Tanya Khovanova's Math Blog

The beautiful Pascal triangle has been around for many years. Can you say something new about it?

Of course you can. Mathematicians always find new way to look at things. In 2012 RSI student, Kevin Garbe, did some new and cool research related to the triangle. Consider Pascal’s triangle modulo 2, see picture which was copied from a stackexchange discussion.

A consecutive block of m digits in one row of the triangle modulo 2 is called an m-block. If you search the triangle you will find that all possible binary strings of length 2 are m-blocks. Will this trend continue? Yes, you can find any possible string of length 3, but it stops there. The blocks you can find are called accessible blocks. So, which blocks of length 4 are not accessible?

There are only two strings that are not accessible: 1101 and 1011. It is not surprising that they are reflections of each other. Pascal’s triangle respects mirror symmetry and the answer should be symmetric with respect to reflection.

You can’t find these blocks on the picture, but how do we prove that they are not accessible, that is, that you can’t ever find them? The following amazing property of the triangle can help. We call a row odd/even, if it corresponds to binomial coefficients of n choose something, where n is an odd/even number. Every odd row has every digit doubled. Moreover, if we take odd rows and replace every double digit with its single self we get back Pascal’s triangle. Obviously the two strings 1101 and 1011 can’t be parts of odd rows.

What about even rows? The even rows have a similar property: every even-indexed digit is a zero. If you remove these zeros you get back Pascal’s triangle. The two strings 1101 and 1011 can’t be part of even rows. Therefore, they are not accessible.

The next question is to count the number of inaccessible blocks of a given length: a(n). This and much more was done by Kevin Garbe for his RSI 2012 project. (I was the head mentor of the math projects.) His paper is published on the arxiv. The answer to the question can be found by constructing recurrence relations for odd/even rows. It can be shown that a(2r) = 3a(r) + a(r+1) − 6 and a(2r+1) = 3a(r) + 2a(r+1) − 6. As a result the number of inaccessible blocks of length n is n² − n + 2. I wonder if there exists a direct proof of this formula without considering odd and even rows separately.

This RSI result was so pretty that it became a question at our entrance PRIMES test for the year 2013. In the test we changed the word accessible to admissible, so that it would be more difficult for applicants to find the research. Besides, Garbe’s paper wasn’t arxived yet.

The pretty picture above is from the stackexchange, where one of our PRIMES applicants tried to solicit help in solving the test question. What a shame.

I picked four problems that I liked from the Moscow Math Olympiad 2016:

Problem 1. Ten people are sitting around a round table. Some of them are knights who always tell the truth, and some of them are knaves who always lie. Two people said, “Both neighbors of mine are knaves.” The other eight people said, “Both neighbors of mine are knights.” How many knights might be sitting around the round table?

Problem 2. Today at least three members of the English club came to the club. Following the tradition, each member brought their favorite juice in the amount they plan to drink tonight. By the rules of the club, at any moment any three members of the club can sit at a table and drink from their juice bottles on the condition that they drink the same amount of juice. Prove that all the members can finish their juice bottles tonight if and only if no one brings more than the third of the total juice brought to the club.

Problem 3. Three piles of nuts together contain an even number of nuts. One move consists of moving half of the nuts from a pile with an even number of nuts to one of the other two piles. Prove that no matter what the initial position of nuts, it is possible to collect exactly half of all the nuts in one pile.

Problem 4. N people crossed the river starting from the left bank and using one boat. Each time two people rowed a boat to the right bank and one person returned the boat back to the left bank. Before the crossing each person knew one joke that was different from all the other persons’ jokes. While there were two people in the boat, each told the other person all the jokes they knew at the time. For any integer k find the smallest N such that it is possible that after the crossing each person knows at least k more jokes in addition to the one they knew at the start.

Spoiler for Problem 2. I want to mention a beautiful solution to problem 2. Let’s divide a circle into n arcs proportionate to the amount of juice members have. Let us inscribe an equilateral triangle into the circle. In a general position the vertices of the triangle point to three distinct people. These are the people who should start drinking juices with the same speed. We rotate the triangle to match the drinking speed, and as soon as the triangle switches the arcs, we switch drinking people correspondingly. After 120 degree rotation all the juices will be finished.

I already posted a funny true story that Smullyan told me when I last visited him. Raymond Smullyan died recently at the age of 97 and my mind keeps coming back to this last visit.

The year was 2012 and I was about to drive back to Boston after my talk at Penn State. Smullyan’s place in the Catskills was on the way—sort of. I wanted to call him, but I was apprehensive. Raymond Smullyan had a webpage on which his email was invisible. You could find his email address by looking at the source file or by highlighting empty space at the bottom of the page. Making your contact information invisible sends a mixed message.

While this was a little eccentric, it meant that only people who were smart enough to find it, could access his email address. I already knew his email because he had given it to me along with his witty reply to my blog post about our meeting at the Gathering for Gardner in 2010.

In our personal interactions, he always seemed to like me, so I called Raymond and arranged a visit for the next day around lunch time. When I knocked on his door, no one answered, but the door was open, and since Smullyan was expecting me, I walked right in. “Hello? Anyone there? Hello? Hello?” As I wandered around the house, I saw an open bedroom door and inside Smullyan was sleeping. So I sat down in his library and picked up a book.

When he woke up, he was happy to see me, and he was hungry. He told me that he didn’t eat at home, so we should go out together for lunch. I was hungry too, so I happily agreed. Then he said that he wanted to drive. I do not have a poker face, so he saw the fear in me. My only other trip with a nonagenarian driver flashed in front of my eyes. The driver had been Roman Totenberg and it had been the scariest drive I have ever experienced.

I said that I wanted to drive myself. Annoyed, Raymond asked me if I was afraid of him taking the wheel. I told him that I have severe motion sickness and always prefer to drive myself. Raymond could see that I was telling the truth. I got the impression that he was actually relieved when he agreed to go in my car.

We went to Selena’s Diner. He took out playing cards with which he showed me magic tricks. I showed him some tricks too. This was probably a bad move as he abandoned me to go to the neighboring table to show his magic tricks to a couple of young girls. They were horrified at first”his unruly hair, his over-the-top energy, his ebullient behavior”but between me and the waitress, we quickly reassured them. The girls enjoyed the tricks, and I enjoyed my visit.

Like many people, I was appalled by Trump’s immigration ban. On the Internet I found many essays that explained that he did not include in the ban those majority-Muslim countries in which he has business interests. See for example, an article at Forbes with a nice map, and an article at NPR.

Now the countries that are excluded are motivated to continue to support Trump’s businesses, and to offer him bribes and good deals in exchange for staying out of the ban. The countries on the list are also motivated to approach Trump and offer him a sweet business deal.

So even if the courts stopped the ban, he has already succeeded in showing every country in the world that to be on his good side requires that they pay up. And China got the hint and granted Trump a trademark he’s been seeking for a decade.

Looks like Trump’s vision of a great America is a very rich Mr Trump.

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.