Tanya Khovanova's Math Blog

I am reading the book Eat to Live by Joel Fuhrman. It contains a formula that as a math formula doesn’t make any sense. But as an idea, it felt like a revelation. Here it is:

HEALTH = NUTRIENTS/CALORIES

The idea is to choose foods that contain more nutrients per calorie. The formula doesn’t make sense for many reasons. Taken to its logical conclusion, the best foods would be vitamins and tea. The formula doesn’t provide bounds: it just emphasizes that your calories should be nutritious. However, too few calories — nutritious or not — and you will die. And too many calories — even super nutritious — are still too many calories. In addition the formula doesn’t explain how to balance different types of nutrients.

Let’s see why it was a revelation. I often crave bananas. I assumed that I need bananas for some reason and my body tells me that. Suppose I really need potassium. As a result I eat a banana, which contains 800 milligrams of potassium and adds 200 calories as a bonus. If I ate spinach instead, I would get the same amount of potassium at a price of only 35 calories.

The book suggests that if I start eating foods that are high in nutrients, I will satisfy my need for particular nutrients, and my cravings will subside. As a result I will not want to eat that much. If I start my day eating spinach, that might eliminate my banana desire.

I’ve been following an intuitive eating diet. I am trying to listen to my body hoping that my body will tell me what is better for it. It seems that my body sends me signals that are not precise enough. It’s not that my body isn’t communicating with me, but it is telling me “potassium” and all I hear is “bananas.” What I need to do is use my brain to help me decipher what my body really, really wants to tell me.

As Dr. Fuhrman puts it, we are a nation of overfed and malnourished people. But Fuhrman’s weight loss plan is too complicated and time-consuming for me, so I designed my own plan based on his ideas:

I will start every meal with vegetables, as they are the most nutritious. I hope that vegetables will provide the nutrients I need. That in turn will make me less hungry by the next meal, at which time I’ll take in fewer calories. I will report to my readers whether or not my plan works. I’m off to shop for spinach. Will I ever love it as much as bananas?

by Olivier Bernardi and Tanya Khovanova

The Cookie Monster is a peculiar creature that appeared in The Inquisitive Problem Solver (Vaderlind, Guy & Larson, MAA, P34). Presented with a set of cookie jars, the Cookie Monster will try to empty all the jars with the least number of moves, where a move is to select any subset of the jars and eat the same number of cookies from each jar in the subset.

Even an untalented Cookie Monster would be able to empty n jars in n moves: to fulfill this strategy the Monster can devour all the cookies of one jar at a time. If the Monster is lucky and some jars have the same number of cookies, the Monster can apply the same eating process to all these identical jars. For example, if all the jars have the same number of cookies, the Monster can gulp down all of them in one swoop.

Now, let us limit our discussion to only cases of n non-empty jars that contain distinct numbers of cookies. If indeed all the numbers are distinct, can the Monster finish eating faster than in n moves?

The answer depends on the actual number of cookies in each jar. For example, if the number of cookies in jars are different powers of 2, then even the most talented Monsters can’t finish faster than in n steps. Indeed, suppose the largest jar contains 2^N cookies. That would be more than the total number of cookies in all the other jars together. Therefore, any strategy has to include a step in which the Monster only takes cookies from the largest jar. The Monster will not jeopardize the strategy if it takes all the cookies from the largest jar in the first move. Applying the induction process, we see that we need at least n steps.

On the other hand, sometimes the Monster can finish the jars faster. If 2^k−1 jars contain respectively 1, 2, 3, …, 2^k−1 cookies, the Cookie Monster can empty them all in k steps. Here is how. For its first move, the Monster eats 2^k-1 cookies from each of the jars containing 2^k-1 cookies or more. What remains are 2^k-1−1 pairs of identical non-empty jars containing respectively 1, 2, 3, …, 2^k-1−1 cookies. The Monster can then continue eating cookies in a similar fashion, finishing in k steps. For instance, for k=3 the sequences of non-empty jars are: 1,2,3,4,5,6,7 → 1,1,2,2,3,3 → 1,1,1,1 → all empty.

Now we would like to prove a theorem that shows that the example above is the lowest limit of moves even for the most gifted Cookie Monsters.

Theorem. If n non-empty jars contain distinct numbers of cookies, the Cookie Monster will need at least ⌈log₂(n+1)⌉ steps to empty them all.

Proof. Suppose that n jars contain distinct numbers of cookies and let f(n) be the number of distinct non-empty jars after the first move of the Cookie Monster. We claim that n ≤ 2f(n)+1. Indeed, after the first move, there will be at least n − 1 non-empty jars, but there cannot be three identical non-empty jars. That means, the number of jars plus 1 can’t decrease faster than twice each time.

Now here is something our readers can play with. Suppose a sequence of numbers represents the number of cookies in the jars. Which sequences are interesting, that is, which can provide interesting solutions for the Cookie Monster problem?

I read an interesting article on the paradoxes involved in allocating seats for the Congress. The problem arises because of two rules: one congressperson has one vote, and the number of congresspeople per state should be proportional to the population of said state.

These two rules contradict each other, because it is unrealistic to expect to be able to equally divide the populations of different states. Therefore, two different congresspeople from two different states may represent different sizes of population.

Let me explain how seats are divided by using as an example a country with three states: New Nevada (NN), Massecticut (MC) and Califivenia (C5). Suppose the total number of congresspeople is ten. Also suppose the population distribution is such that the states should have the following number of congresspeople: NN — 3.33, MC — 3.34 and C5 — 3.33. As you know states generally do not send a third of a congressperson, so the situation is resolved using the Hamilton method. First, each state gets an integer portion of the seats. In my example, each state gets three seats. Next, if there are seats left they are allocated to states with the largest remainders. In my example, the remainders are 0.33, 0.34 and 0.33. As Massecticut has the largest reminder it gets the last seat.

This is not fair, because now each NN seat represents a larger population portion than each MC seat. Not only is this not fair, but it can also create some strange situations. Suppose there have been population changes for the next redistricting: NN — 3.0, MC — 3.4 and C5 — 3.6. In this case, NN and MC each get 3 seats, while C5 gets the extra seat for a total of 4. Even though MC tried very hard and succeeded in raising their portion of the population, they still lost a seat.

Is there any fair way to allocate seats? George Szpiro in his article suggests adding fractional congresspersons to the House of Representatives. So one state might have three representatives, but one of those has only a quarter of a vote. Thus, the state’s voting power becomes 2 1/4.

We can take this idea further. We can use the Hamilton method to decide the number of representatives per state, but give each congressperson a fractional voting power, so the voting power of each state exactly matches the population. This way we lose one of the rules that each congressperson has the same vote. But representation will be exact. In my first example, NN got three seats, when they really needed 3.33. So each congressperson from New Nevada will have 1.11 votes. On the other hand MC got four seats, when they needed 3.34. So each MC representative gets 0.835 votes.

Continuing with this idea, we do not need congresspeople from the same state to have the same power. We can give proportional voting power to a congressperson depending on the population in his/her district.

Or we can go all the way with this idea and lose the districts altogether, so that every congressperson’s voting power will be exactly proportionate to the number of citizens who voted for him/her. This way the voting power will reflect the popularity — rather than the size of the district — of each congressperson.

Lionel Levine invented a new hat puzzle.

The sultan decides to torture his hundred wise men again. He has an unlimited supply of red and blue hats. Tomorrow he will pile an infinite, randomly-colored sequence of hats on each wise man’s head. Each wise man will be able to see the colors of everyone else’s hats, but will not be able to see the colors of his own hats. The wise men are not allowed to pass any information to each other.
At the sultan’s signal each has to write a natural number. The sultan will then check the color of the hat that corresponds to that number in the pile of hats. For example, if the wise man writes down “four,” the sultan will check the color of the fourth hat in that man’s pile. If any of the numbers correspond to a red hat, all the wise men will have their heads chopped off along with their hats. The numbers must correspond to blue hats. What should be their strategy to maximize their chance of survival?

Suppose each wise man writes “one.” The first hat in each pile is blue with a probability of one-half. Hence, they will survive as a group with a probability of 1 over 2¹⁰⁰. Wise men are so wise that they can do much better than that. Can you figure it out?

Inspired by Lionel, I decided to suggest the following variation:

This time the sultan puts two hats randomly on each wise man’s head. Each wise man will see the colors of other people’s hats, but not the colors of his own. The men are not allowed to pass any info to each other. At the sultan’s signal each has to write the number of blue hats on his head. If they are all correct, all of them survive. If at least one of them is wrong, all of them die. What should be their strategy to maximize their chance of survival?

Suppose there is only one wise man. It is clear that he should write that he has exactly one blue hat. He survives with the probability of one-half. Suppose now that there are two wise men. Each of them can write “one.” With this strategy, they will survive with a probability of 1/4. Can they do better than that? What can you suggest if, instead of two, there is any number of wise men?

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Today I saw an ad — “A printer for sale” — handwritten. Hmm.

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What do you call a motherboard on your spouse’s computer?
The motherboard-in-law.

In a puzzle book by Mari Berrondo (in Russian), I found the following logic problem:

Alfred, Bertran and Charles are asked about their profession. One of them always lies; another one always tells the truth; and the third one [who I will refer to as a “half-liar”] sometimes lies and sometime tells the truth. Here are their answers:

Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?

The solution in the book is the following. As Alfred is right about what Bertran would say, Alfred can’t be a liar. If Alfred is a half-liar then the other two people would give the opposite statements, since one will be a truth-teller and the other a liar. But they both say that Alfred is a piano-tuner, therefore Alfred must be a truth-teller. Hence, Alfred’s statement about everyone’s profession must be the truth. Now we know that Charles is an insurance agent. As Charles confirms that, thus telling the truth in this instance, we recognize that he must be a half-liar. The answer to the problem is that the half-liar is an insurance agent.

But I have a problem with this problem. You see, a liar can say many things. He can say that he is a conductor, a mathematician, a beekeeper or whatever. So there is no way of knowing what a person who decides to lie can say. Let’s just analyze the statement by Alfred: “Concerning Bertran, if you ask him, he will tell you that he is a painter.”

If Alfred tells the truth about what Bertran would say, he needs to know for sure that Bertran will say that he is a painter. Hence, Bertran must be a truth-teller and a painter. If Alfred lies, he needs to be sure that Bertran won’t say that he is a painter. So Bertran must be either a truth-teller and not a painter, or a liar and a painter. Bertran can’t be a half-liar, because a half-liar can say that he is a painter as well as he can say something else, no matter what his real profession.

There is one interesting aspect of this that many people overlook. There are different types of people who are half-liars. In some books half-liars are introduced as people who, before making a statement, flip a coin to decide whether to lie or to tell the truth. Such a person needs to know in advance exactly what other people are saying, in order to construct a statement about what those people might say that corresponds to the coin flip. On the other hand, other types of half-liars exist. One half-liar can say something and then see later whether it is true. If Alfred is a half-liar who doesn’t care in advance about the truth of his statement, he can say that Bertran will claim that he is a painter.

I leave it to my readers to finish my analysis and see that the problem doesn’t have a solution. To end my essay on a positive note, I decided to slightly change the problem, so that there is no contradiction. In the same setting:

Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is not a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?

The data from annual surveys carried out by the American Mathematical Society shows the same picture year after year: the percentage of females in different categories decreases as the category level rises. For example, here is the data for 2006:

The high percentage of female math majors means that a lot of women do like mathematics. Why aren’t women becoming professors of mathematics? In the picture to the left, little Sanya fearlessly took her first integral. I hope, even as an adult, she will never be afraid of integrals.

I am one of the organizers of the Women and Mathematics Program at the Institute for Advanced Study at Princeton In 2009 we had a special seminar devoted to discussing this issue. Here is the report of our discussion based on the notes that Rajaa Al Talli took during the meeting.

Many of us felt, for the following three reasons, that the data doesn’t represent the full picture.

First, the different stages correspond to women of different ages; thus, the number of tenured faculty should be compared, not to the number of current math majors, but rather to women who majored in math many years ago. The percentage of female PhDs in mathematics has been increasing steadily for the past several years. As a result, we expect an eventual increase in the number of full-time female faculty.

Second, international women mathematicians might be having a great impact on the numbers. Let’s examine a hypothetical situation. If many female professors come to the US after completing their studies in other countries, it would be logical to assume that they would raise the numbers. But since the numbers are falling, we might be losing more females than we think. Or, it could be the opposite: international graduate students complete a PhD in mathematics in the USA and then go back to their own countries. In this case we would be losing fewer females to professorships than the numbers seem to suggest. Unfortunately, we can’t really say which case is true as we do not know the data on international students and professors.

Third, many women who major in mathematics also have second majors. For example, the women who have a second major in education probably plan to become teachers instead of pursuing an academic career. It would be interesting to find the data comparing women who never meant to have careers in science with those women who left because they were discouraged. If we are losing women from the sciences because they decide not to pursue scientific careers, then at least that is their choice.

It is also worth studying why so few women are interested in careers in mathematics in the first place. Changing our culture or applying peer pressure in a different direction might change the ambitions of a lot of people.

We discussed why the data in the table doesn’t represent the full picture. On the other hand, there are many reasons why women who can do mathematics and want to do mathematics might be discouraged from pursuing an academic career:

• Marriage and children distract from mathematics.
• The lack of legal protections for pregnant women, of required maternity leave and of childcare provision.
• The cultural skepticism that women can do math on a high level.
• An educational system that tends to tell students that math is very difficult, thus discouraging women from the early stages of their academic life.
• Boys tend to be more competitive than girls.
• The lack of job opportunities.
• A career in math often requires moving.

Our group proposed many solutions to help retain women in mathematics:

1. Find a way to get men pregnant as well.
2. Incorporate ideas from other countries (like Portugal), where they don’t have this problem.
3. Increase the level of social care for pregnant women and young children.
4. Create new laws to protect the rights of pregnant women.
5. Educate secondary, high school and college math teachers how to present math — such as through games — as an interesting subject, not as a difficult one.

At the end of our meeting, everyone accepted Ingrid Daubechies‘ proposal that we do the following:

Each woman in mathematics should take as her responsibility the improvement of the mathematical environment in which she works. If every woman helps change what’s going on in her university or the school where she teaches, that will help solve the problem on the larger scale.

The book How to Drive Your Man Wild in Bed by Graham Masterton has a chapter on how to choose a lover. It highlights red flags for men who need to be approached with caution. There is a whole list of potentially bad signs, including neglecting to shower in the previous week and talking only about himself.

The list of bad features also includes professions to avoid. Can you guess the first profession on the list? OK, I think you should be able to meta-guess given the fact that I am writing about it. Indeed, the list on page 64 starts:

Avoid, on the whole, mathematicians…

I am an expert on NOT avoiding mathematicians: in fact, I’ve married three of them and dated x number of them. That isn’t necessarily because I like mathematicians so much; I just do not meet anyone else.

When I was a student I had a theory that mathematicians are different from physicists. My theory was based on two conferences on mathematical physics I attended in a row. The first one was targeted for mathematicians and the second for physicists. The first one was very quiet, and the second one was all boozing and partying. So I decided that mathematicians are introverts and physicists are extroverts. I was sure then that my second husband chose a wrong field, because he liked booze and parties.

By now, years later, I’ve met many more mathematicians, and I have to tell you that they are varied. It is impossible and unfair to describe mathematicians as a type. One mathematician even became the star of an erotic movie. I write this essay for girls who are interested in dating mathematicians. I am not talking about math majors here, I am talking about mathematicians who do serious research. Do I have a word of advice?

I do have several words of caution. While they don’t apply to all mathematicians, it’s worth keeping them in mind.

First, there are many mathematicians who, like my first husband, are very devoted to mathematics. I admire that devotion, but it means that they plan to do mathematics on Saturday nights and prefer to spend vacation at their desks. If they can only fit in one music concert per year, it is not enough for me. Of course, this applies to anyone who is obsessed by his work.

Second, there are mathematicians who believe that they are very smart. Smarter than many other people. They expand their credibility in math to other fields. They start going into biology, politics and relationships with the charisma of an expert, when in fact they do not have a clue what they are talking about.

Third, there are mathematicians who enjoy their math world so much that they do not see much else around them. The jokes are made about this type of mathematician:

What is the difference between an extroverted mathematician and an introverted one? The extroverted one looks at your shoes, rather than at his own shoes.

Yes, I have met a lot of mathematicians like that. Do you think that their wives complain that their husbands do not notice their new haircuts? No. Such triviality is not worth mentioning. Their wives complain that their husbands didn’t notice that the furniture was repossessed or that their old cat died and was replaced by a dog. My third husband was like that. At some point in my marriage I discovered that he didn’t know the color of my eyes. He didn’t know the color of his eyes either. He wasn’t color-blind: he was just indifferent. I asked him as a personal favor to learn the color of my eyes by heart and he did. My friend Irene even suggested creating a support group for the wives of such mathematicians.

While you need to watch out for those traits, there are also things I like about mathematicians. Many mathematicians are indeed very smart. That means it is interesting to talk to them. Also, I like when people are driven by something, for it shows a capacity for passion.

Mathematicians are often open and direct. Many mathematicians, like me, have trouble making false statements. I stopped playing —Mafia— because of that. I prefer people who say what they think and do not hold back.

There is a certain innocence among some mathematicians, and that reminds me of the words of the Mozart character in Pushkin’s poetic drama, Mozart and Salieri: —And genius and villainy are two things incompatible, aren’t they?— I feel this relates to mathematicians as well. Many mathematicians are so busy understanding mathematics, they are not interested in plotting and playing games.

Would I ever date a mathematician again? Yes, I would.

33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?

The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.

Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.

First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.

As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.

Let’s look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.

Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.

We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.

Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.

If we continue this, we’ll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.

We get two interesting new sequences out of this. The sequence of the fastest rider’s speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.

The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.

I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.

One day we may all face the necessity of hiring a lawyer. If the case is tricky the lawyer must be smart and inventive. I am collecting puzzles to give to a potential lawyer during an interview. The following puzzle is one of them. It was given at the second Euler Olympiad in Russia:

At a local Toyota dealership, you are allowed to exchange brand new cars. You can exchange three Camrys for one Prius and one Avalon, and three Priuses for two Camrys and one Avalon. Assuming an unlimited supply of cars at the dealership, can collector Vasya exchange 700 Camrys for 400 Avalons?

The beauty of this puzzle is that the answer I may find acceptable from a mathematician is not the same as I want from my future lawyer.

Have I intrigued you? Get to work and send me your solutions.