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08 December 2016 @ 04:15 pm

* * *

—Honey, we are like two parallel lines.
—Why do you say that?
—The intersection of our life paths was a mistake.

* * *

—Q: Why did the obtuse angle go to the beach?
—A: Because it was over 90 degrees.

* * *

Ancient Roman in a clothing store: How come XL is larger than L?

* * *

—Which is the odd one out: one, three, six, seven?
—Well, three of them are odd ones.

* * *

When I am with you, I solve integrals in my head, so that blood can come back to my brain.

* * *

There are two kinds of people in this world: Those that can extrapolate from incomplete data.

* * *

Seven has the word even in it, which is odd.

 
 
08 December 2016 @ 04:03 pm

I have always wanted to be an honest person and have followed my honor code. Soviet Russia had two honor codes. One code was for dealing with people and the other for dealing with businesses and the government. I remind you that in Soviet Russia, all businesses were owned by the government. The money paid for work didn't have any relation to the work done. The government paid standard salaries and the businesses did whatever. Generally, that meant that they were doing nothing. Meanwhile, the government got its income from selling oil.

All people were being screwed by the government, so they had no motivation to play fair. Just as American workers do, Soviet Russians might use the work copier to make personal copies. The difference was that we didn't feel guilty at ripping off the government. We wouldn't just make a few necessary copies; we would make copies for our friends, our family, for strangers—as many copies as possible.

I moved to the United States in 1990. Several of my friends took time to explain to me the difference between Soviet Russia and the US. One of the friends, let's call her Sarah, was working as staff at Harvard University. She told me the story of a recent visit by a famous Russian professor. After he left, the department received a bill. It appears that the professor, in a short visit, used several months-worth of the department's budget allocation for copying and phone calls. I was impressed, in a good way, by the professor who I assume spent a good deal of his time making a lot of copies of papers unavailable in Russia, presumably not only for himself but also for students and colleagues. On the other hand, it was clearly wrong.

My new friend Sarah told me that in the US money does not come from nowhere, and I should include Harvard University in my honor code for people. Actually, not only Harvard, but also any place of work and the government too. Sarah also told me that since that budget problem, she was asked to talk to every incoming Russian visitor and explain to them how capitalism works. Most Russian visitors were ready to accept the rules. I too was delighted with that. It is much easier to follow one honor code than two.

I was also very happy for my son, Alexey. He was eight when we moved to Boston. Before our move I had a dilemma. Should I tell him that Lenin and Stalin were bad guys and killed millions of people? If I gave him a truthful explanation, I would also have to teach him to lie. Otherwise, if he mentioned this at school or on the street, we would be at risk of going to prison. If I didn't teach him the truth, he might become brainwashed and grow up believing in communism, which would be very, very bad.

I was so lucky that I moved to the US: I didn't have to teach my son to lie.

 
 
20 November 2016 @ 02:05 pm
Centaurs, manticores, and minotaurs roam their planet. Their society is very democratic: any two animals can become sex partners. When two different species mate, their orgasm is so potent that they merge into one creature of the third species. For example, once one centaur and one manticore make love, they are reborn as one minotaur. At the beginning of the year 2016 there were 2016 centaurs, 2017 manticores, and 2018 minotaurs. They mated non-stop and at the end of the year only one creature was left on the planet. Which one?

This is one of those puzzles that I love-hate. I hate it because it is easy to answer this puzzle by inventing a specific mating pattern that ends with one animal. It is possible to get the correct answer without seeing the beauty. I love it because there is beauty in the explanation of why, if the mating ends with one animal, it has to be a specific animal.

The solution is charming, but being a mathematician this problem makes me wonder if ii is always possible to end with one animal. So I add another puzzle.

Describe the sets of parameters for which it is impossible to end up with one creature.
 
 
09 November 2016 @ 06:21 pm

Today I have two new coin puzzles that were inspired by my son, Alexey Radul:

You have N > 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have a balance scale which might or might not be faulty. A faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of weighings you need in order to figure out whether the scale is faulty?

If you think about it, this problem is isomorphic to a known problem I wrote about before:

You have N ≥ 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is either lighter than the real ones or heavier than the real ones. You also have a normal balance scale. What is the smallest number of weighings you need in order to figure out whether the fake coin is lighter or heavier?

To make things more interesting let's mix the problems up.

You have N > 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have M > 1 identical-looking balance scales. All but one of them are normal and one is faulty. The faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of total weighings needed to figure out which scale is faulty?
 
 
22 September 2016 @ 03:07 pm

The following problem was at a 2016 entrance test for the MIT PRIMES STEP program.

I drew several triangles on a piece of paper. First I showed the paper to Lev and asked him how many triangles there were. Lev said 5 and he was right. Then I showed the paper to Sasha and asked him how many triangles there were. Sasha said 3 and he was right. How many triangles are there on the paper? Explain.

The intended answer was 8: there were 5 triangles on one side of the paper and 3 on the other.

Most of the students didn't think that the paper might be two-sided, but they came up with other inventive ideas. Below are some of their pictures, and I leave it to you to explain why they work. All the students who submitted these pictures got a full credit for this problem on the test.

Example 7Example 5Example 4Example 3Example 1Example 2Example 6
 
 
 
10 September 2016 @ 07:00 pm

If you are a high-school student who wants to conduct research in mathematics, you should check out the MIT PRIMES program. If you enjoy solving the problems in our entrance test, that's the first indication that you might want to apply. But to determine if the program is right for you, and you are right for the program, please read the following questions and answers which have been prepared for you by Tanya Khovanova, the PRIMES Head Mentor. (This only addresses applications to PRIMES Math, and only to the research track)

Question: I do not like math competitions. Should I apply?
Answer: Math competitions are completely separate from research in mathematics. If you enjoy thinking about mathematics for long periods of time and are fascinated by our test questions, you should apply.

Question: I am good at math, but I really want to be a doctor. Should I apply?
Answer: No. PRIMES requires a huge time commitment, so math should really be your most significant interest.

Question: I want to get into Harvard, and PRIMES looks good on a resume. Should I apply?
Answer: PRIMES does look good on a resume. But if you are more passionate about, say, climate change than math, what would Harvard's admission committee see? Our experience in the program is that if math isn't your top interest, your math student may not be sufficiently impressive to be accepted at Harvard as a math researcher. At the same time, you will not be accepted as the top climate change student as you didn't invest your time in that. Math research is a hard way to earn points for college. See also, the essay, Thoughts on research by Simon Rubinstein-Salzedo.

Question: My parents want me to apply. Should I apply?
Answer: Your parents will not be accepted to the program. Do not apply if you do not really, really want to.

Question: Your website suggests that I should spend ten hours a week on the PRIMES project. I can only spend five. But I am a genius and faster than other people.
Answer: We already assume that you are a genius and faster than anyone else you know. Five hours a week are not enough for a successful project.

Question: I looked at the past PRIMES projects and nothing excites me as much as my current interest in Pascal's triangle. I doubt I should apply.
Answer: When you start working on a project, you will learn a lot about it. You will understand why, for example, Cherednik algebras are cool. The excitement comes with knowledge and invested time. Not yet being excited about Cherednik algebras is not a good reason not to apply. Besides a lot of exciting mathematics is done between several different fields.

Question: I really want to do nothing else than study Pascal's triangle.
Answer: We try to match our projects to students' interests as much as we can. But we almost never can fulfill a specific request as above. You might get a project related to Young diagrams, which are connected to quantum Pascal's triangle. If this connection doesn't excite you, you shouldn't apply.

Question: I think I will be better positioned for research if I spend five more years studying.
Answer: There is nothing wrong with this approach. For many years the standard was to start research in graduate school. Our program is innovative. At PRIMES we are trying a different model. It may sound scary, but you will learn everything you need to know in order to do your project. If the project is in representation theory, for example, you will only learn what you need—not the whole theory. Our hope is that eventually you will take a course in representation theory and expand your grasp of it and see the bigger picture behind your project. We have a reading track for people like you who reside in Boston area.

Question: I love math, but I am not sure that I want to be a mathematician. Should I apply?
Answer: Many people start loving math early in life and then discover that there are many other things that require a similar kind of brain: computer science, cryptography, finance, and so on. We do not require from our students a commitment to become mathematicians. If you want to try research in math, you should apply. If students decide that they do not want to do research in math after finishing our program, we do not consider that a negative result. One way or another, the experience of PRIMES will help you understand better what you want to do with your life.

Question: I want to get to the International Math Olympiad. I am afraid that the time the research project takes prevents me from preparing for competitions. Should I apply?
Answer: People who are good at Olympiads often have fantastic brain power that helps in research. On the other hand, research requires a different mind set and the transition might be painful. It is possible, but not trivial to succeed in both. It is up to you to decide how you want to spend your time.

Question: I like number theory, but I do not see past PRIMES projects in number theory.
Answer: Doable number theory projects are hard to come by and we have fewer number theory projects than students who want to do number theory. There are many high-school programs that teach number theory including PROMYS and Ross programs. Our applicants like number theory because they were exposed to it. During PRIMES you will be exposed to something else and might like it as much.

Question: I found a local professor to work with on a research project. Should I apply to PRIMES?
Answer: PRIMES requires that you devote 10 hours a week to research for a year. It is unrealistic to do two research projects in parallel. Choose one. Working with someone in person may be better than by Skype at PRIMES. Also, usually our mentors are not professors, but rather graduate students. On the other hand, they are MIT grad students and projects are often suggested by professors. Our program is well structured. We guarantee weekly meetings in the Spring, we give extra help with your paper, and we have a conference. It is up to you to decide.

 
 
07 September 2016 @ 05:33 pm

Alexander Shapovalov is a prolific puzzle writer. He has a special webpage of his river-crossing puzzles (in Russian). Here is one of these puzzles.



Three swindlers have two suitcases each. They approach a river they wish to cross. There is one boat that can carry three objects, where a person or a suitcase counts as one object. No swindler can trust his suitcase to his swindler friends when he is away, but each swindler doesn't mind his suitcases left alone at the river shore. Can they cross the river?
 
 
09 August 2016 @ 07:02 pm

In my paper with Joshua Xiong, Nim Fractals, we produced a bijection between P-positions in the three-pile Nim and a three-branch Ulam-Warburton automaton. We also defined a parent-child relationship on games that is induced by this bijection. Namely, two consecutive P-positions in a longest optimal game of Nim are the ones that correspond to a parent-child pair in the automaton. A cell in the Ulam-Warburton automaton has exactly one parent. That means, if (a,b,c) is a Nim P-position, then exactly one of (a − 1,b − 1,c), (a − 1,b,c − 1), and (a,b − 1,c − 1) must be a P-position and a parent of (a,b,c). (See our paper for more details.)

Now I want to explicitly write out the rules of an automaton which will generate the Nim P-positions in 3D.

Let me restrict the evolution of the automaton to the non-negative octant. That is, we consider points (a,b,c) in 3D, where each coordinate is a non-negative integer. We define the neighbors of the point (a,b,c) to be the points that differ from (a,b,c) in two coordinates exactly by 1. So each point strictly inside the octant has 12 neighbors. (There are three ways to choose two coordinates, and after that four ways to choose plus or minus 1 in each of them.

There is a geometric interpretation to this notion of neighborhood. Let us correspond a unit cube to a point with integer coordinates. The center of the cube is located at the given point and the sides are parallel to the axes. Then two points are neighbors if and only if the corresponding cubes share one edge. Now it becomes more visual that a cube has 12 neighbors, as it has 12 edges.

Here is the rule of the automaton. Points never die. We start with the patriarch, (0,0,0), one point being alive. The non-alive point is born inside the non-negative octant if it has exactly 1 alive neighbor that is closer to the patriarch. In other words the point (a,b,c) is born if and only if exactly one out of three points (a − 1,b − 1,c), (a − 1,b,c − 1), and (a,b − 1,c − 1) is alive. It follows that the points that are born at the n-th step has a coordinate sum 2n.

Consider for example the starting growth. At the first step the points (0,1,1), (1,0,1) and (1,1,0) are born. At the next step the points (0,2,2) and (2,0,2) and (2,2,0) are born. while the (1,1,2) will never be born as starting from the second step it has at least two live neighbors: (0,1,1) and (1,0,1) that are closer to the patriarch.

Theorem. In the resulting automaton, the points that are born at step n are exactly the P-positions of Nim with the total of 2n tokens.

Proof. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious. Suppose we proved that at step n exactly P-positions with 2n tokens are born. Consider a P-position of Nim: (a,b,c) such that a + b + c = 2n + 2. Remember, that bitwise XOR of a, b, and c is zero. Consider the 2-adic values of a, b, and c (aka the smallest powers of 2 dividing a, b, and c). There should be exactly two out of these three integers that have the smallest 2-adic value. Suppose these are a and b. Then (a − 1,b − 1,c) is a P-position, while (a − 1,b,c − 1) and (a,b − 1,a − 1) are not. That means by the inductive hypothesis (a,b,c) has exactly one alive neighbor. So the position (a,b,c) is born at time n + 1.

Now we need to proof that nothing else is born. For the sake of contradiction suppose that (a,b,c) is the earliest N-position to be born. That means it has a live neighbor that is a P-position closer to the patriarch. WLOG we can assume that this neighbor is (a − 1,b − 1,c).

If a − 1 and b − 1 are both even, then (a,b,c) is a P-position, which is a contradiction. Suppose a − 1 and b − 1 are both odd. Then their binary representations can't have the same number of ones at the end. Otherwise, (a,b,c) is a P-position. That is a and b have different 2-adic values. Suppose a has a smaller 2-adic value, Then, for (a − 1,b − 1,c) to be a P-position a and c has to have the same 2-adic value. That means (a,b − 1,c − 1) is a P-position too. Now suppose a − 1 and b − 1 are of different parities. Without loss of generality suppose a − 1 is odd and b − 1 is even, then c is odd. Then (a − 1,b,c − 1) is a P-position too. Thus we can always find a second neighbor with the same number of tokens. That is, both neighbors are alive at the same time; and the N-position (a,b,c) is never born. □

One might wonder what happens if we relax the automaton rule by removing the constraint on the distance to the patriarch. Suppose a new point is born if it has exactly one neighbor alive. This will be a different automaton. Let us look at the starting growth, up to a permutation of coordinates. At step one, positions (0,1,1) are born. At the next step positions (0,2,2) are born. At the next step positions (0,1,3), (1,2,3) and (0,3,3) are born. We see that (0,1,3) is not a P-positions. What will happen later? Will this N-position mess up the future positions that are born? Actually, this automaton will still contain all the P-positions of Nim.

Theorem. In the new automaton, the points that are born at step n and have total of 2n tokens are exactly the P-positions of Nim with the total of 2n tokens.

Proof. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious. The birth of the points that have total of 2n tokens and are born at step n depend only on the points with the total of 2n − 2 tokens that are born at step n − 1. By the inductive hypothesis, those are the P-positions with 2n − 2 tokens. So the points have total of 2n tokens and are born at step n match exactly the first automaton described above. To reiterate, N-positions with 2n tokens are born after P-positions with 2n tokens, so they do not influence the birth of P-positions with 2n + 2 tokens. □

 
 
07 August 2016 @ 02:17 pm

In the game of Nim you have several piles with tokens. Players take turns taking several tokens from one pile. The person who takes the last token wins.

The strategy of this game is well-known. You win if after your move the bitwise XOR of all the tokens in all the piles is 0. Such positions that you want to finish your move with are called P-positions.

I play this game with my students where the initial position has four piles with 1, 3, 5, and 7 tokens each. I invite my students to start the game, and I always win as this is a P-position. Very soon my students start complaining that I go second and want to switch with me. What should I do? My idea is to make the game last long (to have many turns before ending) to increase the chances of my students making a mistake. So what is the longest game of Nim given that it starts in a P-position?

Clearly you can't play slower then taking one token at a time. The beauty of Nim is that such an optimal game starting from a P-position is always possible. I made this claim in several papers of mine, but I can't find where this is proven. One of my papers (with Joshua Xiong) contains an indirect proof by building a bijection to the Ulam-Warburton automaton. But this claim is simple enough, so I want to present a direct proof here. Actually, I will prove a stronger statement.

Theorem. In an optimal game of Nim that starts at a P-position the first player can take one token at each turn so that the second player is forced to take one token too.


Proof. Consider a P-position in a game of Nim. Then find a pile with the lowest 2-adic value. That is the pile such that the power of two in its factorization is the smallest. Suppose this power is k. Notice that there should be at least two piles with this 2-adic value.


The first player should take a token from one of those piles. Then the bitwise heap-sum after the move is 2k+1−1. Then the Nim strategy requires the second player to take tokens from a pile such that its value decreases after bitwise XORing with 2k+1−1. All piles with the 2-adic value more than k will increase after xoring with 2k+1−1. That means the second player has to take tokens from another pile with 2-adic value k. Moreover, the second player is forced to take exactly one token to match the heap-sum. □

In the position (1,3,5,7) all numbers are odd, so I can take one token from any pile for my first move, then the correct move is to take one token from any other pile. My students do not know that; and I usually win even as the first player. Plus, there are four different ways I can start as the first player. This way my students do not get to try several different options with the same move I make. After I win several times as the first player, I convince my students that I win anyway and persuade them to go back to me being the second player. After that I relax and never lose. I am evil.

 
 
06 August 2016 @ 08:02 pm

Replace question marks with letters in the following sequence:



Y, Y, ?, ?, Y, ?, Y, ?, R, R, R, R.