Tanya Khovanova's Math Blog

I wonder what the largest number is that can be represented with one character. Probably 9. How about two characters? Is it 9⁹? What about three or four?

I guess I should define a character. Let’s have two separate cases. In
the first one you can only use keyboard characters. In the second one
you can use any Unicode characters.

I’m awaiting your answers to this.

The last time I talked to John H. Conway, he taught me to walk up the stairs. It’s not that I didn’t know how to do that, but he reminded me that a nerd’s goal in climbing the steps is to establish the number of steps at the end of the flight. Since it is boring to just count the stairs, we’re lucky to have John’s fun system.

His invention is simple. Your steps should be in a cycle: short, long, long. Long in this case means a double step. Thus, you will cover five stairs in one short-long-long cycle. In addition, you should always start the first cycle on the same foot. Suppose you start on the left foot, then after two cycles you are back on the left foot, having covered ten stairs. While you are walking the stairs in this way, it is clear where you are in the cycle. By the end of the staircase, you will know the number of stairs modulo ten. Usually there are not a lot of stairs in a staircase, so you can easily estimate the total if you know the last digit of that number.

I guess I am not a true nerd. I have lived in my apartment for eight years and have never bothered to count the number of steps. That is, until now. Having climbed my staircase using John’s method, I now know that the ominous total is 13. Oh dear.

The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year’s Olympiad:

Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?

Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, “My age is equal to the sum of the digits of my birth year.” “Mine too,” replied Paul. By how many years is Paul younger than Peter?

Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?

Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?

Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.

Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.

A mathematician is someone who pauses when asked “How much is two and two?”

Indeed, the answer might be:

• 0 — modulo 4.
• 1 — in characteristic 3.
• 2 — if AND is considered as a logical operation.
• 4 ± ε — for approximate values of 2.
• 4 — (almost forgot this case).
• 10 — in base 4.
• 11 — in base 3.
• 22 — for string concatenations.
• MSC — (this one is your homework).

I already wrote about the research of my friend Olga Amosova who studied the sickle-cell anemia mutation. She and her colleagues needed to store short fragments of hemoglobin genes for their experiments. All the fragments were identical. They noticed that with time the fragments always broke down in the same place. It was a mystery. When good scientists stumble on a mystery, they start digging.

They found that one of the nucleotides rips off the DNA fragment at the site of the Sickle-cell mutation. That place on the DNA becomes fragile and later breaks down. These sites need to be repaired. The repair is very error-prone and often leads to a mutation.

When DNA strands are left unattended, they want to pair up. There are four types of nucleotides: A, C, G and T. So mathematically the fragment of DNA is a string in the alphabet A, C, G, T. These nucleotides are matched to each other. When two DNA strands pair up, A on one strand always matches T and C matches G. So it is logical that if there are two complementary DNA pieces on the same fragment, they will find each other and pair up. They form a hydrogen bond. For example, a piece AACGT matches perfectly another piece TTGCA. Suppose a substring of DNA consists of a piece AACGT and somewhere later the reverse of the match: ACGTT. Such a string is called an inverted repeat. The DNA fragment I mentioned contains a string AACGT****ACGTT. Two pieces AACGT and ACGTT are complementary and not too far from each other in space. So it is easy for them to find each other and to bond to form a so-called stem-loop or a hairpin structure. The site of Sickle-cell mutation falls into the loop.

Olga and her colleagues discovered that for some particular loops the orientation in space becomes awkward and one of the nucleotides rips off. Such a rip off is called depurination. In further investigation, Olga found examples of when depurination happens. The first sequence of the pair that will bond later has to have at least five nucleotides and has to end in T. Correspondingly the second part in the pair has to begin with A. In the middle there needs to be four nucleotides GTGG. The first G flies away. Enzymes rush like a first aid squad to repair it and introduce mistakes that lead to mutation and diseases like cancer.

DNA was thought to be simply a passive information storage system, not capable of any action. Now we see that DNA is capable of action. DNA can damage itself. Damage provokes a mutation. For all practical purposes it is self-mutilation. Olga and her colleagues scanned the human genome for other sequences that are capable of self-mutilation. They found that such sequences are overwhelmingly present. They are present in much higher numbers than would be expected statistically. The pieces that are capable of damaging themselves occur 40 times more often than would occur if the nucleotides were distributed randomly. They are especially overrepresented in genes linked to cancer.

Self-damaging shouldn’t happen in normal situations. It can be provoked by the environment, for example, the chemistry of the cell. That means, that our cancers are not only in our genes but also in our life-style. There was, for example, a suggestion in a recent NY Times article, Is Sugar Toxic?, that too much sugar in a diet might provoke cancer. If the rate of mutation depends on the environment, we can influence it and prolong our lives.

It is not clear why the ability to self-mutilate survives in the evolutionary process. It is quite possible that if something very bad happens to our planet, we need our genes to be able to mutate very fast in order to adjust to the environment so that humans can survive.

Though I never tried to donate my sperm to a sperm bank, because of my inability to produce it, I know that sperm banks look for people who have ancestors who lived for a very long time. Such sperm is in bigger demand as everyone wants their children to live longer. I wonder if this tendency is a mistake. Global warming is upon us. People with longevity genes might not be flexible enough for their children to survive the changing of the Earth.

As you might have guessed from the title, this essay is about domino tilings.

Suppose a subset of a square grid has area N, and the number of possible domino tilings is T. Let’s imagine that each cell is contributing a factor of x tilings to the total independently of the others. Then we get that x^N = T. This mental exercise suggests a definition: we call the nth root of T the degree of freedom per square for a given region.

Let’s consider a 1 by 2k rectangle. There is exactly one way to tile it with dominoes. So the degree of freedom per square of such a rectangle is 1. Now consider a 2 by k rectangle. It has the same area as before, and we know that there should be more than one tiling. Hence, we expect the degree of freedom to be larger than the one in the previous example. The number of tilings of a 2 by k rectangle is F_k-2, where F_k is k^th Fibonacci number. So the degree of freedom for large k will be approximately the square root of the golden ratio, which is about 1.272.

You might expect that squares should give larger degrees of freedom than rectangles of the same area. The degree of freedom for a large square is about 1.3385. You can find more information in the beautiful paper Tilings by Federico Ardila and Richard P. Stanley.

Let’s move from rectangles to Aztec diamonds. They are almost like squares but the side of the diamond is aligned with diagonals of the dominoes rather than with their sides. See the sample diamonds in the picture above, which Richard Stanley kindly sent to me for this essay.

It is easier to calculate the degree of freedom for Aztec diamonds than for regular squares. The degree is the fourth root of 2, or 1.1892…. In the picture below created by James Propp’s tiling group you can see a random tiling of a large Aztec diamond.

Look at its colors: horizontal dominoes are yellow and blue; vertical ones are red and aquamarine. You might wonder what rule decides which of the horizontal dominoes are yellow and which are blue. I will not tell you the rule; I will just hint that it is simple.

Back to freedom. As you can see from the picture, freedom is highly non-uniform and depends on where you live. Freedom is concentrated inside a circle called the arctic circle, perhaps because the areas outside it are frozen for lack of freedom.

Now I would like to expand the notion of freedom to give each cell its own freedom. For a large Aztec diamond, I will approximate freedom with a function that is one outside the arctic circle and is uniform inside. The Aztec diamond AZ(n) consists of 2n(n+1) squares, shaped like a square with side-length n√2. So the area of the circle is πn²/2. Hence we can calculate the freedom inside the circle as the π^th root of 2, which is about 1.247. This number is still much less than the degree of freedom of a cell in a large square.

Jorge Tierno sent me a link to the following puzzle:

There is a certain country where everybody wants to have a son. Therefore, each couple keeps having children until they have a boy, then they stop. What fraction of the children are female?

If we assume that a boy is born with probability 1/2 and children do not die, then every birth will produce a boy with the same probability as a girl, so girls will comprise half of all children.

Now, I wonder why everyone would want a boy? Y-chromosomes are much shorter than X-chromosomes. If a man wants to pass his genes to the next generation, a daughter should be preferable as she keeps more genes from the father. I am a mother of two boys, so my granddaughters will have my X-chromosome while my grandsons will have my ex-husband’s Y-chromosome, so to keep my genes in the pool I should be more interested in granddaughters.

But I digress. I started writing this essay because in the original puzzle link the answer was different from mine. Here is how the other argument goes:

Half of all families have zero girls, a quarter have 1/2 girls, 1/8 have 2/3 girls, and so on. If we sum this up the expected ratio of girls to boys is (1/2)0 + (1/4)(1/2) + (1/8)(2/3) + (1/16)(3/4) + … which adds to 1 − ln 2, which is about 30%.

What’s wrong with this solution?

I found this cute problem in the Russian book Sharygin Geometry Olympiad by Zaslavsky, Protasov and Sharygin.

Find numbers p and q that satisfy the equation: x² + px + q = 0.

The book asks you to find a mistake in the following solution:

By Viète’s formulae we get a system of equations p + q = − p, pq = q. Solving the system we get two solutions: p = q = 0 and p = 1, q = −2.

What is wrong with this solution?

I wrote how the written entrance exam was used to keep Jewish students from studying at Moscow State University, but the real brutality happened at the oral exam. Undesirable students were given very difficult problems. Here is a sample “Jewish” problem:

Solve the following equation for real y:

Here is how my compatriots who studied algebra in Soviet high schools would have approached this problem. First, cube it and get a 9th degree equation. Then, try to use the Rational Root Theorem and find that y = 1 is a root. Factoring out y − 1 gives an 8th degree equation too messy to deal with.

The most advanced students would have checked if the polynomial in question had multiple roots by GCDing it with its derivative, but in vain.

We didn’t study any other methods. So the students given that problem would have failed it and the exam.

Unfortunately, this problem is impossible to appeal, because it has an elementary solution that any applicant could have understood. It goes like this:

Let us introduce a new variable: x = (y³ + 1)/2. Now we need to solve a system of equations:

This system has a symmetry which we can exploit. The graphs of the functions x = (y³ + 1)/2 and y = (x³ + 1)/2 are reflections of each other across the line x = y. As both functions are increasing, the solution to the system of equations should lie on the line x = y. Hence, we need to solve the cubic y = (y³ + 1)/2, one of whose roots we already know.

Now I offer you another problem without telling you the solution:

Four points on a plane used to belong to four different sides of a square. Reconstruct the square by compass and straightedge.

Recently I asked my readers to look at the 1976 written math exam that was given to applicants wishing to study at the math department of Moscow State University. Now it’s time to reveal the hidden agenda. My readers noticed that problems 1, 2, and 3 were relatively simple, problem 4 was very hard, and problem 5 was extremely hard. It seems unfair and strange that problems of such different difficulty were worth the same. It is also suspicious that the difficult problems had no opportunity for partial credit. As a result of these characteristics of the exam, almost every applicant would get 3 points, the lowest passing score. The same situation persisted for many years in a row. Why would the best place to study math in Soviet Russia not differentiate the math abilities of its applicants?

In those years the math department of Moscow State University was infamous for its antisemitism and its efforts to exclude all Jewish students from the University. The strange structure of the exam accomplished three objectives toward that goal.

1. Protect the fast track. There was a fast track for students with a gold medal from their high school who got 5 points on the written exam. The structure of the exam guaranteed that very few students could solve all 5 problems. If by chance a Jewish student solved all 5 problems, it was not much work to find some minor stylistic mistake and not count the solution.

2. Avoid raising suspicion at the next exams. The second math entrance exam was oral. At such an exam different students would talk one-on-one with professors and would have to answer different questions. It was much easier to arrange difficult questions for undesirable students and fail all the Jewish students during the oral exam than during the written exam. But if many students with perfect scores on the written exam had failed the oral exam, it might have raised a lot of questions.

3. Protect appeals. Despite these gigantic efforts, there were cases when Jewish students with a failing score of 2 points were able to appeal and earn the minimum passing score of 3. If undesirable students managed to appeal all the exams, they would only get a half-passing grade at the end and would not be accepted because the department was allowed to choose from the many students that the exams guaranteed would have half-passing scores.

I have only heard about one faculty member who tried to publicly fight the written exam system. It was Vladimir Arnold, and I will tell the story some other time.