Tanya Khovanova's Math Blog

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

• A group of citizens of country A emigrated to B. Show that the rating of both countries can go up.
• After that a group of citizens of B (which may include former citizens of A) emigrated to A. Is it possible that the ratings of both countries go up again?
• A group of citizens of A emigrated to B, and a group of citizens of B emigrated to C. As a result, the ratings of each country increased. After that the migration went the opposite way: some citizens of C moved to B, and some citizens from B moved to A. As a result, the ratings of all three counties went up once more. Is this possible? If yes, then how? If no, then why not?

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

The children A, B, C and D made the following assertions.

• A: B, C and D are girls.
• B: A, C and D are boys.
• C: A and B are lying.
• D: A, B and C are telling the truth.

How many of the children were telling the truth?
A) 0   B) 1   C) 2   D) 3   E) Impossible to determine

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

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

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