Tanya Khovanova's Math Blog

I recently published the following coin puzzle:

There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

My readers, David Reynolds and ext_1973756, wrote to me that I am missing a coin of 4 grams. Indeed, the same puzzle with five coins—1, 2, 3, 4, and 5—is a more natural and a better puzzle.

David Reynolds also suggested to go all the way up to 9 coins:

There are nine silver coins marked 1, 2, 3, 4, 5, 6, 7, 8, and 9. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

It is impossible to resolve this situation with more than nine coins as two weighings provide nine different answers to differentiate coins. But indeed it is possible to solve this problem for nine coins. It is even possible to suggest a non-adaptive algorithm, that is to describe the weighings before knowing the results.

To find such a strategy we need to satisfy two conditions. First, we have to weigh groups of coins of the same supposed weight, otherwise we do not get any useful information. Second, there shouldn’t be any two coins together (in or out of the pan) in both weighings, because it would then be impossible to differentiate between them.

Here is one possible solution of the problem:

• The first weighing: 1, 5, and 9, against 2, 6, and 7
• The second weighing: 1, 6, and 8, against 2, 4, and 9

David Reynolds also suggested a problem in which we do not know whether the fake coin is heavier or lighter:

There are four silver coins marked 1, 2, 3, and 4. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

Again, four coins is the best we can do when in addition to find it, we also want to determine if it is heavier or lighter. Indeed, if there were five coins we would have needed to cover ten different answers, which is too many for two weighings.

Here is the solution for four coins:

The two weighings are 1+3=4, and 1+2=3. If the first weighing balances, then the fake coin is 2 and the second weighing shows if it is heavier or lighter than it should be. Similarly, if the second weighing balances, then the fake coin is four and we can see whether it is heavier or lighter than it should be. If the left pan is lighter/heavier for both weighings, then the fake coin is 1 and is lighter/heavier. But if one pan is heavier on the first of two weighings and the other pan is heavier on the second weighing, then the fake coin is 3. In both cases it is easy to determine whether the fake coin is heavier or lighter.

Now David is missing a coin. If we just want to find the fake coin without determining whether it is heavier of lighter, we can do it with five coins:

There are five silver coins marked 1, 2, 3, 4, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is either lighter or heavier than it should be. Find the fake coin using the balance scale twice.

We can use the same solution as the previous (four coins) problem. If the scale balances both times, then the fake coin is 5. However, in this case we will not know whether the coin is heavier or lighter.

We can’t extend this problem to beyond five coins. Suppose we have six coins. We can’t use more than three coins in the first weighing. This is because if the scale unbalances, we can’t resolve more than three coins in one remaining weighing. Suppose the first weighing balances; then we have at least three leftover coins we know nothing about and one of them is fake. These three coins should be separated for the next weighing. That means one of the coins needs to be on the left pan and one on the right pan. We can add real coins any way we need. But if the second weighing unbalances we do not know if the fake coin is on the left and lighter or on the right and heavier.

Two interesting research results about male homosexuality are intertwined. The first one shows that the probability of homosexuality in a man increases with the number of older brothers. That is, if a boy is the third son in a family, the probability of him being a homosexual is greater than the probability of a first son in a family being homosexual. The second research result shows that the probability of homosexuality increases with the number of children the mother has. So if a woman is fertile and has many children, the probability that each of her sons is a homosexual is greater than the probability that an only child is a homosexual.

Many people conclude from the first result that a woman undergoes hormonal or other changes while being pregnant with boys that influence the probability of future boys being homosexual. Looking at the second result, researchers conclude that homosexuality has a genetic component. Moreover, that component is tied up with the mother’s fecundity. The same genes are responsible for both the mother having many children and for her sons being homosexual. This assumption explains why homosexuality is not dying out in the evolution process.

In one of my previous essays I showed that the first results influences the second result. If each next son is homosexual with higher probability, then the more children a mother has the more probable it is that her sons are homosexuals. That means that the second result is a mathematical consequence of the first result. Therefore, the conclusion that the second result implies a genetic component might be wrong. The correlation between homosexuality and fecundity could be the consequence of hormonal changes.

Now let’s look at this from the opposite direction. I will show that the first result is the mathematical consequence of the second result: namely, if fertile women are more probable to give birth to homosexuals, then the probability that the second sons are is higher than the probability that the first sons are gay.

For simplicity let’s only consider mothers with one or two boys. Suppose the probability of a son of a one-son mother to be a homosexual is p₁. Suppose the probability of a son of a two-sons mother to be a homosexual is p₂. The data shows that p₂ is greater than p₁. What is the consequence? Suppose the number of mothers with one son is m₁ and the number of mothers with two sons is m₂. Then in the whole population the probability of a boy who is the first son to be gay is (p₁m₁+p₂m₂)/(m₁+m₂) and the probability of a boy who is the second son to be gay is p₂. It is easy to see that the first probability is smaller than the second one.

Let me create an extreme hypothetical example. Suppose mothers of one son always have straight sons, and mothers of two sons always have gay sons. Now consider a random boy in this hypothetical setting. If he’s the second son, he is always gay, while if he is the first son he is not always gay.

We can conclude that if the probability of having homosexual sons depends on fecundity, then the higher numbered children would be gay with higher probability than the first-born. This means that if the genetics argument is true and being a homosexual depends on the mother’s fecundity gene, then it would follow mathematically that the probability of homosexuality increases with birth order. The conclusion that homosexuality depends on hormonal changes might not be valid.

So what is first, chicken or egg? Is homosexuality caused by fecundity, while birth order correlation is just the consequence? Or vice versa? Is homosexuality caused by the birth order, while correlation with fecundity is just the consequence?

What do we do when the research results are so interdependent? To untangle them we need to look at the data more carefully. And that is easy to do.

To show that homosexuality depends on the order of birth independently of the mother’s fertility, we need to take all the families with two boys (or the same number of boys) and show that in such families the second child is more probable to be homosexual than the first child.

To show the dependence on fertility, without the influence of the birth order, we need to take all first-born sons and show that they are more probable to be homosexuals if their mothers have more children.

It would be really interesting to look at this data.

My new favorite hat puzzle was invented by Konstantin Knop and Alexander Shapovalov. It appeared (in a different wording) in March 2013 at the Tournament of the Towns:

A sultan decides to give 100 of his sages a test. The sages will stand in line, one behind the other, so that the last person in the line sees everyone else. The sultan has 101 hats, each of a different color, and the sages know all the colors. The sultan puts all but one of the hats on the sages. The sages can only see the colors of the hats on people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. They are not allowed to repeat a color that was already announced. Each person who guesses his color wrong will get his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

I loved it so much that I wrote a paper about it. You can find the solution there.

I particularly like these two problems that I gave my AMSA students for homework:

Athos, Porthos, and Aramis were rewarded with six coins: three gold and three silver. Each got two coins. Athos doesn’t know what kind of coins the others got, but he knows his own coins. Ask him one question to which he can answer “Yes,” “No,” or “I do not know,” so that you will be able to figure out his coins.

There are four silver coins marked 1, 2, 3, and 5. They are supposed to weigh the number of grams that is written on them. One of the coins is fake and is lighter than it should be. Find the fake coin using the balance scale twice.

We’ve all been hearing about parallel computing, and now it has turned up in a coin-weighing puzzle invented by Konstantin Knop.

“We have N indistinguishable coins. One of them is fake and it is not known whether it is heavier or lighter, but all genuine coins weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts a minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?”

This puzzle reminds me of another coin-weighing problem, where in a similar situation you need to find a fake coin by using one scale with four pans. The answer in this variation would be 5⁵ = 3125. We can divide coins in five groups with the same number of coins and put four groups on the scale. If one of the groups is different (heavier or lighter), then this group contains the fake coin. Otherwise, the leftover group contains the fake coin. This way each weighing reduces the pile with the fake coin by a factor of five. One scale with four pans gives you more information than two scales with two pans used in parallel. We can conclude that Knop’s puzzle should require at least the same number of weighings as the four-pan puzzle for the same number of coins. So we can expect the answer to Knop’s puzzle will not be bigger than 3125. But what will it be?

Once when I was working at Telcordia, I received a phone call from my doctor’s office. Here is how it went:

— Are you Tanya Khovanova?
— Yes.
— You should come here immediately and redo your blood test ASAP.
— What’s going on?
— Your blood count shows that you are dead.
— If I’m dead, then what’s the hurry?
Given that I wasn’t dead, the conclusion was that there had been a mistake in the test. If there had been a mistake, the probability that something was wrong after the test was the same as it was before the test. There was no hurry.

I started my Yellow Road plan on February 9 when I was 245.2 pounds.

I decided that my first target weight would be my actual weight on February 9: 245.2. Every day this target weight goes down by 0.1 pounds. I weigh myself every morning and compare my actual weight to my target weight. My actions depend on the difference.

My Yellow Zone is plus or minus one pound of my target weight. My Green Zone means I am doing even better: my weight is less than my target weight minus one pound. My Red Zone means that I am not doing so great: my weight is more than my target weight plus a pound.

If I am within the Yellow corridor, I continue building my healthy habits as I have been doing. If I am in the Green Zone, I can afford to digress from healthy habits and indulge myself a bit. If I am in the Red Zone, I have to reduce my evening meals to apples only, which I do not particularly like. The Red Zone has different shades: if I am one pound over my target weight I have to start my apple restrictions after 8:00 pm. If I am two pounds over, then after 6:00 pm, and so on.

Today I am ten pounds lighter. In the process, I have made these discoveries:

Writing down my weight daily is very important. When I look at yesterday’s number and today’s number, I start thinking about what caused the increase or decrease. Now I have more clarity about which foods are better for me.

I have a better picture of how much I should eat. One day I held a party and I didn’t eat much. In fact, I had only one small desert. I didn’t feel full and went to bed feeling proud of myself. The next morning I weighed myself and was surprised to find I had gained three pounds. The amount of food I should be eating is much smaller than I expected. I think it may be three times less than what I was used to eating. My plan might not be aggressive enough. Currently when I am in my lightest Red Zone, I have to eat just apples after 8:00 pm. I discovered that I can still gain weight with this regime.

A half-empty stomach is not such a bad feeling. I was so afraid of starting a plan where I might feel hungry. Now I discovered that there are several hours between my first signal that I should eat and real hunger. My first sense that I should eat something might not be actual hunger at all. I do experience a light feeling in my stomach, but now I am starting to learn to enjoy it.

The system works for me. That’s the bottom line. For the first time in my life, I found a way to lose weight. All my friends ask me about this system. I explain that this Yellow Road plan is not a panacea. My plan is based on many other things that I did before. If it continues working, I promise to discuss it further: to analyze what exactly works and why.

My next step is to adjust my plan in light of my discoveries. From now on, I’ll eat only apples after 8:00 pm—not as an exception, but as a rule.

Here is a problem from the 2012 Moscow Olympiad:

There were n people at a meeting. It appears that any two people at the meeting shared exactly two common acquaintances.

• Prove that all the people have exactly the same number of common acquaintances at this meeting.
• Show that n can be greater than 4.

My question is: Why 4? I can answer that myself. If in a group of four people any two people share exactly two common acquaintances, then all four people know each other. So in this Olympiad problem, the author wanted students to invent a more intricate example.

Let’s take this up a notch and work on a more difficult problem.

There were n people at a meeting. It appears that any k people at the meeting shared exactly k common acquaintances.

• Prove that all the people have exactly the same number of common acquaintances at this meeting.
• Is it possible that n can be greater than 2k?

I stumbled upon an article, Winners Live Longer, that says:

“When 524 nominees for the Nobel Prize were examined and compared to the actual winners from 1901 to 1950, the winners lived longer by 1.4 years. Why? It seems just having won and knowing you are on top gives you a boost of 1.8% to your life expectancy.”

This goes on top of the pile of Bad Conclusions From Statistics. With any kind of awards where people can be nominated several times, winners on average would live longer. The reason is that nominees who die early lose their chance to be nominated again and to win.

I wonder what would happen if we were to compare Fields medal nominees and winners. There is a cut off age of 40 for receiving a Fields medal. If we compare the life span of Fields medal winners and nominees who survived past 40, we might get a better picture of how winning affects life expectancy.

Living a long life increases your chances of getting a Nobel Prize, but doesn’t help you get a Fields medal.

I wish I could write four papers in three weeks. The title just means that I submitted four papers to the arXiv in the last three weeks—somehow, after the stress of doing my taxes ended, four of my papers converged to their final state very fast. Here are the papers with their abstracts:

• On k-visibility graphs (with Matthew Babbitt and Jesse Geneson). We examine several types of visibility graphs in which sightlines can pass through k objects. For k ≥ 1 we improve the upper bound on the maximum thickness of bar k-visibility graphs from 2k(9k−1) to 6k, and prove that the maximum thickness must be at least k+1. We also show that the maximum thickness of semi-bar k-visibility graphs is between the ceiling of 2(k+1)/3 and 2k. Moreover we bound the maximum thickness of rectangle k-visibility graphs. We also bound the maximum number of edges and the chromatic number of arc and circle k-visibility graphs. Furthermore we give a method for finding the number of edges in semi-bar k-visibility graphs based on skyscraper puzzles.
• Skyscraper Numbers (with Joel Brewster Lewis). We introduce numbers depending on three parameters which we call skyscraper numbers. We discuss properties of these numbers and their relationship with Stirling numbers of the first kind, and we also introduce a skyscraper sequence.
• Connected Components of Underlying Graphs of Halving Lines (with Dai Yang). In this paper we discuss the connected components of underlying graphs of halving lines’ configurations. We show how to create a configuration whose underlying graph is the union of two given underlying graphs. We also prove that every connected component of the underlying graph is itself an underlying graph.
• Efficient Calculation of Determinants of Symbolic Matrices with Many Variables (with Ziv Scully). Efficient matrix determinant calculations have been studied since the 19th century. Computers expand the range of determinants that are practically calculable to include matrices with symbolic entries. However, the fastest determinant algorithms for numerical matrices are often not the fastest for symbolic matrices with many variables. We compare the performance of two algorithms, fraction-free Gaussian elimination and minor expansion, on symbolic matrices with many variables. We show that, under a simplified theoretical model, minor expansion is faster in most situations. We then propose optimizations for minor expansion and demonstrate their effectiveness with empirical data.