Tanya Khovanova's Math Blog

I stumbled upon a TED-Ed video with a frog puzzle:

You’re stranded in a rainforest, and you’ve eaten a poisonous mushroom. To save your life, you need an antidote excreted by a certain species of frog. Unfortunately, only the female of the species produces the antidote. The male and female frogs occur in equal numbers and look identical. There is no way to distinguish between them except that the male has a distinctive croak. To your left you spot a frog on a tree stump. You hear a croak from a clearing in the opposite direction, where you see two frogs. You can’t tell which one made the sound. You feel yourself starting to lose consciousness, and you realize that you only have time to run in one direction. Which way should you go: to the clearing and lick both frogs or to the tree stump and lick the stump frog?

My first thought was that male frogs croak to attract female frogs. That means the second frog in the clearing is probably an already-attracted female. The fact that the stump frog is not moving means it is male. I was wrong. This puzzle didn’t assume any knowledge of biology. The puzzle assumes that each frog’s gender is independent from other frogs. Thus this puzzle is similar to two-children puzzles that I wrote so much about. I not only blogged about this, but also wrote a paper: Martin Gardner’s Mistake.

As in two-children puzzles, the solution depends on why the frog croaked. It is easy to make a reasonable model here. Suppose the male frog croaks with probability p. Now the puzzle can be solved.

Consider the stump frog before the croaking:

• It is a female with probability 1/2.
• It is a croaking male with probability p/2.
• It is a silent male with probability (1-p)/2.

Consider the two frogs in the clearing before the croaking:

• Both are female with probability 1/4.
• One is a female and another is a croaking male with probability p/2.
• One is a female and another is a silent male with probability (1-p)/2.
• Both are silent males with probability (1-p)²/4.
• Both are croaking males with probability p²/4.
• One is a silent male and another is a croaking male with probability p(1-p)/2.

The probabilities corresponding to our outcome—a non-croaking frog on the stump and one croaking frog in the clearing—are in bold. Given that the stump frog is silent, the probability that it’s a female is 1/(2-p). Simillarly, given that one clearing frog croaked, the probability that one of them is a female is 1/(2-p). The probabilities are the same: it doesn’t matter where you go for the antidote.

The TED-Ed’s puzzle makes the same mistake that is common in the two-children puzzles. I don’t want to repeat their incorrect solution. The TED-Ed’s frog puzzle is wrong.

(The calculation in the second to last paragraph was corrected on Nov 13, 2021.)

When I receive a bank statement, I review all the transactions. The problem is my failing memory. I do not remember when and where I last took money from an ATM, and how much. So I decided to create a pattern. I used to take cash in multiples of 100. Probably most people do that. A better idea is to always take the amount that has a fixed remainder modulo 100, but not 0. For example, let’s say I take one of the following amounts: 40, 140, 240, or 340. Sometimes I need more money, sometimes less. This set of numbers covers all of my potential situations, but my pattern is that all the numbers end in 40. This way if someone else gets access to my account, they will almost surely take a multiple of 100. I will be able to discover a fraud without remembering the details of my last withdrawal.

In addition, when I first started doing this, I was hoping I wouldn’t need to wait until I review my own statement to discover problems. My hope was that if a thief tried to take cash from my account, my bank would notice a change in the pattern and notify me immediately. Now I realize that this was wishful thinking. I doubt that banks are as smart as I am.

One day I should try an experiment. I should go to an ATM I never use and withdraw 200 dollars. I wonder if my bank would notice.

One of my jobs is giving linear algebra recitations at MIT. The most unpleasant aspect of it is dealing with late homework. Students attempt to submit their homework late for lots of different reasons: a sick parent needing help, stress, a performance at Carnegie Hall, a broken printer, flu, and so on. How do I decide which excuse is sufficient, and which is not? I do not want to be a judge! Moreover, my assumption is that people tell the truth. In the case of linear algebra homework, this assumption is unwise. As soon as students discover that I trust everyone, there’s a sharp increase in the number of sick parents and broken printers. So I have the choice of being either a naive idiot or a suspicious cynic. I do not like either role.

Our linear algebra course often adopts a brilliant approach: We announce that late homework is not allowed for any reason. To compensate for emergencies, we drop everyone’s lowest score. That is, we allow the students to skip one homework out of ten. They are free to use this option for oversleeping the 4 pm deadline. If all their printers work properly and they do not get so sick that they have to skip their homework, they can forgo the last homework. Happily, this relieves me from being a judge.

Or does it? Unfortunately, this policy doesn’t completely resolve the problem. Some students continue trying to push their late homework on me, despite the rules. In order to be fair to other students and to follow the rules, I reject all late homework. The students who badger me nonetheless waste my time and drain my emotions. This is very unpleasant.

From the point of view of those students, such behavior makes sense. They have nothing to lose and they might get some points. There is no way to punish a person who tries to hand in homework late and from time to time they stumble onto a soft instructor who accepts the homework against the official rules. Because such behavior is occasionally rewarded, they continue doing it. I believe that wrong behavior shouldn’t be rewarded. As a responsible adult, I think it is my duty to counteract the rewards of this behavior. But I do not know how.

What should I do? Maybe I should…

1. Persuade our course leader, or MIT in general, to introduce a punishment for trying to hand in late homework. We might, for example, subtract points from their final score.
2. Promote the value of honorably following the rules through discussions with the students.
3. Growl at any student who submits their homework late.
4. Explain to them what other people think about them, when they do not follow the rules.

Maybe I should just do number 4 right now and explain what I feel. A student who persists in handing in homework late feels to me that s/he is entitled and is better than everyone else, and shows that s/he doesn’t care about the rules and honor. Again, this wastes my time, puts me in a disagreeable position, and reduces my respect for that student.

Earlier I suggested that students don’t have anything to lose by such behavior, but in fact, they do pay a price, even though they may not understand that.

I found the following puzzle on a facebook page of Konstantin Knop (in Russian):

Seven astronauts landed on a small spherical asteroid. They wanted to explore it and walked in different directions starting from the same location. All used the same walking algorithm: walk x kilometers forward, turn 90 degrees left and walk another x kilometers, turn 90 degree left again and walk the last x kilometers. The value of x was different for different astronauts and was one of 30, 40, 50, 60, 70, 80, and 90.
All but one astronaut finished in the same location. What was the value of x for the astronaut who finished alone? What is the size of the asteroid?

I’ve been collecting math jokes for many years. I thought I’ve seen them all. No. Inventive people continue to create them. I was recently sent a link to a math joke website that features many jokes that are new to me. Here are my favorites:

* * *

With massive loss of generality, let $n=5$.

* * *

How do you prove a cotheorem? Using rollaries.

* * *

$0\to A\to B \to C \to 0$. Exactly.

* * *

Let $\varepsilon\to 0$. There goes the neighborhood!

* * *

Take a positive integer $N$. No, wait, $N$ is too big; take a positive integer $k$.

* * *

Calculus has its limits.

* * *

There is a fine line between a numerator and a denominator.

* * *

There’s a marked difference between a ruler and a straightedge.

* * *

Suppose there is no empty set. Then consider the set of all empty sets.

* * *

Q: Why is it an insult to call someone “abelian”?
A: It means they only have a 1-dimensional character, and are self-centered.

* * *

Q: What’s a polar bear?
A: A rectangular bear after a coordinate transform.

* * *

A logician rides an elevator. The door opens and someone asks:
—”Are you going up or down?”
—”Yes.”

Russians cherish the issues of Kvant, a famous Soviet monthly magazine for high-school students devoted to math and physics. At its peak its circulation was about 300,000, which is unparalleled for a children’s science journal. I still have my old childhood issues somewhere in my basement. But one issue is very special: it has a prominent place in my office. I didn’t receive it by subscription; I received it as a gift from my brother Misha.

Strictly speaking Misha is not my brother, but rather a half-brother. His full name is Mikhail Khovanov and he is a math professor at Columbia. The signed issue he gave me contains his math problem published in the journal. He invented this problem when he was a 10th grader. Here it is:

In a convex n-gon (n > 4), no three diagonals are concurrent (intersect at the same point). What is the maximum number of the diagonals that can be drawn inside this polygon so that all the parts they divide into are triangles?

He designed other problems while he was in high school. All of them are geometrical in nature. The journal is available online, and a separate document with all the math problems is also available (in Russian). His problems are M1038, M1103, M1108 (above), M1119, M1153, M1204. I like his other problems too. M1153 (below) is the shortest problem on his list: as usual I am guided by my laziness.

What’s the greatest number of turns that a rook’s Hamiltonian cycle through every cell on an 8 by 8 chessboard can contain?

I wanted to accompany this post with a picture of my brother at the age he was when he invented these problems—about 16. Unfortunately, I don’t have a quality picture from that period. I do have a picture that is slightly off: by about ten years.

What are the Ford circles? A picture is worth a thousand words, so here is a picture.

We draw a circle for any rational number p/q between 0 and 1 inclusive. We assume that p/q is the representation of the number in the lowest terms. Then the center of the circle is located at (p/q,1/q) and the radius is 1/q. The number inside a circle is q.

Here’s the game. We start with any circle on the picture, except for the two largest circles corresponding to integers 0 and 1. In one move we can switch to a larger circle that touches our circle. The person who ends up at the two largest circles corresponding to integers 0 or 1 loses. Equivalently, the person who ends in the central circle marked “2” wins.

There are other ways to describe moves in this game in terms of rational numbers corresponding to circles, that is, the x-coordinates of their centers. Two circles corresponding to numbers p/q and r/s touch each other iff one of the following equivalent statements is true:

• The cross-determinant of two numbers p/q and r/s that is defined as |ps-qr| equals to 1;
• p/q and r/s are neighbors in some Farey sequence;
• One of the numbers is the parent of the other in the Stern-Brocot tree.

Let me explain the last bullet. Given two rational numbers in their lowest terms a/b and c/d, we generate their mediant as: (a+c)/(b+d). We call the two numbers a/b and c/d the parents of the mediant (a+c)/(b+d). The Stern-Brocot tree starts with two parents 0/1 and 1/1. Then their mediant is inserted between them to create a row: 0/1, 1/2, and 1/1. Then all possible mediants of two consecutive numbers are inserted in a given row to get a new row. The process repeats ad infinitum. The famous theorem states that any rational number between 0 and 1 will appear in the process.

What I like about this game is a simple and beautiful description of P-positions (These are the positions you want to end your move at in order to win.) P-positions are numbers with even denominators in their lowest terms.

In the picture above P-positions are blue, while other positions are red. All circles touched by blue are red. And if we look at the larger neighbors of every red circle, one of them is blue and one is red.

Let’s prove that the numbers with even denominators satisfy the conditions for P-positions. First, two circles corresponding to numbers with even denominators can’t touch each other. Indeed, the cross-determinant of two such fractions is divisible by 2. Second, each red circle has to touch one blue and one red circle with larger radii. Indeed, the circles with larger radii touching a given circle are exactly the parents of the circle. If the mediant has an odd denominator, then one of the parents must have an even denominator and the other an odd denominator.

Let us start with regular (non-weighted) mediants. Suppose we have two fractions a/b and c/d, the mediant of these two numbers is the wrong way a child might sum these two fractions: (a+c)/(b+d). There is nothing wrong with this childish way of summing, it is just not a sum of two numbers, but rather an operation the result of which is called a mediant. One must be careful: the result doesn’t depend on the initial numbers chosen, but on their representations. The way to deal with this is to assume that a/b, c/d, and (a+c)/(b+d) are in their lowest terms.

The numerical value of the mediant is always in between given numbers. This is probably why it is called a mediant.

Suppose you start with two rational numbers 0/1 and 1/1, and insert a mediant between them. If you continue doing it recursively, you can reach any rational number between 0 and 1. This is a well-known property of the mediants. For example, after three rounds of inserting mediants into the initial sequence (0/1, 1/1), we get the sequence: 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1.

The mediants are easy to generalize if we assign weights to initial fractions. Suppose the first fraction has weight m and the second n, then their weighted mediant is (am+cn)/(bm+dn). The simplest generalization happens when the total weight, m+n is 3. In this case, given two rational numbers a/b and c/d, the left mediant is (2a+c)/(2b+d) and the right one is (a+2c)/(b+2d). Obviously the inequality property still holds. If a/b ≤ c/d, then a/b ≤ (2a+c)/(2b+d) ≤ (a+2c)/(b+2d) ≤ c/d.

James Propp suggested the following question for our PRIMES project. Suppose the starting numbers are 0/1 and 1/1. If we recursively insert two weighted mediants in order between two numbers we will get a lot of numbers. For example, after two rounds of inserting weighted mediants into the initial sequence (0/1, 1/1), we get the sequence: 0/1, 1/5, 2/7, 1/3, 4/9, 5/9, 2/3, 5/7, 4/5, 1/1. It is easy to see that new fractions must have an odd denominator. Thus unlike the standard case, not every fraction will appear. The question is: will every rational number between 0 and 1, which in reduced form has an odd denominator appear?

I started working on this project with Dhroova Aiylam when he was a high-school junior. We didn’t finish this project during the PRIMES program. But Dhroova finished another project I already wrote about: he analyzed the case of the standard mediants with any starting points. He showed that any rational number in between the starting points appears.

Dhroova became an undergraduate student at MIT and we decided to come back to the initial PRIMES project of weighted mediants. In our paper Stern-Brocot Trees from Weighted Mediants we prove that indeed every fraction with an odd denominator between 0 and 1 appears during the recursive procedure. We also analyzed what happens if we start with any two rational numbers.

In the starting position you see four multi-colored quadrants of a ring. For example, the first picture shows the starting position of level 33.

You can either swipe or touch between the quadrants. A swipe expands one quadrant into two quadrants and compresses two other quadrants into one. You can swipe clockwise or counterclockwise. The second figure shows the result of the clockwise swipe of the North opening. The NW quadrant that was half-red and half-yellow has expanded into two quadrants. The red piece now occupies the entire NW quadrant and the yellow piece—the entire NE quadrant. Two East quadrants got contracted into one SE quadrant. The former blue NE quadrant became the top blue half of the SE quadrant. The former SE half-blue half-green quadrant became the bottom half of the SE quadrant. In general the quadrant where the swiping movement starts expands in the direction of the swipe and the quadrant where the swiping movements ends together with the next quadrant compresses.

You can also touch an opening between quadrants. In this case the neighboring quadrants exchange places. The third figure shows the result of touching the East opening.

The goal of the game is to reach the final position: have each quadrant in one color. The next image shows the end of this particular level. As you can see the game was finished in 7 moves. In this particular case this is the smallest number of moves possible. To tell you a secret, it wasn’t me who finished the game in the smallest number of moves: it was my brother.

There is also a tutorial for this game on youtube.

I was at the 1976 International Mathematics Olympiad as part of the USSR team. There were eight people on the team and I decided to find out what they have achieved in the last 40 years. Here is the picture of our team. From left to right: Sergey Finashin, Yuri Burov, Nikita Netsvetaev, Boris Solomyak, Alexander Goncharov, Tanya Khovanova, Sergei Mironov, our chaperone Z.I. Moiseeva, our team leader A.P. Savin, no clue who this is (probably a translator), Piotr Grinevich.

The list below is ordered by the number of points received at the Olympiad.

Tanya Khovanova, 39 points: a lecturer at MIT. I am interested in a wide range of topics, mostly recreational mathematics and have written 60 papers.

Sergey Finashin, 37 points: a full professor at Middle East Technical University in Turkey, who is interested in topology. He wrote 40+ papers.

Alexander Goncharov, 37 points: a full professor at Yale University. He is the highest achiever on the team. He won the EMS Prize in 1992 and was an Invited Speaker at the 1994 International Congress of Mathematicians. He is interested in geometry, representation theory, and mathematical physics. He published 75 papers in refereed journals.

Nikita Netsvetaev, 34 points: a full professor at Saint-Petersburg University in Russia. He is interested in topology and algebraic geometry and wrote 20 papers.

Boris Solomyak, 31 points: a full professor at Bar-Ilan University. He is interested in fractal geometry and dynamical systems and wrote 90 papers.

Piotr Grinevich, 26 points: a leading scientific researcher at the Landau Institute for Theoretical Physics. He also teaches at Moscow State University. He is interested in integrable systems and wrote 80 papers.

Sergei Mironov, 24 points. Sergey became very ill while he was an undergrad. He stopped doing mathematics.

Yuri Burov, 22 points. Yuri wrote two papers, but quit mathematics after graduate school. He died several years ago from multiple sclerosis.

Six out of eight people on our team became mathematicians. Or more precisely five an a half. I consider myself a mathematician and am grateful for my position at MIT, where I run innovative programs for young mathematicians. But in the research world, a lecturer is a nobody. This makes me sad. I had to take breaks from research in order to raise my two children. And then I had to work in the private sector in order to support them. I was the best on my team and now I am the only one who is not a full professor. If you are looking for an example of how gender affects a career in academia, this is it.