Tanya Khovanova's Math Blog

I recently posted the following puzzle:

Puzzle. We have 3²ⁿ identical-looking coins. One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Here is my son Sergei’s solution. Divide the coins into nine groups of equal size and number the groups in ternary: 00, 01, 02, 10, 11, 12, 20, 21, and 22. On each scale we put three groups versus three groups. On the first scale we compare the three groups that start with 1 with the three groups that start with 2. For the second scale we do the same using the last digit instead of the first one, and for the third scale we use the sum of two digits modulo 3. Any pair of scales, if they are assumed to be normal, would point to one out of nine groups as the group containing the fake coin.

If all three pairs of scales agree on one group, then this is the group containing the fake coin. Thus in three weighings, we reduce the number of groups of coins by a factor of nine. If the pairs of scales do not agree, then the random scale produced a wrong weighing and thus can be found out. How do we do that? We have three out of nine groups of coins each of which might contain the fake coin. We compare two of the groups on all three scales. This way we know exactly which group contains the fake coin and, consequently, which scale generated a wrong weighing. If we know the random scale, we can speed up the rest of the process of finding the fake coin. Thus in the worst case we require 3n+3 weighings.

The big idea here is that as soon as the random scale shows a wrong weighing result it can be found out. So in the worst case, the random scale behaves as a normal scale and messes things up at the very end. Sergei’s solution can be improved to 3n+1 weighings. Can you do that?

The improved solution is written in a paper Взвешивания на «хитрых весах» (in Russian) by Konstantin Knop, that is published in Математика в школе 2009-2. The paper contains an even stronger solution that provides a better asymptotics.

The first time I visited the US was in 1990 at the invitation of an old friend, Joseph Bernstein. After my arrival Joseph proposed and I accepted, but my essay is not about that.

Joseph reintroduced me to his daughter, Mira, who was then in her late teens. I was struck by Mira’s charm. I had never before met teenagers like her. Of course, Joseph got points for that as I was hoping to have a child with him. When I moved to the US I met some other kids who were also incredibly charming. It was too late to take points away from Joseph, but it made me realize what a huge difference there was between Soviet and American teenagers. American teenagers were happier, more relaxed, better mannered, and less cynical than Soviet ones.

My oldest son, Alexey, was born in the USSR and moved to the US when he was eight. One unremarkable day when he was in middle school (Baker public school in Brookline), the principle invited me for a chat. I came to the school very worried. The principal explained to me that there was a kid who was bugging Alexey and Alexey pushed him back with a pencil. While the principal proceeded to explain the dangers of a pencil, I tuned out. I needed all my energy to conceal my happy smile. This was one of the happiest moments of my life in the US. What a great country I live in where the biggest worry of a principal in a middle school is the waving of a pencil! I remembered Alexey’s prestigious school in Moscow. They had fights every day that resulted in bloody noses and lost teeth. When I complained to his Russian teacher, she told me that it was not her job to supervise children during big breaks. Plus the children needed to learn to be tough. No wonder American children are happier.

I was wondering if there were any advantages to a Soviet upbringing. For one thing, Soviet kids grow up earlier and are less naive. They are more prepared for harsh realities than those American kids who are privileged.

Naive children grow up into naive adults. Naive adults become naive presidents. I watched with pain as naive Bush (“I looked the man in the eye. I found him to be very straightforward and trustworthy.”) and naive Obama (Russian reset) misunderstood and underestimated Putin.

Putin is (and, according to Forbes Magazine, has been for the last four years) the most powerful person in the world. Even though the US kept its distance from Russia, he was able to manipulate us from afar. Now that Trump wants to be close to Putin, the manipulation will be even easier. Putin is better at this game. He will win and we will lose.

May 5 of 1955 can be written as 5/5/55. How many times during the 20th century the date in the format month day and the last two digits of the year can be written with the same digit?

I had a distant relative Alla, who was brought up by a single mother, who died in a car crash when the girl was in her early teens. Alla was becoming a sweet and pleasant teenager; she was taken in by her aunt after the accident. Very soon the aunt started complaining that Alla was turning into a cheater and a thief. The aunt found a therapist for Alla, who explained that Alla was stealing for a reason. Because the world had unfairly stolen her mother, Alla felt entitled to compensation in the form of jewelry, money, and other luxuries.

I was reminded of Alla’s story when I was reading The (Honest) Truth About Dishonesty: How We Lie to Everyone—Especially Ourselves by Dan Ariely. Ariely discusses a wide range of reasons why honest people cheat. But to me he neglects to look at the most prominent reason. Often honest people cheat when they feel justified and entitled to do so.

One of Ariely’s experiments went like this. One group was asked to write a text avoiding letters x and z. The other group was asked to write a text avoiding letters a and n. The second task is way more difficult and requires more energy. After the tasks were completed the participants were given a test in which they had a chance to cheat. For this experiment, the participants were compensated financially according to the number of questions they solved. Not surprisingly, the second group cheated more. The book concludes that when people are tired, their guard goes down and they cheat more. I do not argue with this conclusion, but I think another reason also contributes to cheating. Have you ever tried to write a text without using the letters a and n? I did:

I should try it here. But this is so difficult. I give up.

My son, Alexey, was way better than me:

First, God brought forth the sky with the world. The world existed without form. Gloom covered the deep. The Spirit of God hovered over the fluids. Quoth God: let there be light. Thus light existed.

Fun as it is, this is cruel and unusual punishment. The request is more difficult than most people expect at an experiment. It could be that participants cheated not only because their capacity for honesty was depleted, but because they felt entitled to more money because the challenge was so difficult.

In another experiment, the participants received a high-fashion brand of sunglasses before the test. Some of them were told that the sunglasses were a cheap imitation of the luxury brand (when they really were not). This group cheated more than the group who thought that they got a real thing. The book concludes that wearing fake sunglasses makes people feel that they themselves are fake and so they care less about their honor. Unfortunately, the book doesn’t explain in detail what was actually promised. It looks like the participants were promised high-fashion sunglasses. In this case, the fake group would have felt deceived and might have felt more justified to cheat.

Dear Dan Ariely: May I suggest the following experiment. Invite people and promise them some money for a 15-minute task. Pay them the promised minimum and give them a test through which they can earn more. Construct it so that they can earn a lot more if they cheat. Then make the non-control group wait for half an hour. If I were in this group, I would have felt that I am owed for the total of 45 minutes—three times more than what I was promised. I do not know if I would cheat or just leave, but I wouldn’t be surprised that in this group people would cheat more than in the control group.

If like me, you fancy Raymond Smullyan and his books, then you’ve heard about knights and knaves. Knights always tell the truth and knaves always lie. In addition to knights and knaves, there are normal people who sometimes tell the truth and sometimes lie. Here is a puzzle.

Puzzle. How, in one sentence, can a normal person prove that they are normal?

We can draw a parallel between people and coins. We can say that knights correspond to real coins, and knaves to fake coins that are lighter than real ones. Inspired by normal people, my coauthor Konstantin Knop invented chameleon coins. Chameleon coins can change their weight and behave like real or fake coins. I just wrote a post about chameleon coins.

Normal people are too unpredictable: they can consistently pretend to be knights or knaves. So logicians invented a simpler type of person, one who switches from telling the truth in one sentence to a lie in the next and then back to the truth. Such people are called alternators. Here is another puzzle:

Puzzle. You meet a person who is one of the three types: a knight, a knave, or an alternator. In two questions, find out which type they are.

Continuing a parallel between people and coins we can define alternator coins: the coins that switch their weight each time they are on the scale from weighing as much as real ones to weighing as much as fake ones. For the purposes of this essay, we assume that the fake coins are lighter than real ones. Unlike the chameleon coin, which might never reveal itself by always pretending to be real, the alternators can always be found. How do you find a single alternator among many real coins? There is a simple strategy: repeat every weighing twice. This strategy allows us to find an alternator among 9 coins in four weighings. Can we do better?

I used the alternator coins as a research project for my PRIMES STEP program where we do math research with students in seventh and eighth grade. The students started the alternator project and immediately discovered the strategy above. The next step is to describe a better strategy. For example, what is the maximum number of coins containing one alternator such that the alternator can always be found in four weighings?

But first we count possible outcomes. Suppose there is a strategy that finds an alternator. In this strategy we can’t have two unbalanced weighings in a row. To prove that, let us suppose there was an unbalanced weighing. Then the alternator switches its weight to a real coin and whether or not the alternator is on the scale, the next weighing must balance. The beauty of it is that given a strategy each outcome has to point to a particular coin as an alternator. That means the number of outcomes bounds the total number of coins that can be processed.

Counting the number of possible outcomes that do not have two unbalances in row is a matter of solving a recurrence, which I leave to the readers to find. The result is Jacobshtal numbers: the most beautiful sequence you might never have heard of. For example, the total number of possible outcomes of four weighings is 11. Since each outcome of a strategy needs to point to a coin, the total number of coins that can be processed in four weighings is not more than 11. But 11 is better than 9 in our previous strategy. Can we process 11 coins in four weighings? Yes, we can. I will describe the first part of the strategy.

So we have 11 coins, one of which is an alternator. In the first weighing we compare 5 coins against 5 coins. If the weighing unbalances, the alternator is on a lighter pan. Our problem is reduced to finding the alternator among five coins when we know that it is in the real state. If the weighing balances, then we know that if the alternator is among the coins on the scale it must now be in the light state. For the second weighting, we pick two sets of three coins out of this ten coins and compare them against each other. Notice that 3 is a Jacobsthal number, and 5, the number of coins outside the scale, is also a Jacobsthal number. If the second weighing balances, the alternator must be among 5 coins outside the scale. All but one of these coins are in the light state, and I leave it to the readers to finish the strategy. If the weighing unbalances, we need to find the alternator among 3 coins that are in the real state now. This can be done in two weighings, and again the readers are to the rescue.

It appears that Jacobsthal numbers provide the exact lower bound of the number of coins that can be processed. This is what my middle-schoolers discovered and proved. We wrote a paper on the subject. The strategy in the paper is adaptive. That means it changes depending on the results of the previous weighings. Can we find an oblivious strategy? I will tell you in later posts.

Suppose we have 3ⁿ identical-looking coins. One of the coins is fake and lighter than the other coins which all are real. We also have a random scale. That is a scale that at each weighing behaves randomly. Find the fake coin in the smallest number of weighings. Oops! That won’t work! It is impossible to find the fake coin. The scale can consistently misbehave in such a way as to blame a specific real coin for being fake.

Let’s try something else. Suppose we have two scales: one normal and one random. Find the fake coin.

What am I thinking? The normal scale can point to one coin and the random scale can point to another coin and we are in a “she said, he said” situation which we can’t resolve.

Now, in my final try, I’ll make it right. We actually have three scales, one of which is random. So here we go, with thanks to my son Sergei for giving me this puzzle:

Puzzle. We have 3ⁿ identical-looking coins. One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Let’s start with this strategy: repeat every weighing on all three scales and have a majority vote. At least two of the scales will agree, thus pointing to the true result. This way we can use a divide-into-three-equal-groups strategy for one scale to find the fake coin. It will require 3n weighings.

Can we do better? Of course, we can. We can repeat every weighing on two scales. If they agree we do not need the third scale. If they do not agree, one of the scales is random and lying, and we can repeat the weighing on the third scale to “out” the random scale. After we identify one normal scale, the process goes faster. In the worst case we will need 2n + 1 weighings.

Can we do even better? Yes, we can. I will leave it to the readers to find a beautiful solution that is asymptotically better than the previous one.

Update on Dec 24, 2016. The total number of coins should be 3²ⁿ, not 3ⁿ. We are looking at the worst case scenario, when the random scale is adversarial.

We all have played with problems in which we had real coins and fake (counterfeit) coins. For this post I assume that the fake coins are always lighter than the real coins. My coauthor Konstantin Knop invented a new type of a coin: a chameleon coin. This coin can mimic a fake or a real coin. It can also choose independently which coin to mimic for each weighing on a balance scale.

You cannot find the chameleon coin in a mix with real coins if it does not want to be found, because it can consistently behave as a real coin. Let’s add classic fake coins to the mix, the ones that are lighter. Still the task of identifying the chameleons using a balance scale cannot be achieved: the chameleons can pretend to be fake coins. We can’t identify the fake coins either, as the chameleons can mess things us up by consistently pretending to be fake.

What we can do is to find a small number of coins some of which are guaranteed to be fake. Consider the simplest setup, when we have one fake coin and one chameleon in our mix of N coins. That is we have N − 2 real coins. Our task now is to find TWO coins, one of which has to be fake. As usual we want to do it in the smallest number of weighings that guarantees that we’ll find the two coins. Let me give you a fun problem to solve:

Puzzle. The total number of coins is four. And as above we have one chameleon and one classic fake. In two weighings find two coins so that one of them is guaranteed to be fake.

If you want to learn more, we just wrote a paper titled Chameleon Coins.

* * *

—Honey, we are like two parallel lines.
—Why do you say that?
—The intersection of our life paths was a mistake.

* * *

—Q: Why did the obtuse angle go to the beach?
—A: Because it was over 90 degrees.

* * *

Ancient Roman in a clothing store: How come XL is larger than L?

* * *

—Which is the odd one out: one, three, six, seven?
—Well, three of them are odd ones.

* * *

When I am with you, I solve integrals in my head, so that blood can come back to my brain.

* * *

There are two kinds of people in this world: Those that can extrapolate from incomplete data.

* * *

Seven has the word even in it, which is odd.

I have always wanted to be an honest person and have followed my honor code. Soviet Russia had two honor codes. One code was for dealing with people and the other for dealing with businesses and the government. I remind you that in Soviet Russia, all businesses were owned by the government. The money paid for work didn’t have any relation to the work done. The government paid standard salaries and the businesses did whatever. Generally, that meant that they were doing nothing. Meanwhile, the government got its income from selling oil.

All people were being screwed by the government, so they had no motivation to play fair. Just as American workers do, Soviet Russians might use the work copier to make personal copies. The difference was that we didn’t feel guilty at ripping off the government. We wouldn’t just make a few necessary copies; we would make copies for our friends, our family, for strangers—as many copies as possible.

I moved to the United States in 1990. Several of my friends took time to explain to me the difference between Soviet Russia and the US. One of the friends, let’s call her Sarah, was working as staff at Harvard University. She told me the story of a recent visit by a famous Russian professor. After he left, the department received a bill. It appears that the professor, in a short visit, used several months-worth of the department’s budget allocation for copying and phone calls. I was impressed, in a good way, by the professor who I assume spent a good deal of his time making a lot of copies of papers unavailable in Russia, presumably not only for himself but also for students and colleagues. On the other hand, it was clearly wrong.

My new friend Sarah told me that in the US money does not come from nowhere, and I should include Harvard University in my honor code for people. Actually, not only Harvard, but also any place of work and the government too. Sarah also told me that since that budget problem, she was asked to talk to every incoming Russian visitor and explain to them how capitalism works. Most Russian visitors were ready to accept the rules. I too was delighted with that. It is much easier to follow one honor code than two.

I was also very happy for my son, Alexey. He was eight when we moved to Boston. Before our move I had a dilemma. Should I tell him that Lenin and Stalin were bad guys and killed millions of people? If I gave him a truthful explanation, I would also have to teach him to lie. Otherwise, if he mentioned this at school or on the street, we would be at risk of going to prison. If I didn’t teach him the truth, he might become brainwashed and grow up believing in communism, which would be very, very bad.

I was so lucky that I moved to the US: I didn’t have to teach my son to lie.

Centaurs, manticores, and minotaurs roam their planet. Their society is very democratic: any two animals can become sex partners. When two different species mate, their orgasm is so potent that they merge into one creature of the third species. For example, once one centaur and one manticore make love, they are reborn as one minotaur. At the beginning of the year 2016 there were 2016 centaurs, 2017 manticores, and 2018 minotaurs. They mated non-stop and at the end of the year only one creature was left on the planet. Which one?

This is one of those puzzles that I love-hate. I hate it because it is easy to answer this puzzle by inventing a specific mating pattern that ends with one animal. It is possible to get the correct answer without seeing the beauty. I love it because there is beauty in the explanation of why, if the mating ends with one animal, it has to be a specific animal.

The solution is charming, but being a mathematician this problem makes me wonder if ii is always possible to end with one animal. So I add another puzzle.

Describe the sets of parameters for which it is impossible to end up with one creature.