Tanya Khovanova's Math Blog

Alexander Shapovalov is a prolific puzzle writer. He has a special webpage of his river-crossing puzzles (in Russian). Here is one of these puzzles.

Three swindlers have two suitcases each. They approach a river they wish to cross. There is one boat that can carry three objects, where a person or a suitcase counts as one object. No swindler can trust his suitcase to his swindler friends when he is away, but each swindler doesn’t mind his suitcases left alone at the river shore. Can they cross the river?

In my paper with Joshua Xiong, Nim Fractals, we produced a bijection between P-positions in the three-pile Nim and a three-branch Ulam-Warburton automaton. We also defined a parent-child relationship on games that is induced by this bijection. Namely, two consecutive P-positions in a longest optimal game of Nim are the ones that correspond to a parent-child pair in the automaton. A cell in the Ulam-Warburton automaton has exactly one parent. That means, if (a,b,c) is a Nim P-position, then exactly one of (a − 1,b − 1,c), (a − 1,b,c − 1), and (a,b − 1,c − 1) must be a P-position and a parent of (a,b,c). (See our paper for more details.)

Now I want to explicitly write out the rules of an automaton which will generate the Nim P-positions in 3D.

Let me restrict the evolution of the automaton to the non-negative octant. That is, we consider points (a,b,c) in 3D, where each coordinate is a non-negative integer. We define the neighbors of the point (a,b,c) to be the points that differ from (a,b,c) in two coordinates exactly by 1. So each point strictly inside the octant has 12 neighbors. (There are three ways to choose two coordinates, and after that four ways to choose plus or minus 1 in each of them.

There is a geometric interpretation to this notion of neighborhood. Let us correspond a unit cube to a point with integer coordinates. The center of the cube is located at the given point and the sides are parallel to the axes. Then two points are neighbors if and only if the corresponding cubes share one edge. Now it becomes more visual that a cube has 12 neighbors, as it has 12 edges.

Here is the rule of the automaton. Points never die. We start with the patriarch, (0,0,0), one point being alive. The non-alive point is born inside the non-negative octant if it has exactly 1 alive neighbor that is closer to the patriarch. In other words the point (a,b,c) is born if and only if exactly one out of three points (a − 1,b − 1,c), (a − 1,b,c − 1), and (a,b − 1,c − 1) is alive. It follows that the points that are born at the n-th step has a coordinate sum 2n.

Consider for example the starting growth. At the first step the points (0,1,1), (1,0,1) and (1,1,0) are born. At the next step the points (0,2,2) and (2,0,2) and (2,2,0) are born. while the (1,1,2) will never be born as starting from the second step it has at least two live neighbors: (0,1,1) and (1,0,1) that are closer to the patriarch.

Theorem. In the resulting automaton, the points that are born at step n are exactly the P-positions of Nim with the total of 2n tokens.

Proof. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious. Suppose we proved that at step n exactly P-positions with 2n tokens are born. Consider a P-position of Nim: (a,b,c) such that a + b + c = 2n + 2. Remember, that bitwise XOR of a, b, and c is zero. Consider the 2-adic values of a, b, and c (aka the smallest powers of 2 dividing a, b, and c). There should be exactly two out of these three integers that have the smallest 2-adic value. Suppose these are a and b. Then (a − 1,b − 1,c) is a P-position, while (a − 1,b,c − 1) and (a,b − 1,a − 1) are not. That means by the inductive hypothesis (a,b,c) has exactly one alive neighbor. So the position (a,b,c) is born at time n + 1.

Now we need to proof that nothing else is born. For the sake of contradiction suppose that (a,b,c) is the earliest N-position to be born. That means it has a live neighbor that is a P-position closer to the patriarch. WLOG we can assume that this neighbor is (a − 1,b − 1,c).

If a − 1 and b − 1 are both even, then (a,b,c) is a P-position, which is a contradiction. Suppose a − 1 and b − 1 are both odd. Then their binary representations can’t have the same number of ones at the end. Otherwise, (a,b,c) is a P-position. That is a and b have different 2-adic values. Suppose a has a smaller 2-adic value, Then, for (a − 1,b − 1,c) to be a P-position a and c has to have the same 2-adic value. That means (a,b − 1,c − 1) is a P-position too. Now suppose a − 1 and b − 1 are of different parities. Without loss of generality suppose a − 1 is odd and b − 1 is even, then c is odd. Then (a − 1,b,c − 1) is a P-position too. Thus we can always find a second neighbor with the same number of tokens. That is, both neighbors are alive at the same time; and the N-position (a,b,c) is never born. □

One might wonder what happens if we relax the automaton rule by removing the constraint on the distance to the patriarch. Suppose a new point is born if it has exactly one neighbor alive. This will be a different automaton. Let us look at the starting growth, up to a permutation of coordinates. At step one, positions (0,1,1) are born. At the next step positions (0,2,2) are born. At the next step positions (0,1,3), (1,2,3) and (0,3,3) are born. We see that (0,1,3) is not a P-positions. What will happen later? Will this N-position mess up the future positions that are born? Actually, this automaton will still contain all the P-positions of Nim.

Theorem. In the new automaton, the points that are born at step n and have total of 2n tokens are exactly the P-positions of Nim with the total of 2n tokens.

Proof. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious. The birth of the points that have total of 2n tokens and are born at step n depend only on the points with the total of 2n − 2 tokens that are born at step n − 1. By the inductive hypothesis, those are the P-positions with 2n − 2 tokens. So the points have total of 2n tokens and are born at step n match exactly the first automaton described above. To reiterate, N-positions with 2n tokens are born after P-positions with 2n tokens, so they do not influence the birth of P-positions with 2n + 2 tokens. □

In the game of Nim you have several piles with tokens. Players take turns taking several tokens from one pile. The person who takes the last token wins.

The strategy of this game is well-known. You win if after your move the bitwise XOR of all the tokens in all the piles is 0. Such positions that you want to finish your move with are called P-positions.

I play this game with my students where the initial position has four piles with 1, 3, 5, and 7 tokens each. I invite my students to start the game, and I always win as this is a P-position. Very soon my students start complaining that I go second and want to switch with me. What should I do? My idea is to make the game last long (to have many turns before ending) to increase the chances of my students making a mistake. So what is the longest game of Nim given that it starts in a P-position?

Clearly you can’t play slower then taking one token at a time. The beauty of Nim is that such an optimal game starting from a P-position is always possible. I made this claim in several papers of mine, but I can’t find where this is proven. One of my papers (with Joshua Xiong) contains an indirect proof by building a bijection to the Ulam-Warburton automaton. But this claim is simple enough, so I want to present a direct proof here. Actually, I will prove a stronger statement.

Theorem. In an optimal game of Nim that starts at a P-position the first player can take one token at each turn so that the second player is forced to take one token too.

Proof. Consider a P-position in a game of Nim. Then find a pile with the lowest 2-adic value. That is the pile such that the power of two in its factorization is the smallest. Suppose this power is k. Notice that there should be at least two piles with this 2-adic value.

The first player should take a token from one of those piles. Then the bitwise heap-sum after the move is 2^k+1−1. Then the Nim strategy requires the second player to take tokens from a pile such that its value decreases after bitwise XORing with 2^k+1−1. All piles with the 2-adic value more than k will increase after xoring with 2^k+1−1. That means the second player has to take tokens from another pile with 2-adic value k. Moreover, the second player is forced to take exactly one token to match the heap-sum. □

In the position (1,3,5,7) all numbers are odd, so I can take one token from any pile for my first move, then the correct move is to take one token from any other pile. My students do not know that; and I usually win even as the first player. Plus, there are four different ways I can start as the first player. This way my students do not get to try several different options with the same move I make. After I win several times as the first player, I convince my students that I win anyway and persuade them to go back to me being the second player. After that I relax and never lose. I am evil.

Replace question marks with letters in the following sequence:

Y, Y, ?, ?, Y, ?, Y, ?, R, R, R, R.

Problem 1. It is true that it is possible to put positive numbers at the vertices of a triangle so that the sum of two numbers at the ends of each side is equal to the length of the side?

This looks like a simple linear algebra question with three variables and three equations. But it has a pretty geometrical solution. What is it?

Problem 2. Prove that it is possible to assign a number to every edge of a tetrahedron so that the sum of the three numbers on the edges of every face is equal to the area of the face.

Again we have six sides and four faces. There should be many solutions. Can you find a geometric one?

Do you know how to cut a cake? I mean, mathematically. There is a whole area of mathematics that studies cake cutting.

Mathematicians usually assume that each person has their own idea of what is the best part of the cake. Suppose three sisters are celebrating the New Year by having a cake. Anna likes only icing, Bella likes only chocolate chips, and Carol likes only pieces of walnuts mixed into the cake. Mathematicians want to cut cakes fairly. But what is fair? Fair is fair, right? Wrong. There are several different notions of fair cake division.

There is a proportionate division. In such a division every sister gets at least one-third of her value of the cake. This seems fair. Let’s see an example. Anna gets one third of the icing, Bella gets one third of the chocolate chips, and Carol gets everything else. This is a fair proportionate division. Each of the sisters believes that she got at least one-third of the cake, in their own value. But it doesn’t seem quite fair.

There is a stronger notion of fairness. It is called envy-free. In this division each sister gets at least one-third of the cake and, in addition, none of the three sisters would improve their value by swapping pieces. That means, if Anna wants only icing, not only does she get at least one-third of the icing, but also no one else gets more icing than Anna. The previous example of the proportionate division is not envy-free. Carol got two-thirds of the icing, so Anna would want to switch with her.

Let’s try a different division. Anna gets one third of the icing, Bella gets the chocolate chips and another third of the icing, and Carol gets all the walnuts and another third of the icing. Formally, this is envy-free cake cutting. But poor Anna. What do you think Anna feels when she sees the smiles of contentment on the faces of her sisters? Whoever invented the name doesn’t understand envy. Anna got one-third of the cake by her value, but the other sisters got the whole cake!

Luckily mathematicians understand this conundrum. So they invented another name for a cake division. They call a division equitable if everyone values all the pieces the same. So the division above is envy-free but not equitable. Let’s try again. Let’s give each sister one-third of all the components of the cake. This division is very good mathematically: it is proportionate, envy free, and equitable. By the way, envy-free division is always proportionate. This division seems fair. But is it a good division?

There is another term here: Pareto-efficient division means that it is impossible to make one person feel better, without making another person feel worse. All divisions above are not Pareto-efficient. Moving some icing from Carol to Anna, doesn’t decrease the value for Carol, but increases the value for Anna.

There is an even better way to divide the cake. We can give the icing to Anna, the walnuts to Bella, and the chocolate chips to Carol. This division is envy-free, equitable, and Pareto-efficient. It is perfect. Mathematicians even have a word for it. They call it a perfect division.

Mathematically this division is perfect. Unfortunately, sisters are not. I know an Anna who would still envy Bella.

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.

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.

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.

I am running a PRIMES-STEP program for middle school students, where we try to do research in mathematics. In the fall of 2015 we decided to study the following topic in logic.

Suppose there is an island where the following four types of people live:

• Absolute truth-tellers always tell the truth.
• Partial truth-tellers tell the truth with one exception: if they are guilty, they will say that they are innocent.
• Absolute liars always lie.
• Responsible liars lie with one exception: if they are guilty, they will say that they are guilty.

See if you can solve this simple logic puzzle about people on this island.

It is known that exactly one person stole an expensive painting from an apartment. It is also known that only Alice or Bob could have done it. Here are their statements:
Alice: I am guilty. Bob is a truth-teller.
Bob: I am guilty. Alice stole it. Alice is the same type as me.
Who stole the painting and what types are Alice and Bob?

My students and I discovered a lot of interesting things about these four types of people and wrote a paper: Who Is Guilty?. This paper contains 11 cute logic puzzles designed by each of my 11 students.

I envied my students and decided to create two puzzles of my own. You have already solved the one above, so here is another, more difficult, puzzle:

A bank was robbed and a witness said that there was exactly one person who committed the robbery. Three suspects were apprehended. No one else could have participated in the robbery.
Alice: I am innocent. Bob committed the crime. Bob is a truth-teller.
Bob: I am innocent. Alice is guilty. Carol is a different type than me.
Carol: I am innocent. Alice is guilty.
Who robbed the bank and what types are the suspects?

The following cute puzzle of unknown origins was sent to me by Martina Balagovic and Vincent van der Noort:

A car travels from A to B and back again. When going uphill it goes 56 km an hour, when going downhill it goes 72 km an hour and while driving on flat surface it goes 63 km an hour. Getting from A to B takes 4 hours and getting back (over the same road) takes 4 hours and 40 minutes.
What is the distance between A and B?

I received a book Really Big Numbers by Richard Schwartz for review. I was supposed to write the review a long time ago, but I’ve been procrastinating. Usually, if I like a book, I write a review very fast. If I hate a book, I do not write a review at all. With this book I developed a love-hate relationship.

Let me start with love. I enjoyed reading the first 80 pages. The pictures are great, and some explanations are very well thought out. Plus, I haven’t thought much about really big numbers, so the book helped me understand them. I was impressed with how this book treats very difficult ideas with simple explanations and illuminating images. I was captivated by it.

Now to hate. I had two problems with this book: one pedagogical and the other mathematical.

The pedagogical issue. The beginning of the book is suitable for small children. Most of the book is suitable for advanced middle-schoolers who like mathematics. The last part is very advanced. Is it a good idea to show children a book that looks like a children’s book, but which soon becomes totally out of their reach? Richard Schwartz understands it and says many times that pieces of this book might be read several years apart. Several years? What child is ready to wait several years to finish a book? How would children feel about the book and about numbers when no matter how hard they try, they cannot understand the end of the book?

As a reviewer, I can’t recommend the full book for kids who are not ready to grasp the notion of the Ackermann function or arrow notation. Even if the child is capable of understanding these ideas, there are mathematical issues that would prevent me from recommending it.

The mathematical issue. Let me start by explaining the notion of plex. We call an n-plex a number that is equal to 10ⁿ. For example, 2-plex means 10² which is 100, and 10-plex means 10¹⁰. The fun part starts when we plex plexes. The number n-plexplex means 10 to the power n-plex which is 10⁽¹⁰¹⁰⁾. We can continue plexing: n-plexplexplex means 10 to the power n-plexplex. When you are hunting for really big numbers, it is easier to write the number of plexes rather than writing plexes after plexes. Richard Schwartz introduces the following notation to help visualize the whole thing. He puts numbers in a square. Number n in a square means 1-plexed n times. For example, 2 inside a square means 10¹⁰. Ten inside a square is 1-plexplexplexplexplexplexplexplexplexplex.

We can start nesting squares. For example, 2 inside a square means 1-plex-plex or 10¹⁰. Let’s add a square around it: 2 inside two squares means 10¹⁰ inside a square, which equals 1 plexed 10¹⁰ times. To denote 10 nested in n squares Richard Schwartz uses the next symbol: n inside a pentagon. For example, 1 inside a pentagon is 10 inside a square. I wrote this number in the previous paragraph: it is 1-plexplexplexplexplexplexplexplexplexplex. Similarly, n inside a hexagon means 10 inside n nested pentagons. We can continue this forever: n inside a k-gon is 10 inside k nested (k-1)-gons.

What bothers me is why a square? Why not a triangle? If we adopt this scheme, what is the meaning of a number in a triangle? Let’s try to unravel this. Following Richard Schwartz’s notation we get that n inside a square is the same as 10 inside n nested triangles. What do we do n times with 10 to get to 1 plexed n times? 1-plexed n times is 10 plexed n-1 times. There is a disconnect in notation here. For example, 10 in two nested triangles should mean 2 inside a square that is 10¹⁰. 10 inside one triangle should mean 1 inside a square which is 10. This doesn’t make any sense.

I started googling around and discovered the Steinhaus–Moser notation. In this notation a number n in a triangle means nⁿ. A number n in a square means the number n inside n nested triangles. A number n in a pentagon means the number n inside n nested squares. And so on. This makes total sense to me. If we move down the number of sides, we can say that the number n inside a 2-gon means n times n and the number n inside a 1-gon means n. This is perfect.

Schwartz changed the existing notation in two places. First he made everything about 10. This might not be such a bad idea except his 10 inside a square doesn’t equal 10 inside a square in Steinhaus–Moser notation. In Steinhaus–Moser notation 10 inside a square means 10 plexed 10 times. The author removed one of the plexes. He made 10 inside the square to mean 1 plexed 10 times, and as a result it stopped working.

Even though the first 83 pages are delightful and the pictures are terrific, the notation doesn’t work. What a pity.