My favorite of the Escher plane tessellations is the one with shells. It is stunning, and the mathematics behind it is beautiful. I want to thank the late John Conway for teaching me the secret of this drawing.
Mathematicians are interested in tessellations because of the symmetries behind them. This tessellation has translational and rotational symmetries. Can you find them?
When I ask my students to find the rotational symmetries, they immediately tell me that they see two different 4-fold points, aka points where 90-degree rotations preserve the drawing. One point, I call G, is where four greenish shells meet, and one point, I call R, is where four reddish shells meet.
As you might have guessed, the students’ answer is not quite correct. There is more to the picture. Look at a dark-brown shell that looks like a curvy rectangle. This shape has markings. Now look at a specific point R and its four closest brown shells. You can see that going around this point R, the brown shells alternate their orientation: the darker side of these shells either faces towards point R or away.
The big secret of this artwork is that it contains TWO symmetry groups: a group and a subgroup. If we ignore the markings on the brown shells and consider them one solid color, then point R is indeed a 4-fold symmetry point. In addition, the center of the brown shape is a 2-fold symmetry point. Thus, the symmetry group of this simplified drawing is 442 in orbifold notation.
If we take the markings of the brown shells into account, then point R is not a 4-fold rotation, it is a 2-fold rotation. Point G keeps the property of being a 4-fold rotation. If you know your symmetry groups, you can conclude that there should be another 4-fold rotation. But where is it?
I will spill my answer. The symmetry point G is not ONE symmetry point anymore. There are two different points where greenish shells meet. The dark side of the brown shells faces one of them and looks away from the other.
The dark secret of this drawing is that it demonstrates two symmetry groups: group 442 and its subgroup 442, with different fundamental regions. To see the secret, you must look closely at a dark brown shell and find its darker side.
Puzzle. Four glasses are placed on the four corners of a square rotating table; each glass is either right-side up or upside down. You need to turn them all in the same direction, either all facing up or all down. You may do so by grasping any two glasses and turning either none, one of them, or both over. There are two catches: 1) You are blindfolded, and 2) the table is spun after each round. Assuming a bell rings when you have them all facing the same way, how do you do it?
When I first heard this problem, the person who tortured me with it forgot to mention the bell. The problem was impossible: there was no feedback. Then, the bell was mentioned. It was an aha moment: you get information, but only a confirmation that the problem is solved. I tried to think about the puzzle backwards. What could be my last step? Assume that I know that the glasses alternate around the table: up, down, up, down. Then, as my last step, I can reach for the diagonal glasses and turn them over.
This is how you might start thinking about this puzzle. You can find the rest of the solution on the puzzle’s Wikipedia page.
Here is a wonderful variation, proposed by Michael Hotiner from Ukraine, that appeared ten years ago at a puzzle competition. This variation is not cheap; the players must pay with their lives, albeit virtual ones.
Puzzle setup. Consider a computer game where at level M, there is an M-sided rotating table with glasses at each corner. The setup is similar to the four-glasses puzzle above. The player is blindfolded, and the table rotates between rounds. As soon as level M is over, if all the glasses face one direction, the bell rings, and the player moves to the next level, M+1. At each round, the player can decide on the number N of glasses to touch. Upon deciding, they have to pay N! lives for that round. Then, the player can touch the glasses one by one, choosing the next glass depending on how the previous glasses are oriented. After all the glasses are touched, the player can decide which ones to turn upside down.
Let’s see what happens at the very beginning of this game. The game starts at level 1, with only one glass. The round is solved before it starts. The cost is 0. The bell rings, and the game immediately goes to level 2.
Consider level 2. If the bell doesn’t ring immediately, the glasses face different directions. The cheapest way to proceed is to choose N = 1 and turn any glass. The cost is 1.
Consider level 3. A player can solve it in one round by touching all three glasses and turning them the same way. The cost is 3! = 6. There is a cheaper method that costs 4 lives in two rounds. In the first round, the player can touch any two glasses and turn both of them up. If the bell doesn’t ring, then the third glass is upside down. In the next round, the player can touch two glasses. If one is upside down, that glass should be turned up. If both glasses are right side up, then the player should turn both of them over to finish the round.
Puzzle tasks.
Find a way to pass level 3 using only three lives.
“Do you really hope to get hired here? We will never hire a woman physicist.” This was my friend’s job interview here in the US, though many years ago.
Another friend had an adviser who insisted that to recommend her for a PhD he needed to get into her pant first.
I heard more stories over the years, but my personal experience with gender bias was not so dramatic. Or, it could be that I didn’t pay attention. Let me start with my most memorable story.
Why did I get an NSF postdoc? Thirty years ago, I had recently arrived in the US, and just had a baby. A friend suggested that I apply for an NSF postdoc. I applied, and I got it. I was elated, but my happiness didn’t last long. It lasted until some bitter guy, who didn’t get into a postdoc, told me I got the position only because I was a woman. This was the first time I heard that gender might play a role in such decisions. I was devastated. To this day, I still do not know whether it was my talents or my gender that got me the postdoc. For many years after, I had impostor syndrome.
I wrote a blog essay, Polite Gender Bias, about some of my other stories. Each individual case might not seem gender-related, but they were repeated too many times, so I became sensitive. I call these stories:
An unbelievably amazing proof.
Simple versus elegant.
Saying “I am dumb” as a defense.
Who generates ideas for Tanya Khovanova’s math blog?
Here is a more recent story. I do not know whether I should be proud, angry, or embarrassed.
Where is the sugar? From time to time, in the MIT math department tea room, I am asked where is the sugar, or some similar things. This often happens when there are several males in the room. Why was I chosen? It could be that I look friendly and have a great smile, and this is why people approach me. Or people might assume that, being a woman, I am a secretary who knows everything about the kitchen. Or, they might assume that, being overweight, I love sugar and know about every secret sugar stash in any kitchen.
Here is a story where I know exactly how I felt: I was angry.
Can I say a word? I was invited to a lunch discussion on gender issues at our MIT math department. I am not faculty, so I was sure the only reason I was there was because they wanted more female participants. People around me enthusiastically praised our department’s handling of gender issues. I tried to speak many times, but some guy would always interrupt me. I was about to start laughing very loudly at the irony of the situation when our department head finally noticed what was going on and gave me a chance to say a word. The situation was resolved then, but I regret that I didn’t start laughing.
As stories pile up, I become more vocal. I am tired of being nice at my own expense. Now people think I am a bitch. So be it.
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.
This puzzle was in a homework assignment I gave to my students.
Puzzle. In the given configuration of matches, move two matches to form three squares.
I assume that the intended solution is the one below. Its appeal is that it has three squares of the same size and no dangling matches.
As usual, my students were inventive and suggested many alternative solutions.
The first picture shows a solution with three squares of different sizes: two small ones and one twice as big.
The next solution has two big squares and one smaller one.
The following picture shows three squares of different sizes.
Since the problem doesn’t forbid forming more than three squares, there are, of course, many solutions with more than three squares. The last picture has six squares.
I recently stumbled upon the following challenge on Facebook.
Puzzle. Type the number of seconds in a week using the smallest number of characters.
I lied. The challenge was to type the same number using the smallest number of clicks on a computer keyboard. There is a slight difference, and I like my version better.
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?
