed the names of the men I did not marry. But there is no point in changing the names of my ex-husbands, as my readers probably know their names anyway.
I received my first marriage proposal when I was 16. As a person who was unable to say "no" to anything, I accepted it. Luckily, we were not allowed to get married until I was 18, the legal marriage age in the USSR, and by that time we broke up.
To my next proposal, from Sasha, I still couldn't say "no", and ended up marrying him. The fact that I was hoping to divorce him before I got married at 19 shows that I should have devoted more effort in learning to say "no". I decided to divorce him within the first year.
My next proposal came from Andrey, I said yes, with every intention of living with Andrey forever. We married when I was 22 and he divorced me when I was 29.
After I recovered from my second divorce, I had a fling with an old friend, Sam, who was visiting Moscow on his way to immigrate to Israel.
Sam proposed to me in a letter that was sent from the train he took from the USSR to Israel. At that point I realized I had a problem with saying "no". The idea of marrying Sam seemed premature and very risky. I didn't want to say yes. I should have said no, but Sam didn't have a return address, so I didn't say anything.
That same year I received a phone call from Joseph. Joseph was an old friend who lived in the US, and I hadn't seen or heard from him for ten years. He invited me to visit him in the US and then proposed to me the day after my arrival. The idea of marrying Joseph seemed premature and very risky, but in my heart it felt absolutely right. I said yes, and I wanted to say yes.
I was very glad that I hadn't promised anything to Sam. But I felt uncomfortable. So even before I called my mother to notify her of my marriage plans, I located Sam in Israel and called him to tell him that I had accepted a marriage proposal from Joseph. I needed to consent to marry someone else as a way of saying "no" to Sam.
After I married Joseph, I came back to Russia to do all the paperwork and pick up my son, Alexey for our move to the US. There I met Victor. I wasn't flirting with Victor and was completely disinterested. So his proposal came as a total surprise. That was the time I realized that I had a monumental problem with saying "no". I had to say "no" to Victor, but I couldn't force myself to pronounce the word. Here is our dialogue as I remember it:
My sincere attempt at saying "no" didn't work. I moved to the US to live with Joseph and I soon got pregnant. Victor was the first person on my list to notify — another rather roundabout way to reject a proposal.
The marriage lasted eight years. Sometime after I divorced Joseph, I met Evan who invited me on a couple of dates. I wasn't sure I wanted to get involved with him. But he proposed and got my attention. I was single and available, though I had my doubts about him.
Evan mentioned that he had royal blood. So I decided to act like a princess. I gave him a puzzle:
I have two coins that together make 15 cents. One of them is not a nickel. What are my coins?
He didn't solve it. In and of itself, that wouldn't be a reason to reject a guy. But Evan didn't even understand my explanation, despite the fact that he was a systems administrator. A systems administrator who doesn't get logic is a definite turn-off.
So I said "no"! That was my first "no" and I have mathematics to thank.
When the Women and Math program at IAS was coming to an end, Ingrid Daubechies invited me to a picnic at her place for PACM (The Program in Applied and Computational Mathematics at Princeton University). I accepted with great enthusiasm for three reasons: I was awfully tired and needed a rest; PACM was my former workplace, so I was hoping to meet old acquaintances; and most of all, I loved the chance to hang out with Ingrid.
Ingrid is a great cook, so she prepared some amazing deserts for the picnic. While I was helping myself to a second serving of her superb lemon mousse, a man asked me if I was Ingrid's sister. Ingrid overheard this and laughingly told him that we are soul-sisters.
I admire Ingrid, so at first I took this as a compliment and felt all warm and fuzzy. But when my critical reasoning returned, I had to ask myself: Why would someone think I am Ingrid's sister with my Eastern European round face, my Russian name and my Russian accent?
I started talking to the man. He asked me what I do. I told him that I am a mathematician. He was stunned. What is so surprising in meeting a mathematician at a math department picnic?
Now I think I understand what happened. It never occurred to him that I was a mathematician. I was clearly unattached, so he couldn't place me as someone's wife. As the picnic was at Ingrid's house, he must have concluded that I had to be Ingrid's relative. Very logical, but very gender biased.
I heard this problem many years ago when I was working for Math Alive.
Three men are fighting in a truel. Andrew is the worst shot; he misses 2/3 of the time. Bob is better; he misses 1/3 of the time. Connor is the best shot; he always hits. Each of the three men have an infinite number of bullets. Each shot is either a kill or a miss. They have to shoot at each other in order until two of them are dead. To make it more fair they decide to start with Andrew, followed by Bob, and then Connor. We assume that they choose their strategies to maximize their probability of survival. At whom should Andrew aim for his first shot?
This is a beautiful probability puzzle, and I will not spoil it for you by writing a solution. Recently, I watched an episode of Numb3rs: The Fifth Season
Imagine a duel between three people. I'm the worst shot; I hit the target once every three trials. One of my opponents [Charlie] is better; hits it twice every three shots. The third guy [Colby] is a dead shot; he never misses. Each gets one shot. As the worst I go first, then Charlie, then Colby. Who do I aim for for my one shot?
This is a completely different problem; it's no longer about the last man standing. Colby doesn't need to shoot since the other two truelists have already expended their only shots. If Charlie believes that Colby prefers nonviolence, all else being equal, then Charlie doesn't need to shoot. Finally, the same is true of Marshall. There is no point in shooting at all.
To make things more mathematically interesting, let's assume that the truelists are bloodthirsty. That is, if shooting doesn't decrease their survival rate, they will shoot. How do we solve this problem?
If he survives, Colby will kill someone. If Charlie is alive during his turn, he has to shoot Colby because Colby might kill him. What should Marshall do? If Marshall kills Colby, then Charlie misses Marshall (hence Marshall survives) with probability 1/3. If Marshall kills Charlie, then Marshall is guaranteed to be killed by Colby, so Marshall survives with probability 0. If he doesn't kill anyone, things look much better: with probability 2/3, Colby is killed by Charlie and Marshall survives. Even if Colby is alive, he does not necessarily shoot Marshall, so Marshall survives with probability at least 2/3. Overall, Marshall's chances of staying alive are much better if he misses. He should shoot in the air!
The sad part of the story is that Charlie Eppes completely missed. That is, he completely missed the solution. In the episode he suggested that Marshall should shoot Charlie. If Marshall shoots Charlie, he will be guaranteed to die.
It is disappointing that a show about math can't get its math right.
From the 1966 Moscow Math Olympiad:
Prove that you can choose six weights from a set of weights weighing 1, 2, …, 26 grams such that any two subsets of the six have different total weights. Prove that you can't choose seven weights with this property.
Let us define the sequence a(n) to be the largest size of a subset of the set of weights weighing 1, 2, …, n grams such that any subset of it is uniquely determined by its total weight. I hope that you agree with me that a(1) = 1, a(2) = 2, a(3) = 2, a(4) = 3, and a(5) = 3. The next few terms are more difficult to calculate, but if I am not mistaken, a(6) = 3 and a(7) = 4. Can you compute more terms of this sequence?
Let's see what can be said about upper and lower bounds for a(n). If we take weights that are different powers of two, we are guaranteed that any subset is uniquely determined by the total weight. Thus a(n) ≥ log2n. On the other hand, the total weight of a subset has to be a number between 1 and the total weight of all the coins, n(n+1)/2. That means that our set can have no more than n(n+1)/2 subsets. Thus a(n) ≤ log2(n(n+1)/2).
Returning back to the original problem we see that 5 ≤ a(26) ≤ 8. So to solve the original problem you need to find a more interesting set than powers of two and a more interesting counting argument.
Fifteen years ago I attended Silvio Micali's cryptography course. During one of the lectures, he asked me to close my eyes. When I did, he wrote a random sequence of coin flips of length six on the board and invited me to guess it.
I am a teacher at heart, so I imagined a random sequence I would write for my students. Suppose I start with 0. I will not continue with zero, because 00 looks like a constant sequence, which is not random enough. So my next step would be sequence 01. For the next character I wouldn't say zero, because 010 seems to promise a repetitive pattern 010101. So my next step would be 011. After that I do not want to say one, because I will have too many ones. So I would follow up with 0110. I need only two more characters. I do not want to end this with 11, because the result would be periodic, I do not want to end this with 00, because I would have too many zeroes. I do not want to end this with 01, because the sequence 011001 has a symmetry: reversing and negating this sequence produces the same sequence.
During the lecture all these considerations happened in the blink of an eye in my mind. I just said: 011010. I opened my eyes and saw that Micali had written HTTHTH on the board. He was not amused and may even have thought that I was cheating.
Many teachers, when writing a random sequence, do not flip a coin. They choose a sequence that looks "random": it doesn't have too many repetitions and the number of ones and zeroes is balanced (that is, approximately the same). When they write it character by character on the board, they often choose a sequence so that any prefix looks "random" too.
As a result, the sequence they choose stops being random. Actually, they're making a choice from a small set of sequences that look "random". So the fact that I guessed Micali's sequence is not surprising at all.
If you have gone to many math classes, you've seen a lot of professors choosing very similar-looking "random" sequences. This discriminates against sequences that do not look "random". To restore fairness to those under-represented sequences, I have decided that the next time I need a random sequence, I will choose 000000.
Two month ago I made a minor rearrangement of my math humor page. The traffic to that page tripled. Would you like to know what I did? I collected all the suggestive jokes in one chapter and named it Dirty Math Jokes.
Mathematics is so far from sex that stumbling on a math sex joke is always a special treat.
Combinatorialists do it discretely.
When those jokes were randomly placed in my joke file, it was easy to miss flirtatious connotations.
She was only a mathematician's daughter, and she sure learned how to multiply using square roots.
So I decided to collect them together in one place.
Math Problem: A mother is 21 years older than her son. In 6 years she will be 5 times as old as her son. Where is the father?
Sergei Bernstein and Nathan Benjamin brought back a variation of the "Rock, Paper, Scissors" game from the Mathcamp. They call it "Rock, Paper, Scissor." In this variation one of the players is not allowed to play Scissors. The game ends as soon as someone wins a turn.
Can you suggest the best strategy for each player?
They also invented their own variation of the standard "Rock, Paper, Scissors." In their version, players are not allowed to play the same thing twice in a row.
If there is a draw, then it will remain a draw forever. So the game ends when there is a draw. The winner is the person who has more points.
They didn't invent a nice name for their game yet, so I am open to suggestions.
Let me describe a variation of Nim that is at the same time a variation of Chomp. Here's a reminder of what Nim and Chomp are.
In the game of Nim, there are several piles of matches and two players. Each of the players, in turn, can take any number of matches, but those matches must come from the same pile. The person who takes the last match wins. Some people play with a different variation in which the person who takes the last match loses.
Mathematicians do not differentiate between these two versions since the strategy is almost the same in both cases. The classic game of Nim starts with four piles that have 1, 3, 5 and 7 matches. I call this configuration "classic" because it is how Nim was played in one of my favorite movies, "Last Year at Marienbad". Recently that movie was rated Number One by Time Magazine in their list of the Top 10 Movies That Mess with Your Mind.
In the game of Chomp, also played by two people, there is a rectangular chocolate bar consisting of n by m squares, where the lowest left square is poisoned. Each player in turn chooses a particular square of the chocolate bar, and then eats this square as well as all the squares to the right and above. The player who eats the poisonous square loses.
Here is my Nim-Chomp game. It is the game of Nim with an extra condition: the piles are numbered. With every move a player is allowed to take any number of matches from any pile, with one constraint: after each turn the i-th pile can't have fewer matches than the j-th pile if i is bigger than j.
That was a definition of the Nim-Chomp game based on the game of Nim, so to be fair, here is a definition based on the game of Chomp. The game follows the rules of Chomp with one additional constraint: the squares a player eats in a single turn must all be from the same row. In other words, the chosen square shouldn't have a square above it.
The game of Nim is easy and its strategy has been known for many years. On the other hand, the game of Chomp is very difficult. The strategy is only known for 2 by m bars. So I invented the game of Nim-Chomp as a bridge between Nim and Chomp.
More and more I stumble upon the claim that the difference in individual personal choices between men and women is one of the main contributors to the gender gap in mathematical careers. Let me tell you some stories that I've heard that illustrate some choices that women made. The names have been changed.
Ann got her PhD in math at the same time as her husband. They both got job offers at places very far from each other. As Ann was pregnant, she decided not to accept her offer and to follow her husband. In two years she was ready to go back to research and she started to search for some kind of position at the University where her husband worked. At that time there was a nuptial rule that prevented two spouses from working at the same place. There was no other college nearby, and Ann's husband didn't yet have tenure and freaked out at the thought of changing his job. They decided to have a second child. After two more years of staying home, she felt completely disqualified and dropped the idea of research.
Olga was very passionate about her children's education. She felt that public education was insufficient and that parents needed to devote a lot of time to reading and playing with their kids. Her husband insisted that the children would be fine on their own and he refused outright to read to them or to participate in time-consuming activities. Olga took upon herself the burden she was hoping to share. At that same time, she started a tenure track position. In addition to all of this, she took her teaching very seriously. Her students loved her, but her paper-writing speed declined. She didn't get tenure and quit academia. Later, she realized that in fact her husband had shared her sense of the importance of their children's education, but he had played a power-game which left her doing all the work.
Maria told her husband Alex that she wanted a babysitter for their two children now that she was ready to get back to mathematics. Alex sat down with her to look at their finances. Before looking for a job Maria needed to finish her two papers. That meant they had to pay a babysitter even though Maria wasn't bringing in any money. Maria agreed to postpone her comeback until the kids went to kindergarten. She somehow finished her papers and found her first part-time job as an adjunct lecturer. Alex sat down with her again to discuss their finances. They calculated how much time she would need to prepare to teach after such a long break. The babysitter would cost more than her income. In addition, they would have to buy a second car and some professional clothes for Maria. They summed everything up: it appeared that they couldn't afford for her to take this job. Maria rejected the offer.
Two years later, Alex was offered an additional job as a part-time mathematical consultant. Instead of accepting it Maria's husband suggested that the company interview Maria, who was longing to return to work. Maria got an offer, but at half the fee her husband was offered. The manager explained this difference by pointing to her husband's superior work record. Maria and Alex sat down together again and calculated that it would more profitable for them as a couple if he took the job and dropped all his household responsibilities. Maria couldn't find a way to argue.
I recently wrote about my own decision to quit academia twelve years ago. Although what I really wanted to do was to work in academia, my family responsibilities took precedence.
Yes, personal choices are a great contributor to the gender gap in mathematical careers. I just do not like when people assume that women chose freely. Some choices we were cornered into making.
I was somewhat taken aback by the popularity of my earlier essay "Divisibility by 7 is a Walk on a Graph." Tanya tells me it got a good number of hits. The graph in that article is rather crude, and takes a bit of care to use, because the arrows go off in random directions from each node. So taking a hint from a commenter on the first graph, I redrew the graph, sacrificing planarity in favor of ease of use. Specifically, I arranged the black arrows in a counterclockwise circle, which makes them easy to follow.
The graph is used in the same way as the first graph. To find the remainder on dividing a number by 7, start at node 0, for each digit D of the number, move along D black arrows (for digit 0 do not move at all), and as you pass from one digit to the next, move along a single white arrow.
For example, let n = 325. Start at node 0, move along 3 black arrows (to node 3), then 1 white arrow (to node 2), then 2 black arrows (to node 4), then 1 white arrow (to node 5), and finally 5 black arrows (to node 3). Finishing at node 3 shows that the remainder on dividing 325 by 7 is 3.
I fancy it to be a little animal face.
I am a milk person. I can easily drink half a gallon of milk a day. The problem with half a gallon of milk a day is that it is about half of my target calorie intake. That is why I switched to the reduced-fat milk. It didn't taste good, but I was very proud of myself. That is, I was proud for about a year until I finally decided to read the labels. One serving of whole milk is 150 calories, while one serving of reduced-fat milk is 130 calories. All this year-long suffering saved me an insignificant 13%. Not to mention that the amount of sugar is the same.
I decided to look into this more closely. I went to my nearest super-market and checked the milk. The low-fat milk is 110 calories per serving, while non-fat milk is 90 calories.
If I had just reduced my milk intake by half, I would have consumed fewer calories than by replacing whole milk with non-fat milk, but I would have enjoyed it so much more. The lesson? Read the labels and do the math!
Forty years ago, it took about 18 months for us to find the rules that eventually became the Game of Life. We thought in terms of birth rules and death rules. Maybe one day's death rule would be a bit too strong compared to its birth rule. So the next day at coffee time we'd either try to weaken the death rule or strengthen the birth rule, but either way, only by a tiny bit. They had to be extremely well-balanced; if the death rule was even slightly too strong then almost every configuration would die off. And conversely, if the birth rule was even a little bit stronger than the death rule, almost every configuration would grow explosively.
What's wrong with that, you might ask. Well, if the "radius" grows by 1 unit per generation, then after 9 or 10 moves, it's off the (19 by 19) Go board. We can probably find more Go boards, of course, but after another 20 or so moves it will outflow the coffee table and then it is awfully hard to keep track. We wanted to be able to study configurations for much longer than that, which meant that we had to disallow rules that might lead to linear growth. Of course, we weren't interested in rules that usually led to collapse.
Who were "we"? Well, I was the chief culprit and had an aim in mind — to find a simple set of rules that would lead to a system able to simulate a universal computer. Von Neumann had already shown that this was possible, but his system had 29 states and a very complicated set of rules. The rest of "us" were mostly graduate students who had no higher aim than amusing themselves. Every now and then some rather older colleagues or visitors took an interest.
So my plan was, first, to find a set of rules that almost always prevented explosive growth and catastrophic collapse. Second, I wanted to study it long enough to learn how it could be "programmed". I hoped to find a system whose rules were much simpler than Von Neumann's, preferably with only two states (on and off) per cell, rather than his 29.
I'll just describe the last few rule fiddles. We had in fact given up on finding a two-state system, in favor of one with three states: 0, A, B. State 0 represented an empty cell, and it was natural to think of A and B as two sexes, but we only found their proper names when Martin Huxley walked by and said, "Actresses and bishops!"
Perhaps I should explain this. There is a British anecdote that starts like this:
"The actress sat on the left side of the bed, and removed her stockings. The bishop, on the right side of the bed, removed his gaiters. Then she unbuttoned her blouse and he took off his shirt…"
You are supposed to be getting excited, but it all ends quite tamely, because it turns out that the bishop was in his palace, while the actress was in her bedsit near the theater. There are lots of stories in England about the actress and the bishop, and if a person says something that has a salacious double meaning, it's standard to respond "as the actress said to the bishop," or "as the bishop said to the actress".
Okay, back to Life! To inhibit explosive growth, we decided to imitate biology by letting death be a consequence of either overcrowding or isolation. The population would only grow if the number of neighbors was neither too large nor too small. Rather surprisingly, this turned out to mean that children had to have three or more parents. Let's see why. If two parents could give birth, then in the figure below, the parents A and B, who are on the border of the population, would produce children A' and B' at the next time step, followed by grandchildren A" and B" and so on, thus giving us linear growth!
So we moved to threesomes. Children were born to three parents, made up of both sexes. Moreover, the sex of the child was determined by the sex of its parents — two bishops and one actress would give birth to a little actress, while two actresses and one bishop would produce a tiny bishop. This was "the weaker-sex birth rule," and it was accompanied by "the sexual frustration death rule," which made death the punishment for not touching somebody of the opposite sex!
However, the weaker-sex birth rule lived up to its name, by being weaker than the death rule. Remember we weren't interested in rules that led to disappointingly swift collapses, as the actress said to the bishop. Therefore, we strengthened the birth rule by allowing same-sex conception, but again by applying the weaker-sex rule — so that three actresses would produce a bishop or three bishops an actress. However this strengthened the birth rule too much, causing us to apply the death penalty more often.
We decided to apply the death penalty to those who weren't touching at least two other people, whatever their sex. At first sight it was not obvious that this was stronger than the sexual frustration rule, but in fact it was, because the weaker-sex rule ensured that the sexes were fairly evenly mixed, so if you were touching at least two other people, there was a good chance that one of them would be of the opposite sex.
According to our new set of rules, the sex of parents played no role except to determine the sex of the children, so we abolished sex. After all, according to the bishop, Life without sex is much cleaner.
This is now called the Game of Life and these rules, at last, turned out to be clean and well-balanced.
The article Daring to Discuss Women in Science by John Tierney in the New York Times on June 7, 2010 purports to present a dispassionate scientific defense of Larry Summers's claims, in particular by reviewing and expanding his argument that observed differences in the length of the extreme right tail of the bell curves of men's and women's test scores indicate real differences in their innate ability. But in fact any argument like this has to acknowledge a serious difficulty: it is problematic to assume without comment that the abilities of a group can be inferred from the tail of a bell curve. We are so used to invoking bell curves to talk about group abilities, we don't notice that such arguments usually use only the mean of the curve. Using the tail is a totally different story.
Think about it: it is reasonable to question whether a single data point — the test score of an individual person — is a true indication of his/her ability. It might not be. Maybe a single test score represents a dunce with hyper-overachieving parents who push him to study all the time. So does that single false reading destroy the validity of the curve? No of course not: because some other kid might have been a super-genius who was drunk last night and can barely keep his eyes open during the test. One is testing above his "true ability" and the other is testing below his "true ability," and the effect cancels out. Thus the means of curves are a good way to measure the ability of large groups, because all the random false readings average out.
But tails are not. On the tail this "canceling out" effect doesn't work. Look at the extreme right tail. The relatively slow but hyper-motivated kids are not canceled out by the hoard of far-above-the-mean super geniuses who had drunken revels the night before. There just aren't that many super-geniuses and they just don't party that much.
Or let's look at it another way: imagine that you had a large group which you divided in half totally at random. At this point their bell curve of test scores looks exactly the same. Lets call one of the group "boys" and the other group "girls". But they are two utterly randomly selected groups. Now lets inject the "boys" with a chemical that gives the ones who are very good already a burning desire to dominate any contest they enter into. And let us inject the "girls" with a chemical that makes the ones who are already good nonetheless unwilling to make anyone feel bad by making themselves look too good. What will happen to the two bell curves? Of course the upper tail of the "boys'" curve will stretch out, while the "girls'" tail will shrink in. It will look like the "boys" whipped the "girls" on the right tail of ability hands down, no contest. But the tail has nothing to do with ability. Remember they started out with the same distribution of abilities, before they got their injections. It is only the effect of the chemicals on motivation that makes it look like the "boys" beat the "girls" at the tail.
So, when you see different tails, you can't automatically conclude that this is caused by difference in underlying innate ability. It is possible that other factors are at play — especially since if we were looking to identify these hypothetical chemicals we might find obvious candidates like "testosterone" and "estrogen".
The possibility of alternative explanations for these findings calls into question Tierney and Summers' claims to superior dispassionate scientific objectivity. Moving from the mean to the tail of a bell curve makes systematic effects on averages irrelevant, true, but it is instead susceptible to systematic effects on deviations, which are irrelevant at the mean. An argument that uses this trick to dodge gender differences in averages cannot claim the mantle of scientific responsibility without accounting for gender differences in deviations. I am deeply disappointed that Tierney and Summers did not accompany their assertions with a suitable reminder of this fact.
I recently published a puzzle about wizards, hats of different colors and rooms. Unfortunately, I was too succinct in my description and didn't explicitly mention several assumptions. Although such assumptions are usual in this type of puzzle, I realize now, from your responses, that I should have listed them and I apologize.
The Sultan decided to test the wisdom of his wizards. He collected them together and gave them a task. Tomorrow at noon he will put hats of different colors on each of the wizard's heads.
The wizards have a list of the available colors. There are enough hats of each color for every wizard. The wizards also have a list of rooms. There are enough rooms to assign a different room for every color.
Tomorrow as the Sultan puts hats on the wizards, they will be able to see the colors of the hats of the other wizards, but not the color of their own. Without communicating with each other, each wizard has to choose a room. The challenge comes when two wizards have the same hat color, for they must choose the same room. On the other hand, if they have different hat colors, they must choose different rooms. Wizards have one day to decide on their strategy. If they do not all complete their task, then all of their heads will be chopped off. What strategy would you suggest for the wizards?
In his comment on the first blog about this problem, JBL beautifully described the intended solution for the finite number of wizards, and any potentially-infinite number of colors. I do not want to repeat his full solution here. I would rather describe his solution for two wizards and two colors.
Suppose the colors are red and blue. The wizards will designate one of the rooms as red and another as blue. As soon as each wizard sees the other wizard's hat color, he chooses the room of the color he sees. The beauty of this solution is that if the colors of hats are different, the color of the rooms will not match the color of the corresponding hats: the blue-hatted wizard will go to the red room and vice versa. But the Sultan's condition would still be fulfilled.
JBL's solution doesn't work if the number of wizards is infinite. I know the solution in that case, but I do not like it because it gives more power to the axiom of choice than I am comfortable with. If you are interested, you can extrapolate the solution from my essay on Countable Wise Men with Hats which offers a similar solution to a slightly different problem.
The International Math Olympiad started in Eastern Europe in 1959. Romania was the first host country. The Olympiad grew and only in 1976 did it move outside the Eastern bloc. The competition was held in Austria.
I was on the Soviet team in 1975 and 1976, so I was able to compare competitions held in Eastern vs. Western countries. Of course, the Austrian Olympiad was much better supported financially, but today I want to write about the differences in how our team was prepped.
Before our travel to Austria the Soviet team members were gathered in a room with strangers in suits for a chat. I assumed that we were talking to the KGB. They gave us a series of instructions. For example, they told us not to leave the campus during the competition, to always walk in groups, and to avoid talking to kids from countries that are enemies of the USSR. They warned us that they would be watching, and I was scared to death.
Now that I am older and wiser, I understand that their goal was to frighten us. Our team traveled with adult supervisors, who were trusted by the KGB. But for several days during the grading period of the competition, our supervisors were not allowed to see us. So the KGB wanted us to be too afraid to be very adventurous when we were left on our own.
In addition, the KGB had a Jewish problem. In general, Jews were not allowed to go abroad. I had many Jewish friends who qualified for the pre-IMO math camp where the team was chosen, but who were not able to get on the IMO because of delays with their travel documents. Some local bureaucrats were eager to impress the KGB and therefore held up visas for Jewish students, preventing them from being on the team. But the team selection process itself wasn't yet corrupt in 1976. So every year despite the efforts of the system, some young Jewish mathematicians would end up on the team.
Before 1976, the Olympiad was in the Eastern bloc, so the KGB wasn't quite so concerned about having Jewish members on the team. But Austria was not only a Western country, it was also the transition point for Jewish refugees from the Soviet Union. The speed with which the IMO moved their competition to a Western country was much faster than the Soviet bureaucratic machine could build a mechanism for completely preventing Jews from joining the team.
One very strong candidate, Yura Pass, didn't get his documents, but two other Jewish boys made it on to the team that was going to Austria. They were joking that they would be the only Soviet Jews who would go to Austria and actually come back. They did come back, only to go forward later: both are now math professors working in the US.
Because we had Jewish members on our team, it gave the KGB a special extra reason to scare us. But the biggest pressure was to win. We were told that 1976 was the most important year for the Soviet team to be the best. We were told that capitalist countries spread rumors that the judges in Eastern bloc countries favored the Soviet team and that the relative success of the Soviet team throughout the years had not been fully deserved. Now that the competition was in Austria, the suits told us, the enemies of the USSR were hoping for the downfall of the Soviet team. Our task was to prove once and for all that the Soviet students were the best at math, and that the rumors were unfounded. We had to win the team competition not only to prove ourselves, but also to clear the name of the Soviet team for all the previous years.
We did have a very strong team. The USSR came out first with 250 points, followed by the UK with 214 points and the USA with 188 points. Out of nine gold medals, we took four.
We could have gotten one more gold medal if Yura Pass had been allowed on the team. Yura was crushed by the machine's treatment of Jews and soon afterwards quit mathematics.
I just received a mass email on how to live longer and it made these points:
Tip 1. Delay your retirement. Studies show that people who retire at 65 live longer than people who retire at 60.
Tip 2. Sex makes you younger. Studies show that older people who have sex twice a week look ten years younger than their peers who do not have sex at all.
People who draw conclusions from such studies usually do not understand statistics. Correlation doesn't mean causality. Let me use the above-mentioned studies to reach different conclusions by reversing the causality assumption of the unknown writer of the mass email. You can compare results and make your own decisions.
Case 1. Studies show that people who retire at 65 live longer than people who retire at 60. Reversed causality: People who live longer are healthier, so they are able to keep working and to retire later in life.
Case 2. Studies show that older people who have sex twice a week look ten years younger than their peers who do not have sex at all. Reversed causality: Older people who look ten years younger than their peers can get laid easier, so they have sex more often.
Sergei just came back from MOP — Mathematical Olympiad Summer Program, where he was a grader. The first question I asked him was, "What was your favorite math problem there?" Here it is:
There are wisemen, hats and rooms. Hats are of different color and there are enough hats of each color for every wisemen. There are enough rooms, so that you can assign a different room for every color. At some moment in time the sultan puts hats on the wisemen's heads, so as usual they see all other hats, but do not see their own hat color. At the same time, each wiseman has to choose a room to go to. If two wisemen have the same hat color, they should go to the same room. If they have different hat colors, they should go to different rooms. What strategy should the wisemen decide upon before this event takes place?
Oh, I forgot to mention the most interesting part of this problem is that you shouldn't assume that the number of wisemen or hats or rooms is finite. You should just assume that they have the power of choice.
My son Alexey Radul and I were discussing the Tuesday's child puzzle:
You run into an old friend. He has two children, but you do not know their genders. He says, "I have a son born on a Tuesday." What is the probability that his second child is also a son?
Here is a letter he wrote me on the subject. I liked it because unlike many other discussions, Alexey not only asserts that different interpretations of the conditions in the puzzle form different mathematical problems, but also measures how different they are.
If you assume that boys and girls are symmetric, and that days of the week are symmetric (and if you have no information to the contrary, assuming anything else would be sheer presumption on your part), then you can be in one of at least two states.
1) You say that "at least one son born on a Tuesday" is all the information you have, in which case your distribution including this information is uniform over consistent cases, in which case your answer is 13/27 boy and your information entropy is
− ∑27 (1/27) log(1/27) = − log(1/27) = 3.2958.
2) You say that the information you have is "The guy might have said
any true thing of the form 'I have at least one {boy/girl} born on a
{day of the week}', and he said 'boy', 'Tuesday'." This is a
different mathematical problem with a different solution. The
solution: By a symmetry argument (see below [*]) we must assign uniform
probability of him making any true statement in any particular
situation. Then we proceed by Bayes' Rule: the statement we heard is
d, and for each possible collection of children
− ∑ p log p = − 1/14 log 1/14 − 26/28 log 1/28 = 3.2827
Therefore that assumed additional structure really is more information, which is only present at best implicitly in the original problem. How much more information? The difference in entropies is 3.2958 - 3.2827 = 0.0131 nats (a nat is to a bit what the natural log is to the binary log). How much information is that? Well, the best I can do is to reproduce an argument of E.T. Jaynes', which may or may not really apply to this situation. Suppose you have some repeatable experiment with some discrete set of possible outcomes, and suppose you assign probabilities to those outcomes. Then the number of ways those probabilities can be realized as frequencies counted over N trials is proportional to eNH, where H is the entropy of the distribution in nats. Which means that the ratio by which one distribution is easier to realize is approximately eN(H1-H2). In the case of N = 1000 and H1 - H2 = 0.0131, that's circa 5x105. For each way to get a 1000-trial experiment to agree with version 2, there are half a million ways to get a 1000-trial experiment to agree with version 1. So that's a nontrivial assumption.
[*] The symmetry argument: We are faced with the following probability assignment problem
Suppose our subject's first child is a boy born on a Tuesday, and his second child is a girl born on a Friday. What probability must we assign to him asserting that he has at least one boy born on a Tuesday?
Good question. Let's transform our coordinates: Let Tuesday' be Friday, Friday' be Tuesday, boy' be girl, girl' be boy, first' be second and second' be first. Then our problem becomes
Suppose our subject's second' child is a girl' born on a Friday', and his first' child is a boy' born on a Tuesday'. What probability must we assign to him asserting that he has at least one girl' born on a Friday'?
Our transformation necessitates p(boy Tuesday) = p(girl' Friday'), and likewise p(girl Friday) = p(boy' Tuesday'). But our state of complete ignorance about what's going on with respect to the man's attitudes about boys, girls, Tuesdays, Fridays, and first and second children has the symmetry that question and question' are the same question, and must, by the desideratum of consistency, have the same answer. Therefore p(boy Tuesday) = p(boy' Tuesday') = p(girl Friday) = 1/2.
My classmate Misha gave me a math problem. Although I liked the math part, I hated the setup. Here is the problem using the new setup:
You are hoping to get very rich one day and you base your hopes on your new invention: a diet pill. The pill works beautifully and doesn't have side effects. Patients simply take a pill when they get hungry. Within one hour they will fall asleep and will be unable to awake for exactly two hours, at which time they will awake by themselves feeling completely sated.
You have just arrived at your lab, when you realize that one of your interns has misplaced your bottle of wonder pills. After some investigation, you come to the conclusion that your bottle is on the shelf with 239 other bottles that contain a placebo. Unfortunately, those placebo bottles were custom-designed for your statistical tests to exactly match your medicine bottle.
You need to bring the medicine bottle to your boss in two hours. While you panic, all your five interns volunteer for experiments, hoping to be mentioned in your paper. Can you find your bottle in two hours?
The problem Misha gave me had 240 barrels of wine, one of which contained deadly poison and five slaves who could be spared.
I do not like killing people even when they are imaginary. But while I was slow in inventing my own setup, the original version of the puzzle started making the rounds on the Internet. So I decided to kill the problem by writing the solution.
The strategy is to give out some pills immediately, wait for one hour and see who falls asleep. The next step is to give some other pills at the beginning of the next hour to some of the interns who are awake.
Let's count the information you can get out of this. Each intern will experience one of three different situations: falling asleep in the first hour, in the second hour, or not falling asleep at all. Thus, you can have a total of 35 = 243 different outcomes.
If you had more bottles than 243, there would be no way to distinguish between them. The fact that you have 240 bottles might mean that 243 will work too, but apparently the designer of the puzzle didn't want to hint into powers of three and picked the largest round number below 243. These considerations should increase your willingness to look into this problem base 3.
Let us label the bottles with different 5-character strings containing three characters 0, 1, and 2. Now we can use the label as instructions. The first character will be associated with the first intern and so on. Suppose the fourth character on the bottle's label is 0, then the fourth intern doesn't need to struggle with digesting a pill from this particular bottle. If the fourth character is 1, then the fourth intern gets the pill in the first hour. If the fourth character is 2, the fourth intern gets the pill in the second hour.
Note this minor detail: Suppose the fourth character on a bottle is 2, but the fourth intern is asleep by the second hour. That means, the bottle doesn't contain the medicine, and we can put it aside.
At the end of two hours you know who fell asleep and when. This data will exactly match the label on the bottle with the medicine.
Once I read a book in Russian that mentioned a study of the children of Soviet military personnel who had to move often. The conclusion was that frequent relocation is very damaging for children's psyche. The children had to build new friendships, which they would lose the next time they had to move. After several moves they would stop making friends; later, as adults, they would be afraid of getting close to anyone.
In September 1996, my husband, my two children and I came to Princeton from Israel for my husband's month-long visit to the Institute for Advanced Study. After the visit we were supposed to go back to Israel, but that didn't happen. My husband returned alone and I stayed in Princeton with my children. That's a long and sentimental story for another time.
Meanwhile, my older son Alexey started going to Princeton High School. By this time he had attended seven schools in three different countries. In light of the evidence presented in that book about the impact on children of moving, I felt very guilty. Alexey was entering 10th grade. Moving him again not only would further damage his ability to make friends, but would also screw up his college chances. He needed a stable environment leading up to college. For example, recommendation letters are better written by people who are involved with kids for several years. I was afraid to mess up his future. I promised myself not to move him again during high school, especially as Princeton High School was one of the best public schools in New Jersey.
At the same time, I got a Visiting Scholar position at Princeton University. Although it didn't pay me any salary, through that position I received university housing, library privileges and an office. I was living on my personal savings and the monthly check my husband was sending me from Israel. My money was running out and I felt completely lost, like so many immigrants. I was new to Princeton; I didn't have friends there; and I was struggling with English. On top of that, I had medical problems, not the least very low energy.
Ingrid Daubechies noticed me at the Princeton math department and approached me. After our conversation, she found some money for me to work on the Math Alive course she was designing. That work was a breath of fresh air. I enjoyed it tremendously, but the part-time salary was not enough. Then I received more help from Ingrid. She appreciated my work on Math Alive a lot, but realized that I needed a different solution. She sacrificed her own interests and started recommending me around. She arranged an interview for me at Telcordia, who offered me a job as a systems engineer.
The decision to accept this job was very painful, because I did not want to leave academia. However, considering that my priority was to keep Alexey in Princeton High School, I didn't feel I had other options. I knew that I couldn't stay much longer at Princeton University and I was aware that getting a University job often requires relocation.
Looking back, I think the reasons behind this decision were more complex than sacrificing my career for my child. If I had known more about social supports for poor families and about other possible research jobs, or if I had been more confident in my research abilities, I might not have left academics.
Alexey triumphed at Princeton High School. The school allowed him to take math courses at Princeton University. He took several, including the course in logic by John Conway and two courses in graph theory by Paul Seymour. Alexey's multi-variable calculus professor complained to me that she couldn't fit her grades into the required curve. If she gave Alexey 100%, the others would have to get less than 20. Luckily, it turned out that because her class was small, she didn't need to bother about making a curve. After three years in Princeton High School, Alexey secured an impressive resume and great recommendation letters and went to MIT to pursue a double major in mathematics and computer science.
Alexey translated from a Russian joke site:
American scientists finally developed a car that runs on water. Unfortunately, at the moment it only runs on water from the Gulf of Mexico.
I've heard about many mathematicians running polymath projects through their blogs. I wasn't planning to do that. It just happened. In this essay, I describe the collaborative effort that was made to solve the following problem that appeared in my blog on July 2009:
Baron Münchhausen has n identical-looking coins weighing 1, 2, …, n grams. The Baron's guests know that he has this set of coins, but do not know which one is which. The Baron knows which coin is which and wants to demonstrate to his guests that he is right. He plans to conduct weighings on a balance scale, so that the guests will be convinced about the weight of every of coin. What is the smallest number of weighings that the Baron must do in order to reveal the weights?
The sequence a(n) of the minimal number of weighings is called the Baron Münchhausen's omni-sequence to distinguish it from the Existential Baron's sequence where he needs the smallest amount of weighings to prove the weight of one coin of his choosing.
In this essay I will describe efforts to calculate a(n). The contributors are: Max Alekseyev, Ilya Bogdanov, Maxim Kalenkov, Konstantin Knop, Joel Lewis and Alexey Radul.
The sequence starts as a(1) = 0, because there is nothing to demonstrate. Next, a(2) = 1, since with only one weighing you can find which coin is lighter.
Next, a(3) = 2. Indeed you can't prove all the coins in one weighing, but in the first weighing you can show that the 1-gram coin is lighter than the 2-gram coin. In the second weighing you can show that the 2-gram coin is lighter than the 3-gram coin. Thus, in two weighings you can establish an order of weights and prove the weight of all three coins.
As you can see in the case of n = 3, you can compare coins in order and prove the weight of all the coins in n − 1 weighings. But this is not at all the optimal number. Let us see why a(4) = 2. In his first weighing the Baron can put the 1- and the 2-gram coins on the left pan of the balance and the 4-gram coin on the right pan. In the future, I will just describe that weighing as 1 + 2 < 4. This way everyone agrees that the coin on the right pan is 4 grams, and the coin that is left out is 3 grams. The only thing that is left to do is to compare the 1-gram and the 2-gram coins in the second weighing.
Later Konstantin Knop sent me a different solution for n=4. His solution provides an interesting example. While looking for solutions, people usually try to have an unbalanced weighing to be "tight". That is, they make it so that the heavier cup is exactly 1 gram heavier than the lighter cup. If you are trying to prove one coin in one weighing, "tightness" is a requirement. But it is not necessary when you have several weighings. Here is the first weighing in Konstantin's solution: 1 + 3 = 4; and his second second weighing is: 1 + 2 < 3 + 4. We see that the second weighing has a weight difference of four between pans.
Next, a(5) = 2. We can have the first weighing the same as before: 1 + 2 < 4, and the second weighing: 1 + 4 = 5. The second weighing confirms that the heavy coin on the right pan in the first weighing can't be the heaviest one, thus it has to be the 4-gram coin. After that you can see that every coin is identified.
Next, a(6) = 2. The first weighing, 1 + 2 + 3 = 6, divides all coins into three groups: {1,2,3}, {4,5} and {6}. We know to which group each coin belongs, but we do not know which coin in the group is which. The second weighing: 1 + 6 < 3 + 5, identifies every coin. Indeed, the only possibility for the left side to weigh less than the right side is when the smallest weighing coin from the first group and 6 are on the left, and the two largest weighing coins from the first two groups are on the right.
When I was writing my essay I suspected that n = 6 is the largest number for which a solution can be established in two weighings, but I didn't have any proof. So I was embarrassed to show my solutions of three weighings n equals 7, 8 and 9.
On the other hand I published the solutions suggested by my son, Alexey Radul, for n = 10 and n = 11. In these cases the theoretical lower bound of log3(n) for a(n) is equal to 3, and finding solutions in three weighings was enough to establish the value of the sequence a(n) for n = 10 and n = 11.
So, a(10) = 3, and here are the weighings. The first weighing is 1 + 2 + 3 + 4 = 10. After this weighing, we can divide the coins into three groups {1,2,3,4}, {5,6,7,8,9} and {10}. The second weighing is 1 + 5 + 10 < 8 + 9. After the second weighing we can divide all coins into groups we know they belong to: {1}, {2,3,4}, {5}, {6,7}, {8,9} and {10}. The last weighing contains the lowest weighing coin from each non-single-coin group on the left and the largest weighing coin on the right, plus, in order to balance them, the coins whose weights we know. The last weighing is 2 + 6 + 8 + 5 = 4 + 7 + 9 + 1.
Similarly, a(11) = 3, and the weighings are: 1 + 2 + 3 + 4 < 11; 1 + 2 + 5 + 11 = 9 + 10; 6 + 9 + 1 + 3 = 8 + 4 + 2 + 5.
After publishing my blog I wrote a letter to the Sequence Fans mailing list asking them to expand the sequence. Max Alekseyev replied with the results of an exhaustive search program he wrote. First of all, he found a counter-intuitive solution for n=6. Namely, the following two weighings: 1 + 3 < 5 and 1 + 2+ 5 < 3 + 6. He also confirmed that it is not possible to identify the coins in two weighings for n=7, n=8 and n=9.
So now I can stop being embarrassed and proudly present my solution for n=7 in three weighings. That is, a(7) = 3 and the first weighing is: 1+2+3 < 7, and it divides all the coins into three groups {1,2,3}, {4,5,6} and 7. The second weighing, 1 + 4 < 6, divides them even further. Now we know the identity of every coin except the group {2,3}, which we can disambiguate with the third weighing: 2 < 3.
In many solutions that I've seen, one of the weighings was very special: every coin on one cup was lighter than every coin on the other cup. I wondered if that was always the case. Konstantin Knop send me a counterexample for n=7. The first weighing is: 1 + 2 + 3 + 5 = 4 + 7. The second is: 1 + 2 + 4 < 3 + 5. The third is: 1 + 3 + 4 = 2 + 6.
Later Max Alekseyev sent me two more special solutions for n=7. The first one contains only equalities: 2 + 5 = 7; 1 + 2 + 4 = 7; 1 + 2 + 3 + 5 = 4 + 7. The second one contains only inequalities: 1 + 3 < 5; 1 + 2 + 5 < 3 + 6; 5 + 6 < 2 + 3 + 7.
Moving to the next index, a(8) = 3 and the first weighing is: 1 + 2 + 3 + 4 + 5 < 7 + 8. The second weighing is: 1 + 2 + 5 < 4 + 6. After that we have identified all coins but two groups {1,2} and {3,4} that can be resolved by 2 + 4 = 6.
Meanwhile my blog received a comment from Konstantin Knop who claimed that he found solutions in three weighings for n in the range between 12 and 17 inclusive and four weighings for n = 53. I had already corresponded with Konstantin and knew that his claims are always well-founded, so I didn't doubt that he had found the solutions.
Later I began to write a paper with Joel Lewis on the upper bound of the omni-sequence, where we prove that a(n) ≤ 2 ⌈log2n⌉. For this paper, we wanted a comprehensive set of examples, so I emailed Konstantin asking him to write up his solutions. He promptly sent me the results and mentioned that he had found the weighings together with Ilya Bogdanov. They used several different ideas in the solutions. First I'll describe their solutions based on ideas we've already seen, namely to compare the lightest coins in the range to the heaviest coins.
Here is the proof that a(13) = 3. The first weighing is: 1 + … + 8 = 11 + 12 + 13, and it identifies the groups {1, 2, 3, 4, 5, 6, 7, 8}, {9, 10} and {11, 12, 13}. The second weighing is: (1 + 2 + 3) + 9 + (11 + 12) = (7 + 8) + 10 + 13, and it divides them further into groups {1, 2, 3}, {4, 5, 6}, {7, 8}, {9}, {10}, {11, 12}, {13}. And the last weighing identifies all the coins: 1 + 4 + 7 + 11 + 9 + 10 = 3 + 6 + 8 + 12 + 13.
Similarly, let us show that a(15) = 3. The first weighing is: 1 + … + 7 < 14 + 15, and it divides the coins into three groups {1, 2, 3, 4, 5, 6, 7}, {8, 9, 10, 11, 12, 13}, and {14, 15}. The second weighing is: (1 + 2 + 3) + 8 + (14 + 15) = (5 + 6 + 7) + (12 + 13), and this divides them further into groups {1, 2, 3}, {4}, {5, 6, 7}, {8}, {9, 10, 11}, {12, 13} and {14, 15}. The third weighing identifies every coin: 1 + 5 + 8 + 9 + 12 + 14 = 3 + 7 + 11 + 13 + 15.
As I mentioned earlier it is not always possible to find the first weighing which will nicely divide the coins into groups. We already discussed an example, n = 5, in which neither of the two weighings divided the coins into groups. Likewise, the same thing happened in the second mysterious solution for n = 6. What these solutions have in common is that the first weighing nearly divides everything nicely. The left pan is almost the set of the lightest coins and the right pan is almost the set of the heaviest coins. But not quite.
That is not our only situation in which the first weighing does not quite divide the coins into groups. For example, here is Konstantin's solution for a(9) = 3. For the first weighing, we put five coins on the left pan and two coins on the right pan. The left pan is lighter. This could happen in three different ways:
The second weighing, 1 + 2 + 3 = 6, in which we took three coins from the left pan and balanced them against one coin – again from the left pan – could only happen in case "C." After the two weighings, the following groups were identified: {1, 2, 3}, {4}, {5, 7}, {6}, {8, 9}. The third weighing, 1 + 4 + 5 + 8 < 3 + 7 + 9, identifies all the coins.
A similar technique is used in the solution that Konstantin sent to us to demonstrate that a(12) = 3. The first weighing is: 1 + 2 + 3 + 4 + 5 + 6 < 10 + 12. The audience which sees the results of the weighings understands that there are three possibilities for the distribution of coins:
The second weighing, (1 + 2 + 3) + (7 + 8) + 10 < (9 + 11 ) + 12, convinces the audience that the left pan must weigh at least 31 if the first weighing was case "A" above (31 = 1 + 2 + 3 + 7 + 8 + 10) or "C" (31 = 1 + 2 + 3 + 6 + 8 + 11), and at least 32 (32 = 1 + 2 + 3 + 7 + 8 + 11) if the first weighing was case "B." At the same time the right pan is not more than 12 + 9 + 11 = 32 for case "A" above, not more than 12 + 9 + 10 = 31 for case "B" and not more than 12 + 9 + 10 = 31 for case "C."
Hence the inequality in the second weighing is only possible when the first weighing was indeed as described by case "A" above. Consequently, the first two weighings together identify groups: {1, 2, 3}, {4, 5, 6}, {7, 8}, {9}, {10}, {11} and {12}. The third weighing, 1 + 4 + 7 + 11 + 12 < 3 + 6 + 8 + 9 + 10, identifies all the coins.
Other cases that Konstantin Knop sent me used a completely different technique. I would like to explain this technique using the mysterious solution for n = 6 found by Max Alekseyev. Suppose we have six coins labeled c1, … c6. The first weighing is: c1 + c3 < c5. The second weighing is: c1 + c2 + c5 < c3 + c6.
Let us prove that these two weighings identify all the coins. Let us replace the two inequalities above with the following: c1 + c3 − c5 ≤ −1, and c1 + c2 + c5 − c3 − c6 ≤ −1. Now we multiply the first inequality by 3 and the second by 2 and sum the results. We get: 5c1 + 2c2 + c3 + 0c4 − c5 − 2c6 ≤ −5. Note that the coefficients for labels are in a decreasing order. By the rearrangement inequality the smallest value the expression 5c1 + 2c2 + c3 + 0c4 − c5 − 2c6 reaches is when the labels on the coins match the indices. This smallest value is −5. Hence, the labels have to match the coins.
The technique that Konstantin and his collaborators are using is to search for appropriate coefficients to multiply the weighings by, rather than searching for the weighings themselves. In lieu of lengthy explanations, I will just list the weighings that he uses together with coefficients to multiply them by for their proof that the weighings differentiate coins.
We will start with showing that a(14) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 < 11 + 13 + 14, and: 1 + 2 + 3 + 8 + 11 + 13 = 7 + 9 + 10 + 12, followed by 1 + 4 + 7 + 10 = 3 + 6 + 13. The coefficients to multiply by are {9, 5, 2}.
Next we will show that a(16) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 8 < 14 + 16, and 1 + 2 + 3 + 7 + 9 + 14 = 8 + 13 + 15, followed by 1 + 4 + 7 + 10 + 13 < 3 + 6 + 12 + 15. The coefficients to multiply by are {11, 5, 2}.
Next we will show that a(17) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 < 15 + 16 + 17, and 1 + 2 + 3 + 8 + 11 + 15 + 16 < 7 + 9 + 10 + 14 + 17, and 1 + 4 + 7 + 8 + 12 + 14 = 3 + 6 + 10 + 11 + 16. The coefficients to multiply by are {11, 5, 2}.
Next we will show that a(53) = 4. The weighings are: (1 + 2 + … + 23) + 25 < 47 + (49 + … + 53), and (1 + … + 9) + 24 + (26+ … + 31) + 47 + (49 + … + 52) < (16 + … + 23) + 25 + (41 + … + 46) + 48 + (51 + 52), and (1 + 2 + 3) + (10 + 11) + (16 + 17 + 18) + 24 + (26 + 27) + (32 + 33 + 34) + (41 + 42 + 43) + 47 + 49 + 53 =(7 + 8 + 9) + 15 + (22 + 23) + 25 + (30 + 31) + (38 + 39) + 40 + (45 + 46) + 48 + (51 + 52), and the last one 1 + 4 + 7 + 10 + 12 + 16 + 19 + 22 + 24 + 28 + 30 + 32 + 35 + 38 + 41 + 45 + 47 + 51 + 53 < 3 + 6 + 9 + 11 + 14 + 18 + 21 + 25 + 27 + 29 + 34 + 37 + 40 + 43 + 48 + 49 + 50 + 52. The coefficients to multiply by are {43, 15, 5, 2}.
When I was working on the paper with Joel Lewis I re-established my email discussions about the Baron's onmi-sequence with Konstantin Knop. At that time Konstantin's colleague, Maxim Kalenkov, got interested in the subject and wrote a computer search program to find other solutions that can be proven with the rearrangement inequality. Thus, we know two more terms of this sequence.
The next known term is a(18) = 3. The weighings are: 1 + 2 + 4 + 5 + 7 + 10 + 12 = 9 + 15 + 17, and 1 + 3 + 4 + 6 + 9 + 11 + 17 = 7 + 12 + 14 + 18, and 2 + 3 + 7 + 8 + 9 + 14 + 15 = 4 + 10 + 11 + 16 + 17. The corresponding coefficients are: {8, 7, 5}.
Similarly, a(19) = 3. The weighings are: 1 + 2 + 3 + 4 + 5 + 7 + 8 + 10 + 13 = 16 + 18 + 19, and 1 + 2 + 3 + 6 + 9 + 11 + 16 = 8 + 10 + 13 + 17, and 1 + 4 + 6 + 8 + 12 + 18 = 3 + 7 + 11 + 13 + 15. The coefficients are {12, 7, 3}.
Maxim Kalenkov continued his search. He didn't find any new solutions in three weighings, but he found a lot of solutions in four weighings, namely for numbers from 20 to 58. Below are his solutions, with multiplier coefficients in front of every weighing:
a(20) ≤ 4
18: 1+2+3+4+5+10+14+16+18 = 6+7+11+12+17+20
19: 1+2+4+5+12+15+17+19 < 6+8+10+14+18+20
21: 2+6+11+17+18 = 4+9+10+15+16
26: 1+6+7+8+9+10+20 = 2+5+17+18+19
a(21) ≤ 4
18: 3+5+6+11+15+17+19 = 7+8+9+13+18+21
19: 4+6+9+13+16+18+20 = 1+7+11+12+15+19+21
21: 1+4+5+7+12+18+19 = 3+9+10+11+16+17
26: 1+2+3+7+8+9+10+11+21 = 4+5+6+18+19+20
a(22) ≤ 4
18: 3+5+6+12+15+17+20 = 7+8+10+13+18+22
19: 1+2+4+10+13+16+18+21 = 7+9+12+15+20+22
21: 1+6+7+18+19+20 = 2+3+10+11+12+16+17
26: 2+3+7+8+9+10+11+12+22 = 6+18+19+20+21
a(22) ≤ 4
18: 1+2+5+6+12+16+18+22 < 3+7+8+10+13+19+23
19: 1+3+4+6+10+17+19+21 = 7+9+12+14+16+23
21: 3+7+13+14+19+20 = 2+6+10+11+12+17+18
26: 2+7+8+9+10+11+12+23 = 19+20+21+22
a(24) ≤ 4
18: 1+3+6+12+17+19+21+23 < 2+4+7+8+10+13+15+20+24
19: 1+2+4+5+6+10+15+18+20+22 < 7+9+12+14+17+21+24
21: 4+7+13+14+20+21 = 3+6+10+11+12+18+19
26: 2+3+7+8+9+10+11+12+24 = 20+21+22+23
a(25) ≤ 4
18: 1+2+3+5+6+8+9+14+17+19+22 < 10+12+16+20+24+25
19: 2+6+7+9+12+16+18+20+23 = 10+11+14+15+17+22+24
21: 1+2+4+7+8+10+15+20+21+22 = 3+6+12+13+14+18+19+25
26: 3+10+11+12+13+14+24+25 = 2+7+8+9+20+21+22+23
a(26) ≤ 4
18: 1+2+3+5+6+8+9+14+18+20+23 = 10+12+15+21+25+26
19: 2+3+4+6+7+9+12+19+21+24 = 11+14+16+18+23+25
21: 7+8+15+16+21+22+23 = 2+6+12+13+14+19+20+26
26: 1+2+10+11+12+13+14+25+26 = 7+8+9+21+22+23+24
a(27) ≤ 4
18: 1+3+4+6+7+8+9+14+19+21+23+25 < 10+11+13+15+17+22+26+27
19: 1+2+3+5+7+9+13+17+20+22+24 < 4+10+12+14+16+19+23+26
21: 2+3+4+8+10+15+16+22+23 = 1+7+13+14+20+21+27
26: 1+10+11+12+13+14+26+27 = 3+8+9+22+23+24+25
a(28) = 4
18: 3+6+8+9+10+15+19+21+24+26 = 1+5+11+13+16+18+22+27+28
19: 1+4+5+7+9+10+13+18+20+22+25 = 3+6+11+12+15+17+19+24+27
21: 5+6+11+16+17+22+23+24 = 4+9+13+14+15+20+21+28
26: 1+2+3+4+11+12+13+14+15+27+28 = 10+22+23+24+25+26
a(29) = 4
18: 1+3+5+6+7+9+10+16+20+22+25+27 < 11+12+14+17+18+23+28+29
19: 4+8+10+14+18+21+23+26 = 2+3+6+11+13+16+20+25+28
21: 1+2+6+8+9+11+17+23+24+25 = 4+5+14+15+16+21+22+29
26: 2+3+4+5+11+12+13+14+15+16+28+29 = 8+9+10+23+24+25+26+27
a(30) = 4
18: 2+8+10+16+21+23+26+27+28 = 5+11+12+14+17+19+24+29+30
19: 2+4+5+7+9+14+19+22+24+28 = 11+13+16+18+21+26+29
21: 1+5+6+9+10+11+17+18+24+25+26 = 4+14+15+16+22+23+28+30
26: 1+3+4+11+12+13+14+15+16+29+30 < 9+10+24+25+26+27+28
a(31) = 4
18: 1+2+6+9+10+16+21+23+26+28+29 < 3+4+7+11+12+14+17+19+24+30+31
19: 1+2+4+7+14+19+22+24+27+29 < 6+9+11+13+16+18+21+26+30
21: 2+3+7+8+9+11+17+18+24+25+26 = 14+15+16+22+23+29+31
26: 1+3+4+5+6+11+12+13+14+15+16+30+31 = 2+24+25+26+27+28+29
a(32) = 4
18: 1+5+8+9+10+12+13+22+24+27+29 = 6+14+16+18+20+25+30+31
19: 4+6+10+11+13+16+20+23+25+28 = 1+2+8+15+19+22+27+30+32
21: 1+2+6+7+8+11+12+18+19+25+26+27 = 4+5+10+16+17+23+24+31+32
26: 1+2+3+4+5+14+15+16+17+30+31+32 < 11+12+13+25+26+27+28+29
a(33) = 4
18: 1+2+6+7+8+9+10+12+13+23+27+29+30 = 3+14+15+17+19+21+25+31+32
19: 1+2+10+11+13+17+21+24+25+28+30 = 4+6+8+14+16+20+23+27+31+33
21: 1+3+4+8+11+12+14+19+20+25+26+27 < 7+10+17+18+24+30+32+33
26: 3+4+5+6+7+14+15+16+17+18+31+32+33 = 11+12+13+25+26+27+28+29+30
a(34) = 4
18: 2+3+4+8+10+12+13+19+24+28+30+31 < 6+14+15+17+20+22+26+32+33
19: 1+2+3+5+6+9+11+13+17+22+25+26+29+31 = 4+8+14+16+19+21+24+28+32+34
21: 1+3+6+7+8+11+12+14+20+21+26+27+28 = 2+5+17+18+19+25+31+33+34
26: 1+2+4+5+14+15+16+17+18+19+32+33+34 = 3+11+12+13+26+27+28+29+30+31
a(35) = 4
18: 1+3+4+6+8+10+11+13+19+24+26+29+31+32 = 14+15+17+20+22+27+33+34+35
19: 1+4+7+9+11+12+17+22+25+27+30+32 = 6+14+16+19+21+24+29+33+35
21: 2+12+13+14+20+21+27+28+29+35 = 4+7+8+11+17+18+19+25+26+32+34
26: 1+2+3+4+5+6+7+8+14+15+16+17+18+19+33+34 = 12+13+27+28+29+30+31+32
a(36) = 4
18: 1+2+3+4+5+9+11+12+14+15+20+25+29+31+32 < 6+16+18+21+23+27+33+34+36
19: 1+2+4+5+6+8+10+12+13+15+18+23+26+27+30+32 < 16+17+20+22+25+29+33+35+36
21: 1+3+5+13+14+16+21+22+27+28+29+36 = 2+8+9+12+18+19+20+26+32+34+35
26: 2+6+7+8+9+16+17+18+19+20+33+34+35 = 5+13+14+15+27+28+29+30+31+32
a(37) = 4
18: 1+2+3+6+8+10+12+13+15+16+21+27+30+32+33 = 4+9+17+19+22+24+28+34+35+37
19: 1+2+3+7+9+11+13+14+16+19+24+26+28+31+33 = 5+6+10+17+18+21+23+30+34+36+37
21: 3+4+5+9+10+14+15+17+22+23+28+29+30+37 = 1+7+8+13+19+20+21+26+27+33+35+36
26: 1+4+5+6+7+8+17+18+19+20+21+34+35+36 = 3+14+15+16+28+29+30+31+32+33
a(38) = 4
18: 2+3+4+7+9+11+13+15+16+21+26+28+31+33+34 = 5+10+17+18+19+22+24+29+35+36+38
19: 1+3+4+8+10+12+13+14+16+19+24+27+29+32+34 = 7+11+17+21+23+26+31+35+37+38
21: 1+2+4+5+10+11+14+15+17+22+23+29+30+31+38 = 8+9+13+19+20+21+27+28+34+36+37
26: 5+6+7+8+9+17+18+19+20+21+35+36+37 = 4+14+15+16+29+30+31+32+33+34
a(39) = 4
18: 1+2+4+7+9+11+13+14+16+22+27+29+32+34+35 = 10+17+18+20+23+25+30+36+38+39
19: 2+3+4+8+10+12+14+15+20+25+28+30+33+35 = 5+7+11+17+19+22+24+27+32+37+38
21: 1+2+3+5+10+11+15+16+17+23+24+30+31+32+38 < 8+9+14+20+21+22+28+29+35+36+37
26: 1+5+6+7+8+9+17+18+19+20+21+22+36+37 = 15+16+30+31+32+33+34+35
a(40) = 4
18: 1+2+3+4+5+8+9+12+14+15+17+18+23+27+29+32+34+35 < 7+10+19+21+24+26+30+36+37+39+40
19: 1+3+4+5+7+10+13+15+16+18+21+26+28+30+33+35 < 6+8+12+20+23+25+27+32+36+38+39
21: 1+5+6+10+11+12+16+17+24+25+30+31+32+39 < 3+9+15+21+22+23+28+29+35+37+38
26: 2+3+6+7+8+9+19+20+21+22+23+36+37+38 = 5+16+17+18+30+31+32+33+34+35
a(41) = 4
18: 1+3+5+6+8+10+12+14+15+17+18+23+28+30+33+35+36 < 7+11+19+21+24+26+31+37+38+40+41
19: 1+2+4+5+6+7+9+11+13+15+16+18+21+26+29+31+34+36 = 8+12+19+20+23+25+28+33+37+39+40
21: 1+4+6+11+12+16+17+19+24+25+31+32+33+40 < 9+10+15+21+22+23+29+30+36+38+39
26: 2+3+7+8+9+10+19+20+21+22+23+37+38+39 = 6+16+17+18+31+32+33+34+35+36
a(42) = 4
18: 1+3+5+6+7+11+13+15+16+18+24+29+31+34+36+37 = 2+8+12+19+20+22+25+27+32+38+40+41
19: 1+2+4+6+7+10+12+14+16+17+18+22+27+30+32+35+37 = 3+13+19+21+24+26+29+34+39+40+42
21: 1+2+3+4+5+7+8+12+13+17+19+25+26+32+33+34+40 = 10+11+16+22+23+24+30+31+37+38+39
26: 2+3+8+9+10+11+19+20+21+22+23+24+38+39 = 7+17+18+32+33+34+35+36+37
a(43) = 4
18: 1+2+5+6+7+10+12+14+16+17+19+20+29+31+34+36+37 = 8+21+23+25+27+32+38+39+41+42
19: 2+4+5+6+7+8+11+15+17+18+20+23+27+30+32+35+37 = 10+14+22+26+29+34+38+40+41+43
21: 1+2+3+7+13+14+18+19+25+26+32+33+34+41 < 5+11+12+17+23+24+30+31+37+39+40
26: 1+3+4+5+8+9+10+11+12+21+22+23+24+38+39+40 < 7+18+19+20+32+33+34+35+36+37
a(44) = 4
18: 1+2+5+6+7+8+10+11+14+16+17+19+20+25+30+32+35+37+38 = 3+12+21+22+23+26+28+33+39+40+42+44
19: 1+2+3+7+8+12+15+17+18+20+23+28+31+33+36+38+44 = 9+10+14+21+25+27+30+35+39+41+42+43
21: 2+3+4+6+8+9+12+13+14+18+19+21+26+27+33+34+35+42 = 11+17+23+24+25+31+32+38+40+41+44
26: 1+3+4+5+9+10+11+21+22+23+24+25+39+40+41 = 8+18+19+20+33+34+35+36+37+38
a(45) = 4
18: 2+4+5+6+11+13+16+17+18+20+21+26+31+33+36+38+39 = 7+9+14+22+24+27+29+34+40+42+43+45
19: 1+2+4+6+9+12+14+18+19+21+24+29+32+34+37+39+45 = 8+11+16+23+26+28+31+36+40+41+42+44
21: 1+2+3+5+7+8+14+15+16+19+20+27+28+34+35+36+42 = 4+12+13+18+24+25+26+32+33+39+41+45
26: 1+3+4+7+8+9+10+11+12+13+22+23+24+25+26+40+41 = 19+20+21+34+35+36+37+38+39
a(46) = 4
18: 1+2+3+5+6+8+9+14+16+17+19+21+22+26+31+33+36+38+39 = 10+12+23+27+29+34+40+41+42+43+45
19: 2+4+6+7+8+9+12+15+18+20+22+29+32+34+37+39+45 = 3+11+14+17+23+24+26+28+31+36+40+42+44
21: 1+3+7+9+10+11+17+20+21+23+27+28+34+35+36+42 = 6+15+16+25+26+32+33+39+41+45+46
26: 1+2+3+4+5+6+10+11+12+13+14+15+16+23+24+25+26+40+41 = 9+20+21+22+34+35+36+37+38+39
a(47) = 4
18: 2+3+5+7+8+9+13+14+16+18+19+21+22+27+32+34+37+39+40 < 11+23+24+28+30+35+41+42+43+44+46
19: 1+3+6+7+8+9+11+17+19+20+22+30+33+35+38+40+46 < 5+10+13+16+23+25+27+29+32+37+41+43+45
21: 1+2+4+5+9+10+15+16+20+21+23+28+29+35+36+37+43 < 7+14+19+26+27+33+34+40+42+46+47
26: 1+2+3+4+5+6+7+10+11+12+13+14+23+24+25+26+27+41+42 < 9+20+21+22+35+36+37+38+39+40
a(48) = 4
18: 3+5+6+7+8+13+17+19+20+22+23+28+33+35+38+40+41 < 9+11+15+24+26+29+31+36+42+44+45+47
19: 1+4+6+8+11+14+15+18+20+21+23+26+31+34+36+39+41+47 < 3+10+13+17+24+25+28+30+33+38+42+43+44+46
21: 1+2+3+7+8+9+10+15+16+17+21+22+24+29+30+36+37+38+44 = 6+14+20+26+27+28+34+35+41+43+47+48
26: 1+2+3+4+5+6+9+10+11+12+13+14+24+25+26+27+28+42+43 = 8+21+22+23+36+37+38+39+40+41
a(49) = 4
18: 2+3+6+7+8+9+10+15+18+20+21+23+24+29+34+38+40+41+49 < 12+16+25+26+30+32+36+42+43+44+45+47
19: 1+3+5+7+9+10+12+14+16+19+21+22+24+32+35+36+39+41+47 < 11+18+25+27+29+31+34+38+42+44+46+49
21: 1+2+3+4+8+10+11+16+17+18+22+23+25+30+31+36+37+38+44 < 7+14+15+21+28+29+35+41+43+47+48+49
26: 1+2+4+5+6+7+11+12+13+14+15+25+26+27+28+29+42+43 = 10+22+23+24+36+37+38+39+40+41
a(50) = 4
18: 1+2+3+4+6+7+9+10+11+15+16+19+20+21+23+24+34+36+39+41+42+50 = 12+17+25+26+28+30+32+37+43+44+45+46+48
19: 2+3+5+7+8+10+11+17+21+22+24+28+32+35+37+40+42+48 = 4+13+15+19+25+27+31+34+39+43+45+47+50
21: 1+3+4+8+9+11+12+13+17+18+19+22+23+25+30+31+37+38+39+45 = 7+16+21+28+29+35+36+42+44+48+49+50
26: 1+2+4+5+6+7+12+13+14+15+16+25+26+27+28+29+43+44 = 11+22+23+24+37+38+39+40+41+42
a(51) = 4
18: 2+3+4+6+8+10+11+15+17+19+20+21+23+24+30+35+37+40+42+43+50 < 12+13+18+25+26+28+31+33+38+44+46+47+49+51
19: 1+3+4+7+9+11+13+16+18+21+22+24+28+33+36+38+41+43+49 < 6+15+19+25+27+30+32+35+40+45+46+48+50
21: 1+2+4+5+6+9+10+11+12+18+19+22+23+25+31+32+38+39+40+46+51 < 16+17+21+28+29+30+36+37+43+44+45+49+50
26: 1+2+3+5+6+7+8+12+13+14+15+16+17+25+26+27+28+29+30+44+45 < 11+22+23+24+38+39+40+41+42+43+51
a(52) = 4
18: 2+5+7+8+10+11+12+17+19+22+25+26+31+35+37+40+42+43+51 = 3+13+15+20+27+28+29+32+38+44+46+47+49+52
19: 1+2+3+6+8+9+11+12+15+18+20+23+24+26+29+36+38+41+43+49 = 5+14+17+22+27+31+33+35+40+45+46+48+51
21: 1+3+4+5+9+10+12+13+14+20+21+22+24+25+27+32+33+38+39+40+46+52 = 8+18+19+29+30+31+36+37+43+44+45+49+50+51
26: 1+2+3+4+5+6+7+8+13+14+15+16+17+18+19+27+28+29+30+31+44+45 = 12+24+25+26+38+39+40+41+42+43+52
a(53) = 4
18: 2+3+4+7+8+9+10+11+12+17+19+21+23+25+26+31+36+38+41+43+44+52 < 5+13+15+27+29+32+34+39+45+46+47+48+50+53
19: 1+3+4+5+9+11+12+15+18+22+23+24+26+29+34+37+39+42+44+50 = 7+14+17+21+27+28+31+33+36+41+45+47+49+52
21: 1+2+4+5+6+7+10+12+13+14+20+21+24+25+27+32+33+39+40+41+47+53 < 9+18+19+23+29+30+31+37+38+44+46+50+51+52
26: 1+2+3+5+6+7+8+9+13+14+15+16+17+18+19+27+28+29+30+31+45+46 = 12+24+25+26+39+40+41+42+43+44+53
a(54) = 4
18: 2+3+4+6+8+9+11+12+13+18+20+23+25+26+28+29+33+37+39+41+42+43+51 = 5+14+16+21+30+31+32+35+44+45+47+48+49+52+54
19: 1+3+4+5+7+9+10+12+13+16+19+21+24+26+27+29+32+35+38+43+49+54 < 6+15+18+23+30+33+34+37+41+44+46+47+51+53
21: 1+2+4+5+6+10+11+13+14+15+21+22+23+27+28+30+34+40+41+47+52+53 < 9+19+20+26+32+33+38+39+43+45+46+49+50+51
26: 1+2+3+5+6+7+8+9+14+15+16+17+18+19+20+30+31+32+33+44+45+46 < 13+27+28+29+40+41+42+43+52+53+54
a(55) = 4
18: 2+3+6+8+9+11+12+13+18+19+22+24+25+27+28+32+37+39+42+44+45+54 < 4+14+16+20+29+31+33+35+40+46+47+49+50+52+55
19: 1+2+3+4+7+9+10+12+13+16+20+23+25+26+28+31+35+38+40+43+45+52 < 6+15+18+22+30+32+34+37+42+46+48+49+51+54
21: 1+3+4+5+6+10+11+13+14+15+20+21+22+26+27+33+34+40+41+42+49+55 = 9+19+25+31+32+38+39+45+47+48+52+53+54
26: 1+2+4+5+6+7+8+9+14+15+16+17+18+19+29+30+31+32+46+47+48 = 13+26+27+28+40+41+42+43+44+45+55
a(56) = 4
18: 2+3+4+7+9+10+12+13+14+18+20+23+25+26+28+34+39+41+44+46+47+55 < 5+15+16+21+29+30+32+35+37+42+48+50+52+53+56
19: 1+3+4+5+8+10+11+13+14+16+19+21+24+26+27+28+32+37+40+42+45+47+52 < 7+18+23+29+31+34+36+39+44+49+51+54+55+56
21: 1+2+4+5+6+7+11+12+14+15+21+22+23+27+29+35+36+42+43+44+53+54 < 10+19+20+26+32+33+34+40+41+47+48+49+52+56
26: 1+2+3+5+6+7+8+9+10+15+16+17+18+19+20+29+30+31+32+33+34+48+49+56 = 14+27+28+42+43+44+45+46+47+53+54+55
a(57) = 4
18: 2+3+4+7+9+10+12+14+18+19+22+24+27+32+35+36+39+43+45+49+51 < 5+13+16+21+26+28+31+34+38+41+42+48+50+53+56
21: 1+3+4+5+8+10+11+14+16+18+20+21+25+29+31+35+38+40+41+46+51+53+57 < 7+15+19+24+26+30+36+37+39+42+45+47+48+52+55+56
25: 1+2+4+5+6+7+11+12+13+15+18+21+23+24+26+29+32+34+37+41+44+45+48+55 = 10+20+22+31+33+36+40+43+47+51+53+54+56+57
33: 1+2+3+5+6+7+8+9+10+13+15+16+17+19+20+22+26+28+30+31+33+36+42+47+56 = 18+29+32+35+41+44+45+46+49+51+55+57
a(58) = 4
17: 2+3+4+8+9+10+12+13+14+21+22+25+26+27+29+30+37+42+43+45+46+47+56 < 5+15+16+20+31+32+33+35+36+41+48+49+51+52+53+55
20: 1+3+4+5+7+10+11+13+14+16+19+20+24+27+28+30+33+36+40+41+44+47+53 = 8+17+21+25+31+37+38+42+45+48+50+51+56+57
21: 1+2+4+5+6+8+11+12+14+15+17+20+23+25+28+29+31+35+38+41+45+51+55+57 < 10+19+22+27+33+34+37+40+43+47+49+50+53+54+56
26: 1+2+3+5+6+7+8+9+10+15+16+17+18+19+21+22+31+32+33+34+37+48+49+50 < 14+28+29+30+41+44+45+46+47+55+57+58
You might ask why this piece is titled George Hart, when the only man in the photo on the left is John Conway. George Hart is related to this picture in three different ways.
First, this picture is of the math department common room at Princeton University. It was taken during a joint event of the WaM and SWIM programs in June, 2009. It shows the Zometool workshop conducted by George Hart that resulted in the construction of the expanded 120-cell, which appears in the photo's foreground.
The second connection to George Hart is that beautiful shiny object under the lights on the far left. The object is the propello-octahedron sculpture that George Hart created out of 150 CDs. The sculpture has been in the common room for many years, and I have always loved it.
Unfortunately, the sculpture was slowly degrading, even losing some of its parts. I visited Princeton in August 2008 and realized that the sculpture was facing a short life expectancy, so I took the picture of it that is below. I couldn't find any angle to shoot the photo that hid the lost pieces. The sculpture survived until my visit in June 2009, as evidenced by the first picture. But unfortunately it wasn't there any more during my last visit in May 2010.
Oops, I almost forgot. I promised you a third way in which George Hart relates to the first picture. He is the one who took it.
My coauthor Konstantin Knop publishes cute math problems in his blog (in Russian). Recently he posted a coin weighing problem that was given at the 2010 Euler math Olympiad in Russia to eighth graders. The author of the problem is Alexander Shapovalov.
Among 100 coins exactly 4 are fake. All genuine coins weigh the same; all fake coins, too. A fake coin is lighter than a genuine coin. How would we find at least one genuine coin using two weighings on a balance scale?
It is conceivable that your two weighings may find more than one genuine coin. The more difficult question that Konstantin and his commentators discuss is the maximum number of genuine coins you can guarantee to identify in two weighings. Konstantin and the others propose 14 as the answer, but do not have a proof yet.
I wonder if one of you can find a bigger number than Konstantin or alternatively a proof that indeed 14 is the largest possible.
You might ask, considering the title of this piece, why I think that coins are scary. No, I am not afraid of coins. It scares me that this problem was given to eighth graders in Russia, because I cannot imagine that it would be given to kids that age in the USA.
By the way, ten eighth grade students in Russia solved this problem during the competition.
I recently wrote two pieces about the puzzle relating to sons born on a Tuesday: A Son born on Tuesday and Sons and Tuesdays. I also posted a beautiful essay on the subject by Peter Winkler: Conditional Probability and "He Said, She Said". Here is the problem:
You run into an old friend. He has two children, but you do not know what their gender is. He says, "I have a son born on a Tuesday." What is the probability that his second child is also a son?
A side note. My son Alexey explained to me that I made an English mistake in the problem in those previous posts. It is better to say "born on a Tuesday" than "born on Tuesday." I apologize.
Despite this error, I was gratified to hear from a number of people who told me that I had converted them from their solution to my solution. To ensure that the conversion is substantial, I've created a new version of the puzzle on which my readers can test out their new-found understanding. Here it is:
You run into an old friend. He has two children, but you do not know what their gender is. He says, "I have a son born on a Tuesday." What is the probability that his second child is born on a Wednesday?
I was terribly shy when I was a teenager. I worked on this problem and overcame it. But when I moved to the US my shyness returned in a strange form. I was fine around Russians but shy around Americans. At first I assumed that it was a language problem.
I became friends with a Russian sexologist and psychotherapist. He pointed out that I never initiated a conversation with Americans and so I realized that my shyness had returned. He prescribed an exercise for me: I had to invite a new American guy to lunch once a week.
Why guys? Maybe because he was a sexologist or maybe because my problems with self-esteem were more pronounced when I was around men. In any case, I decided to do the exercise.
To paint the full picture I need to add some relevant details. At that time I was married, although I didn't wear a ring, and wasn't especially interested in other men. The reason I didn't wear a ring was that Joseph, my husband at the time, did not himself want to wear a ring. As I love symmetry in relationships more than I love rings, I didn't wear one either.
The men I was about to invite to lunch were mere acquaintances, because I had not yet made any American friends. So although I didn't intend to hide it, they may not have realized that I was married.
Two things surprised me in this exercise. First, it was very easy. Most people agreed to do lunch with me.
Second, every man I invited mentioned his girlfriend. This was unexpected. From my experience with Russians, I anticipated that every man would hide his involvement with someone else, even with a wife, at least for some time. At the very least, many Russian men would try to flirt.
The Americans were different. Unclear why I had invited them out, they wanted to be upfront with me from the start, just in case I was interested in them. Since that experience, I admire the way that American men come clean.
I never invited any of these guys out twice: I just needed a supply of new men for my exercise in overcoming my shyness. I wonder if they thought I was put off by their confessions. Perhaps my loss of interest in them after the first lunch confirmed their suspicions that I was attracted to them.
The sexologist's exercise was a success. Today I have no trouble inviting someone to lunch.
John Conway has a T-shirt with his theorem on it. I couldn't miss this picture opportunity and persuaded John to pose for pictures with his back to me. Here is the theorem:
If you continue the sides of a triangle beyond every vertex at the distances equaling to the length of the opposite side, the resulting six points lie on a circle, which is called Conway's circle.
Poor John Conway had to stand with his back to me until I figured out the proof of the theorem and realized which point must be the center of Conway's circle.
Heard somewhere:
Teacher: What's bigger: 22/7 or 3.14?
Student: They are equal.
Teacher: Why do you say that?!
Student: They are both equal to π.
For the last three years I've been coming to the Institute for Advanced Study in Princeton every spring for the Women and Mathematics program. Every year I am assigned to an office in the main building: Fuld Hall.
The problem is that there is a different office that I crave. Every year I go and check on it over in Simonyi Hall, where the Mathematics Department is located. This year I took this photo of the empty name-tag, hoping that one day it will say Tanya Khovanova.
Credit cards often keep your mother's maiden name in their database for security purposes. This so called "security" is based on two assumptions:
Were these assumptions true, only your close relatives would know your mother's maiden name. In reality, if your mother was never married, then your last name is the same as your mother's last name. So, crooks who are trying to steal identities can try to use your last name as your mother's presumed maiden name. Very often they will succeed. Besides, many women do not change their last names. If you have a different last name from your mother, but your mother uses her maiden name, then the bank's security question is not very secure at all.
If you want your identity to be secure you might need to invent a maiden name for your mother. Alternatively, perhaps your parents can tell you a family secret that will help you choose a name that is related to you, but not transparent to the public.
My relative Martin took his wife's last name after their marriage. Before his children apply for credits cards and bank accounts, he needs to explain to them that it is better for them to use his maiden name as their mother's maiden name for banking purposes.
I was driving on MassPike when, for no apparent reason, a car driving in the opposite direction started flashing its headlights. I remembered the Russian tradition of informing the oncoming traffic that the police are nearby. So I adjusted my speed and very soon I saw a police car. I got this warm feeling in my heart because I didn't need to panic or check my speedometer. I mentally thanked that anonymous Russian driver and started wondering why the tradition had not been adopted in the USA. Is it because we are so responsible that we want to punish speeders, or do we think that the police are on our side?
The following problem was sent to me by Joel Lewis.
You have 12 mice, one of which eats faster than all the others. You need to find it. You have a supply of standard cupcakes that you value very much and want to minimize how many of them you have to use. The only way you can find the mouse is to give cupcakes to several groups of mice and see which group is the fastest.
We assume that mice chew at a constant speed and all the mice in one group can attack the cake at the same time. I love this puzzle because I love coin problems. Let me restate the puzzle as a coin problem:
You have 12 coins, one of which is fake and weighs less than all the others. You have a balance scale with multiple pans, that is you can weigh several things at once and order them by weight. You do not care about the total number of weighings as in most classical coin puzzles, instead, this time using a pan is expensive and you want to find the fake coin with as few pan-uses as possible.
Spoiler warning: below I will discuss the solution for n mice.
You can, of course, give a cake to every mouse and see which one finishes first. You can save one cake by giving cakes at the same time to all but one of the mice. If everyone finishes simultaneously, the faster mouse is the unfed one.
It wastes cakes to give them to unequally-sized groups of mice. We can do better by copying the classical way to find a fake coin with the minimum number of weighings. That is, for each test, divide the mice into three groups as evenly as possible and give a cake to each of two equally-sized groups. The number of cakes you use is about 2log3n.
I wouldn't have written this essay if that was the solution. Sometimes you can do even better. For example, you can find the faster mouse out of 12 using only 5 cakes.
First, if you give out k cakes in one test, the test tells you which of k+1 groups the mouse is in. In the worst case, the faster mouse will be in the biggest group, so you should minimize the biggest group. Hence, your groups that get cakes should have ⌈n/(k+1)⌉ mice.
A test with one cake gives no information. I argue that giving out more than three cakes doesn't gain anything. Indeed, suppose we use four cakes in a test. That is, we divide the mice into five groups A, B, C, D and E, of which the first four are the same size. We can simulate the test by two tests in each of which we give out two cakes. In the first test we give cakes to A+B and C+D. If one of the groups is faster, say A+B, then in the second test give cakes to A and B; if not, E has the faster mouse. I leave it as an exercise to simulate a test with more than four cakes.
Thus, in an optimal strategy we can use two or three cakes per test. Also, if you give one test with k − 1 cakes and the next one with m − 1 cakes, you can switch them with the same effect. The largest group after either order of tests will have at most ⌈n/km⌉ mice.
I don't need two tests of three cakes each, which would give me a group of size at least ⌈n/16⌉. I can achieve the same result with three tests of two cakes each, with the faster mouse restricted to a group of size at most ⌈n/27⌉.
That means all my tests use two cakes, except I might use three cakes once. It doesn't matter in what order I conduct the tests, so I can wait until the end to use three cakes. I leave it as an exercise to the reader that the only small number of mice for which we would prefer three cakes is four. From this it follows quickly that for numbers of mice between 3 * 3i + 1 and 4 * 3i, the number of cakes is 2i + 1. For numbers between 4 * 3i + 1 and 3i+2 the answer is 2i + 2.
A week ago I chatted with my son Sergei about memorable math problems. He mentioned problem 5 from USAMO 2007. The problem can be reduced to the following:
Prove that (x7 + 1)/(x + 1) is composite for x = 77n, where n is a non-negative integer.
Perhaps Sergei remembered this problem because as far as he knew, he was the only one in that competition to solve it. That made me curious as to how he solved it. His solution is available as solution 2 at the Art of Problem Solving website. His solution seemed mysterious and impossible to invent on the spot. I became even more curious to understand the thought process underlying his solution.
Here is his recollection:
We need to factor x6 − x5 + x4 − x3 + x2 − x + 1. If such factoring existed it would have been known. Therefore, we need to somehow use the fact that x = 77n. What is the simplest way to factor? We should try to represent the polynomial in question as the difference of squares. Luckily, x is an odd power of 7. We can make it a square if we multiply or divide it by 7 or another odd power of 7. So with a supply of squares on one side, we need to find a match for one of them to build the difference.
Let us simplify the problem and see what happens for (y3 + 1)/(y + 1) for y = 33n, when n = 1. In other words we want to represent 703 as a difference of squares. This can be done: 703 = 282 − 92. Now let us see how we can express 282 and 92 through y which in this case is equal to 27. The first term is (y + 1)2, and the second is 3y.
Now let's go back to 7 and x, and check whether (x + 1)6 − (x6 − x5 + x4 − x3 + x2 − x + 1) is 7x. Oops, no. The difference is 7x5 + 14x4 + 21x3 + 14x2 +7x. On the plus side, it is divisible by 7x which we know is a square. The leftover factor is x4 + 2x3 + 3x2 +2x + 1, which is a square of x2 +x + 1.
The problem is solved, but the mystery remains. The problem can't be generalized to numbers other than 3 and 7.
I decided to take a closer look at the Putnam Competition. I analyzed the results of the three top contenders for the best Putnam teams: Harvard, MIT, Princeton. I looked at the annual number of Putnam Fellows from each of these three schools starting from 1994.
| Year | Harvard | MIT | Princeton |
|---|---|---|---|
| 1994 | 2 | 0 | 1 |
| 1995 | 3 | 0 | 0 |
| 1996 | 2 | 0 | 0 |
| 1997 | 4 | 0 | 0 |
| 1998 | 2 | 0 | 1 |
| 1999 | 2 | 1 | 0 |
| 2000 | 2 | 2 | 0 |
| 2001 | 2 | 1 | 0 |
| 2002 | 2 | 2 | 0 |
| 2003 | 1 | 2 | 1 |
| 2004 | 0 | 3 | 2 |
| 2005 | 2 | 3 | 1 |
| 2006 | 1 | 3 | 0 |
| 2007 | 1 | 2 | 1 |
| 2008 | 1 | 3 | 0 | 2009 | 1 | 2 | 0 |
As you may notice MIT couldn't even generate a Putnam Fellow until 2000, but starting from 2003 MIT consistently had more Putnam Fellows than Harvard or Princeton.
Richard Stanley, the coach of the MIT team, kindly sent me the statistics for the most recent competition, held in 2009.
| Category | Overall | MIT |
|---|---|---|
| Number of participants | 4036 | 116 |
| Mean score | 9.5 | 34.7 |
| Median score | 2 | 31 |
| Geometric mean | 0 | 0 |
| Percent of 0 score | 43.7 | 4.3 |
Furthermore, MIT had 40% in the top 5, 33% in the top 15, 32% in the top 25, and 35% in the top 81. For comparison, in the top 81, MIT had 28 winners — more than the next three schools together: Caltech 11, Harvard 9, Princeton 7.
No comment.
OnlineDegree.net selected the 50 Best Blogs for Math Majors, and I am pleased that Tanya Khovanova's Math Blog is number two. Since they did not explain their criteria, I suspected that it might be according to the number of Google hits. To double check, I Googled "math blog" and once again my blog was number two.
This might be the right moment to acknowledge the others involved with my blog. First, Sue Katz, my writing teacher and editor, corrected the English in most of my posts. Now I do not "do" mistakes in English any more, I make them.
My sons, Alexey and Sergei, are a huge support. Sometimes my poor kids have to listen endlessly to my latest idea, until I am ready to write about it. And then they will even read the final piece, and continue to encourage me.
But the most important motivators are you, my readers. Your comments, your personal emails and your feedback keep me writing.
I am usually disappointed with American math text books. I have had an underwhelming experience with them. Often I open a book and in the next 15 minutes, I find a mistake, a typo, a misguided explanation, sloppiness, a misconception or some other annoyance.
I was pleasantly surprised when I opened the book Introduction to Algebra
Look at problem 7.3.1.
Five chickens can lay 10 eggs in 20 days. How long does it take 18 chickens to lay 100 eggs?
There is nothing wrong with this problem. But the book is accompanied by the Introduction to Algebra Solutions Manual
The number of eggs is jointly proportional to the number of chickens and the amount of time. One chicken lays one egg in 10 days. Hence, 18 chickens will lay 100 eggs in 1000/18 days, which is slightly more than 55 and a half days.
What is wrong with this solution? Richard Rusczyk is human after all.
I like this book for its amazing accuracy and clean explanations. There are also a lot of diverse problems in terms of difficulty and ideas. Richard Rusczyk has good taste. Many of the problems are from different competitions and require inventiveness. I like that there are a lot of challenge problems that go beyond the boring parts of algebra. Also, I like that important points of algebra are chosen wisely and are emphasized.
This book might not be for everyone. It doesn't have pretty pictures. It doesn't have color at all. This is not a flaw for a math book. The book concentrates on ideas and problems, not entertainment. So if you're looking for math entertainment, you'll find it on my blog. For solid study, try Richard Rusczyk's books
I love Raymond Smullyan's books
This happened at the Gathering for Gardner 8. We were introduced and then later that day, the conference participants were treated to a dinner event that included a magic show. In one evening I saw more close-up magic tricks than I had in my whole life. This left me lightheaded, doubting physics and my whole scientific outlook on life.
Afterwards, Raymond Smullyan joined me in the elevator. "Do you want to see a magic trick?" he asked. "I bet I can kiss you without touching you." I was caught off guard. At that moment I believed anything was possible. I agreed to the bet.
He asked me to close my eyes, kissed me on the cheek and laughed, "I lost."
As a writer of books on mathematical puzzles I am often faced with delicate issues of phrasing, none more so than when it comes to questions about conditional probability. Consider the classic "X has two children and at least one is a boy. What is the probability that the other is a boy?"
It is reasonable to interpret this puzzle as asking you "What is the probability that X has two boys, given that at least one of the children is a boy" in which case the answer is unambiguously 1/3—given the usual assumptions about no twins and equal gender frequency.
This puzzle confounds people *legitimately*, however, because most of the ways in which you are likely to find out that X has at least one boy contain an implicit bias which changes the answer. For example, if you happen to meet one of X's children and it's a boy, the answer changes to 1/2.
Suppose the puzzle is phrased this way: X says "I have two children and at least one is a boy." What is the probability that the other is a boy?
Put this way, the puzzle is highly ambiguous. Computer scientists, cryptologists and others who must deal carefully with message-passing know that what counts is not what a person says (even if she is known never to lie) but *under what circumstances would she have said it.*
Here, there is no context and thus no way to know what prompted X to make this statement. Could he instead have said "At least one is a girl"? Could he have said "Both are boys"? Could he have said nothing? If you, the one faced with solving the puzzle, are desperate to disambiguate it, you'd probably have to assume that what really happened was: X (for some reason unconnected with X's identity) was asked whether it was the case that he had at least one son, and, after being warned—by a judge?—that he had to give a yes-or-no answer, said "yes." An unlikely scenario, to say the least, but necessary if you want to claim that the solution to the puzzle is 1/3.
Consider the puzzle presented (according to Alex Bellos) by Gary Foshee at the recent 9th Gathering for Gardner:
I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?
If the puzzle was indeed put exactly this way, and your life depended on defending any particular answer, God help you. You cannot answer without knowing, for example, what the speaker would have said if he had one boy and one girl, and the boy was born on Wednesday. Or if he had two boys, one born on Tuesday and one on Wednesday. Or two girls, both born on Tuesday. Et cetera.
Now, there is nothing mathematically wrong (given the usual assumptions, including X being random) about saying that "The probability that X has two sons, given that at least one of X's two children is a boy born on Tuesday, is 13/27." But if that is to be turned into an unambiguous puzzle attached to a presumed situation, some serious hypothesizing is necessary. For instance: you get on the phone and start calling random people. Each is asked if he or she has two children. If so, is it the case that at least one is a boy born on a Tuesday? And if the answer is again yes, are the children both boys? Theoretically, of the times you reach the third question, the fraction of pollees who say "yes" should tend to 13/27.
Kind of takes the fun out of the puzzle, though, doesn't it? Kudos to Gary for stirring up controversy with a quickie.
I recently discussed the following problem:
You run into an old friend. He has two children, but you do not know their genders. He says, "I have a son born on Tuesday." What is the probability that his second child is also a son?
I had heard this problem at the Gathering for Gardner 9 in a private conversation. My adversary had been convinced that the answer to the problem is 13/27. I came back to Boston from the gathering and wrote my aforementioned essay in which I disagreed with his conclusion.
I will tell you my little secret: when I started writing I substituted Wednesday for Tuesday. Then I checked my sons' birthdays and they were born on Saturday and Tuesday. So I changed my essay back to Tuesday.
After I published it people sent me several links to other articles discussing the same problem, such as those of Keith Devlin and Alex Bellos, both of whom think the answer is 13/27. So I invented a fictional opponent — Jack, and here is my imaginary conversation with him.
Jack: The probability that a father with two sons has a son born on Tuesday is 13/49. The probability that a father with a son and a daughter has a son born on Tuesday is 1/7. A dad with a son and a daughter is encountered twice as often as a dad with just two sons. Hence, we compare 13/49 with 14/49, and the probability of the father having a second son is 13/27.
Me: What if the problem is about Wednesday?
Jack: It doesn't matter. The particular day in question was random. The answer should be the same: 13/27.
Me: Suppose the father says, "I have a son born on *day." He mumbles the day, so you do not hear it exactly.
Jack: Well, as the answer is the same for any day, it shouldn't matter. The probability that his second child will also be a son is still 13/27.
Me: Suppose he says, "I have a son born …". So he might have continued and mentioned the day, he might not have. What is the probability?
Jack: We already decided that it doesn't depend on the day, so it shouldn't matter. The probability is still 13/27.
Me: Suppose he says, "I have a son and I do not remember when he was born." Isn't that the same as just saying, "I have a son." And by your arguments the probability that his second child is also a son is 13/27.
Jack: Hmm.
Me: Do you remember your calculation? If we denote the number of days in a week as d, then the probability of him having a second son is (2d−1)/(4d−1). My point is that this probability depends on the number of days of the week. So, if tomorrow we change a week length to another number his probability of having a son changes. Right?
At this point my imaginary conversation stops and I do not know whether I have convinced Jack or not.
Now let me give you another probability problem, where the answer is 13/27:
You pick a random father of two children and ask him, "Yes or no, do you have a son born on Tuesday?" Let's make a leap and assume that all fathers know the day of the births of their children and that they answer truthfully. If the answer is yes, what is the probability of the father having two sons?
Jack's argument works perfectly in this case.
My homework for the readers is: Explain the difference between these two problems. Why is the second problem well-defined, while the first one is not?
On March 5, 2010 I visited Princeton and had dinner with John Conway at Tiger Noodles. He gave me the second Doomsday lesson right there on a napkin. I described the first Doomsday lesson earlier, in which John taught me to calculate the days of the week for 2009. Now was the time to expand that lesson to any year.
As you can see on the photo of the napkin, John uses his fingers to make calculations. The thumb represents the DoomsDay Difference, the number of days your birthday is ahead of DoomsDay for a given year. To calculate this number you have to go back to my previous post.
The index finger represents the century adjustment. For example, the Doomsday for the year 1900 is Wednesday. Conway remembers Wednesday as We-are-in-this-day. He invented his algorithm in the twentieth century, not to mention that most people who use his algorithm were born in that century. Conway remembers the Doomsday for the year 2000 as Twosday.
The next three fingers help you to calculate the adjustment for a particular year. Every non-leap year has 52 weeks and one day. So the Doomsday moves one day of the week forward in one year. A leap year has one extra day, so the Doomsday moves forward two days. Thus, every four years the Doomsday moves five days forward, and, consequently, every twelve years it moves forward to the next day of the week. This fact helps us to simplify our year adjustment by replacing every dozen of years with one day in the week.
The middle finger counts the number of dozens in the last two digits of your year. It is important to use "dozen" instead of "12" as later we will sum up all the numerals, and the word "dozen" will remind us that we do not need to include it in the sum.
The ring finger represents the remainder of the last two digits of the year modulo 12, and the pinkie finger represents the number of leap years in that remainder.
John made two sample calculations on the napkin. The first one was for his own birthday — December 26, 1937. John was born exactly on Doomsday. I suspect that that is the real reason he called his algorithm the Doomsday Algorithm. The century adjustment is Wednesday. There are 3 dozens in 37, with the remainder 1 and 0 leap years in the remainder. When we add four more days to Wednesday, we get Sunday. So John Conway was born on Sunday.
The second napkin example was the day we had dinner: March 5, 2010. March 5 is 5 days ahead of the Doomsday. The century adjustment is Twosday, plus 0 dozens, 10 years in the remainder and 2 leap years in the remainder. 5 + 0 + 10 + 2 equals 3 modulo 7. Hence, we add three days to Tuesday, demonstrating that we dined out together on Friday. But then, we already knew that.
Do you know that some Russian letters are shaped exactly as some letters in the English alphabet? The shapes are the same, but the sounds of the letters are not. My Russian last name can be completely spelled using English letters: XOBAHOBA.
The adequate translation of my last name into English is Hovanova. You might ask where the first "K" came from. For many years French was considered the language of diplomacy and the USSR used French as an official language for traveling documents.
But "H" in French is silent and "Hovanova" would have been pronounced as "Ovanova." To prevent that, Russians used "kh" for the "h" sound.
Now to my first name. I was born Tatyana, for which Tanya is a nickname. Back in Russia, Tanya is used for children and students and Tatyana for adults and teachers. As I was a student throughout my 30 years of life in Russia, I was always Tanya. When I moved to the US, I decided to keep using Tanya, which I much preferred to Tatyana.
A psychiatrist might think that I wanted to be a student forever or refused to grow up. Or I could be accused of being lazy, as Tanya is shorter. In reality, I was just trying to be considerate. Tanya is easier to write and to spell for Americans. Anyway, I already had enough problems spelling out my last name in this country.
Now that more information is getting translated from Russian into English, I keep stumbling on references to me as to Hovanova or Tatyana. For example, the IMO official website used Russian sources to come up with the names of the Russian participants. They then translated the names directly into English, instead of going through French. As a result, on their website I am Tatyana Hovanova. This is not unique to me: many Russian names on the IMO website differ from those peoples' passport names.
By the way, if you Google my last name you will encounter other Khovanovas. Khovanova is not a particularly unusual name. Only one of the Khovanovas that came up in my search results is a close relative. Elizabeth Khovanova is my father's second wife and a dear friend. She is also an accomplished geneticist.
Khovanova is used only for females in Russia. The male equivalent is Khovanov. Surely you have heard of my half-brother Mikhail Khovanov and his homologies.
I gave you the Wise Men Without Hats puzzle in one of my previous posts:
A sultan decides to check how wise his two wise men are. The sultan chooses a cell on a chessboard and shows it to the first wise man. In addition, each cell on the chessboard either contains a pebble or is empty. The first wise man has to decide whether to remove one pebble or to add one pebble to an empty cell. Next, the second wise man must look at the board and guess which cell was chosen by the sultan. The two wise men are permitted to agree on their strategy beforehand. What strategy can they find to ensure that the second wise man will always guess the chosen cell?
My readers solved it. The solution is the following. Let us assign a number between 0 and 63 to every cell of the board. The second wise man takes numbers corresponding to cells with pebbles, converts them to binary and XORs the result. The answer is the cell number that he is seeking. The first wise man can always add or remove a pebble to make the XORing operation of the remaining pebbles produce any given number from 0 to 63.
This solution only works for boards that have a power of two as the number of cells.
Let's look at the solution more closely. Let us create a graph corresponding to this problem. The vertices of the graph will correspond to the positions of pebbles. That means vertices are in one-to-one correspondence with the subsets of the set of 64 elements. Let us connect two vertices if we can get from one position to another by removing or adding a pebble. That means vertices are connected if two corresponding sets differ by exactly one element. We can see that the resulting graph is regular and each vertex is connected to exactly 64 other vertices.
Let us assign one out of 64 colors to each cell of the chessboard. The second wise man can guess the cell by looking at the chessboard. From this we can conclude that there is a bijection from the vertices of the graph to chessboard cells. In other words, we can color the graph in 64 colors. The existence of the strategy for wise men means that we can color the graph in such a way that each vertex is connected to the vertices of all colors.
As each vertex in our graph has exactly 64 neighbors, the graph has the following property: It can be colored in 64 colors in such a way that every vertex is connected to exactly one vertex of every color.
As soon as I realized that there is such a graph-theoretical object, I started to run around MIT asking everyone if such objects were studied or have a name.
It appears that indeed such an object has a name. A graph that can be colored into k colors in such a way that every vertex has exactly one neighbor of every color is called a rainbow graph.
Andrew Woldar discusses properties of such graphs in his paper. In particular, rainbow graphs are matching graphs. Indeed, every vertex is connected to exactly one vertex of the same color. Hence there is a natural pairing on vertices. From here, we can conclude that the smallest size of a rainbow graph is 2k.
Several MIT students liked the wise men problem and the associated graph object so much that they decided to study them. Hwanchul Yoo, SuHo Oh, and Taedong Yun enumerated all rainbow graphs of size 2k. The number of non-isomorphic rainbow graphs of size 2k equals mitthe number of switching classes of graphs with k vertices. The corresponding sequence A002854 starts as: 1, 1, 2, 3, 7, 16, 54. The paper is soon to appear. It is titled "Rainbow Graphs and Switching Classes."
Just received from Victor Gutenmacher:
Fibonacci salad: For today's salad, mix yesterday's leftover salad with that of the day before.
New additions to my trick problems collection:
* * *
It takes 12 minutes to saw a log into 3 parts. How much time will it take to saw it into 4 parts?
* * *
The Davidsons have five sons. Each son has one sister. How many children are there in the family?
* * *
A caterpillar wants to see the world and decides to climb a 12-meter pole. It starts every morning and climbs 4 meters in half a day. Then it falls asleep for the second half of the day during which it slips 3 meters down. How much time will it take the caterpillar to reach the top?
First Name: David
Last Name: (hidden for privacy protection)
Year of Birth: (hidden for privacy protection)
email: buchanan1985@gmail.com
I remember a math problem from my childhood: divide an L-shaped triomino into four congruent parts. The answer is in the picture on the left. Such division is quite appropriately called a reptile (repetitive tiling). Solomon Golomb invented the name many years ago. He wasn't aware that his definition would end up creating Googling problems, for when you search for such a mathematical object you will stumble upon a lot of amphibians.
Similarly, you can divide the same shape into 9 congruent pieces (see the figure on the left).
Suppose you want to divide a shape into pieces that are similar, but not necessarily of the same size. Such tiling is called an irreptile (irregular reptile).
At the Gathering for Gardner 9 I listened to Carolyn Yackel's talk about the L-reptiles and L-irreptiles. One of the ways to create an irreptile is to start with a reptile, then to make a sub-tiling of one of the existing tiles. This procedure can be repeated many times.
Carolyn brought a ceramic table to the Gathering for Gardner. This table is made of two L-shapes. Both shapes are irreptiles, created by this procedure. In one part of the table she started with a 9-reptile, and in the other with a 4-reptile. She sent me this picture of her table to use in this essay.
After her talk I started wondering how many tiles can an L-irreptile be comprised of. We start with one piece: the L-shape itself. If we divide a tile into four smaller tiles we add three more pieces. If we divide it into nine tiles we add eight more pieces. We can mix sub-dividing into four and nine tiles. The total number of tiles that an L-shape can be comprised of by this procedure is all the numbers you can get from 1 by adding three or eight. The sequence is 1, 4, 7, 9, 10, 12, 13, 15, 16, 17 and so on. Starting from 15 we get all the consecutive numbers.
The numbers that are not represented in the above sequence are 2, 3, 5, 6, 8, 11 and 14. Can we divide an L-shape into such numbers of tiles? Benoît Jubin reminded me that there is an L-reptile with six pieces.
Consequently, we can add 5 more pieces to any L-irreptile. Thus, there exists an L-irreptile made out of 11 (1+5+5) and out of 14 (1+8+5) pieces. The numbers that are left are 2, 3, 5 and 8.
While I was discussing L-irreptiles with fans of sequences, David Wilson suggested a conjecture.
David Wilson's conjecture. If there is an L-irreptile, there is a corresponding square-irreptile with similarly-sized pieces.
If this conjecture is true, then we can see that L-irreptiles with 2, 3 or 5 pieces can't exist as corresponding square-irreptiles do not exist.
For example, to prove that 2 or 3 square-irreptiles can't exist, you need to notice that each corner of the square we are trying to tile should belong to a different small tile.
The question of the existence of an 8-irreptile of the L-shape is more interesting and challenging. The square 8-irreptile exists. If you can prove that the L-shape 8-irreptile doesn't exist, then you will automatically prove that the converse to Wilson's conjecture is not true.
Only at MIT. Room 4-231.
Let n coins weighing 1, 2, … n be given. Baron Münchhausen knows which coin weighs how much, but his audience does not. Define a(n) to be the minimum number of weighings the Baron must conduct on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins.In the paper Baron Münchhausen's Sequence, three of us completely described the Baron's sequence. In particular, we proved that a(n) ≤ 2. Here we would like to outline another proof idea, which is interesting in part because it touches the Riemann hypothesis.
We denote the total weight of coins in some set A as |A|.
Lemma. Numbers n that can be represented as Ti + Tj + Tk = 3n, where i ≤ j < k, such that there is a subset A of coins from j + 1 to k such that n = Tj + |A|, can be done in two weighings.
Proof. Suppose Ti + Tj + Tk = 3n and there is a subset A of coins from j + 1 to k such that n = Tj + |A|. We propose the two weighings
[1…j] + A = n
and
[1…i] + B = n + A,
where B is the complement of A in {j + 1, j + 2, … , k}.
If we sum up twice the first weighing with the second weighing we get
3[1…i] + 2[(i + 1)…j] + 2A + B = 3n + A.
In other words, three times the weight of the coins that were on the left side in both weighings, plus twice the weight of the coins that were on the left side in only the first weighing, plus the weight of the coins that were moved from the left cup to the right cup plus the weight of the coins on the left cup in only the second weighing equals three times the weight of the coin on the right cup in both weighings. Hence three times the weight of the coin on the right cup in both weighings can't be less than the weight of the k other coins that participated plus the weight of the j coins that were on the left cup in the first weighing and weren't moved to the right cup, plus the weight of the i coins that were one the left cup in both the first and the second weighing. But because Ti + Tj + Tk = 3n, then 3n is the smallest possible weight of any set of i plus j plus k coins, the coin on the right cup in both weighings has to be the n-coin.
We checked that any number up to 600,000 except 20 can be represented so as to satisfy the Lemma. To show how to solve 20 coins in two weighings is easy, and, as usual, is left as an exercise for the reader. Next, we want to look at the following lemma.
Lemma. Given a set of consecutive numbers {(j + 1), … , k}, if k > 2j + 2, then it is possible to find a subset in the set that sums up to any number in the range from j + 1 to (j + k + 1)(k – j)/2 – j – 1.
We won't prove the lemma, but it means that if k is about twice larger than j, then we have a lot of flexibility for building our set A in the weighing above. For moderately large n (where 600000 >> "moderately large"), it is not hard to prove that this flexibility is sufficient.
Now the question becomes: can we find such a decomposition into triangular numbers? It is enough to find a representation Ti + Tj + Tk = 3n, where Tk is at least 81% of 3n.
We know that decompositions into triangular numbers are associated with decompositions into squares. Namely, if Ti + Tj + Tk = 3n, then (2i + 1)2 + (2j + 1)2 + (2k + 1)2 = 24n + 3. If the largest square is at least 81% of 24n + 3, then the largest triangular number in the decomposition of 3n is at least 81%.
There is a theorem (W. Duke, Hyperbolic distribution problems and half-integral weight Maass forms, in Inventiones Math 92 (1988) p.73-90) that states that in the limit the decompositions of numbers into three squares are equidistributed. That is, if we take some region on the unit sphere x2 + y2 + z2 = 1 (for example, the region |z| > 0.8) and view decompositions of 24n + 3 into squares as points on the sphere x2 + y2 + z2 = 24n + 3, then, as n grows, decompositions whose projections fall into our chosen region are guaranteed to appear.
This theorem is great, because it tells us that for large enough n we will always be able to find a decomposition of 24n + 3 into triangle numbers where one of the triangle numbers will be much bigger than the others, and it will be possible to prove the weight of the n coin in two weighings. Unfortunately, this summary, as stated, does not tell us how large that n needs to be. So we need some exact estimates.
The number of decompositions of m into sums of three squares is about the square root of m. More precisely, it is possible to compute a number N, such that for any number m > N, with at most one exception, the number of decompositions is at least Cm1/2−1/30, where C is a known constant.
The more specific statement of Duke's theorem is that if the number of solutions to the quadratic x2 + y2 + z2 = 24n + 3 is large, for a computable value of "large", then the solutions are equidistributed. More precisely, let us denote 3n by m and fix an area Ω on the unit sphere. Then the number of solutions (x, y, z) such that the unit vector (x, y, z)/|(x, y, z)| belongs to Ω is
1/(4π) Ωh(8m+3) + E(m),
where h(8m+3) is the total number of solutions of x2 + y2 + z2 = 24n + 3, and E(m) is an error term, which starting from some number satisfies the inequality: E(m) ≤ 1000m1/2-1/7.
That's pretty good, because combining these two lets us, at least in principle, actually calculate an N such that for all n > N except maybe one a(n) = 2. After that we hoped to write a program to exhaustively search smaller numbers by computer.
This situation is still somewhat annoying, because that possible exception must then be propagated into the proof, and if we are not careful, possibly into the final theorem. ("No matter how many coins the Baron has, he can prove the weight of one in at most two weighings, except maybe one number of coins, and we don't know which...") This is where the Riemann Hypothesis comes in. If the Riemann Hypothesis is true, then that exception isn't there, and all is sunlight and flowers.
The beauty of the Baron's puzzle is such that we actually do not need the Riemann hypothesis. As we can use unbalanced weighings, it is enough to find a good decomposition for one out of the four numbers 3n, 3n-1, 3n-2, or 3n-3.
Instead of finding all these exact estimates we found a different elementary proof of our theorem. But we are excited that methods that are used in very advanced number theory can be used to solve a simple math problem that can be described to middle school children.
It would be great if someone decided to finish this proof.
I am used to thinking that a "woman in numbers" means a female number theorist. But not anymore. I just discovered drawings by Svetlana Bogatyr. From now on the expression a "woman in numbers" will convey an additional meaning to me.
I am grateful to Svetlana for permitting me to post several of her drawings. The "Mature Woman" is on the left. "Eurydice", "Girl in Scarf" and "Holland Woman " are below.
Enjoy.
I just invented a two-player game. To start, you have an empty n by n board. When it's your turn you must write an integer between 1 and n into an empty cell on the board. Your integer has to differ from the integers that are already present in the same row or column. If you finish filling up the board, you will get a Latin square and the game will be a tie. The person who doesn't have a move loses. What is the best strategy?
Let's see what happens if n is 2. The first player puts any number in one of the four corners of the 2 by 2 board. The second player wins by placing a different number in the opposite corner.
I played this game with my son Sergei Bernstein on a 3 by 3 board. We discovered that the first player can always win. Since this game is so much fun, I'll leave it to the reader to play it and to find the winning strategy for the first player.
Can you analyze bigger boards? Remember that this game has many symmetries. You can permute rows and columns. Also, you can permute numbers.
While we were playing Sergei invented two theorems.
Sergei's theorem 1. If n is odd the first player can guarantee a tie.
Proof. In the first move the first player writes (n+1)/2 in the center cell. If the second player puts number x in any cell the first player puts number n+1−x into the cell that is rotationally symmetric to the second player's cell with respect to the center. With this strategy the first player will always have a legal move.
Sergei's theorem 2. If n is even the second player can guarantee a tie.
Proof. If the first player puts number x in any cell the second player puts number n+1−x into the cell that is vertically symmetric to the first player's cell with respect to the vertical line of symmetry of the board. With this strategy the second player will always have a legal move.
As you play the game, let me know if you develop any theorems of your own.
Suppose you meet a friend who you know for sure has two children, and he says: "I have a son born on Tuesday." What is the probability that the second child of this man is also a son?
People argue about this problem a lot. Although I've discussed similar problems in the past, this particular problem has several interesting twists. See if you can identify them.
First, let us agree on some basic assumptions:
Now let us consider the first scenario. A father of two children is picked at random. He is instructed to choose a child by flipping a coin. Then he has to provide information about the chosen child in the following format: "I have a son/daughter born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun." If his statement is, "I have a son born on Tuesday," what is the probability that the second child is also a son?
The probability that a father of two daughters will make such a statement is zero. The probability that a father of differently-gendered children will produce such a statement is 1/14. Indeed, with a probability of 1/2 the son is chosen over the daughter and with a probability of 1/7 Tuesday is the birthday.
The probability that a father of two sons will make this statement is 1/7. Among the fathers with two children, there are twice as many who have a son and a daughter than fathers who have two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 1/2 for the second child to also be a son.
Now let us consider the second scenario. A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son. If he has one son, he is instructed to provide information on his son's day of birth. If he has two sons, he has to choose one at random. His statement will be, "I have a son born on Mon/Tues/Wed/Thurs/Fri/Sat/Sun." If his statement is, "I have a son born on Tuesday," what is the probability that the second child is also a son?
The probability that a father of differently-gendered children will produce such a statement is 1/7. If he has two sons, the probability will likewise be 1/7. Among the fathers with two children, twice as many have a son and a daughter as have two sons. Plugging these numbers into the formula for calculating the conditional probability gives us a probability of 1/3 for the second child to also be a son.
Now let us consider the third scenario. A father of two children is picked at random. If he doesn't have a son who is born on Tuesday, he is sent home and another is picked at random until one who has a son that was born on Tuesday is found. He is instructed to tell you, "I have a son born on Tuesday." What is the probability that the second child is also a son?
The probability that a father of two daughters will have a son born on Tuesday is zero. The probability that a father of differently-gendered children will have a son who is born on Tuesday is 1/7. The probability that a father of two sons will have a son born on Tuesday is 13/49. Among the fathers with two children, twice as many have a son and a daughter than two sons. Plugging these numbers into the formula for calculating the conditional probability will give us a probability of 13/27 for the second child to also be a son.
Now let's go back to the original problem. Suppose you meet your friend who you know has two children and he tells you, "I have a son born on Tuesday." What is the probability that the second child is also a son?
What puzzles me is that I've never run into a similar problem about daughters or mothers. I've discussed this math problem about these probabilities with many people many times. But I keep stumbling upon men who passionately defend their wrong solution. When I dig into why their solution is wrong, it appears that they implicitly assume that if a man has a daughter and a son, he won't bother talking about his daughter's birthday at all.
I've seen this so often that I wonder if this is a mathematical mistake or a reflection of their bias.
How to solve the original problem? The problem is under-defined. The solution depends on the reason the father only mentions one child, or the Tuesday.
The funny part of this story is that I, Tanya Khovanova, have two children. And the following statement is true: "I have a son born on Tuesday." What is the probability that my second child is a son?
How many different months' lengths are possible?
For "simplicity" let's stick to the Gregorian calendar.
Lennart Green is an amazing magician who performs card tricks. He is so good that the judges at the competition of the International Federation of Magic Societies didn't believe his tricks. They assumed that he used the help of stooges, and unfairly disqualified him. At the next competition he used a judge to assist him and won first place in the cards category.
I saw his performance twice. Both times he brought a woman to the stage, but each time it was a different woman. It was clear that he talked to each woman before the performance, presumably asking her permission. Furthermore, during his performances, both of the women looked slightly bored, implying that it might be not their first time. My first impression was that the women were a part of the act. I was fooled just as the judges had been fooled.
What can I say? Lennart Green isn't skilled at picking up the right women. Watch his performance at TED, and remember that he proved that the assistants were clueless.
When the Abel Prize was announced in 2001, I got very excited and started wondering who would be the first person to get it. I asked my friends and colleagues who they thought was the greatest mathematician alive. I got the same answer from every person I asked: Alexander Grothendieck. Well, Alexander Grothendieck is not the easiest kind of person to give a prize to, since he rejected the mathematical community and lives in seclusion.
Years later I told this story to my friend Ingrid Daubechies. She pointed out to me that my spontaneous poll was extremely biased. Indeed, I was asking only Russian mathematicians living abroad who belonged to "Gelfand's school." Even so, the unanimity of those responses continues to amaze me.
Now several years have passed and it does not seem that Alexander Grothendieck will be awarded the Prize. Sadly, my advisor Israel Gelfand died without getting the Prize either. I am sure I am biased with respect to Gelfand. He was extremely famous in Soviet Russia, although less well-known outside, which may have affected the decision of the Abel's committee.
I decided to assign some non-subjective numbers to the fame of Gelfand and Grothendieck. On Pi Day, March 14, 2010, I checked the number of Google hits for these two men. All the Google hits in the rest of this essay were obtained on the same day, using only the full names inside quotation marks.
Google hits do not give us a scientific measurement. If the name is very common, the results will be inflated because they will include hits on other people. On the other hand, if a person has different spellings of their name, the results may be diminished. Also, people who worked in countries with a different alphabet are at a big disadvantage. I tried the Google hits for the complete Russian spelling of Gelfand: "Израиль Моисеевич Гельфанд" and got an impressive 137,000.
Now I want to compare these numbers to the Abel Prize winners' hits. Here we have another problem. As soon as a person gets a prize, s/he becomes more famous and the number of hits increases. It would be interesting to collect the hits before the prize winner is announced and then to compare that number to the results after the prize announcement and see how much it increases. For this endeavor, the researcher needs to know who the winner is in advance or to collect the data for all the likely candidates.
John Thompson is way beyond everyone else's range because he shares his name with a famous basketball coach. But my point is that Gelfand and Grothendieck could have been perfect additions to this list.
I have this fun book at home written by Clifford Pickover and titled Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning
The most puzzling thing about this list is that there is no overlap with the Abel Prize winners. Here is the list with the corresponding Google hits.
Since there are other great mathematicians with a lot of hits, I started trying random names. In the list below, I didn't include mathematicians who had someone else appear on the first results page of my search. For example, there exists a film director named Richard Stanley. So here are my relatively "clean" results.
If prizes were awarded by hits, even when the search is polluted by other people with the same name, then the first five to receive them would have been:
If we had included other languages, then Gelfand might have made the top five with his 48,000 English-language hits plus 137,000 Russian hits.
This may not be the most scientific way to select the greatest living mathematician. That's why I'm asking you to tell me, in the comments section, who you would vote for.
I got this problem from my friend, a middle-school math teacher, Tatyana Finkelstein.
We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weight the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale.
Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.
I would like to add an extra twist to the problem above. It is conceivable that there might be several different strategies for finding the direction in which the weight of the fake coin deviates from the real coins. In this case it is better to choose a strategy that can redeem as many coins as possible — that is, to identify the maximum number of coins as real.
The number of coins you identify as real depends on the outcomes of your weighings. Then what is the precise definition of the best strategy?
Let us call a strategy k-redeem if after the weighings you are guaranteed to demonstrate that k coins are real, but you are not guaranteed to demonstrate that k+1 coins are real. Your task is to analyze two-weighing strategies and choose the most profitable one — the strategy that guarantees to redeem the largest possible number of coins, that is, a k-redeem strategy for the largest k.
Sara was born in Boston on February 29, 2008 at 11:00 am. Her parents were quite upset that their calendar-challenged daughter would only be able to celebrate her birthday once in four years. Luckily, science can help Sara's parents. How? Sara can celebrate her birthday every year at the moment when the Earth passes the same point on its orbit around the Sun as when Sara was born.
Assuming that Sara lives her entire life in Boston and that the daylight savings time is not moved earlier into February, your task is to calculate the schedule of Sara's birthday celebrations for 100 years starting from her birth. To simplify your homework, you can approximate one year as 365 days and 6 hours.
Joseph DeVincentis heard my prayers and created an index for MIT mystery hunt puzzles. He created it not because I requested it, but rather because he was on the writing team this year and they needed it. Anyway, finally there is an index.
I have to warn you, though, that this index was created for people who have already solved the puzzles, so the index contains hints for many of the problems and, on rare occasions, solutions.
Now I will do the math index for this year, and I promise that I will avoid big hints.
