Tanya Khovanova's Math Blog

Here is a famous old joke about Russian propaganda. I heard it when Leonid Brezhnev was General Secretary of the Communist Party of the Soviet Union and Ronald Reagan was the president of the US.

Joke. A Russian newspaper wrote an article about a two-person race between the Russian and the US leaders. They wrote, “Brezhnev got an honorable second place, while Reagan was almost last”.

Here is a modern version of this joke I heard going around.

Joke. A Russian TV commentator: Biden cowardly goes to war-torn Kyiv, while Putin heroically hides in his bunker.

Before dividing chores, let’s divide a cake. Suppose I bought a two-layer cake. One layer is chocolate and the other is vanilla. Suppose I love chocolate and hate vanilla, My friend, Joe, is the opposite: he loves vanilla and hates chocolate. It’s very easy to divide the cake. I take all the chocolate, and Joe takes all the vanilla. I think I am lucky to get the full value of the whole cake. Joe feels lucky too.

My point is that people with different tastes can divide things so that both get more than half by their own estimates.

Let’s look at another example. Suppose Alice and Bob are divorcing. Their estate consists of one valuable thing: a portrait of Alice’s grandfather. Alice loved her grandfather and values the portrait at 10,000 dollars. Meanwhile, Bob values it the same as its market value, 2,000 dollars. There exists a division algorithm, called the Knaster procedure, which allows them to divide the portrait so that both of them end up with the same amount of money on top of their perceived half of the estate.

I will skip the calculations. The end result is that Alice gets the portrait and pays Bob 3,000 dollars. In her view, she gets a portrait worth 10,000 and loses 3,000. Her total gain is 7,000 dollars, which is 2,000 more than her estimated half. Bob gets 3,000 dollars. In his view, half of the estate is 1,000 dollars, and he gets 2,000 more than that.

The bigger the difference in perceived value, the more each person gets in addition to their expected half. Suppose this difference is D. If you trust my calculations, the Knaster procedure means each person gets D/4, in addition to one half of their estate’s perceived value.

The same idea can be applied to chores. Suppose I hate shopping while my husband hates doing the dishes. So, I can do the dishes, and he can shop. And we can live happily ever after without doing the things we hate.

So, theoretically, it is very profitable for two people to live together. Have you seen couples where each one thinks that they won the lottery by marrying their partner? Such couples benefit from dividing chores, appreciate their partners’ help, and are happy.

I have seen such couples, though not many. Actually, not many at all. If mathematics says that living together should be profitable, then why are happy couples such a rarity?

I will divide unhappy couples into four categories depending on whether they both benefit from dividing the chores and whether they both appreciate each other.

Benefit and appreciate. In this case, other parameters could affect their happiness: love, sex, children, jobs, and so on. Consider, for example, Alice and Bob. Alice relies on her husband for financial support for her and their small children and appreciates said support. Bob likes how Alice cares for the children and appreciates her for that. However, Alice doesn’t love Bob anymore, and Bob wants something special in his sex life but is afraid to request it from Alice. They are both profoundly unhappy.

Benefit and do not appreciate. It is possible that both people do not appreciate each other, or it could be just one. In addition, it could be that a person underestimates the real value of the partner’s contribution, or it could be that the appreciation is not enough for the partner. This became a more complicated paragraph than I initially expected. As Lev Tolstoy said, “All happy families are alike; each unhappy family is unhappy in its own way.” So, let me have two subcases.

Benefit and do not appreciate. Case 1. Underestimation. Such couples have a good division of labor but underestimate each others’ contribution. For example, Bob thinks that staying at home with children is a walk in the park. He thinks his job is way more difficult than his wife’s daily caring for the house and the children. He assumes that when he returns from work, the house needs to be clean and dinner ready. He is very angry when this doesn’t happen. Alice is very unhappy as she knows how much she actually works.

Benefit and do not appreciate. Case 2. No gratitude. Such couples have a good division of labor but do not express their gratitude sufficiently for the other partner. For example, Alice wants Bob to take her out to dinner as a thank you. Or to say thank you on a regular basis. But Bob brags to his friends that he is lucky in marriage and thinks that this is enough. Everyone but her knows that he feels lucky.

Do not benefit. It is usually one person who is used. There are many ways for people to force their spouses to divide the chores in their favor. There are many types of abusive relationships. I do not even want to give an example. After all, my goal was to discuss the mathematics of chores’ division, not to analyze why people do not divide their labor fairly.

Special cases. Life is complicated. Here is the case that doesn’t quite fit the cases above as the division of chores with time delays. For example, Alice worked and cared for the children while Bob went to medical school. The benefit for Alice was implied in the future. Bob promised to shower her with money when he would get rich. However, as soon as he got rich, he showered someone else.

The mathematics show that living with a partner can be extremely beneficial for both. But people’s emotions are complicated. They do not follow mathematics and often mess it up in more ways than I can imagine.

I got immediately attracted to the puzzle Oleg Polubasov recently posted on Facebook.

Puzzle. A rectangular clearing in a forest is an N-by-M grid, and some of the cells contain a tower. There are no towers in the cells that neighbor the forest. A tower protects its own cell completely and parts of the eight neighboring cells at a depth of half of a cell. Here by neighbors, we mean the cells horizontally, vertically, and diagonally adjacent to the given cell. In particular, if each cell is one square unit, a tower protects four square units. The protected area forms a square with borders that lie in between the grid lines. A tsar knows the towers’ locations and wants to calculate the protected area. Prove that the following formula gives the answer: the number of 2-by-2 subgrids that contain at least 1 tower.

I like this puzzle because it has an elegant solution. But there is more. The puzzle reminds me of one of my favorite novels by Arkady and Boris Strugatsky: The Inhabited Island, also known as Prisoners of Power. This is a science fiction novel where Max Kammerer, a space explorer, ends up on a planet with desolate people who, twice daily, experience sudden bouts of enthusiasm and allegiance to the government. Later, it becomes clear that the love for the rulers comes from towers that broadcast mind-control signals suppressing critical thinking and making people prone to believe propaganda.

The novel was written in 1969 but accurately describes the modern Russian propaganda machine. It appears that there is no need for a secret signal. Just synchronized propaganda on government-controlled TV turns off people’s brains.

A Whitney umbrella is a cool surface I wanted to crochet. The umbrella continues to infinity, and there is no way I want to crochet the whole thing. I wanted to make a finite portion of a Whitney umbrella surrounding the most exciting point.

The result is seen in the picture. Technically, I crocheted not Whitney umbrellas but topologically equivalent surfaces. I am most proud of my secret — and painful — method of crocheting the self-intersecting segment. One day I will spill it.

As Wikipedia defines it: the Whitney umbrella is the union of all straight lines that pass through points of a fixed parabola and are perpendicular to a fixed straight line parallel to the parabola’s axis and lies on its perpendicular bisecting plane. If you look at the picture, the fixed straight line is the self-intersection line, which is a continuation of the line segment where the colors are woven through each other. You can find the parabola as the curve formed by the two-colored edge on either side of the umbrella. Oops, I forgot that only three of these umbrellas are made of two colors.

The Whitney umbrella is a ruled surface, meaning that for every point, there is a straight line on the surface that passes through the point. A ruled surface can be visually described as the set of points swept by a moving straight line.

Oh, look, the stitched rows can pretend to be these straight lines. Actually, if I fold these thingies, the stitched rows ARE straight lines. But, when I make the edges into parabolas, the rows stop being straight. In the real Whitney umbrella, if you look along the intersection line, the straight lines are closer to each other than they are along the parabola. But in crochets, the distances between rows have to be fixed. If my crochets are folded, they become rectangles and ruled surfaces. The real Whitney umbrella does not fold into a plane.

The Whitney umbrella is famous for being the only stable singularity of mappings from R² to R³. I am grateful to Paul Seidel for emailing me the proof. This singularity is so famous it even has two names: pinch point and cuspidal point. Though my crochets are not exactly Whitney umbrellas, they show this singularity. Hooray! I found a secret way to crochet the most famous stable singularity!

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.

Here is one of my all-time favorite problems.

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.

1. Find a way to pass level 3 using only three lives.
2. Find the cheapest way to pass level 4.
3. Find the cheapest way for level 6.
4. Find a way to pass level 5 using only 30 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.

Putin’s bodyguards `collect his excrement on trips abroad and take it back to Russia with them’. This was an article in the Independent. Presumably, Putin is afraid that someone can analyze his waste to get information about his health. Here is a quote from the article.

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.

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