Archive for May 2011

Self-Mutilating DNA

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.

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Freedom and Diamonds

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 xN = 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 Fk-2, where Fk is kth 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.

 

Aztec Diamonds

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.

 

Aztec Diamond

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 πth 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.

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Averaging Averages

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?

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A Wrong 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?

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The Oral Exam

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:

Solve the equation

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:

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.

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The Hidden Agenda Revealed

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.

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Enemies and Friends

by Tanya Khovanova and Alex Ryba

The following problem appeared at the Gillis Math Olympiad organized by the Weizmann Institute:

A foreign government consists of 12 ministers. Each minister has 5 friends and 6 enemies amongst the ministers. Each committee needs 3 ministers. A committee is considered legitimate if all of its members are friends or all of its members are enemies. How many legitimate committees can be formed?

Surprisingly, this problem implies that the answer doesn’t depend on how exactly enemies and friends are distributed. This meta thought lets us calculate the answer by choosing an example. Imagine that the government is divided into two factions of six people. Within a faction people are friends, but members of two different factions dislike each other. Legitimate committees can only be formed by choosing all three members from the same faction. The answer is 40.

We would like to show that actually the answer to the problem doesn’t depend on the particular configuration of friendships and enmities. For this, we will count illegitimate committees. Every illegitimate committee has exactly two people that have one enemy and one friend in the committee. Let’s count all the committees from the point of view of these “mixed” people. Each person participates in exactly 5*6 committees as a mixed person. Multiply by 12 (the number of people), divide by 2 (each committee is counted twice) and you get the total 180. This gives an answer of 40 for the number of legitimate committees without using a particular example.

What interests us is the fact that the number of illegitimate, as well as legitimate, committees is completely defined by the degree distribution of friends. For any set of people and who are either friends or enemies with each other, the number of illegitimate committees can be calculated from the degree distribution of friends in the same way as we did above.

Any graph can be thought of as representing friendships of people, where edges connect friends. This cute puzzle tells us that the sum of the number of 3-cliques and 3-anti-cliques depends only on the degree distribution of the graph.

As a non mathematical comment, the above rule for legitimate committees is not a bad idea. In such a committee there is no reason for two people to gang up on the third one. Besides, if at some point in time all pairs of friends switch to enemies and vice versa, the committees will still be legitimate.

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A Math Exam’s Hidden Agenda

In 1976 I was about to become a student in the math department at Moscow State University. As an IMO team member I was accepted without entrance exams, but all of my other classmates had to take the exams. There were four exams: written math, oral math, physics, and an essay.

The written math exam was the first, and here are the problems. I want my non-Russian readers to see if they notice anything peculiar about this exam. Can you explain what is peculiar, and what might be the hidden agenda?

Problem 1. Solve the equation

Equation

Problem 2. Solve the inequality

Inequality

Problem 3. Consider a right triangle ABC with right angle C. Angle B is 30° and leg CA is equal to 1. Let D be the midpoint of the hypotenuse AB, so that CD is a median. Choose F on the segment BC so that the angle between the hypotenuse and the line DF is 15°. Find the area of CDF. Calculate its numeric value with 0.001 precision.

Problem 4. Three balls, two of which are the same size, are tangent to the plane P, as well as to each other. In addition, the base of a circular cone lies on the plane P, and its axis is perpendicular to the plane. All three balls touch the cone from the outside. Find the angle between a generatrix of the cone and the plane P, given that the triangle formed by the points of tangency of the balls and the plane has one angle equal to 150°.

Problem 5. Let r < s < t be real numbers. If you set y equal to any of the numbers r, s or t in the equation x2 − (9 − y)x + y2 − 9y + 15 = 0, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that −1 < r < 1.

Let me describe some background to this exam. Applicants who solve fewer than two problems fail the exam and are immediately rejected. People who solve two or three problems are given 3 points. Four problems earn 4 points, and five problems earn 5 points.

If you still do not see the hidden agenda, here is another clue. People who get 5 points on the first exam and, in addition, have a gold medal from their high school (that means all As) are admitted right after the first exam. For the others, if they do not fail any of the exams, points are summed up with their GPAs to compute their scores. The so-called half-passing score is then calculated. Scores strictly higher than the half-passing score qualify applicants for admission. However, there are too many applicants for the available openings with at least the half-passing score. As a result only some people with exactly the half-passing score are accepted, at the discretion of the department.

Now my readers have enough information to figure out the hidden agenda behind that particular exam.

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Good Math Research Projects for High School

by Pavel Etingof and Tanya Khovanova

We worked for several years with RSI where we supervised summer math research projects by high school students. Now, we’ve started an additional program at MIT’s math department called PRIMES, where local high school students do math research during the academic year. In this essay we would like to discuss what makes a good math research project for a high school student.

A doable project. The project should not be believed to be extremely difficult to yield at least results. It is very discouraging for an aspiring mathematician not to produce anything during their first project.

An accessible beginning. The student should be able to start doing something original soon after the start of the project. After all, they don’t come to us for coursework, but for research.

Flexibility. It is extremely important to offer them a project that is adjustable; it should go in many directions with many different potential kinds of results. Since we do not know the strength of incoming students in advance, it is useful to have in mind both easier and harder versions of the project.

Motivation. It is important for the project to be well motivated, which means related to other things that have been studied and known to be interesting, to research of other people, etc. Students get more excited when they see that other people are excited about their results.

A computer component. This is not a must for a good project. But modern mathematics involves a lot of computation and young students are better at it than many older professors. Such a project gives young students the opportunity to tackle something more senior people are interested in but might not have enough computer skills to solve. In addition, through computer experiments students get exposed to abstract notions (groups, rings, Lie algebras, representations, etc.) in a more “hands-on” way than when taking standard courses, and as a result are less scared of them.

A learning component. It is always good when a project exposes students to more advanced notions.

The student should like their project. This is very difficult to accomplish when projects are chosen in advance before we meet the students. However, we try to match them to great projects by using the descriptions they give of their interests on their applications. It goes without saying that mentors should like their project too.

Having stated the desired properties of a good project, let us move on to giving examples: bad projects and good projects. We start with a bad one:

Prove that the largest power of 2 that doesn’t contain 0 is 286.

The project satisfies only one requirement: it contains a computer component. Otherwise, it doesn’t have an accessible beginning. It is not very flexible: if the student succeeds, the long-standing conjecture will be proven; if s/he doesn’t, there is not much value in intermediate results. The question is not very interesting. The only motivation is that it has been open for a long time. Also, there is not much to learn. Though, almost any theoretical question can be made flexible. We can start with the question above and change its direction to make it more promising and enticing.

Another bad example is a project where the research happens after the programs are written. This is bad because it is difficult to estimate the programming abilities of incoming students. It doesn’t have an accessible beginning and there is no flexibility until the programming part is finished. If the student can’t finish the programming quickly, s/he will not have time to look at the results and produce conjectures. For example, almost any project in studying social networks may fall into this category:

Study an acquaintance graph for some epic movies or fiction, for example Star Wars or The Lord of the Rings. In this graph people are vertices and two people are connected by an edge if they know each other. The project is to compare properties of such graphs to known properties of other social networks.

Though the networks in movies are much smaller than other networks that people study, the amount of programming might be substantial. This project can be a good project for a person with a flexible time frame or a person who is sure in advance that there will be enough time for him/her to look at the data.

Now on to an example of a good project. Lynnelle Ye and her mentor, Tirasan Khandhawit, chose to analyze the game of Chomp on graphs during RSI 2009.

Given a graph, on each turn a player can remove an edge or a vertex together with all adjacent edges. The player who doesn’t have a move loses. This game was previously solved for complete graphs and forest graphs, so the project was to analyze the game for other types of graphs.

It is clear how to analyze the game for any particular new graph. So that could be a starting point providing an accessible beginning. After that the next step could be to analyze other interesting sets of graphs. The flexibility is guaranteed by the fact that there are many sets of graphs that can be used. In addition, the project entails learning some graph theory and game theory. And the project has a computational component.

Lynnelle Ye successfully implemented this project and provided a complete analysis of complete n-partite graphs for arbitrary n and all bipartite graphs. She also gave partial results for odd-cycle pseudotrees. The paper is available at the arxiv. Not surprisingly, Lynelle got fourth place in the Intel Science Talent Search and second place in the Siemens Competition.

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A Son Named Luigi

Suppose that we choose all families with two children, such that one of them is a son named Luigi. Given that the probability of a boy to be named Luigi is p, what is the probability that the other child is a son?

Here is a potential “solution.” Luigi is a younger brother’s name in one of the most popular video games: Super Mario Bros. Probably the parents loved the game and decided to name their first son Mario and the second Luigi. Hence, if one of the children is named Luigi, then he must be a younger son. The second child is certainly an older son named Mario. So, the answer is 1.

The solution above is not mathematical, but it reflects the fact that children’s names are highly correlated with each other.

Let’s try some mathematical models that describe how the parents might name their children and see what happens. It is common to assume that the names of siblings are chosen independently. In this case the first son (as well as the second son) will be named Luigi with probability p. Therefore, the answer to the puzzle above is (2-p)/(4-p).

The problem with this model is that there is a noticeable probability that the family has two sons, both named Luigi.

As parents usually want to give different names to their children, many researchers suggest the following naming model to avoid naming two children in the same family with the same name. A potential family picks a child’s name at random from a distribution list. Children are named independently of each other. Families in which two children are named the same are crossed out from the list of families.

There is a problem with this approach. When we cross out families we may disturb the balance in the family gender distributions. If we assume that boys’ and girls’ names are different then we will only cross out families with children of the same gender. Thus, the ratio of different-gender families to same-gender families will stop being 1/1. Moreover, it could happen that the number of boy-boy families will differ from the number of girl-girl families.

There are several ways to adjust the model. Suppose there is a probability distribution of names that is used for the first son. If another son is born, the name of the first son is crossed out from the distribution and following that we proportionately adjust the probabilities of all other names for this family. In this model the probability of naming the first son by some name and the second son by the same name changes. For example, the most popular name’s probability decreases with consecutive sons, while the least popular name’s probability increases.

I like this model, because I think that it reflects real life.

Here is another model, suggested by my son Alexey. Parents give names to their children independently of each other from a given distribution list. If they give the same name to both children the family is crossed-out and replaced with another family with children of the same genders. The advantage of this model is that the first child and the second child are named independently from each other with the same probability distribution. The disadvantage is that the probability distribution of names in the resulting set of families will be different from the probability distribution of names in the original preference list.

I would like my readers to comment on the models and how they change the answer to the original problem.

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