Archive for the ‘Math Education’ Category.

Penney’s Game and Groups

For the last year, I’ve been obsessed with Penney’s game. In this game, Alice picks a string of coin tosses, say HHH for three heads. After that, Bob picks his string of tosses of the same lengths, say HTH. Then they toss a fair coin. The person whose string shows up first wins. For example, if the tosses are THTTHHH, then Alice wins after the seventh toss. For these particular choices, Bob wins with probability 3/5.

With my PRIMES student, Sean Li, we looked at this game and asked a different question. Suppose Alice picks a pattern of three tosses in a row that are the same. Suppose after that, Bob chooses a pattern of three alternating tosses. Then they toss a fair coin. Alice is hoping for HHH or TTT, while Bob is hoping for HTH or THT. The person whose pattern shows up first wins. For example, if the tosses are THTTHHH, then Bob wins after the third toss. For these particular choices, Bob wins with probability 1/2.

In this example, what actually happens. We assume that the group of two elements acts on the alphabet of two letters. The group’s non-identity element swaps letters H and T. We assume that two strings are equivalent if they belong to the same equivalency class under the group action. We call such an equivalency class a pattern.

In the new game we invented, we have an alphabet of any size and any group acting on the alphabet. Then Alice and Bob pick their patterns. After that, they play the Penney’s game on these patterns. The answers to all the relevant questions are in our paper, The Penney’s Game with Group Action, posted at the math.CO arxiv 2009.06080.

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Confirming the Labels of Coins in One Weighing

I wrote a paper Confirming the Labels of Coins in One Weighing together with my PRIMES STEP students Isha Agarwal, Paul Braverman, Patrick Chen, William Du, Kaylee Ji, Akhil Kammila, Shane Lee, Alicia Li, Anish Mudide, Jeffrey Shi, Maya Smith, and Isabel Tu. The paper is available at math.HO arxiv:2006.16797. Below my students describe what happens in the paper in their own words.

Tragedy has struck in an iCOINic land known as COINnecticut. One day, while everyone was minding their own business, the vault door of the bank was found to have been forcefully opened. COINcerns spread amongst the COINmoners that someone had tampered with their n sacred COINtainers of coins! The COINtainers are labeled with the integers 1 through n, which usually describe the weight of each of the countless coins inside that corresponding COINtainer. For example, the COINtainer labeled 1 should only COINtain coins that weigh 1 gram, the COINtainer labeled 2 should only COINtain coins that weigh 2 gram, and so on, you get the COINcept.

The acCOINtants COINclude that someone may have switched around the labels on the COINtainers. To resolve this COINplication, aka to check if the labels have been tampered with, they bought a balance scale for a microsCOINpic amount of money. However, they can only use the scale to COINduct one weighing as the angry COINmoners are impatient and wish to withdraw their money ASAP.

The COINfused acCOINtants COINvinced 11 COINspicuous students from Boston’s COINmunity to help them. With their COINbined efforts, they COINcluded that indeed, no matter how many COINtainers there are, their labels can be COINfirmed as correct or incorrect with just one weighing! Unfortunately, sometimes, such a weighing requires the use of many coins or coins with a large COINbined weight, which could potentially break the scale. Seeing this COINundrum, the students wished to be eCOINomical and find the least amount of coins or weight they need to place on the scale.

The acCOINtants and the 11 students COINtinued examining the nature of these weighings and discovered patterns that occur within them. They COINfined their research to special weighings they called downhill. They COINfirmed the effectiveness of such weighings to solve the problem at hand. The students estimated the weight and the number of coins, thus COINpleting their task.

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The No-Flippancy Game

My STEP students invented a coin-flipping game that doesn’t require a coin. It is called The No-Flippancy Game.

Alice and Bob choose distinct strings of length n consisting of the letters H (for heads) and T (for tails). The two players alternate selecting the outcome of the next “flip” to add to the sequence by the rule below.

The “flip” rule: Let i < n be the maximal length of a suffix of the sequence of chosen outcomes that coincides with a prefix of the current player’s string. The player then selects the element of their string with index i + 1 as the next term in the sequence.

Alice goes first, and whoever’s string appears first in the sequence of choices wins. In layman terms, the game rules mean that the players are not strategizing, but rather greedily finishing their strings.

Suppose n = 2 and Alice chose HH. If Bob chooses HT, then Bob wins. Alice has to choose H for the first flip. Then Bob chooses T and wins. On the other hand, if Bob chooses TT for his string, the game becomes infinite. On her turn, Alice always chooses H, while on his turn Bob always chooses T. The game outcome is an alternating string HTHTHT… and no one wins.

Suppose n = 4, Alice chooses HHTT, and Bob chooses THHH. The game proceeds as HTHHTHHH, at which point Bob wins.

This game is very interesting. The outcome depends on how Alice’s and Bob’s chosen strings overlap with each other. We wrote a paper about this game, which is available at math.CO arXiv:2006.09588.

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SET Tic-Tac-Toe

The academic year is over and my junior PRIMES STEP group finished their paper about a classification of magic SET squares. A magic SET square is a 3 by 3 square of SET cards such that each row, column, and diagonal is a set. See an example below. The paper is posted at the arXiv:2006.04764.

A magic SET square

In addition to classifying the magic SET squares, my students invented the game of SET tic-tac-toe. It is played on nine cards that form a magic SET square. Two players take turns picking a card from the square. The first player who has a set wins.

One might think that this game is the same as tic-tac-toe, as a player wins as soon at they have cards from the same row, column, or diagonal. But if you build a magic SET square, you might notices that each magic SET square contains 12 sets. In addition to rows, columns, and diagonals, there are sets that form broken diagonals. The picture below shows all the sets in a magic SET square.

Sets in magic SET squares

There are more ways to win in this game than in a regular tic-tac-toe game. My students proved that ties are impossible in this game. They also showed, that, if played correctly, the first player always wins.

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Anchored Rectangles

Suppose we want to pack a unit square with non-overlapping rectangles that have sides parallel to the axes. The catch is that the lower left corners of all the rectangles are given. By the way, such rectangles are called anchored. Now, given some points in the unit square, aka the lower left corners, we want to find anchored rectangles with the maximum total area.

Imcreasing permutation

When the given points are close to the right upper corner of the square, the total area is small. When a single point is in the bottom left corner of the square, we can cover the whole square. The problem becomes more interesting if we add one extra assumption: one of the given points has to be the bottom left corner of the square. In the 1960’s, it was conjectured by Allen Freedman that any set of points has an anchored rectangle packing with the area of at least one half. The problem is quite resistant. In 2011, Dumitrescu and Tóth showed that every set of points has a packing of area at least 0.09, which was the first constant bound found, and is the best bound currently known.

I gave this problem to my PRIMES student Vincent Bian. He wrote a paper, Special Configurations in Anchored Rectangle Packings, that is now available at the arxiv. When you look at this problem you see that the number of ways to pack depends on the relative coordinates of the points. That means you can view the points as a permutation. Vincent showed that the conjecture is true for several different configurations of points: increasing, decreasing, mountain, split layer, cliff, and sparse decreasing permutations.

An increasing permutation is easy. There are two natural ways to pack the rectangles. One way, when rectangles are horizontal and each rectangle reaches to the right side of the square (see picture above). Another way, when rectangles are vertical. When you take the union of both cases, the square is completely covered, which means at least one of the cases covers at least half of the square. The worst case scenario, that is, the case when the maximum possible area is the smallest is when your points are placed equidistantly on the diagonal.

Decreasing permutation

Other cases are more difficult. For example, Vincent showed that for a decreasing permutation with n points, the worst case scenario is when the points are arranged equidistantly on a hyperbola xy = (1-1/n)n. The picture shows the configuration for 15 points. The total area is more than one half.


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Meta Logic

Here is a logic puzzle.

Puzzle. You are visiting an island where all people know each other. The islanders are of two types: truth-tellers who always tell the truth and liars who always lie. You meet three islanders—Alice, Bob, and Charlie—and ask each of them, “Of the two other islanders here, how many are truth-tellers?” Alice replies, “Zero.” Bob replies, “One.” What will Charlie’s reply be?

The solution proceeds as follows. Suppose Alice is a truth-teller. Then Bob and Charlie are liars. In this situation Bob’s statement is true, which is a contradiction. Hence, Alice is a liar. It follows, that there is at least one truth-teller between Bob and Charlie. Suppose Bob is a liar. Then the statement that there is one truth-teller between Alice and Charlie is wrong. It follows that Charlie is a liar. We have a contradiction again. Thus, Alice is a liar and Bob is a truth-teller. From Bob’s statement, we know that Charlie must be a truth-teller. That means, Charlie says “One.”

But here is another solution suggested by my students that uses meta considerations. A truth-teller has only one possibility for the answer, while a liar can choose between any numbers that are not true. Even if we assume that the answer is only one of three numbers—0, 1, or 2—then the liar still has two options for the answer. If Charlie is a liar, there can’t be a unique answer to this puzzle. Thus, the puzzle question implies that Charlie is a truth-teller. It follows that Alice must be lying and Bob must be telling the truth. And the answer is the same: Charlie says, “One.”

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Playing with Pascal’s Triangle

The beautiful Pascal triangle has been around for many years. Can you say something new about it?

Pascal Triangle Mod 2

Of course you can. Mathematicians always find new way to look at things. In 2012 RSI student, Kevin Garbe, did some new and cool research related to the triangle. Consider Pascal’s triangle modulo 2, see picture which was copied from a stackexchange discussion.

A consecutive block of m digits in one row of the triangle modulo 2 is called an m-block. If you search the triangle you will find that all possible binary strings of length 2 are m-blocks. Will this trend continue? Yes, you can find any possible string of length 3, but it stops there. The blocks you can find are called accessible blocks. So, which blocks of length 4 are not accessible?

There are only two strings that are not accessible: 1101 and 1011. It is not surprising that they are reflections of each other. Pascal’s triangle respects mirror symmetry and the answer should be symmetric with respect to reflection.

You can’t find these blocks on the picture, but how do we prove that they are not accessible, that is, that you can’t ever find them? The following amazing property of the triangle can help. We call a row odd/even, if it corresponds to binomial coefficients of n choose something, where n is an odd/even number. Every odd row has every digit doubled. Moreover, if we take odd rows and replace every double digit with its single self we get back Pascal’s triangle. Obviously the two strings 1101 and 1011 can’t be parts of odd rows.

What about even rows? The even rows have a similar property: every even-indexed digit is a zero. If you remove these zeros you get back Pascal’s triangle. The two strings 1101 and 1011 can’t be part of even rows. Therefore, they are not accessible.

The next question is to count the number of inaccessible blocks of a given length: a(n). This and much more was done by Kevin Garbe for his RSI 2012 project. (I was the head mentor of the math projects.) His paper is published on the arxiv. The answer to the question can be found by constructing recurrence relations for odd/even rows. It can be shown that a(2r) = 3a(r) + a(r+1) − 6 and a(2r+1) = 3a(r) + 2a(r+1) − 6. As a result the number of inaccessible blocks of length n is n2n + 2. I wonder if there exists a direct proof of this formula without considering odd and even rows separately.

This RSI result was so pretty that it became a question at our entrance PRIMES test for the year 2013. In the test we changed the word accessible to admissible, so that it would be more difficult for applicants to find the research. Besides, Garbe’s paper wasn’t arxived yet.

The pretty picture above is from the stackexchange, where one of our PRIMES applicants tried to solicit help in solving the test question. What a shame.Share:Facebooktwitterredditpinterestlinkedinmail

How Many Triangles?

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

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Should You Apply to PRIMES?

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.Share:Facebooktwitterredditpinterestlinkedinmail

The Reuleaux Tetrahedron

Why are manhole covers round? The manhole covers are round because the manholes are round. Duh! But the cute mathematical answer is that the round shapes are better than many other shapes because a round cover can’t fall into a round hole. If we assume that the hole is the same shape as the cover but slightly smaller, then it is true that circular covers can’t fall into their holes. But there are many other shapes with this property. They are called the shapes of constant width.

A circle and a Reuleaux triangle

Given the width, the shape with the largest area is not surprisingly a circle. The shape with the smallest area and a given constant width is a Reuleaux Triangle. Here is how to draw a Reuleaux triangle. Draw three points that are equidistant from each other at distance d. Then draw three circles of radius d with the centers at given points. The Reuleaux triangle is the intersection of these three circles.

Can we generalize this to 3d? What would be an analogue of a Reuleaux Triangle in 3d? Of course, it is a Reuleaux Tetrahedron: Take four points at the vertices of a regular tetrahedron; take a sphere at each vertex with the radius equal to the edge of the tetrahedron; intersect the four spheres.

Is this a shape of the constant width? Many people mistakenly think that this is the case. Indeed, if you squeeze the Reuleaux tetrahedron between two planes, one of which touches a vertex and another touches the opposite face of the curvy tetrahedron, then the distance between them is equal to d: the radius of the circle. This might give you the impression that this distance is always d. Not so. If you squeeze the Reuleaux tetrahedron between two planes that touch the opposite curvy edges, the distance between these planes will be slightly more than d. To create a shape of constant width you need to shave off the edges a bit.

Meissner Bodies

Theoretically you can shave the same amount off every edge to get to a surface of constant width. But this is not the cool way to do it. The cool way is to shave a bit more but only from one edge of the pair of opposite edges. You can get two different figures this way: one that has three shaved edges forming a triangle, and the other, where three shaved edges share a vertex. These two bodies are called Meissner bodies and they are conjectured to be shapes of the constant width with the smallest volume.

On the picture I have two copies of a pair of Meissner bodies. The two left ones have three edges that share a vertex shaved off. The very left shape gives a top view of this vertex and the solid next to it has its bottom with holes looking forward. The two shapes on the right show the second Meissner body in two different positions.

I recently discovered a TED-Ed video about manhole covers. It falsely claims that the Reuleaux tetrahedron has constant width. I wrote to TED-Ed, to the author, and posted a comment on the discussion page. There was no reaction. They either should remove the video or have an errata page for it. Knowingly keeping a video with an error that is being viewed by thousands of people is irresponsible.Share:Facebooktwitterredditpinterestlinkedinmail