According to the report, members of the Russian president’s Federal Protection Service (FPS) are responsible for collecting his bodily waste in specialised packets which are then placed in a dedicated briefcase for the journey home.
Here is my joke on the subject.
Joke. Putin’s guard collecting his waste wrote a book of memoirs. The book has two chapters: Number One and Number Two.
One of my all-time favorite puzzles is about tiling a chessboard with 1-by-2 dominoes. But the chessboard is special: two cells at the opposite corners are removed. A similar elegant chessboard puzzle recently appeared on Facebook.
Puzzle. Our special chessboard has one corner cell removed. On each of the remaining cells, there is an ant. At a signal, each ant moves two cells horizontally or vertically (for example, an ant can move from b3 to b5). Is it possible that after all the ants perform their move, each of the 63 cells will have an ant?
This puzzle is a variation of another puzzle, where all the setup is the same, but each ant moves only one cell horizontally or vertically. You can start with this variation, as it is easier.
The Wythoff array is an array of integers that can be defined through Wythoff’s game. For my purposes, I will skip over Wythoff’s game and define his array using Fibonacci numbers.
You probably heard about binary, ternary, decanary, and other bases. I am sure you know the decanary base: it is our standard decimal system. But, have you ever heard about the Fibonacci base?
Let me define it. Any positive integer can be represented uniquely as a sum of distinct Fibonacci numbers that are not neighbors. For example, 16 is 13 + 3. Now, we just write ones for Fibonacci numbers that we used and zeros for unused Fibonacci numbers, so that the digit corresponding to Fibonacci number 1 is last. (The digit corresponding to Fibonacci number 2 is second to last). Thus, 16 in the Fibonacci base is 100100. The result looks like binary, but it will never have two consecutive ones. Such a representation is called the Zeckendorf representation.
What happens if we add 0 at the end of a number in its Zeckendorf representation? This is like multiplying by 2 in binary. But, in the Fibonacci base, it corresponds to replacing every Fibonacci number in the sum with the next Fibonacci number. The result is called the Fibonacci successor. For example, the Fibonacci successor of 16 has the Zeckendorf representation 1001000 and is equal to 21 + 5 = 26.
Now back to the Wythoff array. The first row consists of the Fibonacci numbers in order. Their Zeckendorf representations look like powers of two in binary. The first column consists of numbers ending in 1 in the Zeckendorf representation, in increasing order. In other words, the Zeckendorf representations of numbers in the first column resemble odd numbers in binary. To finish the definition of the array, each number in the nth column is the Fibonacci successor of a number to its left.
As an example, let’s calculate the first three numbers of the second row. The smallest number that is not a Fibonacci number and looks odd in its Zeckendorf representation is 4, represented as 101 (remember, we are not allowed to have two consecutive ones). We place 4 in the first column of the second row. The next number in the second row has to be the Fibonacci successor of 4, so its Zeckendorf representation has to be 1010. This number equals 5+2 = 7. The Fibonacci successor of 7 is 11, with Zeckendorf representation 10100.
John Conway and Alex Ryba wrote a paper where they studied the Wythoff array. They continued the array to the left and drew walls according to a few rules. Here is a picture of their result. The picture is too small to make out the numbers, but you can see the shape, which looks like the Empire State Building. Oops, I forgot to mention, the paper is called The Extra Fibonacci Series and the Empire State Building.
My last year’s PRIMES STEP senior group (Eric Chen, Adam Ge, Andrew Kalashnikov, Ella Kim, Evin Liang, Mira Lubashev, Matthew Qian, Rohith Raghavan, Benjamin Taycher, and Samuel Wang) studied the Conway-Ryba paper and decided to generalize it to Tribonacci numbers.
Just a reminder. The Tribonacci sequence starts as 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, and continues so that any next term is the sum of the three previous terms. For example, the next term after 44 is 81.
Now we need an analog of the Wythoff array for Tribonacci numbers. My readers might by now guess why I chose the above definition of the Wythoff array. Yes, this definition is perfect for generalizing to Tribonacci numbers.
We start by defining the Tribonacci base. We can represent every integer uniquely as a sum of distinct Tribonacci numbers on the condition that there are no three consecutive Tribonacci numbers in the sum. This is called a Tribonacci representation. We can express it with zeros and ones, and there will never be three ones in a row. For example, 16 can be expressed as a sum of Tribonacci numbers in the following way: 16 = 13 + 2 + 1. Thus, its Tribonacci representation is 10011.
We can define a Tribonacci successor, similarly to the Fibonacci successor, by adding a zero in the Tribonacci representation. For example, the Tribonacci successor of 16 has to have the Tribonacci representation 100110 and is equal to 24 + 4 + 2 = 30.
Now, we define the Tribonacci array similarly to the Wythoff array. The first row of the Tribonacci array consists of distinct Tribonacci numbers in order. The first column consists of integers whose Tribonacci representations end in 1. The number in the nth column is a Tribonacci successor of the number to its left.
Consider any integer sequence such that the next term is a sum of the three previous terms. We call such sequences Tribonacci-like. One of the cool properties is that, in some sense, the Tribonacci array contains all positive Tribonacci-like sequences. More formally, if a Tribonacci-like sequence has three consecutive positive terms, then the tail of such a sequence has to appear in the Tribonacci array as one of the rows.
It follows that multiples of any row in the Tribonacci array have to appear in the array. An awesome fact is that they appear in order.
Moreover, when all the numbers of a row are divisible by n, the row number minus 1 is also divisible by n.
It is easy to extend the Tribonacci array to the left, but this extension doesn’t have the same nice properties as the extension of the Wythoff array. So finding an Empire State Building, or any other building, in the Tribonacci array is futile.
I agreed to review the book, The Raven’s Hat, because of the hats. I love hat puzzles. When I give them to my students, I bring hats to class to reenact the solutions.
The book contains eight awesome puzzles as well as ideas for playing with them. I both loved and hated this book. I loved it because it is great, and hated it because it isn’t perfect. Let me start with three places I didn’t like.
Consider a famous hat puzzle when there are hats of N colors. The sages are in a line, and hats are put on their heads. As usual, they are not allowed to give each other signals. Each of them has to announce their hat’s color, and they want to minimize the number of mistakes.
The big idea is to number the colors. The book suggests that the last sage in line calculates the total number of colors they see modulo N and announces the result to the rest. Then the others, starting from the end of the line, one by one, can calculate and name their hat colors. With this strategy, only the last sage in line might be mistaken.
This is a correct solution, but this is the first place I didn’t like. I prefer a different strategy, where everyone assumes that the total sum of the hat colors is 0 modulo N. In this case, every sage makes the same calculation: each sage sums up everything they see or hear and subtract the result from 0 modulo N. This solution is more elegant, since all the sages follow the same rule.
Then the book extends the same puzzle to an infinite number of sages. My second point of contention is that the authors think that, in this case, two sages might be mistaken. No. The answer is still the same, there is a strategy where not more than one sage is mistaken. See my blog post for the solution.
My third pet peeve happened when the authors introduced ballroom dancing in the puzzle on picture hanging. What is the connection between picture hanging and ballroom dancing? I’ll keep the book’s secret. My beef is with how the roles in ballroom dancing are described. Ballroom dancing is usually danced in pairs with asymmetric roles, which, in the past, were designated for males and females. Gender doesn’t play such a big role anymore; anyone can dance any role.
The authors are afraid to be politically incorrect by calling the dancers male and female. Instead, they say that the dancers dance male and female parts. Though formally, this choice of words might be politically correct, it still sounds awkward and draws attention to gender. If the authors ever talked to any person who has ever danced, they would have known that there is a much simpler way to describe dance roles. The dancers are divided into leaders and followers.
Did I ever tell you that reviewing my students’ writing is part of my job? So I am good at it and like critiquing other people’s writing. Now that my complaints are out, the issues with the book are actually minor.
The book is great. I even bought a second inflatable globe because of this book. The game, described in the book, is to rotate two globes randomly and then find a point on the globes in the same relative position towards the center. The game helped me teach my students that any movement of a sphere is a rotation.
My main goal in this post is to describe the only puzzle in the book that I haven’t seen before.
Puzzle. In a group of opera singers, there are two stars who are either friends or enemies. Surprisingly, only the host, who is not an opera singer, knows who the stars are and the nature of their relationship (the stars do not know that they are stars and whether or not they are friends). The group’s common goal is to identify the stars and to determine whether they are friends or enemies. To do so, they send a few of the singers to sing opera on a stage, which is divided into two halves: left and right. During the opera, the singers do not move between the halves. After the opera is over, the host classifies the opera. If there were no stars or only one star on stage, he classifies it as “neutral”. If both stars were on stage, the opera is a big success or a disaster. If both stars are friends and sing on the same half of the stage, or if they are enemies and sing on different halves, then the opera is a big success. Otherwise, it is a disaster. What is the best strategy for a group of five singers to determine who are the stars and what is their relationship? What is the smallest number of operas they have to sing to guarantee that they can figure everything out?
It is weird that two people do not know whether they are friends. But sacrifices are needed for mathematics. I am excited that there is a nontrivial puzzle related to information theory, and it is ternary based. All other such puzzles I know are about weighing coins on a balance scale. I wrote too many papers about coin weighing. Now I can switch to opera singers with passionate relationships, secret from themselves.
I stumbled on this puzzle on Facebook the day of the World Cup finals. How timely!
Puzzle. Sixteen teams play a single-elimination tournament, where the losing team is immediately out. The teams have different rankings, and a higher-ranking team always wins against a lower-ranking team. The initial seeding is random. In the semifinals, A won against B, and C won against D. Given that B is stronger than D, what is the probability that A will become the champion?
Consider a group of people in which some are friends. We assume that friendship is symmetric: if Alice is Bob’s friend, then Bob is Alice’s friend. That means we can build a friendship graph where vertices are people and edges correspond to friendships. Let’s assume that every person has at least one friend, so the friendship graph doesn’t have isolated vertices.
Darla needs to conduct research by surveying random pairs of friends. But first, she has to find those pairs. To ensure that the pairs are randomly selected, she must pick two random people from the group, contact them, and ask them whether or not they are friends. If they are, she gives them her questionnaire. If not, Darla wasted tons of time and had to keep looking.
The group she is surveying is enormous. So, when she picks two random people, they typically have never even heard of each other. Bother!
Darla decides to speed up the process. She would pick a random person, ask them for a list of friends, and then randomly pick one person from the list. Since every person has at least one friend, Darla always ends up with a filled questionnaire.
Puzzle question. Why is Darla’s method wrong? Can you describe the pairs of friends her method favors?
As my readers know, I run a PRIMES STEP program, where we conduct mathematical research with students in grades 6 through 9. Last year, our junior group wrote the paper, The Struggles of Chessland, which is now posted on the arXiv.
This is the funniest paper ever written at PRIMES STEP. In the paper, the King, with his self-centered Queen and their minions, try to protect their lands in the Bermuda triangle, called the Chessland, which consists of square islands of different sizes.
In the first part of their fairy tale, the King, his lazy Queen, and other chess pieces are trying to survey their islands. The first volunteer for this surveying job is the Knight. The Knight starts somewhere on an island and walks around it by using the knight’s moves in chess. The goal is to see every cell, and a Knight can see all cells that are a knight’s move away from where he is standing. In other words, every cell is either visited by the Knight or is one knight’s move away from a visited cell. In mathematical terms, the Knight walks a path that is a domination set on a knight’s graph.
The students found a lot of such paths. The picture shows a surveying path for the Knight for the island of size 7. The students called this pattern a shoelace pattern. They used similar shoelaces to survey any island of size 7 or larger. However, when I look at the picture, I see a cat.
Other chess pieces survey the land too. Funnily, the King surveys better when he is drunk. Do you know why? Take a look at the paper.
After all the islands had been surveyed, enemy spies started to appear in Chessland, and our band of chess pieces tried to trap them. My students invented the rules for trapping enemies. An example of the black Queen being trapped is in the picture below, and the rules are the following. Wherever the black Queen might move, should be under attack by a white queen. Also, white queens can’t directly attack the black Queen, or each other.
Oh! I forgot to mention: the trapping should use as few white queens as possible. Also, if the enemy is not a queen but another kind of chess piece, it can only be trapped by its own kind.
The trapping can be described in terms of graph theory. The black Queen is located at a vertex of the queen’s graph. The white queens should be positioned at vertices in such a way that they are not neighbors of the black Queen or each other. In addition, any vertex that is a neighbor of the black Queen, has to be a neighbor of at least one white queen.
This year’s PRIMES STEP project was a real chess adventure!
The 2018 International Collegiate Programming Contest (ICPC) had a very hard problem which is of interest to mathematicians. The problem was proposed by super-coder Derek Kisman. Here is the problem as it was presented at the contest.
Problem J. Uncrossed Knight’s Tour. A well-known puzzle is to “tour” all the squares of an 8 × 8 chessboard using a knight, which is a piece that can move only by jumping one square in one direction and two squares in an orthogonal direction. The knight must visit every square of the chessboard, without repeats, and then return to its starting square. There are many ways to do this, and the chessboard size is manageable, so it is a reasonable puzzle for a human to solve.
However, you have access to a computer, and some coding skills! So, we will give you a harder version of this problem on a rectangular m × n chessboard with an additional constraint: the knight may never cross its own path. If you imagine its path consisting of straight line segments connecting the centers of squares it jumps between, these segments must form a simple polygon; that is, no two segments intersect or touch, except that consecutive segments touch at their common end point. This constraint makes it impossible to visit every square, so instead you must maximize the number of squares the knight visits. We keep the constraint that the knight must return to its starting square. Figure J.1 shows an optimal solution for the first sample input, a 6 × 6 board.
The input consists of a single line containing two integers m (1 ≤ m ≤ 8) and n (1 ≤ n ≤ 1015), giving the dimensions of the rectangular chessboard.
Mathematicians know many things about knight tours on a standard 8 × 8 chessboard. But one of the limits in this puzzle is so huge, that the answer to this computational puzzle constitutes a new mathematical discovery. Unsurprisingly, no one solved this puzzle during the contest. A well-written solution summary is available on the ICPC website. The solution requires dynamic programming and a realization that tours have to have repeating patterns.
Since no one solved this problem during the competition, it is useful that, in addition to the solution summary, Derek posted his innovative code online. The two figures below (courtesy of Derek Kisman) show his answer for the longest closed tour on a 6 × 24 board and the longest open tour on a 6 × 26 board.
Derek got interested in uncrossed knight’s tours after reading my blog post, 2014 Math Festival in Moscow, where I presented the following problem given at that festival to 7th graders.
Problem. Inside a 5 × 8 rectangle, Bart draws closed paths that follow diagonals of 1 × 2 rectangles. Find the longest possible path.
The 2014 Math Festival organizers offered an extra point for every diagonal on top of 16. It is funny that the organizers obfuscated that it is useless to try and find 17 diagonals, as the number of diagonals has to be even. The official solution, shown below, had 24 diagonals.
In the solution booklet, the organizers mentioned that they did not know the best answer to their own question. They hoped that the longest possible path matched their official solution.
In Derek’s notation, the above problem is equivalent to finding the largest uncrossed knight’s closed tour on a 6 × 9 grid. And Derek proved that, indeed, 24 is the largest tour. While proving this, he also calculated the answer for gazillions of other boards.
We can go further: beyond gazillions. Let’s consider what happens on m × n boards, where m is fixed, and n is astronomically large. Suppose fm(n) is the largest size of an uncrossed knight’s closed tour on the m × n board and gm(n) is the largest size of an uncrossed open tour.
The answer has repeating patterns everywhere except around the ends of the board. We would expect that the number of possible repeating patterns is finite. Moreover, for large boards, only the densest patterns would appear. This means that asymptotically, both functions f and g are linear functions of n. On top of that, as each pattern is periodic, the behavior at the ends of the long boards should become periodic as well. Hence, the difference functions of f and g for fixed m would eventually become periodic. (Here, the f’s difference function, which I denote D(fm) is defined as follows: D(fm)(n) = fm(n+1) – fm(n).)
That means we can solve a more difficult problem. We do not need the limits for n. It should be possible to calculate these functions for a given m and any n (even if n is way more than a gazillion). I am being over-optimistic here. First, we need some mathematical theory to find the bounds for where the periodic behavior has to start and estimate the size of the period. Suppose we can prove that for D(f6)(n), the eventual period has to start before n reaches A, and the length of the eventual period is not more than B. Then, we need to calculate the A + 2B values of the function f6(n) to know the function for any n.
Derek actually calculated the functions fm(n) and gm(n) for m up to 9 and n up to infinity. More precisely, he found the cycles I am describing above. What a mindblowing achievement!