Archive for the ‘Math in Life’ Category.

Fair-Share Sequences

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 2k+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 Fn + Fn-3 + Fn-5 + Fn-6. Bob will have Fn-1 + Fn-2 + Fn-4. We see that Alice gets more by Fn-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.

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Your Deal is Better than my Deal

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.

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Is My Bank Smart Enough?

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.

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Ringiana

Starting position

My brother, Mikhail Khovanov, has invented a new game: Ringiana. It is now available for iPhone, and soon should be available for Android.

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.

 

After Swipe
After Touch
End Game

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.

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Problems with Problems with Two Children

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.

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The Secretary Problem with a Sliding Window

I love The Secretary Problem. I first heard about it a long time ago with a different narrative. Then it was a problem about the marriage of a princess:

The king announces that it is time for his only daughter to marry. Shortly thereafter 100 suitors line up in a random order behind the castle walls. Each suitor is invited to the throne room in the presence of the princess and the king. At this point, the princess has to either reject the suitor and send him away, or accept the suitor and marry him. If she doesn’t accept anyone from the first 99, she must marry the last one. The princess is very greedy and wants to marry the richest suitor. The moment she sees a suitor, she can estimate his wealth by his clothes and his gifts. What strategy should she use to maximize the probability of marrying the richest person?

The strategy contains two ideas. The first idea is trivial: if the princess looks at a suitor and he is not better than those she saw before, there is no reason to marry him. The second idea is to skip several suitors at the beginning, no matter how rich they might seem. This allows the princess to get a feel for what kinds of suitors are interested in her. Given that we know the strategy, the interesting part now is to find the stopping point: how many suitors exactly does she have to skip? The answer is ⌊N/e⌋. (You might think that this formula is approximate. Surprisingly, it works for almost all small values. I checked the small values and found a discrepancy only for 11 and 30 suitors.)

The problem is called The Secretary Problem, because in one of the set-ups the employer tries to hire a secretary.

In many situations in real life it is a good idea to sample your options. Whether I’m shopping for an apartment or looking for a job, I always remember this problem, which reminds me not to grab the first deal that comes my way.

Mathematically, I try to find variations of the problem that are closer to real life than the classical version. Here’s one of the ideas I had: you can always delay hiring a secretary until you have interviewed several candidates. You can’t wait too long, as that good secretary you interviewed two weeks ago might have already found a job. And of course the king has a small window of time in which he can run out of the castle and persuade a suitor to come back before he saddles up his horse and rides away.

To make the problem mathematical we should fix the window size as an integer w. When you are interviewing the k-th suitor, you are allowed to go w − 1 suitors back. In other words, the latest you can pick the current suitor is after interviewing w − 1 more people. I call this problem: The Secretary Problem with a Sliding Window.

It is easy to extrapolate the standard strategy to the sliding window problem. There is no reason to pick a suitor who is not the best the princess have seen so far. In addition, if she sees the best person, it is better to wait for the last moment to pick him in case someone better appears. So the strategy should be to skip several people at the beginning and then to pick the best suitor at the last moment he is available.

The difficult part after that is to actually calculate the probabilities and find the stopping point. So I suggested this project to RSI 2015. The project was assigned to Abijith Krishnan under the direction of Dr. Shan-Yuan Ho. Abijith is a brilliant and hard-working student. Not only did he (with the help from his mentor) write a formula for the stopping point and winning probabilities during the short length of RSI, he also resolved the case when the goal is to pick one candidate out of the best two.

If you are interested to see what other RSI students did this year, the abstracts are posted here.

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My Sleep Study

I recently had a home sleep study. I was given a small box which I attached to my chest. I also had to attach a thingy to my finger and put small tubes into my nose. It was relatively easy. Now I have my report:

The total time in bed is 468 minutes. Overall AHI is 37 events per hour. The supine AHI is 58 events per hour. The oxygen saturation baseline is 91%. The hypoxemic burden is 58 minutes. The oxygen saturation nadir is 63%. The heart rate ranges from 76-118 beats per minute.

I didn’t have a clue what all that meant so I hit the Internet. AHI means Apnea–Hypopnea Index, and a normal score is below 5. Anything above 30 indicates severe sleep apnea. Because mine is 37, I now have my diagnosis. My 63% oxygen saturation scared me the most. Wikipedia says 65% or less means impaired mental function. I do not need mental function when I sleep, but Wikipedia also says that loss of consciousness happens at 55%. What would happen if I lose consciousness while I sleep? Can I die? Will I wake up?

Overall the sleep study was a great thing. Now I know the diagnosis and there are ways to treat it. So I am looking forward to my improved energy and health.

But there was something in this report that would bother any mathematician. As you can see apnea gets worse when people sleep on their backs. (Thanks to this study I learned a new English word: supine means lying on the back.) The apparatus that I had to attach to my chest prevented me from sleeping on my stomach, one of my favorite sleep positions.

This report doesn’t say anything about my average AHI when I am not supine. If this average is low, then the solution might be to learn to never sleep on the back. It also means that the oxygen saturation nadir number is not very meaningful. It shows how bad it can be if I am forced to sleep on my back. It doesn’t say much about my standard sleep situation.

When I next see my doctor, I hope she’ll have answers to all my questions.

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Puzzling Grades Resolved

This story started when my student asked for an explanation for his grade B in linear algebra. He was slightly above average on every exam and the cut-off for an A was the top 50 percent of the class. I wrote a post in which I asked my readers to explain the situation. Here is my explanation.

The picture below contains histogram for a typical first midterm linear algebra exam.

First Midterm Histogram

The spike in the lowest range indicates zeros for those who missed the exam.

The mean is 74.7 and the median 81.5. As you can see the median is 7 points higher than the mean. That means that if a student performs around average on all the exams, s/he is in the bottom half of the class.

But this is not the whole story. In addition to the above, MIT allows students to drop the class after the second midterm. Suppose 30 students with lower grades drop the class; then the recalculated median for the first midterm for the students who finish the course goes up to 85. This is a difference of more than 10 points from the original average.

If this was a statistics class, then I could have told the puzzled student that he deserves that B. Instead I told him that he didn’t even have the highest score among those with Bs. Somehow that fact made him feel better.

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Linear Algebra on a Mission Impossible

I love making math questions out of the movies. Here is a Mission Impossible III question.

Tom Cruise is cute. He plays Ethan Hunt in Mission Impossible movies. In Mission Impossible III he needs to steal the Rabbit’s Foot from a secure skyscraper in Shanghai. He arrives in Shanghai and studies the skyscraper looking out his window. He decides to break in through the roof. And the way to get to the roof is to use a rope and swing across from another, even taller, skyscraper. 1:21 minutes into the movie, Ethan Hunt calculates the length of the rope he will need by using the projection of a skyline on his window, as seen on the first picture.

MI3 Skyline

Explain why the projection is not enough to calculate the length of the rope. What other data does he need for that? Ethan Hunt does request extra data. But he makes one mistake. He uses his pencil as a compass to draw the end of the rope curve, as seen on the second picture. Explain what his mistake is.

MI3 Rope

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The Virtue of Laziness

My son, Alexey Radul, is a programmer. He taught me the importance of laziness in programming.

One of his rules:

Not to write the same line of code in the same program twice.

If you need the same line of code in the same program, that means you should either use a loop or outsource the line to a function. This style of coding saves time; it makes programs shorter and more elegant. Such programs are easier to debug and understand.

I remember how I copied and pasted lines of code before he taught me this rule. Then I needed to change parameters and missed some of the lines during changing. Debugging was such a headache.

Mathematicians are way lazier than programmers. Consider the system of two equations: x+2y=3 and 4x+5y=6. There are no repeating lines here. Only letters x and y appear twice. Mathematicians invented the whole subject of linear algebra and matrices so that they would not need to rewrite variables.

Mathematicians are driven by laziness. Once ancient mathematicians first solved a quadratic equation, they didn’t want to do it again. So they invented a formula that solves all quadratic equations once and for all.

I try to keep up with tradition. I try to make my theorems as general as possible. When I write my papers, I try to make them short and simple. When I think about mathematics I try to get to the stage where the situation is so clear I can think about it without paper and pencil. I often discover new theorems while I am in bed, about to fall asleep. Sometimes I wake up with a good idea. So I do my job while I sleep.

I love my profession. I get paid for being lazy.

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