Tanya Khovanova's Math Blog

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My Last Visit with John Conway

John Conway came to the 2017 MOVES conference and told me that he wanted to talk to me about subprime fibs. The subprime Fibonacci sequence was invented by John Conway, and I wrote a paper about it. The paper, Conway's Subprime Fibonacci Sequences, wasn't written with John, but rather with Richard Guy and Julian Salazar, and is published in Mathematics Magazine.

I wanted to visit my friend Julia, who lives in Princeton, and this was a good opportunity to discuss the mysteries of subprime fibs with John. On my second day in Princeton, I came to the math department around 3:00 pm carrying some apples. John never goes out for lunch, as he has trouble walking, so he is always hungry by the end of his work day. Thus, each time I go visit him, I come with food. We have very different tastes in apples: unlike me, he likes his apples unwashed.

Anyway, by the time I arrived to the department, John had already left. This was somewhat unusual, so I called him. He sounded weird and not very coherent, as if he wasn't feeling well. Considering also that he had left early, I started to worry. Unfortunately, there was a lot of background noise during our conversation and I only understood that he was at a pizza place. John walks very slowly, so he couldn't have gone too far away from campus. I found him in the second pizza place I checked. It was Tiger's Pizza. He told me that he felt very sleepy and tired. However, I was gratified to see how much having an interested listener gave him energy. He started telling me stories of his trip to Germany a long time ago. He had already eaten, but decided to have some more fries. As a perfect gentleman he offered me some, but I didn't want any.

At some point he dropped a couple of fries on the floor. He tried to reach them and I jumped to help. That was a mistake. I actually know that he likes proving to me and to himself that he can do stuff independently. He accepts my help when I am subtle about it, or when it is unavoidable. Anyway, he looked at me angrily and I backed off. He picked up his fries from the floor and ate them.

John Conway, Aug 11, 2017

I liked his T-shirt and tried to take a picture of it. As you can see, I am no photographer. The T-shirt shows a test question: Name the triangles. Then it features three triangles: an equilateral, isosceles, and right. It also provides someone's answers to this naming test: Geoffrey, Frederick, Eugene.

John asked me if I am more scared of Donald Trump or Kim Jong Un. We agreed that Trump is scarier. At this time he seemed his usual self.

I offered John a ride home, as I do whenever I visit him. He was very glad as he felt very tired. He started to get up. This time, I remembered not to try to help. He couldn't get up, I waited. He tried to push his weight off the table top, but the table was wobbly. I leaned on the table, as if to rest. We often play this sort of game in which he welcomes my help as long as we both pretend that I'm not helping.

My car was a block away and he wanted to walk the block. But after making two steps out of the pizzeria he changed his mind and asked me to bring the car to him. This was the first time ever. This visit he was so much worse than ever before.

On the drive to his place, he gave me a puzzle:

John's puzzle. Given a Mebius strip with a hole, how do you embed it in 3-D so that the two circular borders of the surface are equivalent?

I dropped him off at his house and offered to walk him to the door. He refused. I sat in my car and watched him walking very slowly along his path. I had this sinking feeling in my gut that I was seeing John for the last time. I drove away, once he disappeared behind his door.

On my way back to Boston I visited my friend Vitaly in East Brunswick, and the next day my high school friend Olga in Edison. In Edison, my car started beeping and I panicked. I was far away from home, and didn't want to be stuck in NJ. I started to look for the source of the sound. It was John's phone. As always, my gut feeling deceived me: I had to go back to Princeton.

I drove back to John's apartment. His door was unlocked and I entered. He was resting in bed. He was greatly annoyed at being disturbed. I explained the reason, and gave him his phone. He took the phone and said, "Off you go." I had this sinking feeling in my gut that these words would be the last words that I would hear from John.

A Crypto Word Search for G4G13

Here is the crypto word search I designed as a gift exchange for G4G13 (Gathering for Gardner). The submitted file is here: Crypto Word Search.



Punny Puzzles at the MIT Mystery Hunt

This is a math blog, but from time to time, I write about other things. Today I have something to say about puns, which I adore.

I also like gym, but rarely go there: it doesn't work out. I stopped using stairs, because they are up to something. I wanted to learn how to juggle, but I don't have the balls to do it.

I work at MIT, the work place with the best dam mascot: Tim the Beaver. My salary is not big, and I stopped saving money after I lost interest. I'm no photographer, but I have pictured myself outside of MIT too. I am a mathematician, which is the most spiritual profession: I am very comfortable with higher powers. I praise myself on great ability to think outside the box: it is mostly due to my claustrophobia. I am also a bit of a philosopher: I can go on talking about infinity forever.

I would love to tell you a joke. I recently heard a good one about amnesia, but I forgot how it goes.

My biggest problem is with English. So what if I don't know what apocalypse means? It's not the end of the world!

I never get tired of puns and here is my list of pun puzzles from the MIT Mystery Hunt:

Scam Jokes

* * *

Business plan:

  1. Sign-up for a premium-rate telephone number through which you make money from every call.
  2. Take a loan at the bank.
  3. Do not pay back.
  4. Collection agencies start calling non-stop.

* * *

  1. TMake a full-body selfie.
  2. Eat greedily for a year.
  3. Take a full-body selfie again.
  4. Swap before and after.
  5. Post.
  6. Collect the likes.
  7. Give diet advice.

* * *

A cafe patron ordered a pastry, then changed his mind and replaced it with a cup of coffee. When he finished his coffee, he started leaving without paying. The waiter approached him:
—You didn't pay for coffee!
—But I had it instead of the pastry.
—You didn't pay for the pastry either!
—But I didn't have the pastry.

* * *

At a farmers market stand there is a sign: 1 melon—3 dollars, 3 melons—10 dollars. A client requests one melon and pays 3 dollars, then repeats the procedure two more times. Then he says: "I bought three melons for 9 dollars, while you are trying to sell them for 10 dollars. This is really stupid." The farmer talks to himself: This happens all the time: they buy three melons instead of one, and try to teach me how to make money.

* * *

If the government listens in on my phone conversations, should they be paying half of my phone bill?

* * *

To get to free downloads, please, enter your credit card number.

* * *

The biggest lie of the century, "I have read and agree to the terms of ..."

* * * (submitted by Sam Steingold)

Ignorance: If your poker opponent got lucky cards four times in a row, he must get lousy cards now.
Knowledge: Nope, the deals are independent; prior observations have no bearing on the next deal.
Wisdom: The opponent is cheating; get away from the table now!

Solutions to Geometry Puzzles from the 35-th Lomonosov Tournament

I recently posted two geometry problems. Now is the time for solutions:

Problem 1. Is it 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?

One might guess that the following numbers work: (a+b-c)/2, (b+c-a)/2 and (c+a-b)/2, where a, b, and c are the side lengths. But there exists a geometric solution: Construct the incircle. The tangent points divide each side into two segment, so that the lengths of the segments ending at the same vertex are the same. Assigning this length to the vertex solves the problem. Surprisingly, or not surprisingly, this solution gives the same answer as above.

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.

The problem is under-constrained: there are six sides and four faces. There should be many solutions. But the solution for the first problem suggests a similar idea for the second problem: Construct the inscribed sphere. Connect a tangent point on each face to the three vertices on the same face. This way each face is divided into three triangles. Moreover, the lengths of the segments connecting the tangent points to a vertex are the same. Therefore, two triangles sharing the same edge are congruent and thus have the same area. Assigning this area to each edge solves the problem.

There are many solutions to the second problem. I wonder if for each solution we can find a point on each face, so that the segments connecting these points to vertices divide the faces into three triangles in such a way that triangles sharing an edge are congruent. What would be a geometric meaning of these four points?

Mathy Puzzles at 2018 MIT Mystery Hunt

I was on the writing team for the 2018 MIT Mystery Hunt. I am pleased that the hunt got very positive reviews from the participants. I spent tons of hours working on the hunt and it is good that folks liked it. I edited and tested a lot of puzzles. Here is my review of these year's puzzles that are math-related.

I already posted an essay about the puzzles I wrote myself. Four of my five puzzles are math-related, so I am including them below for completeness. I will mention the topic of each puzzle unless it is a spoiler.

I start with Nikoli-type puzzles. Four elegant Nikoli-type puzzles were written or cowritten by Denis Auroux. In all of them the rules of the logic are stated at the beginning. That means the logic part doesn't contain a mystery and can be solved directly.

There were several puzzles that were very mathematical.

There were also some math-related or computer-sciency puzzles.

There were also several decryption puzzles:

Trump and Pirates

Here is a famous math problem I never before wrote about:

Puzzle. Five pirates discovered a treasure of 100 gold coins. They decide to split the coins using the following scheme. The most senior pirate proposes how to share the coins, and all the pirates vote for or against it. If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the next most senior pirate making a proposal.

As pirates tend to be a bloodthirsty bunch, if a pirate would get the same number of coins whether he votes for or against a proposal, he will vote against so that the pirate who proposed the plan will be thrown overboard. Assuming that all five pirates are intelligent, rational, greedy, and do not wish to die, how will the coins be distributed?

You can find the solution in many places including Wikipedia's Pirate game. The answer is surprising: the most senior pirate gets 98 coins, and the third and the fifth pirates by seniority get one coin each. I always hated this puzzle, but never bothered to think through and figure out why. Now I know.

This puzzle emphasizes the flaws of majority voting. The procedure is purely democratic, but it results in extreme inequality.

That means a democracy needs to have a mechanism to prohibit the president from blatantly benefiting himself. With our current president these mechanisms stopped working. Given that Trump does everything to enrich himself, the pirates puzzle tells us what to expect in the near future.

We, Americans, will lose everything: money, clean air and water, national parks, future climate, health, social security, and so on, while Trump will make money.

My 2018 MIT Mystery Hunt Puzzles

I was on the writing team of this year's hunt, which was based on the movie "Inside Out." One of our goals was to create an easy first round to allow small teams to have a full hunt experience. Our first round consisted of 34 puzzles related to five basic emotions: joy, sadness, disgust, fear, and anger. Each emotion had its own meta puzzle. And the round had a meta-meta puzzle and a runaround. As I tend to write easy puzzles, I contributed three puzzles to this emotions round. The puzzles had references to corresponding emotions that were not needed for the solve path. They were inserted there for flavor.

I also wrote another easy puzzle called A Tribute: 2010-2017 (jointly with Justin Melvin, Wesley Graybill, and Robin Diets ). Though the puzzle is easy, it is useful in solving it to be familiar with the MIT mystery hunt. This is why the puzzle didn't fit the first emotions round.

I also wrote a very difficult puzzle called Murder at the Asylum. This is a monstrosity about liars and truth-tellers.


In mathematics one of the most important questions is why. Let us consider a problem:

Problem. A number has three hundred ones and three hundred zeroes. Can it be a square?

The solution goes like this. Consider divisibility of this number by 9. The sum of the digits is 300. That means the number is divisible by 3, but not by 9. Therefore, it can't be a square.

Why do we consider divisibility by 9? The divisibility by 9 is a very powerful tool, but why was it the first thing that came to my mind? The divisibility by 9 doesn't depend on the order of the digits. Whenever I see a problem where the question talks about digits that can be in any order, the first tool to use is the divisibility by 9.

The why question, is very important in mathematics. But it is also very important in life. It took me many years to start asking why people did this or that. I remember my mom was visiting me in the US. Every time I came back from work, she complained that she was tired. Why? Because she did the laundry in the bath tub. She wouldn't use my washing machine, because she didn't have such a thing in Russia. I promised her that I'd do the laundry myself when there was a sufficient pile. However, she insisted that the dirty clothes annoyed her. I would point that my water bill went up. And so on.

We argued like this every day. We were both frustrated. Then I asked myself why. Why does she do the laundry? The answer was there. She wanted to be helpful. I calmed down and stopped arguing with her. I sucked it up and paid the water bills. Her time with me turned into the most harmonious visit we ever had. Unfortunately, it was the last.

A Scooter Riddle

Puzzle. Alice, Bob, and Charlie are at Alice's house. They are going to Bob's house which is 33 miles away. They have a 2-seat scooter which rides at 25 miles per hour with 1 rider on it; or, at 20 miles per hour with 2 riders. Each of the 3 friends walks at 5 miles per hour. How fast can all three of them make it to Bob's house?

Mathy Jokes

* * * (submitted by Sam Steingold)

I can count to 1023 on my 10 fingers. The rudest number is 132.

* * *

I kept forgetting my password, so I changed it to "incorrect". Now, when I make a mistake during login, my computer reminds me: "Your password is incorrect."

* * *

—You promised me 8% interest, and in reality it is 2%.
—2 is 8—to some degree.

* * * (submitted by Sam Steingold)

Quantum entanglement is simple: when you have a pair of socks and you put one of them on your left foot, the other one becomes the "right sock," no matter where it is located in the universe.

* * *

—I keep telling my students that one half can't be larger or smaller than the other. Still the larger half of my class doesn't get it.

A Gender-Biased Puzzle

This famous trick puzzle is very old:

Puzzle. The professor is watching across a field how the son of the professor's father is fighting with the father of the professor's son. How is this possible?

This puzzle is tricky only because of gender-bias. Most people assume that the professor is male and miss the obvious intended solution, in which a female professor is watching her brother fighting with her husband.

I just gave this problem on a test. Here are other answers that I received.

Years ago people couldn't figure out this puzzle at all. So there has been progress. I was glad that my students suggested so many ideas that work. Nonetheless, many of them revealed their gender-bias by initially assuming that the professor is a man.

I can't wait until this puzzle stops being tricky.

Who Lives in the White House?

Puzzle. There are five houses of different colors next to each other equally spaced on the same road. In each house lives a man of a different profession.

Who lives in the white house?

Correction Nov 11, 2017. Replaced "the same distance from" with "halfway between" to eliminate the possibility of the plumber living in the yellow house. Thank you to my readers for catching this mistake and to Smylers for suggesting a correction.

A Domino-Covering Problem

I do not remember where I saw this problem.

Problem. Invent a connected shape made out of squares on the square grid that cannot be cut into dominoes (rectangles with sides 1 and 2), but if you add a domino to the shape then you can cut the new bigger shape.

This problem reminds me of another famous and beautiful domino-covering problem.

Problem. Two opposite corner squares are cut out from the 8 by 8 square board. Can you cover the remaining shape with dominoes?

The solution to the second problem is to color the shape as a chess board and check that the number of black and white squares is not the same.

What is interesting about the first problem is that it passes the color test. It made me wonder: Is there a way to characterize the shapes on a square grid that pass the color test, but still can't be covered in dominoes?

More Computer Jokes

* * *

Don't anthropomorphize computers: They don't like it.

* * *

I do not have dreams any more. What did I do wrong to make them delete my account?

* * *

How to restore justice: Create a folder named Justice. Delete it. Go to the trash bin and click restore.

* * *

An asocial network: When you sign up, you are friends with everyone. Then you send un-friend requests.

Derek Kisman's Ex Post Facto

I already wrote about two puzzles that Derek Kisman made for the 2013 MIT Mystery Hunt. The first puzzle is now called the Fractal Word Search. It is available on the Hunt website under its name In the Details. I posted one essay about the puzzle and another one describing its solution. The second puzzle, 50/50, is considered one of the most difficult hunt puzzles ever. Unfortunately, the puzzle is not available, but my description of it is.

Today let's look at the third puzzle Derek made for the 2013 Hunt, building on an idea from Tom Yue. This is a non-mathematical crossword puzzle. Derek tends to write multi-layered puzzles: You think you've got the answer, but the answer you've got is actually a hint for the next step.

Often multi-layered puzzles get solvers frustrated, but the previous paragraph is a hint in itself. If you expect the difficulty, you might appreciate the fantastic beauty of this puzzle.

Welcome to Ex Post Facto.

Fair-Share Sequences

Every time I visit Princeton, or otherwise am in the same city as my friend John Conway, I invite him for lunch or dinner. I have this rule for myself: I invite, I pay. If we are in the same place for several meals we alternate paying. Once John Conway complained that our tradition is not fair to me. From time to time we have an odd number of meals per visit and I end up paying more. I do not trust my memory, so I prefer simplicity. I resisted any change to our tradition. We broke the tradition only once, but that is a story for another day.

Let's discuss the mathematical way of paying for meals. Many people suggest using the Thue-Morse sequence instead of the alternating sequence of taking turns. When you alternate, you use the sequence ABABAB…. If this is the order of paying for things, the sequence gives advantage to the second person. So the suggestion is to take turns taking turns: ABBAABBAABBA…. If you are a nerd like me, you wouldn't stop here. This new rule can also give a potential advantage to one person, so we should take turns taking turns taking turns. Continuing this to infinity we get the Thue-Morse sequence: ABBABAABBAABABBA… The next 2n letters are generated from the first 2n by swapping A and B. Some even call this sequence a fair-share sequence.

Should I go ahead and implement this sequence each time I cross paths with John Conway? Actually, the fairness of this sequence is overrated. I probably have 2 or 3 meals with John per trip. If I pay first every time, this sequence will give me an advantage. It only makes sense to use it if there is a very long stretch of meals. This could happen, for example, if we end up living in the same city. But in this case, the alternating sequence is not so bad either, and is much simpler.

Many people suggest another use for this sequence. Suppose you are divorcing and dividing a huge pile of your possessions. A wrong way to do it is to take turns. First Alice choses a piece she wants, then Bob, then Alice, and so on. Alice has the advantage as the first person to choose. An alternative suggestion I hear in different places, for example from standupmaths, is to use the Thue-Morse sequence. I don't like this suggestion either. If Alice and Bob value their stuff differently, there is a better algorithm, called the Knaster inheritance procedure, that allows each of them to think they are getting more than a half. If both of them have the same value for each piece, then the Thue-Morse sequence might not be good either. Suppose one of the pieces they are dividing is worth more than everything else put together. Then the only reasonable way to take turns is ABBBB….

The beauty of the Thue-Morse sequence is that it works very well if there are a lot of items and their consecutive prices form a power function of a small degree k, such as a square or a cube function. After 2k+1 turns made according to this sequence, Alice and Bob will have a tie. You might think that if the sequence of prices doesn't grow very fast, then using the Thue-Morse sequence is okay.

Not so fast. Here is the sequence of prices that I specifically constructed for this purpose: 5,4,4,4,3,3,3,2,2,2,2,1,1,0,0,0. The rule is: every time a turn in the Thue-Morse sequence switches from A to B, the value goes down by 1. Alice gets an extra 1 every time she is in the odd position. This is exactly half of her turns. That is every four turns, she gets an extra 1.

If the prices grow faster than a power, then the sequence doesn't work either. Suppose your pieces have values that form a Fibonacci sequence. Take a look at what happens after seven turns. Alice will have pieces priced Fn + Fn-3 + Fn-5 + Fn-6. Bob will have Fn-1 + Fn-2 + Fn-4. We see that Alice gets more by Fn-3. This value is bigger than the value of all the leftovers together.

I suggest a different way to divide the Fibonacci-priced possessions. If Alice takes the first piece, then Bob should take two next pieces to tie with Alice. So the sequence might be ABBABBABB…. I can combine this idea with flipping turns. So we start with a triple ABB, then switch to BAA. After that we can continue and flip the whole thing: ABBBAABAAABB. Then we flip the whole thing again. And again and again. At the end we get a sequence that I decided to call the Fibonacci fair-share sequence.

I leave you with an exercise. Describe the Tribonacci fair-share sequence.

2015 Coin Problem Solution

A while ago I posted my second favorite problem from the 2015 All-Russian Math Olympiad:

Problem. A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn't know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?

Now it's solution time. First we show that we can do this in 70 weighings. The strategy is to compare one coin against one coin. If the scale balances, we are lucky and can stop, because that means we have found two real coins. If the scale is unbalanced, the heavier coin is definitely fake and we can remove it from consideration. In the worst case, we will do 70 unbalanced weighings that allow us to remove all the fake coins, and we will find all the real coins.

The more difficult part is to show that 69 weighings do not guarantee finding the real coin. We do it by contradiction. Suppose the weights are such that the real coin weighs 1 gram and the i-th fake coin weighs 100i grams. That means whatever coins we put on the scale, the heaviest pan is the pan that has the fake coin with the largest index among the fake coins on the scale.

Suppose there is a strategy to find a real coin in 69 weighings. Given this strategy, we produce an example designed for this strategy, so that the weighings are consistent, but the collector cannot find a real coin.

For the first weighing we assign the heaviest weight, 10070 to one of the coins on the scale and claim that the pan with this coin is heavier. We continue recursively. If a weighing has the coins with assigned weights, we pick the heaviest coin on the pans and claim that the corresponding pan is heavier. If there are no coins with assigned weights on pans, we pick any coin on the pans, assigned the largest available weight to it and claim that the corresponding pan is heavier.

After 69 weighings, not more than 69 coins have assigned weights, while all the weighings are consistent. The rest of the coins can have any of the leftover weights. For example, any of the rest of the coins can weigh 100 grams. That means that there is no coin that is guaranteed to be real.

2014 Math Festival in Moscow

I stumbled upon a couple of problems that I like while scanning the Russian website of Math Festival in Moscow 2014. The problems are for 7 graders.

Problem. Inside a 5-by-8 rectangle, Bart draws closed paths that follow diagonals of 1-by-2 rectangles. Find the longest possible path.

This problem is really very difficult. The competition organizers offered an extra point for every diagonal on top of 16. The official solution has 24 diagonals, but no proof that it's the longest. I'm not sure anyone knows if it is the longest.

Here is another problem:

Problem. Alice and Bob are playing a game. They start with two numbers: 2014 and 2015. In one move a player can do one of two things:
  1. subtract one of the numbers by one of the non-zero digits in any of the two numbers or,
  2. divide one number by two if the number is even.
The winner is the person who is the first to get a one-digit number. Assuming that Alice starts, who wins?

My First Husband with My Third Husband

Bernstein and Goncharov

The year is 1994. The man on the left is my first husband, Alexander Goncharov. Although we were out of touch for a decade, when I married my third husband, Joseph Bernstein (on the right), Goncharov started visiting us. It wasn't me he was interested in: he wanted to talk mathematics with my husband. I found this situation hilarious, so I took this photo.

But that's not all. My second husband, Andrey Radul, is not in the picture. But all four of us were students of Israel Gelfand. In short, my three ex-husbands and I are mathematical siblings — that is, we are all one big happy mathematical family.

The Best Writing on Mathematics 2016

Best Writing on Mathematics 2016

The Best Writing on Mathematics 2016 is out. I am happy that my paper The Pioneering Role of the Sierpinski Gasket is included. The paper is written jointly with my high-school students Eric Nie and Alok Puranik as our PRIMES-2014 project.

At the end of the book there is a short list of notable writings that were considered but didn't make it. The "short" list is actually a dozen pages long. And it includes two more papers of mine: