Tanya Khovanova's Math Blog

On March 5, 2010 I visited Princeton and had dinner with John Conway at Tiger Noodles. He gave me the second Doomsday lesson right there on a napkin. I described the first Doomsday lesson earlier, in which John taught me to calculate the days of the week for 2009. Now was the time to expand that lesson to any year.

As you can see on the photo of the napkin, John uses his fingers to make calculations. The thumb represents the DoomsDay Difference, the number of days your birthday is ahead of DoomsDay for a given year. To calculate this number you have to go back to my previous post.

The index finger represents the century adjustment. For example, the Doomsday for the year 1900 is Wednesday. Conway remembers Wednesday as We-are-in-this-day. He invented his algorithm in the twentieth century, not to mention that most people who use his algorithm were born in that century. Conway remembers the Doomsday for the year 2000 as Twosday.

The next three fingers help you to calculate the adjustment for a particular year. Every non-leap year has 52 weeks and one day. So the Doomsday moves one day of the week forward in one year. A leap year has one extra day, so the Doomsday moves forward two days. Thus, every four years the Doomsday moves five days forward, and, consequently, every twelve years it moves forward to the next day of the week. This fact helps us to simplify our year adjustment by replacing every dozen of years with one day in the week.

The middle finger counts the number of dozens in the last two digits of your year. It is important to use “dozen” instead of “12” as later we will sum up all the numerals, and the word “dozen” will remind us that we do not need to include it in the sum.

The ring finger represents the remainder of the last two digits of the year modulo 12, and the pinkie finger represents the number of leap years in that remainder.

John made two sample calculations on the napkin. The first one was for his own birthday — December 26, 1937. John was born exactly on Doomsday. I suspect that that is the real reason he called his algorithm the Doomsday Algorithm. The century adjustment is Wednesday. There are 3 dozens in 37, with the remainder 1 and 0 leap years in the remainder. When we add four more days to Wednesday, we get Sunday. So John Conway was born on Sunday.

The second napkin example was the day we had dinner: March 5, 2010. March 5 is 5 days ahead of the Doomsday. The century adjustment is Twosday, plus 0 dozens, 10 years in the remainder and 2 leap years in the remainder. 5 + 0 + 10 + 2 equals 3 modulo 7. Hence, we add three days to Tuesday, demonstrating that we dined out together on Friday. But then, we already knew that.

Do you know that some Russian letters are shaped exactly as some letters in the English alphabet? The shapes are the same, but the sounds of the letters are not. My Russian last name can be completely spelled using English letters: XOBAHOBA.

The adequate translation of my last name into English is Hovanova. You might ask where the first “K” came from. For many years French was considered the language of diplomacy and the USSR used French as an official language for traveling documents.

But “H” in French is silent and “Hovanova” would have been pronounced as “Ovanova.” To prevent that, Russians used “kh” for the “h” sound.

Now to my first name. I was born Tatyana, for which Tanya is a nickname. Back in Russia, Tanya is used for children and students and Tatyana for adults and teachers. As I was a student throughout my 30 years of life in Russia, I was always Tanya. When I moved to the US, I decided to keep using Tanya, which I much preferred to Tatyana.

A psychiatrist might think that I wanted to be a student forever or refused to grow up. Or I could be accused of being lazy, as Tanya is shorter. In reality, I was just trying to be considerate. Tanya is easier to write and to spell for Americans. Anyway, I already had enough problems spelling out my last name in this country.

Now that more information is getting translated from Russian into English, I keep stumbling on references to me as to Hovanova or Tatyana. For example, the IMO official website used Russian sources to come up with the names of the Russian participants. They then translated the names directly into English, instead of going through French. As a result, on their website I am Tatyana Hovanova. This is not unique to me: many Russian names on the IMO website differ from those peoples’ passport names.

By the way, if you Google my last name you will encounter other Khovanovas. Khovanova is not a particularly unusual name. Only one of the Khovanovas that came up in my search results is a close relative. Elizabeth Khovanova is my father’s second wife and a dear friend. She is also an accomplished geneticist.

Khovanova is used only for females in Russia. The male equivalent is Khovanov. Surely you have heard of my half-brother Mikhail Khovanov and his homologies.

I gave you the Wise Men Without Hats puzzle in one of my previous posts:

A sultan decides to check how wise his two wise men are. The sultan chooses a cell on a chessboard and shows it to the first wise man. In addition, each cell on the chessboard either contains a pebble or is empty. The first wise man has to decide whether to remove one pebble or to add one pebble to an empty cell. Next, the second wise man must look at the board and guess which cell was chosen by the sultan. The two wise men are permitted to agree on their strategy beforehand. What strategy can they find to ensure that the second wise man will always guess the chosen cell?

My readers solved it. The solution is the following. Let us assign a number between 0 and 63 to every cell of the board. The second wise man takes numbers corresponding to cells with pebbles, converts them to binary and XORs the result. The answer is the cell number that he is seeking. The first wise man can always add or remove a pebble to make the XORing operation of the remaining pebbles produce any given number from 0 to 63.

This solution only works for boards that have a power of two as the number of cells.

Let’s look at the solution more closely. Let us create a graph corresponding to this problem. The vertices of the graph will correspond to the positions of pebbles. That means vertices are in one-to-one correspondence with the subsets of the set of 64 elements. Let us connect two vertices if we can get from one position to another by removing or adding a pebble. That means vertices are connected if two corresponding sets differ by exactly one element. We can see that the resulting graph is regular and each vertex is connected to exactly 64 other vertices.

Let us assign one out of 64 colors to each cell of the chessboard. The second wise man can guess the cell by looking at the chessboard. From this we can conclude that there is a bijection from the vertices of the graph to chessboard cells. In other words, we can color the graph in 64 colors. The existence of the strategy for wise men means that we can color the graph in such a way that each vertex is connected to the vertices of all colors.

As each vertex in our graph has exactly 64 neighbors, the graph has the following property: It can be colored in 64 colors in such a way that every vertex is connected to exactly one vertex of every color.

As soon as I realized that there is such a graph-theoretical object, I started to run around MIT asking everyone if such objects were studied or have a name.

It appears that indeed such an object has a name. A graph that can be colored into k colors in such a way that every vertex has exactly one neighbor of every color is called a rainbow graph.

Andrew Woldar discusses properties of such graphs in his paper. In particular, rainbow graphs are matching graphs. Indeed, every vertex is connected to exactly one vertex of the same color. Hence there is a natural pairing on vertices. From here, we can conclude that the smallest size of a rainbow graph is 2k.

Several MIT students liked the wise men problem and the associated graph object so much that they decided to study them. Hwanchul Yoo, SuHo Oh, and Taedong Yun enumerated all rainbow graphs of size 2k. The number of non-isomorphic rainbow graphs of size 2k equals mitthe number of switching classes of graphs with k vertices. The corresponding sequence A002854 starts as: 1, 1, 2, 3, 7, 16, 54. The paper is soon to appear. It is titled “Rainbow Graphs and Switching Classes.”

Just received from Victor Gutenmacher:

Fibonacci salad: For today’s salad, mix yesterday’s leftover salad with that of the day before.

New additions to my trick problems collection:

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It takes 12 minutes to saw a log into 3 parts. How much time will it take to saw it into 4 parts?

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The Davidsons have five sons. Each son has one sister. How many children are there in the family?

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A caterpillar wants to see the world and decides to climb a 12-meter pole. It starts every morning and climbs 4 meters in half a day. Then it falls asleep for the second half of the day during which it slips 3 meters down. How much time will it take the caterpillar to reach the top?

First Name: David
Last Name: (hidden for privacy protection)
Year of Birth: (hidden for privacy protection)
email: buchanan1985@gmail.com

I remember a math problem from my childhood: divide an L-shaped triomino into four congruent parts. The answer is in the picture on the left. Such division is quite appropriately called a reptile (repetitive tiling). Solomon Golomb invented the name many years ago. He wasn’t aware that his definition would end up creating Googling problems, for when you search for such a mathematical object you will stumble upon a lot of amphibians.

Similarly, you can divide the same shape into 9 congruent pieces (see the figure on the left).

Suppose you want to divide a shape into pieces that are similar, but not necessarily of the same size. Such tiling is called an irreptile (irregular reptile).

At the Gathering for Gardner 9 I listened to Carolyn Yackel‘s talk about the L-reptiles and L-irreptiles. One of the ways to create an irreptile is to start with a reptile, then to make a sub-tiling of one of the existing tiles. This procedure can be repeated many times.

Carolyn brought a ceramic table to the Gathering for Gardner. This table is made of two L-shapes. Both shapes are irreptiles, created by this procedure. In one part of the table she started with a 9-reptile, and in the other with a 4-reptile. She sent me this picture of her table to use in this essay.

After her talk I started wondering how many tiles can an L-irreptile be comprised of. We start with one piece: the L-shape itself. If we divide a tile into four smaller tiles we add three more pieces. If we divide it into nine tiles we add eight more pieces. We can mix sub-dividing into four and nine tiles. The total number of tiles that an L-shape can be comprised of by this procedure is all the numbers you can get from 1 by adding three or eight. The sequence is 1, 4, 7, 9, 10, 12, 13, 15, 16, 17 and so on. Starting from 15 we get all the consecutive numbers.

The numbers that are not represented in the above sequence are 2, 3, 5, 6, 8, 11 and 14. Can we divide an L-shape into such numbers of tiles? Benoît Jubin reminded me that there is an L-reptile with six pieces.

Consequently, we can add 5 more pieces to any L-irreptile. Thus, there exists an L-irreptile made out of 11 (1+5+5) and out of 14 (1+8+5) pieces. The numbers that are left are 2, 3, 5 and 8.

While I was discussing L-irreptiles with fans of sequences, David Wilson suggested a conjecture.

David Wilson’s conjecture. If there is an L-irreptile, there is a corresponding square-irreptile with similarly-sized pieces.

If this conjecture is true, then we can see that L-irreptiles with 2, 3 or 5 pieces can’t exist as corresponding square-irreptiles do not exist.

For example, to prove that 2 or 3 square-irreptiles can’t exist, you need to notice that each corner of the square we are trying to tile should belong to a different small tile.

The question of the existence of an 8-irreptile of the L-shape is more interesting and challenging. The square 8-irreptile exists. If you can prove that the L-shape 8-irreptile doesn’t exist, then you will automatically prove that the converse to Wilson’s conjecture is not true.

Only at MIT. Room 4-231.

by Tanya Khovanova, Konstantin Knop, Alexey Radul and Peter Sarnak

Let n coins weighing 1, 2, … n be given. Baron Münchhausen knows which coin weighs how much, but his audience does not. Define a(n) to be the minimum number of weighings the Baron must conduct on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins.

In the paper Baron Münchhausen’s Sequence, three of us completely described the Baron’s sequence. In particular, we proved that a(n) ≤ 2. Here we would like to outline another proof idea, which is interesting in part because it touches the Riemann hypothesis.We denote the total weight of coins in some set A as |A|.

Lemma. Numbers n that can be represented as T_i + T_j + T_k = 3n, where i ≤ j < k, such that there is a subset A of coins from j + 1 to k such that n = T_j + |A|, can be done in two weighings.

Proof. Suppose T_i + T_j + T_k = 3n and there is a subset A of coins from j + 1 to k such that n = T_j + |A|. We propose the two weighings

[1…j] + A = n

and

[1…i] + B = n + A,

where B is the complement of A in {j + 1, j + 2, … , k}.

If we sum up twice the first weighing with the second weighing we get

3[1…i] + 2[(i + 1)…j] + 2A + B = 3n + A.

In other words, three times the weight of the coins that were on the left side in both weighings, plus twice the weight of the coins that were on the left side in only the first weighing, plus the weight of the coins that were moved from the left cup to the right cup plus the weight of the coins on the left cup in only the second weighing equals three times the weight of the coin on the right cup in both weighings. Hence three times the weight of the coin on the right cup in both weighings can’t be less than the weight of the k other coins that participated plus the weight of the j coins that were on the left cup in the first weighing and weren’t moved to the right cup, plus the weight of the i coins that were one the left cup in both the first and the second weighing. But because T_i + T_j + T_k = 3n, then 3n is the smallest possible weight of any set of i plus j plus k coins, the coin on the right cup in both weighings has to be the n-coin.

We checked that any number up to 600,000 except 20 can be represented so as to satisfy the Lemma. To show how to solve 20 coins in two weighings is easy, and, as usual, is left as an exercise for the reader. Next, we want to look at the following lemma.

Lemma. Given a set of consecutive numbers {(j + 1), … , k}, if k > 2j + 2, then it is possible to find a subset in the set that sums up to any number in the range from j + 1 to (j + k + 1)(k – j)/2 – j – 1.

We won’t prove the lemma, but it means that if k is about twice larger than j, then we have a lot of flexibility for building our set A in the weighing above. For moderately large n (where 600000 >> “moderately large”), it is not hard to prove that this flexibility is sufficient.

Now the question becomes: can we find such a decomposition into triangular numbers? It is enough to find a representation T_i + T_j + T_k = 3n, where T_k is at least 81% of 3n.

We know that decompositions into triangular numbers are associated with decompositions into squares. Namely, if T_i + T_j + T_k = 3n, then (2i + 1)² + (2j + 1)² + (2k + 1)² = 24n + 3. If the largest square is at least 81% of 24n + 3, then the largest triangular number in the decomposition of 3n is at least 81%.

There is a theorem (W. Duke, Hyperbolic distribution problems and half-integral weight Maass forms, in Inventiones Math 92 (1988) p.73-90) that states that in the limit the decompositions of numbers into three squares are equidistributed. That is, if we take some region on the unit sphere x² + y² + z² = 1 (for example, the region |z| > 0.8) and view decompositions of 24n + 3 into squares as points on the sphere x² + y² + z² = 24n + 3, then, as n grows, decompositions whose projections fall into our chosen region are guaranteed to appear.

This theorem is great, because it tells us that for large enough n we will always be able to find a decomposition of 24n + 3 into triangle numbers where one of the triangle numbers will be much bigger than the others, and it will be possible to prove the weight of the n coin in two weighings. Unfortunately, this summary, as stated, does not tell us how large that n needs to be. So we need some exact estimates.

The number of decompositions of m into sums of three squares is about the square root of m. More precisely, it is possible to compute a number N, such that for any number m > N, with at most one exception, the number of decompositions is at least Cm^1/2−1/30, where C is a known constant.

The more specific statement of Duke’s theorem is that if the number of solutions to the quadratic x² + y² + z² = 24n + 3 is large, for a computable value of “large”, then the solutions are equidistributed. More precisely, let us denote 3n by m and fix an area Ω on the unit sphere. Then the number of solutions (x, y, z) such that the unit vector (x, y, z)/|(x, y, z)| belongs to Ω is

1/(4π) Ωh(8m+3) + E(m),

where h(8m+3) is the total number of solutions of x² + y² + z² = 24n + 3, and E(m) is an error term, which starting from some number satisfies the inequality: E(m) ≤ 1000m^1/2-1/7.

That’s pretty good, because combining these two lets us, at least in principle, actually calculate an N such that for all n > N except maybe one a(n) = 2. After that we hoped to write a program to exhaustively search smaller numbers by computer.

This situation is still somewhat annoying, because that possible exception must then be propagated into the proof, and if we are not careful, possibly into the final theorem. (“No matter how many coins the Baron has, he can prove the weight of one in at most two weighings, except maybe one number of coins, and we don’t know which…”) This is where the Riemann Hypothesis comes in. If the Riemann Hypothesis is true, then that exception isn’t there, and all is sunlight and flowers.

The beauty of the Baron’s puzzle is such that we actually do not need the Riemann hypothesis. As we can use unbalanced weighings, it is enough to find a good decomposition for one out of the four numbers 3n, 3n-1, 3n-2, or 3n-3.

Instead of finding all these exact estimates we found a different elementary proof of our theorem. But we are excited that methods that are used in very advanced number theory can be used to solve a simple math problem that can be described to middle school children.

It would be great if someone decided to finish this proof.

I am used to thinking that a “woman in numbers” means a female number theorist. But not anymore. I just discovered drawings by Svetlana Bogatyr. From now on the expression a “woman in numbers” will convey an additional meaning to me.

I am grateful to Svetlana for permitting me to post several of her drawings. The “Mature Woman” is on the left. “Eurydice”, “Girl in Scarf” and “Holland Woman ” are below.

Enjoy.