Tanya Khovanova's Math Blog

Every day I check coronavirus numbers in the US. Right now the number of deaths is 288 and the number of recovered is 171. More people died than recovered. If you are scared about the mortality rate, I can calm you and myself down: our government is incompetent—the testing wasn’t happening—that means the numbers do not show people who had mild symptoms and recovered. The real number of recovered people should be much higher.

Scientists estimated the mortality rate of coronavirus as being between 1 and 3.5 percent. Also, they say that it usually takes three weeks to die. That means three weeks ago the number of infected people in the US was between 8,000 and 29,000. The official number of cases three weeks ago was 68. I am panicking again—our government is incompetent—three weeks ago they detected between 0.25 and 1 percent of coronavirus cases. If this trend continues, then the official 19,383 infected people as of today means, in reality, somewhere between 2 million and 8 million infected people.

I can calm you and myself down: the testing picked up pace. This means, the ratio of detected cases should be more than 1 percent today. Probably the number of infected people today in the US is much less than 8 million. I am not calm.

I already wrote about my first experience crocheting hyperbolic surfaces. In my first surface I added two more stitches per current stitch. It took me hours to crochet the last row: the same hours it took me to crochet the rest.

For my next project, I reduced the ratio. The light blue thingy has ratio 3/2. I continued making my life simpler. The next project, the purple surface on the left, has ratio 4/3. The last project on the right has a ratio of 5/4 and is my favorite. Mostly because I am lazy and it was the fastest to make.

Happy 2019, the first 4 digit number to appear 6 times in the decimal expansion of Pi.

By the way:

2019 = 1⁴ + 2⁴ + 3⁴ + 5⁴ + 6⁴.

Also, 2019 is the product of two primes 3 and 673. The sum of these two prime factors is a square.

This is not all that is interesting about factors of 2019. Every concatenation of these two prime factors is prime. Even more unusual, 2019 is the largest known composite number such that every concatenation of its prime factors is prime. [Oops, the last statement is wrong, Jan 3,2019]

Happy Happy-go-Lucky year, as 2019 is a Happy-go-Lucky number: the number that is both Happy and Lucky.

In case you are wondering, here is the definition of Happy numbers: One can take the sum of the squares of the digits of a number. Those numbers are Happy for which iterating this operation eventually leads to 1.

In case you are wondering, to build the Lucky number sequence, start with natural numbers. Delete every second number, leaving 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, …. The second number remaining is 3, so delete every third number, leaving 1, 3, 7, 9, 13, 15, 19, 21, …. The next number remaining is 7, so delete every 7th number, leaving 1, 3, 7, 9, 13, 15, 21, …. The next number remaining is 9, so delete every ninth number, etc.

I do not like making objects with my hands. But I lived in Soviet Russia. So I knew how to crochet, knit, and sew — because in Russia at that time, we didn’t have a choice. I was always bad at it. The only thing I was good at was darning socks: I had to do it too often. By the way, I failed to find a video on how to darn socks the same way my mom taught me.

Then I came to the US. I suddenly found myself in a rich society, where it was cheaper to buy new stuff than to spend the time doing things with my hands. So I happily dropped my craftsmanship.

After not working with my hands for 28 years, one day I needed hyperbolic surfaces for my classes and I couldn’t find any to buy. Hyperbolic surfaces are famous for providing an example when the Euclid’s Fifth axiom doesn’t work. These hyperbolic surfaces look flat locally, so you can continue a line in any given direction. If you draw a line on such a surface and pick a point that is not on the line, then you can draw many lines through the point that are parallel to the given line.

My students are more important than my dislike of crochet, so I decided to just do it myself. I asked my friend Debbie, who knows how to crochet, for advice, and she gave me more than advice. She gave me a hook and a piece of yarn and reminded me how to work the hook. She started me with a small circle. After that all I had to do was add two stitches for each stitch on the perimeter of the circle. The finished product is this green ballish thing that looks like a brain, as in the photo.

Outside the starting circle, each small surface segment of this “brain” looks the same, making the “brain” a surface of constant curvature.

I chose a ratio of 2 to 1, adding two new stitches for each previous stitch. With this ratio, my flattish surface started looking like a ball very fast. The length of the perimeter doubled for every row. Thus each new row I crocheted took the same total amount of time that I had already spent on the whole thing. All the hours I worked on this “brain,” I kept thinking: darn, it is so unrewarding to do this. Annoying as it was, the thing that kept me going was my initial decision to continue to use up all the yarn Debbie had given me. In the end, with this ratio, half the time I worked was spent making the final row.Share:

This famous trick puzzle is very old:

Puzzle. The professor is watching across a field how the son of the professor’s father is fighting with the father of the professor’s son. How is this possible?

This puzzle is tricky only because of gender-bias. Most people assume that the professor is male and miss the obvious intended solution, in which a female professor is watching her brother fighting with her husband.

I just gave this problem on a test. Here are other answers that I received.

• The professor is gay and is watching his brother fighting with his husband.
• The professor is watching his brother fighting with the father of the professor’s step-son.
• The father of the professor’s son is himself. So he is watching a video of himself fighting with his brother.

Years ago people couldn’t figure out this puzzle at all. So there has been progress. I was glad that my students suggested so many ideas that work. Nonetheless, many of them revealed their gender-bias by initially assuming that the professor is a man.

I can’t wait until this puzzle stops being tricky.Share:

Every time I visit Princeton, or otherwise am in the same city as my friend John Conway, I invite him for lunch or dinner. I have this rule for myself: I invite, I pay. If we are in the same place for several meals we alternate paying. Once John Conway complained that our tradition is not fair to me. From time to time we have an odd number of meals per visit and I end up paying more. I do not trust my memory, so I prefer simplicity. I resisted any change to our tradition. We broke the tradition only once, but that is a story for another day.

Let’s discuss the mathematical way of paying for meals. Many people suggest using the Thue-Morse sequence instead of the alternating sequence of taking turns. When you alternate, you use the sequence ABABAB…. If this is the order of paying for things, the sequence gives advantage to the second person. So the suggestion is to take turns taking turns: ABBAABBAABBA…. If you are a nerd like me, you wouldn’t stop here. This new rule can also give a potential advantage to one person, so we should take turns taking turns taking turns. Continuing this to infinity we get the Thue-Morse sequence: ABBABAABBAABABBA… The next 2ⁿ letters are generated from the first 2ⁿ by swapping A and B. Some even call this sequence a fair-share sequence.

Should I go ahead and implement this sequence each time I cross paths with John Conway? Actually, the fairness of this sequence is overrated. I probably have 2 or 3 meals with John per trip. If I pay first every time, this sequence will give me an advantage. It only makes sense to use it if there is a very long stretch of meals. This could happen, for example, if we end up living in the same city. But in this case, the alternating sequence is not so bad either, and is much simpler.

Many people suggest another use for this sequence. Suppose you are divorcing and dividing a huge pile of your possessions. A wrong way to do it is to take turns. First Alice choses a piece she wants, then Bob, then Alice, and so on. Alice has the advantage as the first person to choose. An alternative suggestion I hear in different places, for example from standupmaths, is to use the Thue-Morse sequence. I don’t like this suggestion either. If Alice and Bob value their stuff differently, there is a better algorithm, called the Knaster inheritance procedure, that allows each of them to think they are getting more than a half. If both of them have the same value for each piece, then the Thue-Morse sequence might not be good either. Suppose one of the pieces they are dividing is worth more than everything else put together. Then the only reasonable way to take turns is ABBBB….

The beauty of the Thue-Morse sequence is that it works very well if there are a lot of items and their consecutive prices form a power function of a small degree k, such as a square or a cube function. After 2^k+1 turns made according to this sequence, Alice and Bob will have a tie. You might think that if the sequence of prices doesn’t grow very fast, then using the Thue-Morse sequence is okay.

Not so fast. Here is the sequence of prices that I specifically constructed for this purpose: 5,4,4,4,3,3,3,2,2,2,2,1,1,0,0,0. The rule is: every time a turn in the Thue-Morse sequence switches from A to B, the value goes down by 1. Alice gets an extra 1 every time she is in the odd position. This is exactly half of her turns. That is every four turns, she gets an extra 1.

If the prices grow faster than a power, then the sequence doesn’t work either. Suppose your pieces have values that form a Fibonacci sequence. Take a look at what happens after seven turns. Alice will have pieces priced F_n + F_n-3 + F_n-5 + F_n-6. Bob will have F_n-1 + F_n-2 + F_n-4. We see that Alice gets more by F_n-3. This value is bigger than the value of all the leftovers together.

I suggest a different way to divide the Fibonacci-priced possessions. If Alice takes the first piece, then Bob should take two next pieces to tie with Alice. So the sequence might be ABBABBABB…. I can combine this idea with flipping turns. So we start with a triple ABB, then switch to BAA. After that we can continue and flip the whole thing: ABBBAABAAABB. Then we flip the whole thing again. And again and again. At the end we get a sequence that I decided to call the Fibonacci fair-share sequence.

I leave you with an exercise. Describe the Tribonacci fair-share sequence.Share:

Do you know how to cut a cake? I mean, mathematically. There is a whole area of mathematics that studies cake cutting.

Mathematicians usually assume that each person has their own idea of what is the best part of the cake. Suppose three sisters are celebrating the New Year by having a cake. Anna likes only icing, Bella likes only chocolate chips, and Carol likes only pieces of walnuts mixed into the cake. Mathematicians want to cut cakes fairly. But what is fair? Fair is fair, right? Wrong. There are several different notions of fair cake division.

There is a proportionate division. In such a division every sister gets at least one-third of her value of the cake. This seems fair. Let’s see an example. Anna gets one third of the icing, Bella gets one third of the chocolate chips, and Carol gets everything else. This is a fair proportionate division. Each of the sisters believes that she got at least one-third of the cake, in their own value. But it doesn’t seem quite fair.

There is a stronger notion of fairness. It is called envy-free. In this division each sister gets at least one-third of the cake and, in addition, none of the three sisters would improve their value by swapping pieces. That means, if Anna wants only icing, not only does she get at least one-third of the icing, but also no one else gets more icing than Anna. The previous example of the proportionate division is not envy-free. Carol got two-thirds of the icing, so Anna would want to switch with her.

Let’s try a different division. Anna gets one third of the icing, Bella gets the chocolate chips and another third of the icing, and Carol gets all the walnuts and another third of the icing. Formally, this is envy-free cake cutting. But poor Anna. What do you think Anna feels when she sees the smiles of contentment on the faces of her sisters? Whoever invented the name doesn’t understand envy. Anna got one-third of the cake by her value, but the other sisters got the whole cake!

Luckily mathematicians understand this conundrum. So they invented another name for a cake division. They call a division equitable if everyone values all the pieces the same. So the division above is envy-free but not equitable. Let’s try again. Let’s give each sister one-third of all the components of the cake. This division is very good mathematically: it is proportionate, envy free, and equitable. By the way, envy-free division is always proportionate. This division seems fair. But is it a good division?

There is another term here: Pareto-efficient division means that it is impossible to make one person feel better, without making another person feel worse. All divisions above are not Pareto-efficient. Moving some icing from Carol to Anna, doesn’t decrease the value for Carol, but increases the value for Anna.

There is an even better way to divide the cake. We can give the icing to Anna, the walnuts to Bella, and the chocolate chips to Carol. This division is envy-free, equitable, and Pareto-efficient. It is perfect. Mathematicians even have a word for it. They call it a perfect division.

Mathematically this division is perfect. Unfortunately, sisters are not. I know an Anna who would still envy Bella.Share:

When I receive a bank statement, I review all the transactions. The problem is my failing memory. I do not remember when and where I last took money from an ATM, and how much. So I decided to create a pattern. I used to take cash in multiples of 100. Probably most people do that. A better idea is to always take the amount that has a fixed remainder modulo 100, but not 0. For example, let’s say I take one of the following amounts: 40, 140, 240, or 340. Sometimes I need more money, sometimes less. This set of numbers covers all of my potential situations, but my pattern is that all the numbers end in 40. This way if someone else gets access to my account, they will almost surely take a multiple of 100. I will be able to discover a fraud without remembering the details of my last withdrawal.

In addition, when I first started doing this, I was hoping I wouldn’t need to wait until I review my own statement to discover problems. My hope was that if a thief tried to take cash from my account, my bank would notice a change in the pattern and notify me immediately. Now I realize that this was wishful thinking. I doubt that banks are as smart as I am.

One day I should try an experiment. I should go to an ATM I never use and withdraw 200 dollars. I wonder if my bank would notice.Share:

In the starting position you see four multi-colored quadrants of a ring. For example, the first picture shows the starting position of level 33.

You can either swipe or touch between the quadrants. A swipe expands one quadrant into two quadrants and compresses two other quadrants into one. You can swipe clockwise or counterclockwise. The second figure shows the result of the clockwise swipe of the North opening. The NW quadrant that was half-red and half-yellow has expanded into two quadrants. The red piece now occupies the entire NW quadrant and the yellow piece—the entire NE quadrant. Two East quadrants got contracted into one SE quadrant. The former blue NE quadrant became the top blue half of the SE quadrant. The former SE half-blue half-green quadrant became the bottom half of the SE quadrant. In general the quadrant where the swiping movement starts expands in the direction of the swipe and the quadrant where the swiping movements ends together with the next quadrant compresses.

You can also touch an opening between quadrants. In this case the neighboring quadrants exchange places. The third figure shows the result of touching the East opening.

The goal of the game is to reach the final position: have each quadrant in one color. The next image shows the end of this particular level. As you can see the game was finished in 7 moves. In this particular case this is the smallest number of moves possible. To tell you a secret, it wasn’t me who finished the game in the smallest number of moves: it was my brother.

There is also a tutorial for this game on youtube.Share:

I have written ad nauseam about the ambiguity of problems with children. Usually a problem with two children is formulated as follows:

Mr. Smith has two children and at least one of them is a boy. What is the probability that he has two boys?

I don’t want to repeat my arguments for why this problem is ambiguous. Today I want to discuss other problematic assumptions about these problems.

Assumption 1: The probability of a child being a boy is 1/2. We know that this is not the case. Usually boys are born more often than girls. In addition to that, when policy interferes, the numbers can change. When China had their one-child policy, 118 boys were born per 100 girls. That makes a probability of a boy 0.54.

Assumption 2: The gender of one child in a family is independent of the gender of the other children. I am not sure where this assumption comes from, but I easily came up with a list of possible influences on this situation.

• A family can have identical twins.
• Families that adopt children can choose the gender of those kids.
• There are studies showing that people (especially men) can have a genetic predisposition to one gender of their children over the other.
• Sex-selective abortion is possible in many countries.
• In vitro fertilization and artificial insemination can use sex-selection techniques.
• People may reject their newborns based on sex.
• The decision to have a second child depends on the gender of the first child.

I would like to discuss how the last bullet point changes the probabilities in two children problems. Let us consider China. Up to now China had a one-child policy with some exceptions. In some cases if the first child was a girl, the family was allowed a second child. For the sake of argument, imagine a county where people are allowed to have a second child only if the first one is a girl. A family with two boys wouldn’t exist in this county. Thus the probability of having two boys is zero.

I tried to find the data about the distribution of children by gender in multi-children families. I couldn’t find any. I would be curious to know what happens in real life, especially in China.Share: