Tanya Khovanova's Math Blog

Today I have my first invited guest blogger, J.B. He is a 2006, 2007, 2008 and 2009 USAMO qualifier. He was also selected to be on the US team at the Romanian Masters in Mathematics competition. Also, he placed 6th at the North American Computational Linguistics Olympiad. Here is his piece:

The analysis is based on the list of 2009 USAMO qualifiers.

There is a rule that if nobody naturally qualifies for the USAMO from a state, then the highest scoring individual will qualify. Unfortunately, this means that we must remove those states with only one USAMO qualifier. We have 33 states remaining. If we sort these strictly by number of USAMO qualifiers, then we find the following result.

States with at least 4 USAMO qualifiers (24 total) voted for Obama, with the following exceptions: Georgia, Texas, South Carolina, and Missouri. In addition, of the two states with 3 USAMO qualifiers, one voted for Obama and one for McCain. The remaining states with 2 qualifiers (5 total) voted Republican.

Now this is not really unexpected. States with very large populations tend to be democratic and also produce more USAMO qualifiers. The most notable exceptions are Georgia and Texas, both of which were indeed exceptions (major outliers, in fact) above. This prompts the following consideration.

States with at least 8 USAMO qualifiers per 10 million residents (25 total) voted for Obama, with the following exceptions: Florida, Wisconsin, South Carolina, Missouri, and Georgia. Of these, all but Georgia fall within 50% of the target 8 USAMO qualifiers per 10 million residents. Georgia has 18 qualifiers per 10 million residents. Note also that the entire USA has 16 qualifiers per 10 million residents.

Furthermore, if USAMO qualifiers had been used instead of population for determining electoral votes, Obama would have won with 86% of the vote rather than 68%. In general, if the Democrat can secure all those states with at least 1 qualifier per million residents (plus DC), he will win with 303 votes. He can even lose the three red states in that category (Georgia, Missouri, and South Carolina) for exactly 269.

USAMO qualifiers per 10 million residents (for states with more than one qualifier) are:

• NH — 122 (all of their qualifiers are from Phillips Exeter Academy)
• MA — 55
• ME — 30
• CA — 29
• NJ — 29
• CT — 29
• VA — 28
• MD — 25
• WA — 21
• IN — 19
• OR — 19
• NY — 18
• GA — 18
• IA — 17
• NM — 15
• MI — 15
• PA — 14
• MO — 14
• SC — 13
• IL — 13
• NC — 13
• OH — 9
• CO — 8
• MN — 8
• UT — 7
• KS — 7
• WI — 7
• KY — 7
• TX — 7
• FL — 6
• LA — 5
• AL — 4
• TN — 3

The states with only one USAMO qualifier are WY, VT, ND, AK, SD, DE, MT, RI, HI, ID, NE, WV, NV, AR, MS, OK, and AZ. The only blue one of these which falls below 8 qualifiers per 10 million is Nevada (we would expect it to have at least 2 qualifiers to fit the expected pattern). Otherwise, it is at least possible that each state fits the pattern of 8 qualifiers per 10 million residents if and only if it votes Democratic.

More than a year ago, when I had my employment benefits with BAE Systems, I called my benefits center with a general question. The customer service representative refused to answer until I gave her my password. I didn’t have a password, so she told me that they would mail my new password to me.

But I needed an answer, so I tried the website, only to be informed that my new password is in the mail and I should wait for its arrival.

In a week, a letter with a password arrived and I called the benefits center again. I happily told them my new password and opened my mouth to ask my question. However, they didn’t accept my password. Obviously, they had changed my password twice, first when I called and then again when I tried their website. Since only ten minutes passed between these two events, both passwords should have arrived on the same day. But that didn’t happen. So my valid password was still in the mail.

In the second it took me to recover from this news, the customer representative told me that they would be sending me a new password and hung up before I could tell her not to.

A new password arrived the next day. I knew that they had already reset that password, and that I’d have to wait a week for the third password to arrive.

I was tempted to call them again and try to create an infinite password resetting loop, but I actually needed to ask my question. So I threw away my freshly arrived, but no-longer-valid password and waited for a week for the next one.

I was lucky to figure it out so quickly, for otherwise my problem could have spiraled out forever. As a professional specifications writer, here are my suggestions to all benefits centers that have that kind of software on what they should do:

• Don’t send an extra password if a password was sent not long ago, for example, in the last two days.
• If two passwords are mailed to a client in the same week, make both of them valid.
• Use email rather than mail.
• Don’t request passwords for general questions.

I had to wait two weeks to ask a simple question. Now I am writing and complaining about it in the hopes that someone who can fix the problem will read this. Maybe it would have been more productive to write a program that clicks on the “I forgot my password” button every second. This would have daily generated thousands of letters with new passwords to me. Maybe then this problem would have drawn attention sooner.

My son Sergei brought back the Flip-Flop game from Canada/USA Mathcamp, and now I teach it to my students. This game trains students in the multiplication table for seven and eight. These are the most difficult digits in multiplication. This game is appropriate for small kids who just learned the multiplication table, but it is also fun for older kids and adults.

This is a turn-based game. In its primitive simplification kids stand in a circle and count in turn. But it is more interesting than that. Here’s what to say and do on your turn, and how the game determines who is next.

First I need to tell you what to say. On your turn, say the next number by default. However, there are exceptions when you have to say something else. And this something else consists of flips and/or flops.

So what are flips? Flip is related to seven. If a number is divisible by seven or has a digit seven, instead of saying this number, we have to say “flip” with multiplicities. For example, instead of 17 we say “flip” because it contains one digit seven. Instead of 14 we say “flip”, because it is divisible by seven once. Instead of 7 we say “flip-flip”, as it is both divisible by seven and has a digit seven. Instead of 49, we say “flip-flip” as 49 is divisible by the square of seven. Instead of 77 we say “flip-flip-flip” as it has two digits seven and is divisible by seven once.

Flop relates to eight the same way as flip relates to seven. Thus, instead of 16 we say “flop” as it is divisible by eight; instead of 18 we say “flop” as it contains the digit eight; and for 48 we say “flop-flop” as it is both divisible by eight and contains the digit eight.

A number can relate to seven and eight at the same time. For example 28 is divisible by seven and contains the digit eight. Instead of 28 we say “flip-flop”. The general rule is that all flips are pronounced before all flops. For example, instead of 788 we will say “flip-flop-flop-flop” as it is divisible by eight and contains the digit seven once and the digit eight twice.

The sequence of natural numbers in the flip-flop version starts as the following: 1, 2, 3, 4, 5, 6, flip-flip, flop-flop, 9, 10, 11, 12, 13, flip, 15, flop, flip, flop, 19, 20, flip, 22, 23, flop, 25, 26, flip, flip-flop, 29, 30, 31, flop, 33, 34, flip, 36, flip, flop, 39, flop, 41, flip, 43, 44, 45, 46, flip, flop-flop, flip-flip, 50, 51, 52, 53, 54, 55, flip-flop, flip, flop, 59, 60, 61, 62, flip, flop-flop, 65, 66, flip, flop, 69, flip-flip, flip, flip-flop, flip, flip, flip, flip, flip-flip-flip, flip-flop, flip, flop-flop, flop, flop, flop, flip-flop, flop, flop, flip-flop, flop-flop-flop, flop, 90, flip, 92, 93, 94, 95, flop, flip, flopflip-flip-flop, 99, 100.

So how does the turn change? Everyone stands in a circle and says their number the way explained above. We start clockwise and move to the next number. For every flip we reverse the direction and for every flop we skip a person. That means that if we have two flips, we don’t change the direction, while for two flops we skip two people. If we have flips and flops together, for example 28 corresponds to “flip-flop”, then first we change the direction and then we skip a person.

On top of that, there is an extra rule for what you do on your turn. If you say something other than a default number, you switch your position from standing to sitting and vice versa. Sometimes I skip this extra feature — not because I am too lazy to exercise, but because I usually conduct this game in a classroom, where all the desks prevent us from fully enjoying such physical activity.

There are two ways to play this game: as a competition or as practice. When we are competing, a person who makes a mistake drops out. If we’re just practicing, no one drops out. Sometimes I am particularly generous and allow my kids one mistake before making them drop out after the second mistake. So far we have played up to 100. I am curious to see if we can ever reach 700 and how long we will be able to continue the game after that.

Testing in the US is dominated by multiple-choice questions. Together with the time limit, this encourages students to stop thinking and go for guessing. I recently wrote an essay AMC, AIME, USAMO Contradiction, in which I complained about the lack of proofs in the first two rounds of math competitions.

Is there a way to improve the situation? I grew up in the USSR, where each round of the math competition had the same format: you were given several hours to write proofs for three or four difficult problems. There are two concerns with organizing a competition in this way. First, the Russian system is much more expensive, whereas the US’s multiple choice tests can be inexpensively checked by a computer. Second, the Russian system is prone to unfairness. You need many math teachers to check all these papers on the highest level. Some of these teachers might not be fully qualified, and it is difficult to ensure uniform checking. This system can’t easily be adopted in the US. I am surprised I haven’t heard of lawsuits challenging USAMO results, but if we were to start having proofs at the AMC level with several hundred thousand participants, we would get into lots of trouble.

An interesting compromise was introduced at the Streamline Olympiad. The problems were multiple choice, but students were also requested to write proofs. Students got two points for a correct multiple choice answer, and if the choice was correct the proof was checked. Students could get up to three points for a correct proof. This idea solves two issues. The writing of proofs is rewarded at an early stage and the work of the judges is not as overwhelming as it would have been, had they needed to check every proof. However, there is one problem that I discussed in previous posts that this method doesn’t solve: with multiple choice, minor mistakes cost you the whole problem, even though you might have been very close to a solution. If we want to reward thinking more than accuracy, the proof system allows us to give credit for partial solutions.

I can suggest another approach. If the Russians require proofs for all problems and the Americans don’t require proofs for any problem, why not compromise and require a proof for one problem out of the set.

But I actually have a bigger idea in mind. I think that current development in artificial intelligence may soon help us to check the proofs with the aid of a computer. Artificial intelligence is still far from ready to validate that a mathematical text a human has produced constitutes a proof. But in this particular case, we have two things working for us. First, we can use humans and computers together. Second, we do not need to check the validity of any random proof; we need to check the validity of a specific proof of a simple problem that we know in advance, thus allowing us to prepare the computers.

Let us assume that we already can convert student handwriting into computer-legible text or that students write directly in LaTeX.

Here is the plan. Suppose for every problem, we create a database of some sample right, wrong and partial solutions with corresponding scores. The computer checks the students’ solutions against the given sample. Hopefully, the computer can recognize small typos and deviations that shouldn’t change the point value. If the computer encounters a solution that is significantly different from the ones in the sample, it sends the solution to human judges. Humans decide how to score the solution and the solution and its score is added to the sample database.

For this system to work, computers should be smart enough not to send too many solutions to humans. So how many is too many? My estimate is based on the idea that we wouldn’t want the budget of AMC to go too much higher than the USAMO budget. Since USAMO has 500 participants, judges check just a few hundred solutions to any particular problem. With several hundred thousand participants in AMC, the computer would have to be able to cluster all the solutions into not more than a few hundred groups. The judges only have to check one solution in each group.

As a bonus, we can create a system where for a given solution that is not in the database, the computer finds the closest solution and highlights the difference, thus simplifying the human’s job.

In order to improve math education, we need to add proofs when teaching math. My idea might also work for SATs and for other tests.

Now that there is more money available for education research, would anyone like to explore this?

A stick has two ends. If you cut off one end, how many ends will the stick have left?

This pre-kindergarten math problem was given to me by Maxim Kazarian who lives in Moscow, Russia. That got me thinking about math education in the US. Actually, just about anything can get me thinking about education in our country. One of our math education patterns is to provide simplified templates and to train kids to plug numbers into them without thinking.

Math education should be about thinking. We need to give kids a lot of math problems that do not fit into standard templates, in order to encourage creative thinking. Here is another puzzle from Maxim:

A square has four corners. If we cut one corner off, how many corners will the remaining figure have?

I invite my readers to invent additional problems that sound as if a subtraction by one is needed, when, in fact, it is not. Here is my contribution:

Anna had two sons. One son grew up and moved away. How many sons does Anna have now?

To get to the national swimming championship, you need to win the state running championships.

What? Is that a joke? Perhaps you’re having the same reaction. Because this is exactly what is happening with math competitions. The official USA math competition has three rounds: AMC, AIME and USAMO.

AMC is a multiple-choice competition with 25 problems in 75 minutes. To be good at it, you need speed, accuracy and the ability to guess well.

AIME is 3 hours long and has 15 problems. The problems are a different level of difficulty and guessing will not help you. Though AIME is also multiple-choice, unlike AMC where you choose out of 5, in AIME you choose out of 1,000. But you still need speed and accuracy. A small arithmetic mistake will cost you the whole problem.

USAMO is a competition that runs for 9 hours and has 6 problems. The problems are much harder and you have to write proofs. Proofs? What proofs? Where are the proofs coming from? It is like you got to the national swimming championship because you are a great runner, but you do not know how to swim.

As the result of this system of selection, the USA team at the International Math Olympiad has diverse skills: these kids are good at all three types of the math competitions. It is like taking an Olympic triathlon team to the Olympic swimming event.

However, the US loses by selecting in this way. There are many kids who are great mathematicians: they may be good at difficult problems but not that good at speed-racing problems. An arithmetic mistake costs you only one point at IMO, but a whole problem at AIME. There are kids who are deep mathematicians prone to small arithmetic mistakes. They could get a gold medal at IMO, but they can’t pass AMC or AIME.

Not only that. As many AMC problems are boring and do not require ideas, AMC might discourage kids from all math competitions at an early stage.

I will write later with my ideas about how to change AMC. Right now I would like to offer a solution to a smaller problem. I am sure that the US math team organizers know many cases in which a non-senior kid does great at USAMO and is potentially a team member for the next year’s US IMO team, but, oops, next year he can’t pass AMC.

I suggest the following: USAMO participants are allowed to go to next year’s AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year’s USAMO no matter what their AIME results are. This way kids who have proved that they are great swimmers do not need to compete in running again.

I have forgotten all the problems from the math Olympiads I participated in, except for one. That problem cost me one point, and as result I didn’t get a perfect score. Here is Problem Number 4 from the 1976 IMO:

Determine the greatest number that is the product of some positive integers, and the sum of these integers is 1976.

The solution goes like this. Consider a partition of 1976. If there is an integer x in the partition greater than 5, then replacing it with two integers x-2 and 2 gives a bigger product. Replacing 4 in the partition by 2 and 2 doesn’t change the product. On the other hand, if there is 1 in the partition, it is profitable to attach it to any other number: to replace 1 and x by x+1.

That means the maximum-product partition of a number greater than one has to consist only of twos and threes. If we have three twos in the partition, we can replace them by two threes, thus increasing the product. That means that our partition should consist of twos and threes, and the number of twos shouldn’t be greater than three.

I lost my point when I made an elementary arithmetic mistake while calculating the remainder of 1976 modulo 3.

Let’s generalize this problem for any number n. If we define the maximum product of partitions a(n) as a function of the number n, we get the sequence A000792 in the Encyclopedia of Integer Sequences. This sequence has other interesting definitions, which appear if we add some structure to partitions of n.

For example, we can suppose that our n is not just a number, but it also represents n objects that are being permuted by the symmetric group S_n. In this case, we can associate some Abelian subgroups of the symmetric group with every partition. Namely, we can divide the set of n elements into disjoint subsets of sizes corresponding to our partition and cycle the elements in every subset. The product of these cycles is an Abelian subgroup, the order of which doesn’t depend on how exactly we partition or how we cycle. This order is the product of the partition numbers. It appears that we can get all maximal Abelian subgroups of the symmetric group in this manner. Thus the sequence a(n) is the maximum size of the maximal Abelian subgroup in a symmetric group S_n.

We can do something else with our n objects. Suppose they are represented by vertices of a graph. We take any partition of n and divide our graph into groups of vertices according to this partition. Next we connect vertices of the graph that belong to different groups. If we take a vertex from every group, then the corresponding subgraph is a clique. We can see that it is a maximal clique as two vertices from the same group are never connected and we can’t add any more vertices. The number of different cliques of this size is the product of the number of elements in every group. We can prove that a(n) is the maximum possible number of maximal cliques in a graph with n vertices.

Even though I can’t return to 1976 to correct my stupid mistake, I love this problem and the corresponding sequence.

A new study is out. Dr. Satoshi Kanazawa blogged about his new discovery. He claims that he can explain why many studies show that men have higher intelligence than women. In his posting Why men are more intelligent than women he posits that intelligence is correlated with height. Taller men on average are smarter than shorter men.

He also claims that given the same height, women are more intelligent than men. As he puts it: “Women who are 5’10” are on average more intelligent than men who are 5’10”.” According to him the only problem women have in the brains department is that they aren’t tall enough.

I couldn’t find the study itself, only his own announcement. My problem with his study is that, according to Dr. Kanazawa’s description, he used data from the National Longitudinal Study of Adolescent Health. If we are talking about adolescents we need to take into account that children grow taller and smarter with age. So any difference in intelligence among the kids of different heights may easily be explained by age. Taller boys may be older and therefore likely to perform better on intelligence tests.

As to intelligence differences between boys and girls of the same height, these girls who are 5’10” may be from higher grades than boys who are 5’10”.

Some IQ tests are designed to take age into account, but what’s totally weird is that a scientist is studying the correlation of intelligence and height, using a population that is in the middle of a physical and mental growth spurt.

As a mathematician, I’ve been around many people of great intelligence, but I’ve never perceived bright people as especially tall.

Here is the picture of the USAMO winners for 2007. These are certainly extremely smart kids and I know their heights. The sixth student on the left is my son, Sergei. I personally met all these kids and compared to their peers they are not — except for one boy — especially tall.

How many disjoint letters “O” can you draw on a plane? Clearly, the answer is infinitely many. Now the real question is whether this number can be uncountable. We assume that the letter “O” is just a circle. In this case, if we draw all possible concentric circles, we get an uncountable number of circles. I call letters for which you can build a disjoint uncountable set on the plane stackable letters.

If we do not allow one letter “O” to be drawn inside another, then we can prove that it is not possible to draw more than a countable number of letters “O”. Indeed, we can always find a special point inside each letter “O” with both coordinates being rational numbers. As circles do not overlap, no two circles can share the same special point. The set of ordered pairs of rational numbers is countable, thus this set of letters “O” has to be countable.

Now let us move to a more interesting question: how many disjoint letters “T” can you draw on a plane? Let us define the letter “T” as two perpendicular line segments, where the end point of one of the segments is the middle point of the other. Prove that the letter “T” is unstackable.

Can you analyze the whole alphabet for stackability? It doesn’t matter which alphabet — you just need to define the drawing of every letter in some reasonable manner.

My real question is, can you define the drawings of English letters in such a way that they are recognizable and the number of stackable letters is maximized? What is the biggest number of letters that you can make stackable?

I brought up this subject to my son Sergei one fine morning. As a result we came up with the following breakfast theorem: Stackable letters are topologically equivalent to a line segment or a circle.

And here is our proof. If the letter in question has a part which is topologically equivalent to “T”, then the letter is not stackable, because “T” is not stackable.

Fortunately, or unfortunately, this is not the whole story. The stackability of each letter depends on how you define its shape. There are two groups of letters in the English alphabet that you can make stackable. The first group consists of letters like “V” that can be described as a function on a segment and stacked as a parallel translation of the same letter. The second group, like letter “O”, can be described as a function of an angle in polar coordinates, and stacked as a dilation of the same letter.

It is clear that this doesn’t cover all cases. For example, a piece of a spiral can be stackable — you stack it by rotating it.

I invite my readers to define two different homeomorphic shapes for the same letter — stackable and unstackable.

I just watched the film Mackenna’s Gold for the first time. I would have liked it, if not for the sunrise. It is a treasure-hunting movie with gold lying around in a hidden canyon. The secret entrance to the canyon is revealed by the shadow of a big rock tower at the moment of sunrise.

The problem with this clue is that the azimuth of the sunrise differs depending on the day of the year. For example, at the latitude of Cairo, Tel Aviv or Los Angeles, the sunrise azimuth changes from 116 degrees to 62 degrees as the seasons change. This is more than a 40 degree span!

In any map where directions include a long shadow at sunrise, it should include the day of the year. To be more precise, the trajectory — of a given spot on Earth — around the Sun is not periodic. That means the Sun never rises at the exact same azimuth twice. If we are talking about the approximate azimuth, then the Sun rises in the same region two time periods per year. These periods are mirror reflections of each other, with the winter solstice as their focal point. That means MacKenna and his companions needed to wait for several months for the right moment.

I wonder whether the filmmakers knew about this mistake. I see two possibilities.

The first one is that no one noticed. This is very bad. It confirms that this country is illiterate. In fact, it is worse than that, since you don’t really need an education to know that sunrise and sunset happen at different points in the sky depending on the time of year. You just need to look out your window from time to time and pay attention.

The second possibility is that someone noticed, but the filmmakers decided to proceed with it anyway, on the assumption that the audience is too dumb to detect the error. Although there are Internet discussions of goofs in this movie, I was disappointed that I didn’t find any complaints about this mistake.

George Lucas borrowed many ideas for the Indiana Jones movies from Mackenna’s Gold. In particular, for Raiders of the Lost Ark he borrowed the sunrise mistake without the sunrise itself. In the movie the Sun is supposed to penetrate the crystal in the headpiece of the Staff of Ra and reveal the location of the Well of Souls. To do this, the Staff must be placed in a specific spot in the map room at a certain time of day.

The fact that the poetic idea of a sunrise was removed probably means that they knew that it was a mistake. It was replaced by a non-poetic certain time of day. But they kept the mistake. The Sun doesn’t repeat its trajectory every day. Like MacKenna, Indiana Jones would have needed to wait in the map room for several months for the Sun to be in the right spot. George Lucas opted for a fast-paced movie that disregards the laws of physics. The mystery of this movie for me is: Does anyone care?