Tanya Khovanova's Math Blog

One of the 2012 PRIMES projects, suggested by Professor Jacob Fox, was about bounds on the number of halving lines. I worked on this project with Dai Yang.

Suppose there are n points in a general position on a plane, where n is an even number. A line through two given points is called a halving line if it divides the rest of n−2 points in half. The big question is to estimate the maximum number of halving lines.

Let us first resolve the small question: estimating the minimum number of halving lines. Let’s take one point from the set and start rotating a line through it. By a continuity argument you can immediately see that there should be a halving line through any point. Hence, the number of halving lines is at least n/2. If the point is on the convex hull of the set of points, then it is easy to see that it has exactly one halving line through it. Consequently, if the points are the vertices of a convex n-gon, then there are exactly n/2 halving lines. Thus, the minimum number of halving lines is n/2.

Finding the maximum number of halving lines is much more difficult. Previous works estimated the upper bound by O(n^4/3) and the lower bound by O(ne^{√log n}). I think that Professor Fox was attracted to this project because the bounds are very far from each other, and some recent progress was made by elementary methods.

Improving a long-standing bound is not a good starting point for a high school project. So after looking at the project we decided to change it in order to produce a simpler task. We decided to study the underlying graph of the configuration of points.

Suppose there is a configuration of n points on a plane, and we are interested in its halving lines. We associate a graph to this set of points. A vertex in the configuration corresponds to a vertex in the graph. The graph vertices are connected, if the corresponding vertices in the set have a halving line passing through them. So we decided to answer as many questions about the underlying graph as possible.

For example, how long can the longest path in the underlying graph be? As I mentioned, the points on the convex hull have exactly one halving line through them. Hence, we have at least three points of degree 1, making it impossible for a path to have length n. The picture below shows a configuration of eight points with a path of length seven. We generalized this construction to prove that there exists a configuration with a path of length n−1 for any n.

We also proved that:

• The largest cycle can have length n − 3.
• The largest clique is of the order O(√n).
• The degrees of two distinct vertices sum to at most n, if they are connected by an edge, and at most n − 2 otherwise.

After we proved all these theorems, we came back to the upper bound and improved it by a constant factor. Our paper is available at arXiv:1210.4959.

Davidson Institute for Talent Development announced their 2012 Winners. Out of 22 students, three were recognized for their math research. All three of them are ours: that is, they participated in our PRIMES and RSI programs:

• David Ding’s project, “Infinitesimal Cherednik Algebras of gl_n,” came out of his participation in the PRIMES program.
• Sitan Chen’s project, “On the Rank Number of Grid Graphs,” came out of his participation in the RSI program.
• Xiaoyu He’ project, “On the Classification of Universal Rotor-Routers,” came out of his participation in the PRIMES program.

I already wrote about Xiaoyu’s project. Today I want to write about Sitan’s project and what I do as the math coordinator for RSI.

RSI students meet with their mentors every day and I meet with students once a week. On the surface I just listen as they describe their projects. In reality, I do many different things. I cheer the students up when they are overwhelmed by the difficulty of their projects. I help them decide whether they need to switch projects. I correct their mistakes. Most projects involve computer help, so I teach them Mathematica. I teach them the intricacies of Latex and Beamer. I explain general mathematical ideas and how their projects are connected to other fields of mathematics. I never do their calculations for them, but sometimes I suggest general ideas. In short, I do whatever needs to be done to help them.

I had a lot of fun working with Sitan. His project was about the rank number of grid graphs. A vertex k-ranking is a labeling of the vertices of a graph with integers from 1 to k so that any path connecting two vertices with the same label passes through a vertex with a greater label. The rank number of a graph is the minimum possible k for which a k-ranking exists for that graph. When Sitan got the project, the ranking numbers were known for grid graphs of sizes 1 by n, 2 by n, and 3 by n. So Sitan started working on the ranking number of the 4 by n graph.

His project was moving unusually fast and my job was to push him to see the big picture. I taught him that the next step, once he finishes 4 by n graphs is not to do 5 by n graphs, as one might think. After the first step, the second step should be bigger. He should use his insight and understanding of 4 by n graphs to try to see what he can do for any grid graphs.

This is exactly what he did. After he finished the calculation of the rank number of the 4 by n graphs, he found a way to improve the known bounds for the ranking number of any grid graph. His paper is available at the arXiv.

I just looked at my notes for my work with Sitan. The last sentence: “Publishable results, a potential winner.”

Consider central symmetry: squares and circles are centrally symmetric, while trapezoids and triangles are not. But if you have two trapezoids, which of them is more centrally symmetric? Can we assign a number to describe how symmetric a shape is?

Here is what I suggest. Given a shape A, find a centrally symmetric shape B of the largest area that fits inside. Then the measure of central symmetry is the ratio of volumes: B/A. For centrally symmetric figures the ratio is 1, and otherwise it is a positive number less than 1.

The measure of symmetry is positive. But how close to 0 can it be? The picture on the left is a shape that consists of five small disks located at the vertices of a regular pentagon. If the disks are small enough than the largest symmetric subshape consists of two disks. Thus the measure of symmetry for this shape is 2/5. If we replace a pentagon with a regular polygon with a large odd number of sides, we can get very close to 0.

What about convex figures? Kovner’s theorem states that every convex shape of area 1 contains a centrally symmetric shape of area at least 2/3. It is equal to 2/3 only if the original shape is a triangle. That means every convex shape is at least 2/3 centrally symmetric. It also means that the triangle is the least centrally symmetric convex figure. By the way, a convex shape can have only one center of symmetry.

After I started writing this I discovered that there are many ways in which people define measures of symmetry. The one I have defined here is called Kovner-Besicovitch measure. The good news is that the triangle is the least symmetric planar convex shape with respect to all of these different measures.

Sid Dhawan was one of our RSI 2011 math students. He was studying interlocking polyominoes under the mentorship of Zachary Abel.

A set of polyominoes is interlocked if no subset can be moved far away from the rest. It was known that polyominoes that are built from four or fewer squares do not interlock. The project of Dhawan and his mentor was to investigate the interlockedness of larger polyominoes. And they totally delivered.

They quickly proved that you can interlock polyominoes with eight or more squares. Then they proved that pentominoes can’t interlock. This left them with a gray area: what happens with polyominoes with six or seven squares? After drawing many beautiful pictures, they finally found the structure presented in our accompanying image. The system consists of 12 hexominoes and 5 pentominoes, and it is rigid. You cannot move a thing. That means that hexominoes can be interlocked and thus the gray area was resolved.

You can find the proofs and the details in their paper “Complexity of Interlocking Polyominoes”. As you can guess by the title, the paper also discusses complexity. The authors proved that determining interlockedness of a a system that includes hexominoes or larger polyominoes is PSPACE hard.

My son Sergei invented the following game a couple of years ago. Two people, Alice and Bob, agree on a number, say, four. Alice takes a clean Rubik’s cube and secretly makes four moves. Bob gets the resulting cube and has to rotate it to the initial state in not more than four moves. Bob doesn’t need to retrace Alice’s moves. He just needs to find a short path back, preferably the shortest one. If he is successful, he gets a point and then it is Alice’s turn.

If they are experienced at solving Rubik’s cube, they can increase the difficulty and play this game with five or six moves.

By the way, how many moves do you need to solve any position on a Rubik’s cube if you know the optimal way? The cube is so complicated that people can’t always know the optimal way. They think that God can, so they called the diameter of the set of all possible Rubik’s cube positions, God’s Number. It was recently proven that God’s Number is 20. If Alice and Bob can increase the difficulty level to 20, that would mean that they can find the shortest path back to the initial state from any position of the cube, or, in short, that they would master God’s algorithm.

by Andrew Buchanan, Tanya Khovanova, and Alex Ryba

We got this problem from Rados Radoicic:

A 7 by 7 board is covered with 38 dominoes such that each covers exactly 2 squares of the board. Prove that it is possible to remove one domino so that the remaining 37 still cover the board.

Let us call a domino covering of an n by n board saturated if the removal of any domino leaves an uncovered square. Let d(n) be the number of dominoes in the largest saturated covering of an n by n board. Rados’ problem asks us to prove that d(7) < 38.

Let’s begin with smaller boards. First we prove that d(2) = 2. Suppose that 3 dominoes are placed on a 2 × 2 board. Let us rotate the board so that at least two of the dominoes are horizontal. If they coincide, then we can remove one of them. If not, they completely cover the board and we can remove the third one. Similarly, you can check all the cases and show that d(3) = 6.

Now consider a saturated domino covering of an n × n board. We can view the dominoes as vertices of a graph, joining two if they share a cell of the board. No domino can share both cells with other dominoes, or we could remove it. Hence, each domino contains at most one shared cell. This means that all the dominoes in a connected component of the graph must overlap on a single shared cell. Hence, the only possible connected components must have the following shapes:

The largest shape in the picture is the X-pentomino. We can describe the other shapes as fragments of an X-pentomino, where the center and at least one more cell is intact. We call these shapes fragments.

A saturated covering by D dominoes corresponds to a decomposition of the n × n board into F fragments. Note that a fragment with k cells is made from k − 1 dominoes. Summing over the dominoes gives: D = n² − F. Thus, in order to make D as large as possible, we should make F as small as possible. Let f(n) be the minimal number of fragments that are required to cover an n by n board without overlap. Then d(n) = n² − f(n).

Consider the line graph of the n by n board. The vertices of the line graph correspond to cells in the original board and the edges connect vertices corresponding to neighboring cells. Notice that in the line graph our fragments become all star graphs formed by spokes coming out from a single central node. Thus a decomposition of a rectangular board into fragments corresponds to a covering of its line graph by star graphs. Consider an independent set in the line graph. The smallest independent set has the same number of elements as the smallest number of stars that can cover the graph. This number is called a domination number.

Now let’s present a theorem connecting domino coverings with X-pentomino coverings.

Theorem. f(n) equals the smallest number of X-pentominoes that can cover an n by n board allowing overlaps and tiles that poke outside, which is the same as the domination number of the corresponding line graph.

The proof of this theorem and the solution to the original puzzle is available in our paper: “Saturated Domino Coverings.” The paper also contains other theorems and discussions of other boards, not to mention a lot of pictures.

The practical applications of star graph coverings are well-known and widely discussed. We predict a similar future for saturated domino coverings and its practical applications, two examples of which follow:

First, imagine a party host arranging a plate of cookies. The cookies must cover the whole plate, but to prevent the kids sneaking a bite before the party, the cookies need to be placed so that removal of just one cookie is bound to expose a chink of plate. This means the cookies must form a saturated covering of the plate. Of course the generous host will want to use a maximal saturated covering.

For the second application, beam yourself to an art museum to consider the guards. Each guard sits on a chair in a doorway, from where it is possible to watch a pair of adjacent rooms. All rooms have to be observed. It would be a mistake to have a redundant guard, that is, one who can be removed without compromising any room. Such a guard might feel demotivated and then who knows what might happen. This means that a placement of guards must be a saturated domino covering of the museum. To keep the guards’ Union happy, we need to use a maximal saturated covering.

We would welcome your own ideas for applications of saturated coverings.

I recently posted an essay Binary Bulls without Cows with the following puzzle:

The test Victor is taking consists of n “true” or “false” questions. In the beginning, Victor doesn’t know any answers, but he is allowed to take the same test several times. After completing the test each time, Victor gets his score — that is, the number of his correct answers. Victor uses the opportunity to re-try the test to figure out all the correct answers. We denote by a(n) the smallest numbers of times Victor needs to take the test to guarantee that he can figure out all the answers. Prove that a(30) ≤ 24, and a(8) ≤ 6.

There are two different types of strategies Victor can use to succeed. First, after each attempt he can use each score as feedback to prepare his answers for the next test. Such strategies are called adaptive. The other type of strategy is one that is called non-adaptive, and it is one in which he prepares answers for all the tests in advance, not knowing the intermediate scores.

Without loss of generality we can assume that in the first test, Victor answers “true” for all the questions. I will call this the base test.

I would like to describe my proof that a(30) ≤ 24. The inequality implies that on average five questions are resolved in four tries. Suppose we have already proven that a(5) = 4. From this, let us map out the 24 tests that guarantee that Victor will figure out the 30 correct answers.

As I mentioned earlier, the first test is the base test and Victor answers every question “true.” For the second test, he changes the first five answers to “false,” thus figuring out how many “true” answers are among the first five questions. This is equivalent to having a base test for the first five questions. We can resolve the first five questions in three more tests and proceed to the next group of five questions. We do not need the base test for the last five questions, because we can figure out the number of “true” answers among the last five from knowing the total score and knowing the answers for the previous groups of five. Thus we showed that a(mn) ≤ m a(n). In particular, a(5) = 4 implies a(30) ≤ 24.

Now I need to prove that a(5) = 4. I started with a leap of faith. I assumed that there is a non-adaptive strategy, that is, that Victor can arrange all four tests in advance. The first test is TTTTT, where I use T for “true” and F for “false.” Suppose for the next test I change one of the answers, say the first one. If after that I can figure out the remaining four answers in two tries, then that would mean that a(4) = 3. This would imply that a(28) ≤ 21 and, therefore, a(30) ≤ 23. If this were the case, the problem wouldn’t have asked me to prove that a(30) ≤ 24. By this meta reasoning I can conclude that a(4) ≠ 3, which is easy to check anyway. From this I deduced that all the other tests should differ from the base test in more than one answer. Changing one of the answers is equivalent to changing four answers, and changing two answers is equivalent to changing three answers. Hence, we can assume that all the other tests contain exactly two “false” answers. Without loss of generality, the second test is FFTTT.

Suppose for the third test, I choose both of my “false” answers from among the last three questions, for example, TTFFT. This third test gives us the exactly the same information as the test TTTTF, but I already explained that having only one “false” answer is a bad idea. Therefore, my next tests should overlap with my previous non-base tests by exactly one “false” answer. The third test, we can conclude, will be FTFTT. Also, there shouldn’t be any group of questions that Victor answers the same for every test. Indeed, if one of the answers in the group is “false” and another is “true,” Victor will not figure out which one is which. This uniquely identifies the last test as FTTFT.

So, if the four tests work they should be like this: TTTTT, FFTTT, FTFTT, FTTFT. Let me prove that these four tests indeed allow Victor to figure out all the answers. Summing up the results of the last three tests modulo 2, Victor will get the parity of the number of correct answers for the first four questions. As he knows the total number of correct answers, he can deduce the correct answer for the last question. After that he will know the number of correct answers for the first four questions and for every pair of them. I will leave it to my readers to finish the proof.

Knop and Mednikov in their paper proved the following lemma:

If there is a non-adaptive way to figure out a test with n questions by k tries, then there is a non-adaptive way to figure out a test with 2n + k − 1 questions by 2k tries.

Their proof goes like this. Let’s divide all questions into three non-overlapping groups A, B, and C that contain n, n, and k − 1 questions correspondingly. By our assumptions there is a non-adaptive way to figure out the answers for A or B using k tries. Let us denote subsets from A that we change to “false” for k − 1 non-base tests as A₁, …, A_k-1. Similarly, we denote subsets from B as B₁, …, B_k-1.

Our first test is the base test that consists of all “true” answers. For the second test we change the answers to A establishing how many “true” answers are in A. In addition we have k − 1 questions of type Sum: we switch answers to questions in A_i ∪ B_i ∪ C_i; and type Diff: we switch answers to (A ∖ A_i) ∪ B_i. The parity of the sum of “false” answers in A − A_i + B_i and A_i + B_i + C_i is the same as in A plus C_i. But we know A‘s score from the second test. Hence we can derive C_i. After that we have two equations with two unknowns and can derive the scores of A_i and B_i. From knowing the number of “true” answers in A and C, we can derive the same for B. Knowing A and A_i gives all the answers in A. Similarly for B. QED.

This lemma is powerful enough to answer the original puzzle. Indeed, a(2) = 2 implies a(5) ≤ 4, and a(3) = 3 implies a(8) ≤ 6.

Let’s call a projection of a body L onto a hyperplane a shadow. Here is a mathematical way to hide behind. An object K can hide behind an object L if in any direction the shadow of K can be moved by a translation to be inside the corresponding shadow of L. If K can hide inside L, then obviously K can hide behind L. Dan Klain drew my interest to the following questions. Is the converse true? If K can hide behind L can it hide inside L? If not, then if K can hide behind L, does it follow that the volume of K is smaller than that of L?

We can answer both questions for 2D bodies by using objects with constant width. Objects with constant width are ones that have the same segment as their shadow in every direction. The two most famous examples are a circle and a Reuleaux triangle:

Let’s consider a circle and a Reuleaux triangle of the same width. They can hide behind each other. Barbier’s Theorem states that all objects of the same constant width have the same perimeter. We all know that given a fixed perimeter, the circle has the largest area. Thus, the circle can hide behind the Reuleaux triangle which has smaller area and, consequently, the circle can’t hide inside the Reuleaux triangle. By the way, the Reuleaux triangle has the smallest area of all the objects with the given constant width.

To digress. You might have heard the most famous Microsoft interview question: Why are manhole covers round? Presumably because round manhole covers can’t fall into slightly smaller round holes. The same property is true for manhole covers of any shape of constant width. On the picture below (Flickr original) you can see Reuleaux-triangle-shaped covers.

Let’s move the dimensions up. Dan’s questions become both more difficult and more interesting, because the shadows are not as simple as segments any more.

Before continuing, I need to introduce the concept of “Minkowski sums.” Suppose we have two convex bodies in space. Let’s designate the origin. Then a body can be represented as a set of vectors from the origin to the points in the body. The Minkowski sum of two bodies are all possible sums of two vectors corresponding to the first body and the second body.

Another way to picture the Minkowski sum is like this: Choose a point in the second body. Then move the second body around by translations so that the chosen point covers the first body. Then the area swept by the second body is the Minkowski sum of both of them.

Suppose we have two convex bodies K and L. Their Minkowski interpolation is the body tK + (1-t)L, where 0 ≤ t ≤ 1 is a scaling coefficient. The picture below made by Christina Chen illustrates the Minkowski interpolation of a triangle and an inverted triangle.

If two bodies can hide behind L, then their Minkowski interpolation can hide behind L for any value of parameter t. In particular if K can hide behind L, then the Minkowski interpolation tK + (1-t)L can hide behind L, for any t.

In my paper co-authored with Christina Chen and Daniel Klain “Volume bounds for shadow covering”, we found the following connection between hiding inside and volumes. If L is a simplex, and K can hide behind it, but can’t hide inside L, then there exists t such that the Minkowski interpolation tK + (1-t)L has a larger volume than the volume of L.

In the paper we conjecture that the largest volume ratio V(K)/V(L) for a body K that can hide behind another body L is achieved if L is a simplex and K is a Minkowski interpolation of L and an inverted simplex. The 3D object that can hide behind a tetrahedron and has 16% more volume than the tetrahedron was found by Christina Chen. See her picture below.

The main result of the paper is a universal constant bound: if K can hide behind L, then V(K) ≤ 2.942 V(L), independent of the dimension of the ambient space.

Recently I stumbled on a cute xkcd comic with the hidden message:

Wikipedia trivia: if you take any article, click on the first link in the article text not in parentheses or italics, and then repeat, you will eventually end up at “Philosophy”.

Naturally, I started to experiment. The first thing I tried was mathematics. Here is the path: Mathematics — Quantity — Property — Modern philosophy — Philosophy.

Then I tried physics, which led me to mathematics: Physics — Natural science — Science — Knowledge — Fact — Information — Sequence — Mathematics.

Then I tried Pierre de Fermat, who for some strange reason led to physics first: Pierre de Fermat — French — France — Unitary state — Sovereign state — State — Social sciences — List of academic disciplines — Academia — Community — Living — Life — Objects — Physics.

The natural question is: what about philosophy? Yes, philosophy goes in a cycle: Philosophy — Reason — Rationality — philosophy.

The original comic talks about spark plugs. So I tried that and arrived at physics: Spark plug — Cylinder head — Internal combustion engine — Engine — Machine — Machine (mechanical) — Mechanical system — Power — Physics.

Then I tried to get far away from philosophy and attempted sex, unsuccessfully: Sex — Biology — Natural science. Then I tried dance: Dance — Art — Sense — Physiology — Science.

It is interesting to see how many steps it takes to get to philosophy. Here is the table for the words I tried:

Mathematics wins. It thoroughly beats all the other words I tried. For now. Fans of sex might be disappointed by these results, and tomorrow they might change the wiki essay about sex to start as:

Modern philosophy considers sex …

I recently stumbled upon some notes (in Russian) of a public lecture given by Vladimir Arnold in 2006. In this lecture Arnold defines a notion of complexity for finite binary strings.

Consider a set of binary strings of length n. Let us first define the Ducci map acting on this set. The result of this operator acting on a string a₁a₂…a_n is a string of length n such that its i-th character is |a_i − a_(i+1)| for i < n, and the n-th character is |a_n − a₁|. We can view this as a difference operator in the field F₂, and we consider strings wrapped around. Or we can say that strings are periodic and infinite in both directions.

Let’s consider as an example the action of the Ducci map on strings of length 6. Since the Ducci map respects cyclic permutation as well as reflection, I will only check strings up to cyclic permutation and reflection. If I denote the Ducci map as D, then the Ducci operator is determined by its action on the following 13 strings, which represent all 64 strings up to cyclic permutation and reflection: D(000000) = 000000, D(000001) = 000011, D(000011) = 000101, D(000101) = 001111, D(000111) = 001001, D(001001) = 011011, D(001011) = 011101, D(001111) = 010001, D(010101) = 111111, D(010111) = 111101, D(011011) = 101101, D(011111) = 100001, D(111111) = 000000.

Now suppose we take a string and apply the Ducci map several times. Because of the pigeonhole principle, this procedure is eventually periodic. On strings of length 6, there are 4 cycles. One cycle of length 1 consists of the string 000000. One cycle of length 3 consists of the strings 011011, 101101 and 110110. Finally, there are two cycles of length 6: the first one is 000101, 001111, 010001, 110011, 010100, 111100, and the second one is shifted by one character.

We can represent the strings as vertices and the Ducci map as a collection of directed edges between vertices. All 64 vertices corresponding to strings of length 6 generate a graph with 4 connected components, each of which contains a unique cycle.

The Ducci map is similar to a differential operator. Hence, sequences that end up at the point 000000 are similar to polynomials. Arnold decided that polynomials should have lower complexity than other functions. I do not completely agree with that decision; I don’t have a good explanation for it. In any case, he proposes the following notion of complexity for such strings.

Strings that end up at cycles of longer length should be considered more complex than strings that end up at cycles with shorter length. Within the connected component, the strings that are further away from the cycle should have greater complexity. Thus the string 000000 has the lowest complexity, followed by the string 111111, as D(111111) = 000000. Next in increasing complexity are the strings 010101 and 101010. At this point the strings that represent polynomials are exhausted and the next more complex strings would be the three strings that form a cycle of length three: 011011, 101101 and 110110. If we assign 000000 a complexity of 1, then we can assign a number representing complexity to any other string. For example, the string 111111 would have complexity 2, and strings 010101 and 101010 would have complexity 3.

I am not completely satisfied with Arnold’s notion of complexity. First, as I mentioned before, I think that some high-degree polynomials are so much uglier than other functions that there is no reason to consider them having lower complexity. Second, I want to give a definition of complexity for periodic strings. There is a slight difference between periodic strings and finite strings that are wrapped around. Indeed, the string 110 of length 3 and the string 110110 of length 6 correspond to the same periodic string, but as finite strings it might make sense to think of string 110110 as more complex than string 110. As I want to define complexity for periodic strings, I want the complexity of the periodic strings corresponding to 110 and 110110 to be the same. So this is my definition of complexity for periodic strings: let’s call the complexity of the string the number of edges we need to traverse in the Ducci graph until we get to a string we saw before. For example, let us start with string 011010. Arrows represent the Ducci map: 011010 → 101110 → 110011 → 010100 → 111100 → 000101 → 001111 → 010001 → 110011. We saw 110011 before, so the number of edges, and thus the complexity, is 8.

The table below describes the complexity of the binary strings of length 6. The first column shows one string in a class up to a rotation or reflections. The second column shows the number of strings in a class. The next column provides the Ducci map of the given string, followed by the length of the cycle. The last two columns show Arnold’s complexity and my complexity.

As you can see, for examples of length six my complexity doesn’t differ much from Arnold’s complexity, but for longer strings the difference will be more significant. Also, I am pleased to see that the sequence 011010, the one that I called The Random Sequence in one of my previous essays, has the highest complexity.

I know that my definition of complexity is only for periodic sequences. For example, the binary expansion of pi will have a very high complexity, though it can be represented by one Greek letter. But for periodic strings it always gives a number that can be used as a measure of complexity.