Tanya Khovanova's Math Blog

Davidson Institute for Talent Development announced their 2012 Winners. Out of 22 students, three were recognized for their math research. All three of them are ours: that is, they participated in our PRIMES and RSI programs:

• David Ding’s project, “Infinitesimal Cherednik Algebras of gl_n,” came out of his participation in the PRIMES program.
• Sitan Chen’s project, “On the Rank Number of Grid Graphs,” came out of his participation in the RSI program.
• Xiaoyu He’ project, “On the Classification of Universal Rotor-Routers,” came out of his participation in the PRIMES program.

I already wrote about Xiaoyu’s project. Today I want to write about Sitan’s project and what I do as the math coordinator for RSI.

RSI students meet with their mentors every day and I meet with students once a week. On the surface I just listen as they describe their projects. In reality, I do many different things. I cheer the students up when they are overwhelmed by the difficulty of their projects. I help them decide whether they need to switch projects. I correct their mistakes. Most projects involve computer help, so I teach them Mathematica. I teach them the intricacies of Latex and Beamer. I explain general mathematical ideas and how their projects are connected to other fields of mathematics. I never do their calculations for them, but sometimes I suggest general ideas. In short, I do whatever needs to be done to help them.

I had a lot of fun working with Sitan. His project was about the rank number of grid graphs. A vertex k-ranking is a labeling of the vertices of a graph with integers from 1 to k so that any path connecting two vertices with the same label passes through a vertex with a greater label. The rank number of a graph is the minimum possible k for which a k-ranking exists for that graph. When Sitan got the project, the ranking numbers were known for grid graphs of sizes 1 by n, 2 by n, and 3 by n. So Sitan started working on the ranking number of the 4 by n graph.

His project was moving unusually fast and my job was to push him to see the big picture. I taught him that the next step, once he finishes 4 by n graphs is not to do 5 by n graphs, as one might think. After the first step, the second step should be bigger. He should use his insight and understanding of 4 by n graphs to try to see what he can do for any grid graphs.

This is exactly what he did. After he finished the calculation of the rank number of the 4 by n graphs, he found a way to improve the known bounds for the ranking number of any grid graph. His paper is available at the arXiv.

I just looked at my notes for my work with Sitan. The last sentence: “Publishable results, a potential winner.”

Baron Münchhausen is famous for his tall tales. My co-author Konstantin Knop wants to rehabilitate him and so invents problems where the Baron is proven to be truthful from the start. We already wrote a paper about one such problem. Here is a new problem by Konstantin:

Kostya has a black box, such that if you put in exactly 3 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 5 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than 3 times?

Actually, Konstantin designed a more complicated problem that was given at the Euler Olympiad, 2012 in Russia.

Let n be a fixed integer. Kostya has a black box, such that if you put in exactly 2n+1 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 4n+1 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than n+2 times?

Note that Kostya can’t just put 4n+1 coins in the box. The box accepts exactly 2n+1 coins. The problem that I started with is for n = 1. Even such a simple variation was a lot of fun for me to solve. So, have fun.

I recently published my new favorite math problem:

A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic’s goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic’s assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.

If the psychic is only allowed to look at the backs of the cards, then the amount of transmitted information is 2³⁶, which is the same amount of information as suits for 18 cards. This number of guesses is achievable: the backs of every two cards can clue in the suit of the second card in the pair. This way the psychic can guess the suits of all even-numbered cards in the deck. So the problem is to improve on that. Using the info from the cards that the psychic is permitted to turn over can help too.

The problem is from the book Moscow Mathematical Olympiads, 2000-2005. The book and Russian blog discussions provide many different ideas on how to guess more than half of the deck.

Here is the list of ideas.

Idea 1. Counting cards. If you count cards you will know the suits of the last cards.

Idea 2. Trading. As we discussed before, the psychic can correctly guess the suits of even-numbered cards. By randomly guessing the odd-numbered cards she can correctly guess on average the suits of 4.5 additional cards. Unfortunately, this is not guaranteed. But wait. What if we trade the knowledge of the second card’s suit for the majority suit among odd-numbered cards?

Idea 3. Three cards. Suppose we have three cards. Three bits can provide the following knowledge: the majority color, plus the suit of the first and of the second cards in the majority color. Thus, three bits of information will allow the psychic to guess the suits of two cards out of three.

Idea 4. Which card. Suppose the assistant signals the suits of even-numbered cards. With no loss, the psychic can guess the even-numbered card and repeat the same suit for the next card. If this is the plan, the assistant can choose which of the two cards to describe. Which card of the two matches the psychic’s guess provides an additional bit of information.

Idea 5. Surprise. Suppose we have a strategy to inform the psychic about some cards. Suppose the assistant deliberately fails on one of the cards. Then the index of this card provides info to the psychic.

I leave it to my readers to use these ideas to find the solution for 19, 23, 24 and maybe even for 26 cards.

Students should use a different strategy for the AIME than for the AMC. So students who are approaching the AIME for the first time need to question the habits they have developed after years of doing multiple choice tests. Here are some suggestions.

Checking. I’ve noticed that the accuracy level of students who take the AIME for the first time drops significantly. It seems that they are so used to multiple choice questions that they rely on multiple choices as a confirmation that they are right. So when someone solves a problem, they compare their answer to the given choices and if the answer is on the list they assume that the answer must be correct. Their pattern is broken when there are no choices. So they arrive at an answer and since there is no way to check it against choices, they just submit it. Because of this lack of confirmation, checking their answer in other ways becomes more important.

Timing. At the AMC we have 3 minutes per problem. At the AIME — 12. That means the timing strategies need to be different. Indeed, the AMC is so fast-paced that it is reasonable to save time by not reading a problem twice. If you read it, you either solve it or skip it and go on. The student who is not trying to achieve a perfect score can decide in advance not to read those final, highly-difficult problems.
For the AIME it is not expensive, in relative terms of time, to read all the problems. The student can read the problems and choose the most promising ones to start with, knowing that if there is time they can always come back to other problems.

Guessing. Guessing at the AMC is very profitable if you can exclude three choices out of the given five. Guessing for the AIME is a waste of time because the answers are integers between 000 and 999. So the probability of a random guess is one in a thousand. Actually, this is not quite right, because the problem writers are human and it is much easier to write a problem with an answer of 10 than one with an answer of 731. But the AIME designers are trying very hard to make answers that are randomly distributed. So the probability of a random guess is not one in a thousand, but it is very close. You can improve your chances by an intelligent guess. For example, you might notice that the answer must be divisible by 10. But guessing is still a waste of time. Thinking about a problem for two minutes in order to increase the probability of a correct guess to one in a 100 means that your expected gain is 1/200 points per minute. Which is usually much less than the gain for checking your answers. You can play the guessing game if you have exhausted your other options.

What saddens me is that the students who are not trained in checking use their first guess to make their life choices. But this is a subject for a separate discussion.

I have already written about how American math competition are illogically structured, for the early rounds do not prepare students for the later rounds. The first time mathletes encounter proofs is in the third level, USAMO. How can they prepare for problems with proofs? My suggestion is to look East. All rounds of Russian math Olympiads — from the local to the regional to the national — are structured in the same way: they have a few problems that require proofs. This is similar to the USAMO. At the national All-Russian Olympiad, the difficulty level is the same as USAMO, while the regionals are easier. That makes the problems from the regionals an excellent way to practice for the USAMO. The best regional Olympiad in Russia is the Moscow Olympiad. Here is the problem from the 1995 Moscow Olympiad:

We start with four identical right triangles. In one move we can cut one of the triangles along the altitude perpendicular to the hypotenuse into two triangles. Prove that, after any number of moves, there are two identical triangles among the whole lot.

This style of problems is very different from those you find in the AMC and the AIME. The answer is not a number; rather, the problem requires proofs and inventiveness, and guessing cannot help.

Here is another problem from the 2002 Olympiad. In this particular case, the problem cannot be adapted for multiple choice:

The tangents of a triangle’s angles are positive integers. What are possible values for these tangents?

The problems are taken from two books: Moscow Mathematical Olympiads, 1993-1999, and Moscow Mathematical Olympiads, 2000-2005. I love these books and the problems they present from past Moscow Olympiads. The solutions are nicely written and the books often contain alternative solutions, extended discussion, and interesting remarks. In addition, some problems are indexed by topics, which is very useful for teachers like me. But the best thing about these books are the problems themselves. Look at the following gem from 2004, which can be used as a magic trick or an idea for a research paper:

A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic’s goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic’s assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.

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.

Tanya Khovanova and Richard Stanley

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.

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.

1. Match the formulae with their names. Explain your solution.
2. Write the names of the elements with the following formulae: C₂H₄, C₂H₆, C₇H₁₂.
3. Write the formulae for the following elements: Propene, Butane.

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.

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.