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I promised to explain my job situation to my readers. I currently have six part-time jobs, which I describe below in the order that I started them.
The job I have had the longest is coaching math for competitions at AMSA Charter School. I started there in the spring of 2008 and I am still teaching there every week. I post my homework assignments on my webpage for other teachers and students to use.
The next longest has been as the Head Mentor at RSI, which I started in the summer of 2009. RSI is a great program where high school students from all over the world come to MIT to do research during summer vacations. It is amazing how much an ambitious student can do during a short five-week period. My first year I saw some great research done, but I was surprised that RSI papers were generally not available online. Beginning in 2012, at my recommendation, our Director Slava Gerovitch now posts the RSI papers at the math department RSI webpage.
As good a program as it is, RSI is too short for a research project. So we decided to develop our own program called PRIMES (Program for Research in Mathematics, Engineering, and Science for High School Students). I am also the Head Mentor in this program. I provide general supervision of all the projects and review conference slides and final papers. In addition to being the Head Mentor, I mentor my own students, which is great fun. I usually have three projects, but if something happens to a project or a mentor, I take it over.
Together with Pavel Etingof, PRIMES Chief Research Advisor, and Slava Gerovitch, PRIMES Director, I wrote an article Mathematical Research in High School: The PRIMES Experience that appeared in Notices.
My job #4, since the fall of 2013, is as a Teaching Assistant for the MIT linear algebra course.
This year I took on two more jobs at MIT. I am the Head Mentor at Mathroots, a summer program for high school students from underrepresented backgrounds. I am also teaching at PRIMES STEP, the program for gifted middle-school students in the Boston area. The goal of the program is to teach students mathematical thinking, have fun, and prepare them for PRIMES.
Overall, I have reached the stage in my career when I do not have time to breath and I still do not make enough money to pay my bills. The good news: my jobs bring me satisfaction. I just hope that I will not be too tired to notice that I am satisfied.
I already wrote about the sliding-window variation of the Secretary Problem. In this variation, after interviewing a candidate for the job, you can pick him or any out of w − 1 candidates directly before him. In this case we say that we have a sliding window of size w. The strategy is to skip the first s candidates, then pick the person who is better than anyone else at the very last moment. I suggested this project to RSI and it was picked up by Abijith Krishnan and his mentor Shan-Yuan Ho. They did a good job that resulted in a paper posted at the arXiv.
In the paper they found a recursive formula for the probability of winning. The formula is very complicated and not explicit. They do not discuss the most interesting question for me: what is the advantage of a sliding window? How much better the probability of winning with the window as opposed to the classical case without the window?
Let us start with a window of size 2, and n applicants. We compare two problems with the same stopping point. Consider the moment after the stopping point when we see a candidate who is better than everyone else before. Suppose this happens in position b. Then in the classic problem we chose this candidate. What is the advantage of a window? When will we be better off with the window? We will be better if the candidate at index b is not the best, and the window allows us to actually reach the best. This depends on where the best secretary is, and what happens in between.
If the best secretary is the next, in position b + 1, then the window gives us an advantage. The probability of that is 1/n. Suppose the best candidate is the one after next, in position b + 2. The window gives us an advantage only if the person in position b + 1 is better than the person in position b. What is the probability of that? It is less than 1/2. From a random person the probability of the next one being better is 1/2. But the person in position b is not random, he is better than random, so the probability of getting a person who is even better decreases and is not more than 1/2. That means the sliding window wins in this case with probability not more than 1/2n.
Similarly, if the best candidate is in position b + k, then the sliding window allows us to win if every candidate between b and b + k is better than the previous one. The probability of the candidate being better at every step is not more than 1/2. That means, the total probability of getting to the candidate in position b + k is 1/2^{k-1}. So our chances to win when the best candidate is at position b + k are not more than 1/2^{k-1}n. Summing everything up we get an advantage that is at least 1/n and not more than 2/n.
The probability of winning in the classical case is 1/e. Therefore, the probability of winning in the sliding window case, given that the size of the window is 2, is between 1/e + 1/n and 1/e + 2/n.
Let us do the same for a window of any small size w. Suppose the best secretary is in the same window as the stopping candidate and after him, that is, the best candidate is among the next w − 1 people. The probability of this is (w − 1)/n. In this case the sliding window always leads to the best person and gives an advantage over the classical case. When else does the sliding window help? Let us divide the rest of the applicants into chunks of size w − 1. Suppose the best applicant is in the chunk number k. For the sliding window to allow us to get to him, the best candidate in every chunk has to be better than the best one in the previous chunk. The probability of that is not more than 1/2^{k-1}. The probability that we get to this winner is not more that (w-1)/2^{k-1}n. Summing it all up we get that the advantage of the window of size w is between (w − 1)/n and 2(w − 1)/n.
I already wrote about my favorite problem from the 2015 All-Russian Math Olympiad that involved tanks. My second favorite problem is about coins. I do love almost every coin problem.
A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn't know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?
My name is Tanya and here's why. When my mom was pregnant, she was discussing the child's name with my father, Gueliy Khovanov. They decided that she could choose a boy's name while he could choose a girl's name. She wanted me to be Andrey in honor of her half-brother who died in World War II. He wanted me to be Tanya in honor of the unrequited love of his life. In an act of incomprehensible generosity, my mother agreed to name me after this woman. My parents were not even married when I was born.
After the decision was made, I was born on my own Name Day. In Russian culture, different names are celebrated on different dates. The a holiday for Tanyas is especially big and it falls on January 25, the day I was born. This serendipity led me to be very attached to my given name.
When I was a child, I believe that my father introduced me to his love, Tanya, although I do not have any clear memory of her. I doubt they were ever actually together. I do remember how my father loved her all his life. Even today I am sure that his love for Tanya brushed off on me, being her namesake.
I especially remember one day, when I was still a kid, my father agonizing about Tanya and calling to offer her help. My mom grimaced when he offered to babysit her children. My parents were divorced by then, and my father showed no interest in babysitting his own children. Now I understand that watching Tanya's children was the least he could do. This was happening in 1968 when the USSR invaded Czechoslovakia, and Tanya's husband, Konstantin Babitsky, was one of the few people in the USSR who risked their lives to openly protest against it. Tanya's life and freedom were also at risk. Offering any help was the right and only call.
Right before my father died of an illness in 1980, he asked me for one last favor: to tell Tanya that he had died. I was 20 years old and didn't have a clue how to find her; I only knew that her name was Tanya Velikanova. She was a famous Soviet human rights activist and dissident. That made it dangerous for me and for the people I might ask for assistance.
I finally asked my father's best friend to do it for me. He said that Tanya was in exile, but promised to pass a message to her. I am not sure he did it and I still feel guilty that I didn't do it myself before she died.
I forgot to tell you how my dad met the love of his life. They met as teenagers at a math Olympiad, where she beat him.
I already wrote how I build a magic SET hypercube with my students. Every time I do it, I can always come up with a new question for my students. This time I decided to flip over two random cards, as in the picture. My students already know that any two cards can be completed to a set. The goal of this activity is to find the third card in the set without trying to figure out what the flipped-over cards are. Where is the third card in the hypercube?
Sometimes my students figure this out without having an explicit rule. Somehow they intuit it before they know it. But after several tries, they discover the rule. What is the rule?
Another set of questions that I ask my students is related to magic SET squares that are formed by 3 by 3 regions in the hypercube. By definition, each magic SET square has every row, column, and diagonal as a set. But there are four more sets inside a magic SET square. We can call them super and sub-diagonal (anti-diagonal) wrap-arounds. Can you prove that every magic SET square has to have these extra four sets? In addition, can you prove that a magic SET square is always uniquely defined by any three cards that do not form a set, and which are put into places that are not supposed to from a set?
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.
In my paper with Joshua Xiong on Nim Fractals we explained how to build an evolution graph corresponding to an impartial combinatorial game. The vertices of the graph are P-positions. And two vertices are connected if the two corresponding positions are consecutive P-positions in a longest possible optimal game.
What types of graphs do we get? An evolution graph should have at least one sink: these are our terminal P-positions. Also there are no directed loops: the game is finite. In addition, the distance from a vertex to sinks is uniquely defined: the number of directed edges that is needed to move from this vertex to a sink. This number is equal to half of the number of moves in the longest game, starting from the corresponding P-position.
A natural question arises: Can we build a game from a graph? The graph needs to satisfy the properties above. But other than that, can we? We can consider the graph itself as a game. Players can start at a vertex or an edge. From an edge, they can only move to a vertex where this edge ends. From a vertex they can move to any out-edge. The corresponding evolution graph is the graph itself. Vertices correspond to P-positions and edges to N-positions.
There is an equivalent game with a more uniform description. We put a vertex in the middle of every edge in the evolution graph. The new graph becomes bipartite. Players can start at any vertex. A move is allowed from a vertex following an out-edge to the vertex, where this edge ends. Vertices that are in the same part as terminal positions are P-positions. Other vertices are N-positions.
I explained to my AMSA students the idea of meta-solving multiple choice. Sometimes by looking at the choices without knowing what the problem is, it is possible to guess the correct answer. Suppose your choices are 2k, 2, 2/k, 10k, −2k. What is the most probable answer? There are several ideas to consider.
Both these considerations suggest a rule of thumb: the answers that are odd ones out are probably wrong. In the given example (2k, 2, 2/k, 10k, —2k), the second choice is an odd one out because it's the only one that does not contain k. The third choice is an odd one out because it's the only one in which we divide by k. The fourth choice is an odd one out because it's the only one with 10 instead of 2. The last one is an odd one out because of the minus sign. Thus the most probable answer is 2k.
So I explained these ideas to my students and gave them a quiz, in which I took the 2003 AMC 10A test, but only gave them the choices without the problems. I was hoping they would do better than randomly guessing.
Luckily for me, I have six classes in a row doing the same thing, so I can make adjustments as I go along. Looking at the results of the first two groups of students, I discovered that they were worse than random. What was going on?
I took a closer look, and what do you know? Nobody marked the first or the last choice. The answers are in an increasing order, so the first is the smallest and the last is the largest. So these two numbers are odd ones out, in a way.
It is a good idea to consider 189 as an odd one out in the list 1, 2, 4, 5, 189. In many other cases, the fact that the number is either the smallest or largest is insufficient reason to consider it as the odd one out. For example, there is no reason to consider 1 to be an odd one out in the list 1, 2, 3, 4, 5. And the designers of AMC are good: a lot of problems have an arithmetic progression as a list of choices, where none of the numbers are obviously odd ones out.
To correct the situation of worse than random results, I discussed it with my students in my next classes. Problem designers cannot have a tradition in which the first answer is never the correct answer. If such a tradition existed, and people knew about it, that knowledge would help them guess. So the first answer should be correct approximately five times, which is a fifth of the total number of questions (25).
And we came up with a strategy. Use the odd one out method except for arithmetic progressions. Then add the choices to balance out the total number of the first answers, the second answers, etc.
That method worked. In my next four classes my students were better than random.
Professor Bock came to his office at the Math Department of Deys University and discovered that someone had broken in. Luckily he had a lunch scheduled with his friend Detective Radstein. Bock complained to the detective about the break-in and the detective agreed to investigate.
Nothing was stolen from the office. It looked like somebody had just slept on Professor Bock's couch. The couch was bought recently and it was positioned so that it was impossible to see it through the tiny window of the office door, which had been locked. Interestingly, Professor Bock's office was the only one with a couch. The detective concluded that the person who broke in knew about the couch and thus was from the Math Department. In addition, the couch was very narrow and couldn't possibly sleep more than one person.
Investigating crimes at the Math Department of Deys University was very easy. This was due to a fact discovered by Detective Radstein on a previous case: every member of the department was one of three types:
Thus solving a crime at the department is very easy. Detective Radstein just needs to ask every person two questions, "How much is 2+2?" and "Are you guilty?" Only a guilty dodger would answer "4" and dodge. Only a guilty truth-teller would answer "4" and "yes". Only a guilty liar would answer something other than "4" and "no".
Detective Radstein decided to enjoy himself by making this investigation more challenging. He asked everyone only one question "Are you guilty of breaking-in?" These are the first three replies:
—Albert: I did it.
—Bill: Albert didn't do it.
—Connie: Albert did it.
It was amusing how many people knew where Albert slept, but these answers were enough for Detective Radstein to figure out the culprit. After this issue was resolved he asked Professor Bock, "While I am here, what other problems do you have at the department?"
"Well," said the professor, "In fact, I would be grateful if you could help us resolve two more issues. This semester our expenditure for tea increased three times. It is clear someone is stealing tea. It's also clear that this could only be someone who started working at the department this semester. There are two people who fit this description: David and Eve. So Detective Radstein asked each of them, "Are you stealing the tea?" He got the following answers:
—David: Eve steals the tea.
—Eve: Only one person steals the tea.
Having solved the Math Department's tea problem Detective Radstein then asked Professor Bock, "What else?" "Well," said Professor Bock, "one more thing. We have a lot of blackboards around the department. Every day a famous three-letter Russian curse word appears on the blackboards. The handwriting is always the same, so it is one person. Luckily the word starts with XY, so most people assume that this is some math formula. Anyway, I want to look into the eyes of the person who does this. I left the department late yesterday; only Fedor, Grisha and Harry were here. The blackboards were clean. This morning when I opened up the Math Department, the vulgar swearword had been written on the blackboards. I don't suspect someone of sneaking into the department at night to scribble on our blackboards. I am certain it is one of the three people I mentioned."
Once again Detective Radstein asked the suspects whether they did it.
—Fedor: Grisha did it.
—Grisha: Fedor did it.
—Harry: I do not speak Russian.
And again Detective Radstein solved the case. He was surprised that everyone in the department knew what everyone else was doing. Only his friend Professor Bock seemed clueless.
If you were the detective, would you be able to help Professor Bock?
My mom used to tell me "You are born to teach." My mom was a very shy person with no desire to be a teacher. Somehow she ended up teaching chemistry all her life. She also told me that teaching was our family business. All my ancestors were teachers or priests. Given that I lived in the Soviet Union, priesthood was not an option. (Given that I'm a woman, priesthood was not an option anywhere else.)
After my postdocs, I ended up quitting academia and working for industry. As a result, I didn't teach much. Until 2008 my only teaching experience was a course in Discrete Mathematics that I taught at Bar-Ilan University as part of my postdoc. Somehow Bar-Ilan University wanted five people to teach this same course in parallel and wanted me to teach it in Russian to attract fresh Russian immigrants. By the end of my course, I had way more people coming to my class than at the beginning, many of whom had transferred from the other parallel courses.
I had another clue that I was doing a good job. During one of my classes, the door was left open. A guy was walking down the hall, but when he heard me, he stopped. He stood there listening intently until the end of my class.
At the end of 2007 I resigned my industry job to go back to mathematics. I needed some financial support to do so, so my friends tried to arrange a math coach position for me at AMSA (Advanced Math and Sciences Academy Charter School). I had my interview in April, 2008, for a position for the following academic year. My interviewers were skeptical: they had had a series of PhDs who had proven incapable of teaching. I told them that deep down in my heart I knew that I would be a great teacher. And I made them an offer: I will start working in April and work until the end of the school year in June. Hiring me for two months was not a great risk, but gave them a basis on which to decide whether to give me a contract for the following year. I've been working at AMSA ever since.
My class is optional and there are no grades, but most of my students stick with me for all five years. (I teach grades 8 to 12.) My students show that they like me in many ways. I'm especially thrilled when they tell me that I am the best teacher they ever had.
Informally they call my class Tanya Time.
One evening Detective Radstein visited Professor Bock. He was hanging out in the kitchen and overheard a conversation between Professor Bock and his wife. It appeared that Mrs. Bock had discovered that all the cutlets she had prepared for the next day were missing. She asked her husband:
—Did you eat the cutlets?
—I ate soup, the professor replied.
—Oh well, the children were probably very hungry.
Detective Radstein smiled to himself. Many mathematicians have trouble making false statements. Some of them adjust to social situations by learning how to lie while formally telling the truth. Radstein calls them dodgers. Fortunately, dodgers only dodge a question when they are threatened. It was obvious that Professor Bock was such a dodger. He implied that he didn't eat the cutlets, because he had soup for lunch. The detective was ready to bet $10,000 that, in truth, the soup was just an appetizer to Professor Bock's meal of cutlets.
Detective Radstein talked to his friend Professor Bock about this obsession mathematicians have to make true statements. Professor Bock agreed that many mathematicians are like that. In fact, all the faculty members in his department are either dodgers or pathological truth-tellers. Ironically, all other staff members at Professor Bock's math department are liars.
A pathological truth-teller is another term that Detective Radstein uses to describe people who tell the truth no matter what. They never dodge. They answer a question exactly, often disregarding the context and purpose of the question. For example, when someone enters an elevator and asks a pathological truth-teller, "Is this elevator going up or down?" the answer s/he gets is "Yes."
One day Professor Bock asked Detective Radstein for help, following a series of laptop thefts at his department. It was clear that the thefts were committed by someone working at the department and that the criminal acted alone.
Detective Radstein decided that the easiest starting point would be to ask everyone the same question: Did you steal the laptops? If a pathological truth-teller is the perpetrator, s/he would admit to the crime. A dodger would evade the question, but only if they are guilty. A liar is flexible: s/he might either answer the question with a lie or dodge with a lie. These are the first three answers the detective got from members of the department:
—Alice: No, I didn't steal the laptops.
—Bob: Alice stole the laptops.
—Clara: Alice didn't steal the laptops.
Who stole the laptops?
Yesterday I went online to check out the problems at the 2015 All-Russian Math Olympiad. I was stunned to discover a problem about tanks and fighter jets. I have never seen such a militarized problem before. The problem setup tells me something about the mood of people in Russia. It tells me that the mood has changed—alarmingly for the worse.
On the other hand, mathematically it was my favorite problem. So here it is.
The field is a 41 by 41 square made of square cells. A tank is camouflaged and hidden in one of the cells. A pilot flying a fighter jet above knows that the tank is there and wants to destroy it. If there is a tank in a cell, it will be hit by the shot directed at this cell. The pilot also knows that two hits are needed to destroy the tank. The tank will move to a neighboring cell immediately upon being hit, without breaking its camouflage. Other than that, the tank won't move. What is the smallest number of shots required to guarantee that the tank is destroyed?
My post with Kvantik's 2012 problems for middle school was a success. So I scanned the 2013 issues and found 7 more problems that I liked. Here are two cute problems I've seen before:
Problem 1. In the equation 30 − 33 = 3 move one digit to make it correct.
Problem 2. A patient got two pills for his headache and two pills for his cough. He was supposed to use one of each type of pill today and do the same tomorrow. The pills all looked the same. By mistake, the patient mixed up the pills. How should he use them so that he follows the prescription exactly?
Now logic and information-theoretical problems:
Problem 3. One strange boy always tells the truth on Wednesdays and Fridays and always lies on Tuesday. On other days of the week he can lie or tell the truth. He was asked his name seven days in a row. The first six answers in order are Eugene, Boris, Vasiliy, Vasiliy, Pete, Boris. What was his answer on the seventh day?
Problem 4. Ten people are suspected of a crime. It is known that two of them are guilty and eight are innocent. Suspects are shown to a psychic in groups of three. If there is a guilty person in the group the psychic points him out. If there are two guilty people in the group, the psychic points to one of them. If all of them are innocent, the psychic points at one of the three.
- How would you find at least one guilty person in four séances?
- How would you find both criminals in six séances?
Problem 5. There are four balloons: red, blue, green and black. Some of the balloons might be magical. There is also a detector box that can say how many balloons out of the ones put inside are magical. How can you find all the magical balloons using the detector box not more than three times?
I conclude with two miscellaneous problems.
Problem 6. Three runners started their loops at the same time at the same place on the same track. After some time they ended at their starting point together. The fastest runner passed the slowest runner 23 times. Assuming each runner has a constant speed, how many times was one runner passed by another runner in total?
Problem 7. Given a point inside a circular disk, cut the disk into two parts so that you can put them back together into a new disk such that the given point is the center.
I recently posted a set of problems from a Russian magazine for middle school children. Now it's time for solutions.
Problem 1. There are 6 glasses on the table in a row. The first three are empty, and the last three are filled with water. How can you make it so that the empty and full glasses alternate, if you are allowed to touch only one of the glasses? (You can't push one glass with another.)
Solution. Pour the water from the fifth glass to the second glass.
Problem 2. If it is raining at midnight, with what probability will there be sunshine in 144 hours?
Solution. In 144 hours it be midnight. Assuming we are not close to the earth's poles, there will be no sunshine.
Problem 3. How can you fill a cylindrical pan exactly half-full of water?
Solution. Fill it with water, then tilt so that the water slowly runs out until you are about to see the rim of the floor of the pan.
Problem 4. The Jackal always lies; the Lion always tells the truth. The Parrot repeats the previous answer—unless he is the first to answer, in which case he babbles randomly. The Giraffe replies truthfully, but to the previous question directed to him: his first answer he chooses randomly.
The Wise Hedgehog in the fog stumbled upon the Jackal, the Lion, the Parrot, and the Giraffe, although the fog prevented him from seeing them clearly. He decided to figure out the order in which they were standing. After he asked everyone in order, "Are you the Jackal?" he was only able to figure out where the Giraffe was. After that he asked everyone, "Are you the Giraffe?" in the same order, and figured out where the Jackal was. But he still didn't have the full picture. He started the next round of questions, asking everyone, "Are you the Parrot?" After the first one answered "Yes", the Hedgehog understood the order. What is the order?
Solution. On the question "Are you the Jackal?" the Jackal and the Lion have to answer no. The Giraffe has to say "yes." The Parrot could not have said "yes," because in this case the Hedgehog, knowing where Giraffe is, could have figured out where the Parrot is. That means, in the first round there was only one "yes." We also can conclude that the Parrot is not after the Giraffe. On the question "Are you the Giraffe?" the Lion would have said "no," and the Jackal would have said "yes." The argument here is similar to the previous discussion: the Parrot couldn't have said "yes." We also know that the Parrot is not after the Jackal. So the Parrot is either the first or is after the Lion. After the Hedgehog's last question, we know that the Lion can't be the first. That means the Parrot is the first. Which means the previous animal said "yes," implying that the Jackal is the last. So there are two possibilities: The Parrot, the Lion, the Giraffe, the Jackal, or the Parrot, the Giraffe, the Lion, the Jackal. But in the second arrangement, the Hedgehog would have differentiated the Lion from the Parrot before the last question, as the Parrot couldn't have been after the Giraffe. The answer is: the Parrot, the Lion, the Giraffe, the Jackal.
Problem 5. There are 12 cards with the statements "There is exactly one false statement to the left of me," "There are exactly two false statements to the left of me," ..., "There are 12 false statements to the left of me." Pete put the cards in a row from left to right in some order. What is the largest number of statements that might be true?
Solution. Suppose there are more than six statements that are true. That means one of the cards with a number more than six is true, meaning that there are at least 7 false statements. This is a contradiction. Suppose there are 6 true cards. They have to be cards with numbers one through six. In addition, no pair of adjacent cards can both be true. Arranging the cards in the order 7, 1, 8, 2, 9, 3, 10, 4, 11, 5, 12, 6 works. Notice that we can permute the cards with numbers more than 6 and it will still work.
Problem 6. Olga Smirnov has exactly one brother, Mikhail, and one sister, Sveta. How many children are there in the Smirnov family?
Solution. I shouldn't have posted this problem. In Russian it is clear that Olga is a girl and the answer is 3 children. In English this problem might be confusing.
Problem 7. Every next digit of number N is strictly greater than the previous one. What is the sum of the digits of 9N?
Solution. Remember that 9N = 10N − N. That means the i-th digit of 9N is the (i+1)-th digit of N minus the the i-th digit of 9N. The next to last digit of 9N is the last digit of N minus the next to last digit of N minus 1 (because of the carry). The last digit of 9N is 10 minus the last digit of N. Summing this up, the answer is 9.
Problem 8. Nine gnomes stood in the cells of a three-by-three square. The gnomes who were in neighboring cells greeted each other. Then they re-arranged themselves in the square, and greeted each other again. They did this one more time. Prove that there is at least one pair of gnomes who didn't get a chance to greet each other.
Solution. Color the square in checkerboard colors. There are 8 different ways to assign colors to gnomes for the three rounds. That means there are two gnomes who stood on the same colors all three times. They couldn't have greeted each other.
John Conway told me a story about the book Winning Ways for Your Mathematical Plays that he wrote jointly with Elwyn Berlekamp and Richard Guy. The book includes the birthdays of each of the book's three authors. The book has short author bios, each of which mentions that author's birthday. But in addition, each birthday is hidden inside the book.
Elwyn Berlekamp was born on September 6, 1940, and this date is pictured on a tombstone on page 318 of Volume 2. That chapter is about suiciding moves.
John Conway was born on December 26, 1937. The Doomsday algorithm to calculate the day of the week for a given date is discussed on page 903 of Volume 4. Boxing Day in 1937 is chosen as an example. Boxing Day is a British holiday that originated when wealthy people gave gift boxes to servants the day after Christmas. John Conway was born on Boxing Day. Hidden behind this calculation that it was a Sunday, is the fact that it was Conway's day of birth.
Richard Guy was born on September 30, 1916. He often uses his initials RKG for Richard Kenneth Guy. Maybe because of that he got the nickname Archangel. If you look into the index pages of Volumes 3 and 4, you will find an entry for Archangel that refers to pages 9, 30, 1916. Not only are there no mentions of "Archangel " on pages 9 and 30, the book only has 1004 pages.
The trio of Berlecamp, Conway, and Guy will be the topic of an upcoming MOVES (Mathematics Of Various Entertaining Subjects) conference at the Museum of Mathematics.
The last time I visited John Horton Conway, he showed me a book, Genius At Play: The Curious Mind of John Horton Conway by Siobhan Roberts. It had nothing to do with him wanting to brag about the book. Nothing, nothing at all.
He just wanted to show me a picture from the book. While he was browsing this book about himself, I took a photo of him (featured on the left). The picture he was looking for was on page 314. I am reproducing it here (Photo by Michael A. Stecker, courtesy of Stephen D. Miller).
When we finally found the picture, he asked me how many tennis balls were in it. I smelled a trick question right away and didn't bite. I shrugged. It appears that one of the balls at the far corner at the base of the pyramid had rolled away. So there was one less ball than I would have calculated. So, how many balls are there?
Without using a calculator, can you tell how much 75^{2} is? If you are reading my blog, with high probability you know that it is 5625. The last two digits of the square of a number ending in 5 are always 25. So we only need to figure out the first two digits. The first two digits equals 7 times 8, or 56, which is the first digit of the original number (7) multiplied by the next number (8).
I was good at mental arithmetic and saved myself a lot of money back in the Soviet Union. Every time I shopped I calculated all the charges as I stood at the cash register. I knew exactly how much to pay, which saved me from cheating cashiers. To simplify my practice, the shelves were empty, so I was never buying too many items.
When I moved to the US, the situation changed. Salespeople are not trying to cheat, not to mention that they use automatic cash registers. In addition, the US sales taxes are confusing. I remember how I bought three identical chocolate bars and the total was not divisible by 3. So I completely dropped my at-the-cash-registers mental training.
Being able to calculate fast was somewhat useful many years ago. But now there is a calculator on every phone.
John H. Conway is a master of mental calculations. He even invented an algorithm to calculate the day of the week for any day. He taught it to me, and I too can easily calculate that July 29 of 1926 was Thursday. This is not useful any more. If I google "what day of the week is July 29, 1926," the first line in big letters says Thursday.
One day a long time ago I was watching a TV show of a guy performing mental tricks: remembering large numbers, multiplying large numbers, and so on. At the end, to a grand fanfare, he showed his crowned trick. A member from the audience was to write down a 2-digit number, and to raise it to the fifth power using a calculator. The mental guy, once he is told what the fifth power was, struggled with great concentration and announced the original number.
I was so underwhelmed. Anyone knows that the last digit of a number and its fifth power are the same. So he only needs to guess the first digit, which can easily be estimated by the size of the fifth power.
I found the description of this trick online. They recommend memorizing the fifth powers of the numbers from 1 to 9. After that, you do the following. First, listen to the number. Next, the last digit of the number you hear is the same as the last digit of the original number. To figure out the first digit of the original number, remove the last 5 digits of the number you hear. Finally, determine between which fifth powers it fits. This makes sense. Or, you could be lazy and remember only meaningful digits. For example, 39^{5}=90,224,199, and 40^{5}=102,400,000. So if the number is below 100,000,000, then the first digit is less than 4.
You do not have to remember the complete fifth powers of every digit. You just need to remember significant ranges. Here they are:
first digit | range |
---|---|
1 | between 100,000 and 3,000,000 |
2 | between 3,000,000 and 24,000,000 |
3 | between 24,000,000, 100,000,000 |
4 | between 100,000,000 and 300,000,000 |
5 | between 300,000,000 and 750,000,000 |
6 | between 750,000,000 and 1,600,000,000 |
7 | between 1,600,000,000 and 3,200,000,000 |
8 | between 3,200,000,000, and 5,900,000,000 |
9 | above 5,900,000,000 |
Besides, for this trick the guy needed to guess one out of one hundred number. He had a phenomenal memory, so he could easily have remembered one hundred fifth powers. Actually he doesn't need to remember all the numbers, only the differentiated features. On this note, there is a cool thing you can do. You can guess the number before the whole fifth power is announced. One caveat: a 1-digit number x and 10x when taken to the fifth power, both begin with the same long string of digits. So we can advise the audience not to suggest a 1-digit number or a number divisible by 10, as that is too easy (without telling the real reason).
Now we can match a start of the fifth power to its fifth root. Three first digits are almost always enough. Here is the list of starting digits and their match: (104,16), (107,64), (115,41), (116,65), (118,26), (12,66), (130,42), (135,67), (141,17), (143,27), (145,68), (147,43), (156,69), (161,11), (164,44), (17,28), (180,71), (184,45), (188,18), (19,72), (2051,29), (2059,46), (207,73), (221,74), (229,47), (23,75), (247,19), (248,12), (253,76), (254,48), (270,77), (282,49), (286,31), (288,78), (30,79), (33,32), (345,51), (348,81), (370,82), (371,13), (38,52), (391,33), (393,83), (40,21), (4181,53), (4182,84), (44,85), (454,34), (459,54), (470,86), (498,87), (50,55), (51,22), (525,35), (527,88), (53,14), (550,56), (558,89), (601,57), (604,36), (62,91), (64,23), (656,58), (659,92), (693,37), (695,93), (71,59), (73,94), (75,15), (77,95), (792,38), (796,24), (81,96), (84,61), (85,97), (902,39), (903,98), (91,62), (95,99), (97,25), (99,63).
Or, you can come to the mental guy's performance and beat him by shouting out the correct answer before he can even finish receiving the last digit. This would be cool and cruel at the same time.
* * *
Do you know a statistics joke?
Probably, but it's mean!
* * *
Twelve different world statisticians studied Russian roulette. Ten of them proved that it is perfectly safe. The other two scientists were unfortunately not able to join the final discussion.
* * *
A statistician bought a new tool that finds correlations between different fields in databases. Hoping for new discoveries he ran his new tool on his large database and found highly correlated events. These are his discoveries:
* * *
Scientists discovered that the main cause of living 'till old age is an error on the birth certificate.
* * *
Scientists concluded that children do not really use the Internet. This is proven by the fact that the percentage of people saying 'No' when asked 'Are you over 18?' is close to zero.
* * *
— Please, close the window, it is cold outside.
— Do you think it will get warmer, after I close it?
Some of my friends collect volumes of The Best Writing on Mathematics published by Princeton University Press. The first annual volume appeared in 2010. In 2014, one of my papers, Conway's Wizards, was chosen to be included in the volume The Best Writing on Mathematics 2014.
All my life I have hated writing. My worst grades in high school were for writing. I couldn't write in Russian, and I was sure I would never be able to write in English. I was mistaken. I am a better writer than I gave myself credit for being.
Writing in English is easier for me than writing in Russian because when I make a mistake, I have an excuse. As my written English gets better and better, maybe one day I'll get the courage to try writing in Russian.
I would like to use this opportunity to thank Sue Katz, my friend, English teacher, and editor. Sue edits most of my blog essays. Originally I wrote about Conway's wizards on my blog. When I had three essays on the subject I combined them into a paper. Sue edited the essays and the paper. I am honored to be included in this respected volume and I want to share this honor with Sue.
Kvant was a very popular science magazine in Soviet Russia. It was targeted to high-school children and I was a subscriber. Recently I discovered that a new magazine appeared in Russia. It is called Kvantik, which means Little Kvant. It is a science magazine for middle-school children. The previous years' archives are available online in Russian. I looked at 2012, the first publication year, and loved it. Here is the list of the math puzzles that caught my attention.
The first three problems are well known, but I still like them.
Problem 1. There are 6 glasses on the table in a row. The first three are empty, and the last three are filled with water. How can you make it so that the empty and full glasses alternate, if you are allowed to touch only one of the glasses? (You can't push one glass with another.)
Problem 2. If it is raining at midnight, with what probability will there be sunshine in 144 hours?
Problem 3. How can you fill a cylindrical pan exactly half-full of water?
I like logic puzzles, and the next two seem especially cute. I like the Parrot character who repeats the previous answer: very appropriate.
Problem 4. The Jackal always lies; the Lion always tells the truth. The Parrot repeats the previous answer—unless he is the first to answer, in which case he babbles randomly. The Giraffe replies truthfully, but to the previous question directed to him—his first answer he chooses randomly.
The Wise Hedgehog in the fog stumbled upon the Jackal, the Lion, the Parrot, and the Giraffe, although the fog prevented him from seeing them clearly. He decided to figure out the order in which they were standing. After he asked everyone in order, "Are you the Jackal?" he was only able to figure out where the Giraffe was. After that he asked everyone, "Are you the Giraffe?" in the same order, and figured out where the Jackal was. But he still didn't have the full picture. He started the next round of questions, asking everyone, "Are you the Parrot?" After the first one answered "Yes", the Hedgehog understood the order. What is the order?
Problem 5. There are 12 cards with the statements "There is exactly one false statement to the left of me," "There are exactly two false statements to the left of me." ..., "There are 12 false statements to the left of me." Pete put the cards in a row from left to right in some order. What is the largest number of statements that might be true?
The next three problems are a mixture of puzzles.
Problem 6. Olga Smirnov has exactly one brother, Mikhail, and one sister, Sveta. How many children are there in the Smirnov family?
Problem 7. Every next digit of number N is strictly greater than the previous one. What is the sum of the digits of 9N?
Problem 8. Nine gnomes stood in the cells of a three-by-three square. The gnomes who were in neighboring cells greeted each other. Then they re-arranged themselves in the square, and greeted each other again. They did this one more time. Prove that there is at least one pair of gnomes who didn't get a chance to greet each other.
As I told you before I was recently diagnosed with severe sleep apnea. My doctor ordered me to use a CPAP machine&mdasha Continuous Positive Airway Pressure machine&mdashthat blows air into my nose and improves the quality of overnight breathing.
I was elated, hoping for a miracle. I haven't had a good night's sleep in 20 years. I was so looking forward to waking up rested.
When I went to bed, I put on my nose mask and turned on the machine. The mask was uncomfortable. It gave me headaches and I had to sleep on my back which I do not like. It also had this annoying plastic smell. It took me some time to fall asleep, but eventually I did.
To my disappointment I didn't wake up rested; I woke up tired as usual. Still, I decided not to give up and continued trying the machine.
Several days passed and I found myself mopping the kitchen floor. I never ever mop floors. I have my cleaning crew do that. That mopping was my first positive sign. Then my relatives started telling me that my voice had changed and had become more energetic. A week later I invited my son and his family for my weekly family dinner, which I have been canceling every previous week for several months.
A month after I began sleeping with the machine, I started having night dreams, something I forgot even existed. After two months, I would wake up and my first thought would be, "What day is it today?" Prior to using the machine, my first waking thought was, "My alarm clock must be broken. It's impossible that I have to get up now. I feel too weak to stand up."
My machine is very smart. It records the data from my sleep and uploads it to a website, which my doctor and I can access. Instead of the 37 apnea episodes per hour I had during my sleep study, now I have 1 episode per hour. The quality of my life has changed gradually. I am not yet ready to conquer the world, but I have so much more energy. I've even accepted two more job offers, and now I have six part-time jobs. (I'll tell you all about this some other time.)
There have been some side effects. For some reason I gained 10 pounds during the first two weeks of using the machine. And I feel like an elephant. Not because of my heavy frame, but because my nose mask with its pipe looks like a trunk.
But I prefer feeling like an elephant to feeling like a zombie. This indeed was a miracle, though a very slow-acting miracle.
One day you meet your friend Alice enjoying a nice walk with her husband Bob and their son Carl. They are excited to see you and they invite you to their party.
Alice: Please, come to our party on Sunday at our place at 632 Elm St. in Watertown.
Bob: My wife likes exaggerating and multiplies every number she mentions by 2.
Carl: My dad compensates for my mom's exaggerations and divides every number he mentions by 4.
Alice: Our son is not like us at all. He doesn't multiply or divide. He just adds 8 to every number he mentions.
Where is the party?
Since I was a child I prided myself on knowing arithmetic. I knew how to add fractions. I knew that 2/3 plus 1/4 is 11/12. Some kids around me were struggling and often summed it up the wrong way by adding numerators and denominators separately and getting 3/7 as a result.
As I grew older I found more reasons to ignore the wrong way. For example, the result of such addition depends on the representation of a fraction, not on the fraction itself, and this was bad.
Oh well. I was growing older, but not wiser. Now mathematicians study this wrong addition of fractions. They call such a sum a mediant of two rational numbers. To avoid the dependency on the representation of fractions, the fractions are assumed to be in the lowest terms.
Let us start with the sequence of fractions: 0/1 and 1/0. This sequence is called the Stern-Brocot sequence of order 0. The Stern-Brocot sequence of order n is generated from the Stern-Brocot sequence of order n − 1 by inserting mediants between consecutive elements of the sequence. For example, the Stern-Brocot sequence of order 2 is 0/1, 1/2, 1/1, 2/1, 1/0.
Where are the trees promised in the title? We can build a portion of this binary tree out of the sequence of order n in the following manner. First, ignore the starting points 0/1 and 1/0. Then assign a vertex to every number in the sequence. After that, connect every mediant to one of the two numbers it was calculated from. More precisely, if a number first appeared in the i-th sequence, its only parent is the number that first appeared in the sequence i−1.
There is beautiful theorem that states that every non-negative rational number appears in the Stern-Brocot sequences. The proof is related to continued fraction. Suppose a rational number r is represented as a continued fraction [a_{0};a_{1},a_{2},…,a_{k}], where a_{k} is assumed to be greater than 1 for uniqueness. Then this number first appears in the Stern-Brocot tree of order a_{0} + a_{1} + a_{2} + … + a_{k} + 1, and its parent is equal [a_{0};a_{1},a_{2},…,a_{k} − 1].
My PRIMES student, Dhroova Ayilam, was working on a project suggested by Prof. James Propp. The goal was to find out what happens if we start with any two rational numbers in lowest terms. Dhroova proved that as with the classical Stern-Brocot trees, any rational number in a given range appears in the tree. His paper Modified Stern-Brocot Sequences is available at the arXiv.
I was afraid of my advisor Israel Gelfand. He used to place unrealistic demands on me. After each seminar he would ask his students to prove by the next week any open problems mentioned by the speaker. So I got used to ignoring his requests.
He also had an idea that it is good to learn mathematics through problem solving. So he asked different mathematicians to compile a list of math problems that are important for undergraduate students to think through and solve by themselves. I still have several lists of these problems.
Here I would like to post the list by Andrei Zelevinsky. This is my favorite list, partially because it is the shortest one. Andrei was a combinatorialist, and it is surprising that the problems he chose are not combinatorics problems at all. This list was compiled many years ago, but I think it is still useful, just keep in mind that by calculating, he meant calculating by hand.
Problem 1. Let G be a finite group of order |G|. Let H be its subgroup, such that the index (G:H) is the smallest prime factor of |G|. Prove that H is a normal subgroup.
Problem 2. Consider a procedure: Given a polygon in a plane, the next polygon is formed by the centers of its edges. Prove that if we start with a polygon and perform the procedure infinitely many times, the resulting polygon will converge to a point. In the next variation, instead of using the centers of edges to construct the next polygon, use the centers of gravity of k consecutive vertices.
Problem 3. Find numbers a_{n} such that 1 + 1/2 + 1/3 + … + 1/k = ln k + γ + a_{1}/k + … + a_{n}/k^{n} + …
Problem 4. Let x_{1} not equal to zero, and x_{k} = sin x_{k-1}. Find the asymptotic behavior of x_{k}.
Problem 5. Calculate the integral from 0 to 1 of x^{−x} over x with the precision 0.001.
I regret that I ignored Gelfand's request and didn't even try to solve these problems back then.
I didn't have any photo of Andrei, so his widow, Galina, sent me one. This is how I remember him.
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.
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.
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.
Here is my new logic puzzle.
I thought of a positive integer that is below 100 and is divisible by 7. In addition to the public knowledge above, I privately tell the units digit of my number to Alice and the tens digit to Bob. Alice and Bob are very logical people, but their conversation might seem strange:
Alice: You do not know Tanya's number.
Bob: I know Tanya's number.
What is my number?
I recently posted a cute puzzle about the emperor and his wizards from 2015 Moscow Math Olympiad. It is time for the solution and two new variations. But first let me repeat the puzzle.
The emperor invited 2015 of his wizards to a carnival. Some of the wizards are good and others are evil. The good wizards always tell the truth, whereas the evil ones are free to say anything they want. The wizards know who is who, but the emperor does not.
During the carnival, the emperor asks every wizard a yes-or-no question. Then he expels one of the wizards from his kingdom. The expelled wizard leaves through a magic door, which allows the emperor to discover what kind of wizard s/he was. After that the emperor starts the next round of questions and expels another wizard. He continues the rounds until he decides to stop.
Prove that it is possible to expel all the evil wizards, while expelling not more than one good wizard.
Solution: Suppose the emperor knows one good wizard. Then he can create a chain that leads him to an evil wizard, as follows: Suppose Alice is the known good wizard. The emperor chooses some other wizard, say Bob, and asks Alice "Is Bob evil?" (which question Alice, being good, will answer truthfully). If Bob turns out to be evil, the emperor can expel him, and repeat this (starting with Alice) next round. If Bob turns out to be good, the emperor can continue, asking Bob about Carl, etc, until he either reaches an evil wizard or determines that all remaining wizards are good.
The above means that, if the emperor can find a good wizard sacrificing at most one (other) good wizard, the emperor will succeed. Here is one way to do this: Let the emperor pick Anne and ask everyone else whether Anne is good. Suppose at least one wizard, say Bill, says that Anne is good. The emperor expels Bill. If Bill is revealed to be evil, then nothing is lost, and the emperor can try again next round. If Bill is revealed to be good, then the emperor knows for sure that Anne is good and can proceed to expel all the remaining evil wizards with the chain method above. If, on the other hand, no one says that Anne is good, then the emperor expels Anne. If Anne proves evil, the emperor didn't lose anything and can conduct another trial next round. If Anne proves good, then everyone else is evil, and the emperor can expel them all without asking any more questions.
I like the mathematical part of the puzzle, but I hate when innocent people are punished. So I couldn't stop thinking about the puzzle until I found a variation where no good wizard need be expelled (the magic properties of the gate are redundant now, since the emperor only ever sends evil wizards through it):
The setting is the same as before, except the emperor knows how many evil wizards there are. He wants to expel all the evil wizards without expelling a good one. For which numbers of evil wizards can he do that?
In addition, my reader Leo Broukhis couldn't get through my CAPTCHAs to post a comment (I think there's something wrong with the plugin) but sent me a variation of the original puzzle by email:
There is again an emperor with a magic gate plagued with a superfluity of evil wizards, but this time the carnival is not very long, so the emperor does not have the luxury of asking the wizards many questions. In fact, he is restricted to asking all of them the same single question, after which he will conduct a series of expulsions that must rid the empire of evil wizards while expelling at most one good one. The one saving grace to this difficult situation is that the question need not be limited to "Yes" or "No" answers—an unbounded (single) integer is permissible.
My favorite problem at the 2015 Moscow Olympiad was about an emperor and his wizards.
8-10th grade. Designed by I.V. Mitrofanov. The emperor invited 2015 of his wizards to a carnival. Some of the wizards are good and others are evil. The good wizards always tell the truth, whereas the evil ones are free to say anything they want. The wizards know who is who, but the emperor does not.
During the carnival, the emperor asks every wizard a yes-or-no question. Then he expels one of the wizards from his kingdom. The expelled wizard leaves through a magic door, which allows the emperor to realize what kind of wizard s/he was. After that the emperor starts the next round of questions and expels another wizard. He continues the rounds until he decides to stop.
Prove that it is possible to expel all the evil wizards, while expelling not more than one good wizard.
Two other problems at the Olympiad were noteworthy—because no competitor solved them:
11th grade. Designed by O.N. Kosuhin. Prove that it is impossible to put the integers from 1 to 64 (using each integer once) into an 8 by 8 table so that any 2 by 2 square considered as a matrix has a determinant that is equal to 1 or −1.
9th grade. Designed by A.Y. Kanel-Belov. Do there exist two polynomials with integer coefficients such that each of them has a coefficient with absolute value exceeding 2015, but no coefficient of their product has absolute value exceeding 1?
I am scared of that old German gentleman. Forgot his name. Oh, yeah. Alzheimer.
I started to lose my brain processing speed a long time ago. By my estimates I am about 100 times slower than I used to be. From time to time someone gives me a puzzle I remember I solved in 30 seconds years ago, but now it takes me 30 minutes. And it is getting worse. When I moved to my new apartment recently, it took me a year to remember my own phone number.
On the positive side, I feel quite famous, according to one of the signs of success. I am often greeted by people I don't know.
My moment of action came when I was in my basement doing my laundry and couldn't remember how to turn on my washing machine. This is after 10 years of heavily using this damn machine. I started to look for the button to push, but there were no buttons. There were only knobs, and I couldn't find the word "on" anywhere. After struggling with my memory and with those knobs, I pulled out one of the knobs, and the machine started.
I panicked. I am afraid of Alzheimer's Disease. I do not want to become demented. I do not want to forget how to count or to stop recognizing my children. I do not want to become a drain on my children's time, emotions and money.
I had complained about my memory to my doctor before, but the only thing he ever found was anemia. This time I was more insistent. I had an MRI that ruled out tumors. I had more extensive blood tests that confirmed anemia and showed a Vitamin D deficiency. But then he sent me to a neurologist who suggested a sleep study. Finally I got a diagnosis of severe sleep apnea. I am so happy now. I might not have Alzheimer's Disease. In the worst case scenario, I might die in my sleep. In comparison, this doesn't sound so bad.
I want to come back to a middle-school Olympiad problem I posted a while ago.
Streamline School Olympiad 2000 (8th grade). You have six bags of coins that look the same. Each bag has an infinite number of coins and all the coins in the same bag weigh the same amount. Each different bag contains coins of a different weight, ranging from 1 to 6 grams exactly. There is a label (1, 2, 3, 4, 5, 6) attached to each bag that is supposed to correspond to the weight of the coins in that bag. You have only a balance scale. What is the least number of times you need to weigh the coins in order to confirm that the labels are correct?
The answer is unpretentious: one weighing is enough. We can take one 5-gram coin, two 4-gram coins, three 3-gram coins, four 2-gram coins and five 1-gram coins for the total of 35 grams. This number is not divisible by 6, so we can add one more 1-gram coin and weigh all of them against six 6-gram coins. I leave it to the reader to show that this solution works and to extrapolate the solution for any number of bags.
My new challenge is to find a weighing for the above problem using the smallest number of coins. What is the number of coins in such a weighing for a given number of bags?
I manually calculated this number for a small number of bags, but I would like to get a confirmation from my readers. Starting from 6 bags, I don't know the answer. Can you help me?
The 2015 Intel Science Talent Search results are out. This year they divided the prizes into three categories: basic research, global good, and innovation. All three top prizes in basic research were awarded to our PRIMES students:
PRIMES' success in this year's Siemens competition is even more impressive. Unlike Intel, Siemens didn't divide the projects into three groups. We took the first and second overall individual prizes.
PRIMES is the place for high school math research. Congratulations to all our students—and to me (and my colleagues) for a job well done!
I lead recitations for a Linear Algebra class at MIT. Sometimes my students are disappointed with their grades. The grades are based on the final score, which is calculated by the following formula: 15% for homework, 15% for each of the three midterms, and 40% for the final. After all the scores are calculated, we decide on the cutoffs for A, B, and other grades. Last semester, the first cutoff was unusually low. The top 50% got an A.
Some students who were above average on every exam assumed they would get an A, but nonetheless received a B. The average scores for the three midterm exams and for the final exam were made public, so everyone knew where they stood relative to the average.
The average scores for homework are not publicly available, but they didn't have much relevance because everyone was close to 100%. However, a hypothetical person who is slightly above average on everything, including the homework, should not expect an A, even if half the class gets an A. There are two different effects that cause this. Can you figure them out?
One day I received a call on my home line. I do not like calls from strangers, but the guy knew my name. So I started talking to him. I assumed that it was some official business. He told me that their company monitors Internet activities, and that my computer is emitting viruses into the Internet traffic degrading Internet performance. All I need to do is to go to my computer and he will instruct me how to get rid of my viruses.
While he was saying all this, I covered my phone's microphone and made a call to the police from my cell phone. I was hoping the police could trace the call and do something while I kept the line to the guy open. The police told me to hang up. They said there is nothing they can do.
Meanwhile, the guy on the phone kept directing me to my Start button while I kept telling him that I can't find it. After talking to the police, I got so angry that I told the guy that I wasn't actually looking for the Start button, but talking to the police. So the guy asks, "What does the police say?"
These people are laughing at us. They know that the police do nothing. And then continued instructing me about my Start button.
Have you ever solved a CalcuDoku puzzle, or a MathDoku puzzle? Maybe you have, but you do not know it. Many incarnations of this puzzle are published under different names. The MIT's Tech publishes it as TechDoku. What distinguishes this puzzle type from most others is that it is trademarked. The registered name is KenKen. So anyone can invent and publish a KenKen puzzle as long as they do not call it KenKen.
In this variously named puzzle you need to reconstruct a Latin square, where cells of a square are grouped into regions called cages. Each cage has a number and an operation (addition, subtraction, multiplication, division) in the upper-left corner of the cage. The operation applied to the numbers in the cage must result in a given value. For non-commutative operations (subtraction and division), the operation is applied starting from the largest number in the cage.
These are my two NOT KenKen puzzles. I will call them TomToms, the name for this puzzle used by Tom Snyder in his The Art of Puzzle LINK blog. In the TomTom variation, cages without a number in a corner are allowed and the operation might be missing, but it has to be one of the standard four. The first puzzle I call Three Threes and the second is a minimalistic version where only one number without the operation is given.
But my goal today is not to discuss KenKen or its ekasemans. Ekaseman is the reverse of the word namesake. My son Alexey invented the term ekaseman to denote a different name for the same thing. My real goal is to discuss a new type of puzzle that can be called Crypto KenKen. In this puzzle the digits in the corner are encrypted using a substitution cipher: each digit corresponds to its letter. I first saw this puzzle at Tom Snyder's blog, where it is called TomTom (Cipher). I think the crypto version of this puzzle deserves its own name. So I suggest PamPam: it is an encryption of KenKen as well as TomTom. And it would be nice to have a female name for a change.
It is time to report on my weight loss progress. Unfortunately, the report is very boring; I am still stuck at the same weight: 225. What can I do? Let's laugh about it. Here are some jokes on the subject.
* * *
After the holidays I stepped on my scale. After an hour I tried again and had a revelation: tears weigh nothing!
* * *
I am on a miracle diet: I eat everything and hope for a miracle.
* * *
Ideas to lose weight: A glass of water three days before your meal.
* * *
I wanted to lose five pounds by this summer, now I have only ten pounds to lose to reach my goal.
I wrote a paper with my son, Alexey Radul, titled (Not so) Much Ado About Nothing. As the title indicates, nothing is discussed in this paper. It's a silly, humorous paper full of puns about "nothing." We submitted the paper to the arXiv two months ago, and it has been on hold since then.
This reminds me of an earlier paper of mine that the arXiv rejected because it didn't have journal references. (Not so) Much Ado About Nothing is done in proper style. It follows all the formal rules of math papers, and contains references, acknowledgements, an introduction with motivation, and results. However, the results amount to nothing. The fact that this paper is not accepted is a good sign. It means the arXiv doesn't just look at the papers formally; they look at the content as well.
On the other hand, the paper is submitted as a paper in recreational mathematics, and it is humorous, so it could have been accepted, since nothing is more recreational than nothing.
Neither rejection nor acceptance would have surprised me. The only thing I do not understand is why it is on hold. Why hold on to nothing?
I know how to walk. Everyone knows how to walk. Or so I thought. Now I am not sure any more.
I've been taking ballroom dance lessons on and off for many years. But at some point I stopped progressing. I got stuck at the Silver level. I know many steps and am a good follower, but I often lose balance and my steps are short.
Then I met Armin Kappacher, an unusual dance coach for the MIT Ballroom dance team. I would like to share some of his wisdom with you, but Armin doesn't have much presence on the web. He only wrote one article for Dance Archives: A Theoretical and Practical Approach to "Seeing The Ground of a Movement."
Although I've been attending his classes for several years, I haven't been able to understand a word. He might say, "Your right arm is disconnected from your chest center." But what does that mean? Others seemed to understand him, because they greatly improved under his guidance. But I was so out of touch with my body that I couldn't translate his words into something my body could understand. Being a mathematician, I lived my whole life in my brain. I never tried to listen to my body. I was never aware of whether my forehead was relaxed or tensed, or if my pelvic floor was collapsed. I grasped that it was my own fault that I failed to understand Armin. I stuck with his classes.
After three years of group classes, I asked Armin for a private lesson with an emphasis on the basics. He instructed me to take three steps and quickly discovered my issues, which included:
So now I am taking walking classes from Armin. I am slowly starting to feel what Armin means when he says that my pelvic floor is collapsed. I feel better now. Whenever I pay attention to how I am walking, my posture improves. As a result, I feel more confident, my mood approves, and I feel like more oxygen is getting to my brain. My friends have noticed a change. For example, my son Sergei got married recently, and I was sitting under the Chuppah during the ceremony (see photo). Afterwards, several friends told me that I looked like a queen.
I have to give some credit to my earrings. They were too long and were getting caught on my dress. So I was constantly trying to wiggle my head up—using Armin's techniques.
I proudly brought this photo to Armin to show him my queen-like posture. He told me that I look okay above the chest center. But below the chest center my spine is still collapsed. Next time I will take sitting lessons with Armin.
Skyscraper puzzles are one of my favorite puzzles types. Recently I discovered a new cute variation of this puzzle on the website the Art of Puzzles. But first let me remind you what the skyscraper rules are. There is an n by n square grid that needs to be filled as a Latin square: each number from 1 to n appears exactly once in each row and column. The numbers in the grid symbolize the heights of skyscrapers. The numbers outside the grid represent how many skyscrapers are seen in the corresponding columns/row from the direction of the number.
The new puzzle is called Skyscrapers (Sum), and the numbers outside the grid represent the sum of the heights of the skyscrapers you see from this direction. For example, if the row is 216354, then from the left you see 8(=2+6); and from the right you see 15(=4+5+6).
Here's an easy Skyscrapers (Sum) puzzle I designed for practice.
The Art of Puzzles has four Skyscrapers (Sum) puzzles that are more difficult than the one above:
* * *
A button of unknown functionality should be pressed an even number of times.
* * *
When I tell you that I am closer to 30 than to 20, I mean to tell you that I am 42.
* * *
If a car with a student-driver sign gets its windshield wipers turned on, then the car is about to turn.
* * *
I always learn from the mistakes of people who followed my advice.
* * *
A traffic policeman stops a car:
—You're going 70 in a 35 miles-per-hour zone.
—But there are two of us!
* * *
The most popular tweet, "Live your life so that you do not have time for social networks."
I recently wrote a post about the ApSimon's Mints problem:
New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material. Therefore, fake coins weigh the same no matter where they've been minted. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings in order to deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?
The post was accompanied by my paper Attacking ApSimon’s Mints.
Unfortunately, both the post and the paper contain wrong information. They both state that the number of coins as a sequence of the number of mints is 1, 2, 4, 8, 15, 38, 74. This is wrong. I took this data from the sequence A007673 in the OEIS database. The sequence had incorrect data lying dormant for 20 years. I believe that the sequence was generated from the paper of R. K. Guy and R. J. Nowakowsky, ApSimon's Mints Problem, published in Monthly in 1994. To the credit of Guy and Nowakowsky, they never claimed to find the best solution: they just found a solution, thus providing a bound for the sequence. Someone mistook their solution for the optimal one and generated the sequence in the database.
After my post, my readers got interested in the problem and soon discovered the mistake. First Konstantin Knop found a solution for 6 mints with 30 coins and for 7 mints with 72 coins. Konstantin is my long-time collaborator. I trust him so I was sure the sequence was flawed. Then someone located a reference to a paper in Chinese A New Algorithm for ApSimon's Mints Problem. Although none of my readers could find the paper itself nor translate the abstract from Chinese. But judging from the title and the formulae it was clear that they found better bounds than the sequence in the database. My readers got excited and tried to fix the sequence. David Reynolds improved on Konstantin's results with a solution for 6 mints with 29 coins and for 7 mints with 52 coins. David did even better on his next try with 28 and 51 coins correspondingly. He also found a solution with 90 coins for 8 mints. Moreover, his exhaustive search proved that these were the best solutions.
Now the sequence in the database is fixed. It starts 1, 2, 4, 8, 15, 28, 51, 90.
For future generations I would like to support each number of the sequence by an example. I use the set P(Q) to represent the sequence of how many coins are taken from each mint for the first(second) weighing. For one mint, only one coin and one weighing is needed. ApSimon himself calculated the first five values, so they were not in dispute.
There is a solution for nine mints using 193 coins that is not confirmed to be optimal. It was found by David Reynolds: P=(1,2,4,12,5,4,20,39,43) and Q=(0,1,3,3,25,33,34,18,27). In addition, David Reynolds provided a construction that reduces the upper bound for n mints to (3^{(n+1)}−2n−3)/4. The following set of coins work: P=(1,3,7,15,…,2^{n}−1) and Q=(1,4,13,40,…,(3^{n}−1)/2).
Recently I wrote that my blog is under attack by spam comments. Most of the comments were caught by my spam-filter Akismet, the best-known filter for WordPress. I was receiving about 50,000 comments a day and 200 of them were sneaking through this filter. I had to moderate those and delete them. This was an extreme waste of my time. But I can understand that the bots achieved some goal. None of their comments made it to my website, but at least I myself was made aware of opportunities for hair removal in Florida.
The comments crashed my server and I had to install CAPTCHAs. I was happy that the number of comments that I had to moderate went down, but the total number of comments was still so high that my server kept crashing. Now that the comments are blocked from human view, why are they still pouring in? One software package is trying to inform the other software package about weight-loss wonder drugs. I am convinced that Akismet is not interested.
My hosting provider couldn't handle the traffic and asked me to upgrade to a more advanced hosting package. It's annoying that I have to pay a lot of money for these bots' attempts to sell Akismet fake Louis Vuitton bags.
The upgrade was too expensive, so I tried a different solution. I closed comments for older posts. It didn't help. The bad software continued trying to leave comments that can't be left. They especially like my post about Cech cycles, called A Mysterious Bracelet. My weblog tells me that every second someone downloads this page and tries to leave a comment. But no one will ever see these comments. Even Akismet will never know what it is missing: it might have had a chance to make $5,000 a day from home.
I recently posted the following old Olympiad problem:
Prove that you can choose 2^{k} numbers from the set {1, 2, 3, …, 3^{k}−1} in such a way that the chosen set contains no averages of any two of its elements.
Let me show how to find 2^{k} − 1 such numbers. We can pick all the numbers that have 0 or 1 in their ternary representation. Let me prove that this set doesn't contain averages. Summing two such numbers doesn't involve carry, and the sum will contain a 1 in each place where the digits differ. On the other hand, a double of any number in this set doesn't contain ones.
This solution is pretty, but it is not good enough: we need one more number. We can add the number 3^{k}−1. I will leave it to the reader to prove that the largest number in the group whose binary representation consists only of twos can be added without any harm.
There are other ways to solve this problem. It is useful to notice that multiplying a no-average set by a constant or adding a constant to it, doesn't change the no-average property.
If we were allowed to use 0, then the problem would have been solved. As zero doesn't belong to the initial range, we can add zero and shift everything by 1. The resulting sequence is sequence A3278. This sequence is the lexicographically first non-averaging sequence.
Another solution was suggested by devjoe in my livejournal mirror blog site. If we multiply our non-averaging set (the one that doesn't have twos in their ternary representation) by 2, we get a set of all numbers that do not contain ones in their ternary representation. By linearity, such a set doesn't contain averages either. We can add 1 to this set.
The Intel International Science and Engineering Fair announced 2014 Grand Awards. I worked with three out of the top five mathematical award winners. Now I can brag that I've got my finger in more than half of the world's best high-school math research.
To be clear: I wasn't actually mentoring these projects, but I supervised two of the projects and I trained the third student for several years. So I'm proud to list the award-winning papers:
How interesting that each of these three students is from a different part of my present career. It certainly feels that I am in all the right places.
* * *
A mathematical tragedy: two parallel lines fall in love.
* * *
Life is not fair, even among gadgets: the desktop misbehaves, the monitor gets smacked.
* * *
An amazing magic trick! Think of a number, add 5 to it, then subtract 5. The result is the number you thought of!
* * *
—How can you distinguish a mathematician from a physicist?
—Ask for an antonym for the word parallel.
—And?
—A mathematician will answer perpendicular, and a physicist serial.
* * *
—How can you distinguish a physicist from a mathematician?
—Ask the person to walk around a post.
—And?
—A physicist will ask why, and a mathematician clockwise or counter-clockwise?
* * *
—Some bike thief managed to open my combination lock. How could they possibly guess that the combo was the year of the canonization of Saint Dominic by Pope Gregory IX at Rieti, Italy?
—What year was that?
—1234.
* * *
—Hello? Is this the anonymous FBI tip-line?
—Yes, Mr. Benson.
* * *
—My five-year-old son knows the first 20 digits of Pi.
—Wow!
— I use it as the password on my laptop, where I keep all the games.
* * *
I learned three things in school: how to rite and how to count.
I recently published a crypto word search puzzle: a word search puzzle where all the letters are encrypted by a substitution cypher. The answer to such a puzzle is a word or a phrase formed by those decrypted letters that are not in the hidden words.
The puzzle I posted was easy. It can be solved by analyzing the repeated letters in the hidden words. The new puzzle is more difficult. No hidden word has repeated letters.
The hidden words are: FUN, HUNT, SOLVE, STARK, TABLE, THINK.
The puzzle below can be defined as a Crypto Word Search. Guess what needs to be done in this puzzle. The answer is a word.
The hidden words are: DUKE, EYES, RUSE, WORD, WUSS.
The idea of a crypto word search came to me from a beautiful, but devilishly difficult, puzzle In the Details, designed by Derek Kisman. In the Details can be described as a fractal word search; it contains a crypto word search as one of the simpler steps.
Let us call a natural number awesome if it can be represented as a^{b} + b^{a}, where a and b are natural numbers. For example, number 57 is awesome as 57 = 2^{5} + 5^{2}. Is 2014 awesome?
What parking spot number is the car parked in?
John Conway likes playing with the Fibonacci sequence. He invented many new sequences using the following trick. The next number in the sequence is the sum of the two previous number adjusted in some way. Free Fibonacci sequences were invented this way. Here is the recurrence for an n-free Fibonacci sequence: the next number in the sequence is the sum of the previous two numbers divided by the highest possible power of n.
Let us calculate a 2-free Fibonacci sequence starting with 5 and 4: 5, 4, 9, 13, 11, 3, 7, 5, 3, 1, 1, 1, …. I leave it to the reader to show that any 2-free sequence ends with a cycle of length one.
Let us try a 3-free Fibonacci sequence starting with 5 and 6: 5, 6, 11, 17, 28, 5, 11, 16, 1, 17, 2, 19, 7, 26, 11, 37, 16, 53, 23, 76, 11, 29, 40, 23, 7, 10, 17, 1, 2, 1, 1, 2, and so on. We are now in the cycle of length 3. Is this always the case? Not quite. If there is a 1-1-2 cycle there should be a 2-2-4 cycle, or any cycle k-k-2k, where k is coprime with 3. But the question remains: does it always end in a cycle of length 3?
I published a paper Free Fibonacci Sequences with Brandon Avila. We conjecture that a 3-free Fibonacci sequence always ends in a cycle and support this conjecture with a probabilistic argument. We were amused by how the behavior changes when we move to 4-free Fibonacci sequences. It seems that in this case sequences never cycle. We were even more amused when we moved to 5-free Fibonacci sequences and discovered that the behavior changes again.
When n equals 5 there are some sequences that cycle. Can you find the cycles? There are also sequences that grow indefinitely and we do not need a probabilistic argument to prove that. Consider Lucas numbers: 2, 1, 3, 4, 7, 11, and so on. This is a Fibonacci-like sequence that never has a term divisible by 5. Thus Lucas numbers form a 5-free Fibonacci sequence. We made a probabilistic argument that most of the starting terms converge eventually to a Lucas-like sequence that grows indefinitely because there are no terms divisible by 5.
What happens for larger n? We didn't manage to find any cycles there. Would you like to try?
My weight loss progress is progress no more. I am stuck at 225.
I have my morning routine. I wake up and jog to the facilities; then I weigh myself. Why do I do this in this order? Because I do not use an alarm-clock. I depend on my own hydro-alarm that wakes me up every morning very reliably. The downside is that this alarm doesn't have a snooze button. I just have to get up, and get up very fast. So I have to relieve myself first; then I can weigh myself calmly. As a collateral benefit, by availing myself first, I get a slightly better result.
While weighing myself every day, I noticed some patterns. My weight is slightly better if I get a long night of sleep. I guess my availing is more voluminous then.
Another pattern that I noticed is that if I do not eat after 5:00 pm, I do not gain weight and sometimes I lose a bit; if I do eat after 5:00 pm, even a very light dinner, I can gain up to 3 pounds. This becomes a struggle and sometimes hunger wins. So I stick to the plan for three weeks, and then one moment of tiredness and hunger, one dinner, and, boom, I am back where I was.
This is even affecting my social life. I stopped inviting friends and family for dinner and I am tense whenever I visit my friends in the evening.
The surprising thing is that my system worked at the beginning. I keep asking myself what the difference is. I think I was strongly motivated and excited about my weight loss plan. It's impossible to sustain that level of excitement, so I've adopted a new element in my plan. I decided to replace my initial excitement by external motivation. I am now listening to motivational tapes in my car. We'll see.
Whenever I am under stress, I turn to jokes. My recent problems with spam attacks on my blog led me to surf the web for new math jokes. Here are some of my recent translations from Russian.
* * *
Two is the same thing as eight, to some degree.
* * *
A girl to her mathematician boyfriend:
— Let's do something that is forbidden tonight.
— Divide by zero?
* * *
If thoughts converge, they are bounded.
* * *
A mathematician's son:
— Dad, how do I write the number 8?
— That's easy: rotate the infinity symbol by pi over 2.
* * *
My student couldn't take an integral from my book. So he took the book together with all the integrals there.
* * *
Archimedes, Pascal and Newton play hide and seek. Archimedes is the seeker. Pascal hides, but Newton draws a 1-meter square around himself. Archimedes opens his eyes and shouts:
— I see Newton!
— Oh, no! One newton per square meter is the pascal.
* * *
What a pleasure to smoke an e-cigarette after cybersex...
* * *
Russians were the first in the world to create a computer program that passes the Turing test. Scientists tested the program using several Russians with a variety of questions, and each time the program gave the same answer as the people. The reply to every question was, "Go f*ck yourself!"
* * *
There are two types of people: those who know nothing about fractals and those who think that there are two types of people: those who know nothing about fractals and those who think that there are two types people…
My readers noticed that my blog disappeared several times. Here's what's been happening. Spammers were sending tens of thousands of comments a day, which crashed the server several times. My hosting provider doteasy.com couldn't handle it and took down my blog. They asked me to install CAPTCHAs.
Installing CAPTCHAs became a big issue. Since I started my blog in 2008 I've never bothered to update the Wordpress software. As a result, all the CAPTCHA plugins I tried to install wouldn't work. I had no choice but to update Wordpress. After so many years, this would surely not be a simple matter. So I decided to hire someone to help. It took me a month to find Brett Mellor, but finally he updated my Wordpress software.
Updating such an old database starts with a backup. We didn't want to back up all the tens of thousands of spam comments, but we couldn't sort through them, because my hosting provider was blocking the user-friendly access to my blog. We were forced to delete all the new comments. In the process, I've lost some of the legitimate comments as well. I apologize for that. If you do not see your comments, please resubmit.
In addition, we had to re-number all the categories, so the old category links no longer work.
I was paralyzed by all the stress of not only losing access to my blog, but also not knowing how to solve the problem. This prevented me from writing for three months, and for this I apologize as well.
Now my blog is updated and CAPTCHAs are installed. I've gone from 50,000 spam comments per day to 500. However, my hosting provider still complains about unusual activity and spam comments.
Your brilliant suggestions are very welcome.
I always wanted to be a person of the world. I wanted my genes to be a mixture of everything. I was glad that I had a great-grandfather from Poland and a great-great-great-grandmother from France. I was also thrilled when my mom told me that her Asian students think she is one of theirs. So I decided to send my DNA to 23andMe and really see what I have.
To my surprise, my world is not as mixed as I expected: I am 99.5% European. My Asian part is minuscule: 0.2%, out of which only 0.1% is assigned as Yakut. My African part is also 0.2%.
My European part is a mixture of mostly eastern and northern European. I am 2.8% Ashkenazi.
In addition to my genetic profile, 23andMe sent me the list of a thousand of my distant relatives. They also sent me a report about the most common last names among my relatives. The list starts with Cohen and continues with Levine, Levin, Goldberg, and Rubin.
You might be surprised by this list of Jewish names when I am only 2.8% Jewish. But the list is based on people who decided to send their DNA to 23andME and provided their last names. All my Russian relatives remained in Russia. Russia has its own company, I-gene, that provides a similar service, and the two databases are not shared.
Only my distant relatives who moved to the US and who are curious about their ancestry and who are willing to share their last names will appear on this list. So maybe this list is not surprising.
Here is an old Olympiad problem:
Prove that you can choose 2^{k} numbers from the set {1, 2, 3, …, 3^{k}−1} in such a way that the chosen set contains no averages of any two of its elements.
* * *
—Describe yourself in three words.
—Lazy.
* * *
Internet forum:
—Tell me about yourself.
—I am lazy and I like to eat.
—Tell me some more.
—I am tired of typing. I'll go grab a snack.
* * *
—Why do you want to divorce your wife?
—She nags too much. For the last half six month, she's been bugging me everyday to throw away the Christmas tree.
* * *
Yesterday I realized that I'm not the laziest person in the world. I saw my neighbor walking the dog on a leash through his window.
* * *
The list of symptoms of laziness:
1)
This is a variation on an old quiz. Can you answer the last question?
—An airplane carries 500 bricks. One of the bricks falls out. How many bricks are left in the airplane?
—This is easy: 499!
—Correct. Next question. How do you put a giraffe into a refrigerator?
—Open the refrigerator, put in the giraffe, and close the refrigerator door.
—Good, next. How do you put an elephant into a refrigerator?
—Open the refrigerator, take out the giraffe, put in the elephant and close the door.
—Correct. The Lion King is hosting his birthday party. All the animals come to congratulate him—except one. Why?
—The elephant couldn't come because it is in the refrigerator.
—Fantastic, next. A man needs to cross a river inhabited by crocodiles and he doesn't have a boat. What should he do?
—He can just swim: all the crocodiles are attending Lion King's birthday party.
—Amazing! The last question: The man swims across the river, and dies. What happened?
Hugh ApSimon described the following coin puzzle in his book Mathematical Byways in Ayling, Beeling and Ceiling.
New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material: fake coins weigh the same independently of the mint. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings so that you can deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?
I will follow ApSimon's notation. Suppose P_{r} and Q_{r} is the number of coins from the mint r used in the first and the second weighing correspondingly. That is, we are minimizing Σmax(P_{r},Q_{r}). (All my summations are over all the mints. I skip the summation limits because it is difficult to write math in html.) Let us denote by W the weight of the genuine coin and by W(1 + ε) the weight of the fake coin. We do not know ε, except that it is not zero.
Let d_{r} be either 0 or 1, depending on what material the r-th mint uses. Thus, the coin from the r-th mint weighs W(1 + d_{r}ε). We know the results of these two weighings and the weight of the genuine coin. Therefore, we can calculate the following two values: a = ΣP_{r}d_{r}ε and b = ΣQ_{r}d_{r}ε.
It is clear that we need to request at least one coin from each mint and use it in at least one weighing: P_{r} + Q_{r} > 0. If both sums a and b are zero, then all the mints are producing genuine coins. Neither of the two values gives us much information as we do not know ε. We can get rid of ε by dividing a by b.
There are 2^{n} − 1 combinations of possible answers: these are subsets of the set of mints producing fake coins given that there is at least one. Thus we need to select numbers P_{r} and Q_{r}, so that a/b produces 2^{n} − 1 possible answers for different sets of values of d_{r}.
Let us consider cases in which the total number of mints is small. If there is one mint we can take one coin and we won't even need a second weighing. For two mints we need one coin from each mint for a total of 2. For three mints, one coin from each mint is not enough. I leave this statement as an exercise. It is possible to test three mints with four coins: one each from the first and second mints and two from the third mint. The coins from each mint for the first and second weighings are (0,1,2) and (1,1,0) respectively.
To prove that this works we need to calculate (d_{2} + 2d_{3})/(d_{1} + d_{2}) for seven different combinations of d_{r}. I leave this as an exercise.
This puzzle seems to be very difficult. We only know the answer if the number of mints is not more than seven. The corresponding sequence A007673 in the OEIS is: 1, 2, 4, 8, 15, 38, 74. It is possible to give bounds for this sequence, but they are so far apart. The lower bound is n. And the ApSimon's book offers a construction for two weighings were P_{r} = r! and Q_{r} = 1.
You can try to find a better construction, or you can try calculating more terms of the sequence. You can also read more about this problem in my short paper Attacking ApSimon's Mints.
I do not want to leave the readers with the puzzle that might end up being intractable. So I suggest the following easy puzzle. Solve the ApSimon's Mints problem assuming that the weight of the fake coin is known.
I once took an accent reduction course, to modify my Russian accent in English. In the first class the teacher explained that the biggest reason people have strong accents is that they stop learning and trying to improve their speech as soon as they can be understood. I promised myself to never stop learning and to continue working on my accent reduction forever.
Once I was giving a lecture on probability and statistics at the IAP mathematical series. My last slide was about the research on the correlation between masturbation male habits and prostate cancer. Their interpretation of the data had been wrong and a very good example of what not to do.
So I looked directly into the eyes of the course coordinator, who was observing my lecture, and without realizing what I was saying, asked, "Do we have time for masturbation?"
Everyone started laughing and I had to present my slide in order to explain myself.
The news of my double entendre spread. Soon after that I was asked to give a lecture at the Family Weekend at MIT. I wonder if that is why the lecture coordinator asked me not to discuss masturbation as small children might be present.
Luckily that was the only fallout from my blooper. Anyway, I decided to stop working on my accent. When people understand that English is not my first language they forgive more readily my slips of tongue.
* * *
—Honey, have you blocked our computer?
—Yes.
—What's the password?
—Our wedding date.
—%?#!!
* * *
—What's the pin on our card?
—We're on a public chat, honey. Why don't I sms it?
—But I forgot my phone. Please tell me, cupcake!
—Okay. By digit: the second digit of our apartment number, the fourth digit of your phone number, the month of my birthday, and the number of our children.
—Got it. How clever! 8342, right?
* * *
—Where is the report?
—We are stuck. The tech people took our monitor with passwords.
—What!?!
—Our monitor got broken so the techs took it for repair. Our passwords were written on the stand.
Here is a setup that works for the several puzzles that follow it:
The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat—from his inexhaustible supply—on every wizard's head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan's signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide together on a strategy.
I wrote about puzzles with this setup before in my essay The Wizards' Hats. My first request had been to maximize the number of wizards who are guaranteed to survive. It is easy to show that you cannot guarantee more than 50 survivors. Indeed, each wizard will be right with probability 0.5. That means whatever the strategy, the expected number of wizards guessing correctly is 50. My second request had been to maximize the probability that all of them will survive. Again, the counting argument shows that this probability can't be more than 0.5.
Now here are some additional puzzles, including the first two mentioned above, based on the same setup. Suggest a strategy—or prove that it doesn't exist—in which:
As I mentioned, I already wrote about the first two questions. Below are the solutions to those questions. If you haven't seen my post and want to think about it, now is a good time to stop reading.
To guarantee the survival of 50 wizards, designate 50 wizards who will assume that the total number of red hats is odd, and the rest of the wizards will assume that the total number of red hats is even. The total number of red hats is either even or odd, so one of the groups is guaranteed to survive.
To make sure that all of them survive together with probability 0.5, they all need to assume that the total number of red hats is even.
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.
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.
I mentor three PRIMES projects. One of them, with Joshua Xiong from Acton-Boxborough Regional High School, is devoted to impartial combinatorial games. We recently found a connection between these games and cellular automata. But first I need to remind you of the rules of Nim.
In the game of Nim there are several piles with counters. Two players take turns choosing a pile and removing several counters from it. A player loses when he or she who doesn't have a turn. Nim is the most famous impartial combinatorial game and its strategy is well known. To win, you need to finish your move in a so called P-position. Nim P-positions are easy to calculate: Bitwise XOR the number of counters in all the piles, and if the result is zero then it's a P-position.
The total number of counters in a P-position is even. So we calculated the sequence a(n): the number of P-positions in the game of Nim with three piles with the total number of counters equal to 2n. As soon as we got the sequence we plugged it into the OEIS, and voilà it was there: The sequence A130665 described the growth of the three branches of the Ulam-Warburton cellular automaton.
The first picture shows the automaton after 6 generations. The automaton consists of cells that never die and it grows like this: start with a square on a square grid. In the next generation the squares that share exactly one side with the living squares are born. At the end remove the Southern branch.
Everything fell into place. We immediately realized that the language of the automata gives us the right words to describe what we know about the game of Nim.
Now we want to describe the automaton related to any impartial combinatorial game. Again, the cells never die and the initial cells correspond to terminal P-positions. People who write programs for calculating P-positions will find a notion of the next generation very natural. Indeed, the program usually starts with the terminal P-positions: they are generation 0. Then we can proceed by induction. Suppose we have found P-positions up to generation i. Denote the positions that are one move away from generation i and earlier as N_{i}. Then the P-positions that do not belong to generation i and earlier and from which all moves belong to N_{i} are the P-positions from generation i + 1.
This description explains the generations, but it doesn't explain who is the parent of a particular P-position. The parent-child relations are depicted as edges on the cellular automaton graph. The parent of position P_{1} from generation i + 1 is a P-position P_{2} in generation i that can be reached from P_{1} in the game.
The parent-child relationship in the game of Nim is especially easy to explain. A P-position P_{1} is a parent of a P-position P_{2} if P_{1} differs from P_{2} in exactly two piles and it has one fewer counter in each of these piles. For example, a P-position (1,3,5,7) has six parents, one of them is (1,3,4,6). In the game with thee piles a P-position always has exactly one parent.
A position in the game of Nim with three piles is naturally depicted as a triple of numbers, that is as a point in 3D. The picture below shows the Nim automaton in 3D at generation 6.
Our paper, Nim Fractals, about sequences enumerating P-positions and describing the automaton connection in more detail is posted at the arXiv:1405.5942. We give a different, but equivalent definition of a parent-child relationship there. A P-position P_{1} is a parent of P_{2} if there exists an optimal game such that P_{1} is achieved from P_{2} in exactly two moves in a game which takes the longest number of possible moves.
There are too many Olympiads. Now there is even a special undergraduate Olympiad in probability, called Kolmogorov Student Olympiad in Probability. It is run by the Department of Probability Theory of Moscow State University. I just discovered this tiny Olympiad, though it has been around for 13 years.
A small portion of the problems are accessible for high school students. These are the problems that I liked. I edited them slightly for clarity.
Second Olympiad. Eight boys and seven girls went to movies and sat in the same row of 15 seats. Assuming that all the 15! permutations of their seating arrangements are equally probable, compute the expected number of pairs of neighbors of different genders. (For example, the seating BBBBBBBGBGGGGGG has three pairs.)
Third Olympiad. One hundred passengers bought assigned tickets for a 100-passenger railroad car. The first 99 passengers to enter the car get seated randomly so that all the 100! possible permutations of their seating arrangements are equally probable. However, the last passenger decides to take his reserved seat. So he arrives at his seat and if it is taken he asks the passenger in his seat to move elsewhere. That passenger does the same thing: she arrives at her own seat and if it is taken, she asks the person to move, and so on. Find the expected number of moved passengers.
Third Olympiad. There are two 6-sided dice with numbers 1 through 6 on their faces. Is it possible to "load" the dice so that when the two dice are thrown the sum of the numbers on the dice are distributed uniformly on the set {2,...,12}? By loading the dice we mean assigning probabilities to each side of the dice. You do not have to "load" both dice the same way.
Sixth Olympiad. There are M green and N red apples in a basket. We take apples out randomly one by one until all the apples left in the basket are red. What is the probability that at the moment we stop the basket is empty?
Seventh Olympiad. Prove that there exists a square matrix A of order 11 such that all its elements are equal to 1 or −1, and det A > 4000.
Twelfth Olympiad. In a segment [0,1] n points are chosen randomly. For every point one of the two directions (left or right) is chosen randomly and independently. At the same moment in time all n points start moving in the chosen direction with speed 1. The collisions of all points are elastic. That means, after two points bump into each other, they start moving in the opposite directions with the same speed of 1. When a point reaches an end of the segment it sticks to it and stops moving. Find the expected time when the last point sticks to the end of the segment.
Thirteenth Olympiad. Students who are trying to solve a problem are seated on one side of an infinite table. The probability that a student can solve the problem independently is 1/2. In addition, each student will be able to peek into the work of his or her right and left neighbor with a probability of 1/4 for each. All these events are independent. Assume that if student X gets a solution by solving or copying, then the students who had been able to peek into the work of student X will also get the solution. Find the probability that student Vasya gets the solution.
The Russian website problems.ru has a big collection of math problems. I use it a lot in my work as a math Olympiad coach. Recently I was giving a statistics lesson. While there was only one statistics problem on the website, it was a good one.
Assume that every person in every country was tested for IQ. A country's IQ rating is the average IQ of the population. We also assume that for the duration of this puzzle no one is born and no one dies.
- A group of citizens of country A emigrated to B. Show that the rating of both countries can go up.
- After that a group of citizens of B (which may include former citizens of A) emigrated to A. Is it possible that the ratings of both countries go up again?
- A group of citizens of A emigrated to B, and a group of citizens of B emigrated to C. As a result, the ratings of each country increased. After that the migration went the opposite way: some citizens of C moved to B, and some citizens from B moved to A. As a result, the ratings of all three counties went up once more. Is this possible? If yes, then how? If no, then why not?
My AMSA students loved the following puzzle from the 2003 Math Kangaroo contest for grades 7-8:
The children A, B, C and D made the following assertions.How many of the children were telling the truth?
- A: B, C and D are girls.
- B: A, C and D are boys.
- C: A and B are lying.
- D: A, B and C are telling the truth.
A) 0 B) 1 C) 2 D) 3 E) Impossible to determine
I was given this puzzle at the last Gathering for Gardner.
Use arithmetic operations to express 6 using three identical digits. For each digit from 0 to 9 find at least one way to express 6.
For example, I can express 6 using three twos in many ways: 2 + 2 + 2, or 2 · 2 + 2, or 2^{2} + 2. But the problem doesn't ask for many ways. One way is enough, but you need to do it for every digit. So nine more cases to go: 0, 1, 3, 4, 5, 6, 7, 8, and 9.
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.
This is one of my favorite jokes:
Three logicians walk into a bar. The waitress asks, "Do you all want beer?"
The first logician answers, "I do not know."
The second logician answers, "I do not know."
The third logician answers, "Yes."
This joke reminds me of hat puzzles. In the joke each logician knows whether or not s/he wants a beer, but doesn't know what the others want to drink. In hat puzzles logicians know the colors of the hats on others’ heads, but not the color of their own hats.
This is a hat puzzle which has the same answers as in the beer joke. Three logicians walk into a bar. They know that the hats were placed on their heads from the set of hats below. The total number of available red hats was three, and the total number of available blue hats was two.
Three logicians walk into a bar. The waitress asks, "Do you know the color of your own hat?'"
The first logician answers, "I do not know."
The second logician answers, "I do not know."
The third logician answers, "Yes."
The puzzle is, what is the color of the third logician's hat?
This process of converting jokes to puzzles reminds me of the Langland's Program, which tries to unite different parts of mathematics. I would like to unite jokes and puzzles. So here I announce my own program:
Tanya's Program: Find a way to convert jokes into puzzles and puzzles into jokes.
Each time I see John Conway he teaches me something new. At the Gathering for Gardner he decided to quiz me on how well I know a regular six-sided die. I said with some pride that the opposite sides sum up to 7. He said, "This is the first level of knowledge." So much for my pride. I immediately realized that the next level would be to know how all the numbers are located relative to each other. I vaguely remembered that in the corner where 1, 2, and 3 meet, the numbers 1, 2, and 3 are arranged in counter-clockwise order.
Here's how John taught me to remember every corner. There are two types of corners. In the first type numbers form an arithmetic progression. John calls such numbers counters. He chose that name so that it would be easy to remember that counters are arranged in counter-clockwise order. The other numbers he calls chaos: their increasing sequence goes clockwise.
Once I grasped that, I relaxed thinking that now I know dice. "What about the third level?" he asked. "What third level?" "Now that you know which number goes on which side, you need to know how the dots are arranged." Luckily, there are only three sides on which the dots are not placed with rotational symmetry: 2, 3, and 6. And they all meet in a corner, which John calls the home corner. The rule is that the diagonals formed by the dots on the sides with 2, 3, and 6, meet in the home corner. You might argue that 6 doesn't have a diagonal. But if you look at 6, you can always connect the dots to form the letters N or Z, depending on the orientation of the die. When you lay the letter N on its side, it becomes the letter Z. Thus they define the same diagonal. This diagonal has to meet the diagonals from 2 and 3 in the corner.
When I came home from the conference I picked up a die and checked that the rules work. There are 8 corners. It is enough to remember one corner of numbers to recover the other numbers by using the opposite sum rule. But it is nice to have a simple rule that allows us to bypass the calculation. Four of the corners have numbers in arithmetic progression: 1:2:3, 1:3:5, 2:4:6, and 4:5:6. They are counters and they are arranged counter-clockwise. The other four corners are: 1:2:4, 1:4:5, 2:3:6, and 3:5:6, and they are arranged clockwise.
I wanted to provide a picture of a die for this post and went online to see if I could grab one. Many of the graphic images of dice, as opposed to photographs, were arranged incorrectly. Clearly these visual artists did not study dice with John Conway.
Then I decided to check my own collection of dice. Most of them are correct. The ones that are incorrect look less professional. Here is the picture. The ones on the right are correct.
I started my Yellow Road a year ago on February 9, 2013, when my weight was 245.2 pounds. My system worked for eight months. I lost 25 pounds. Then I went to two parties in a row and gained four pounds. According to my plan, I was supposed to eat only apples after lunch. It was too difficult to stick to that, and I got off-target. My target weight continued decreasing daily, as per my plan, while I got stuck. The growing difference between my real weight and my target weight was very discouraging, so I lost my momentum.
I decided to reset the target weight and restart the plan. I changed my plan slightly to incorporate the lessons I had learned about myself.
On February 9, 2014, I started my New Yellow Road. I weighed 223.2 pounds. So I reset my target weight to be 223.2 on February 9. Each day my target weight goes down by 0.1 pounds. I weigh myself each morning. If I am within one pound of my target weight, I am in the Yellow zone and I will eat only fruits and vegetables after 5:00 pm. If I am more than one pound over my target weight, I am in the Red Zone and will eat only apples after 5:00 pm. If I am more than one pound below my target weight, I can eat anything.
I want to tell you the story of my "successful" salary negotiation. The year was 2003 and I had a temporary visiting position at Princeton University. I wanted to move to Boston and my friend showed my resume to Alphatech. The interview went well and they offered me a position as Lead Analyst with a salary of $110.000. This was particularly good news considering that the tech job market was very weak in 2003.
At that time, I was working very hard on building my self-esteem. I read lots of books and was in therapy for two years. I decided to practice what I had learned to try to negotiate a bigger paycheck. While speaking to the HR guy on the phone, I was standing up, as I was taught, and projecting my voice firmly with my chest opened up. I could hardly believe it when I heard myself ask for a $10,000 increase.
Despite my wonderful posture, the human resource person refused. However, he remembered that they had forgotten to give me a moving bonus. He asked me about my living conditions. I told him that I lived in a four-bedroom house. I didn't elaborate: it was a tiny four-bedroom house made out of a garage. I made a counter offer: forget the moving bonus, but give me my salary increase as I asked. He agreed.
For the first year, there was no real difference, because the salary increase was equal to the moving bonus. But I was planning to stay with the company for a long time, so by the second year, my clever negotiation would start to pay off. My negotiations were a success. But were they?
Things change. The boss who hired me and appreciated me stopped being my boss. The company was bought by BAE Systems who were not interested in research. To my surprise, I started getting non-glowing performance reviews. Luckily, by that time I had made a lot of friends at work, and one of them not only knew what was going on, but was willing to tell me. My salary was higher than that of other employees in the same position. Salary increases were tied to performance. They wanted to minimize my increases to bring my salary into the range of others at my level. So to justify it, they needed a negative performance review. After one negative review it is difficult to change the trend. A negative review stays on the record and affects the future reputation.
In a long run I am not sure that my salary negotiations were a success.
There is a famous puzzle about three light bulbs, that is sometimes given at interviews.
Suppose that you are standing in a hallway next to three light switches that are all off. There is another room down the hall, where there are three incandescent light bulbs—each light bulb is operated by one of the switches in the hallway. You can't see the light bulbs from the hallway. How would you figure out which switch operates which light bulb, if you can only go to the room with the light bulbs one and only one time?
This puzzle worked much better in the past when we only had incandescent light bulbs and so didn't need to specify the type of bulbs. Unfortunately, the standard solution only works with incandescent bulbs and the word "incandescent" nowadays needs to be stated. But the use of "incandescent" is a big hint. Indeed, incandescent light bulbs generate heat when they are on, so the standard solution is to turn on the first light switch, to keep the second switch off, and to turn the third switch on for five minutes before turning it off. In the room, the light bulb corresponding to the first switch will be lit, and out of the two unlit bulbs, the one corresponding to the third switch will be warm.
It's a cute solution, but there could be so many other approaches:
I invite my readers to invent other methods to solve this problem. Be creative. After all, if I were to interview you for a job, I would be more impressed by a new solution than the one that is all over the Internet.
You arrive at an archipelago of many islands. On each island there are two villages. In one village truth-tellers live, and they always tell the truth. In the other village liars live, and they always lie. The islanders all know each other.
On the first island you stumbled upon three islanders and you ask each of them your question:
How many truth-tellers are there among you?
Here are their answers:
A: One.
B: A is wrong.
C: A and B are from the same village.
Can you determine who is a truth-teller and who is a liar?
This island is called a classic island, where all behave as if they were in a standard logic puzzle. It is a perfectly nice puzzle but B and C didn't answer the question: B ratted on A, and C went on a tangent. When I was younger, I never cared about the motivations of A, B, or C. Their answers are enough to solve the puzzle. But now that I am older, I keep wondering why they would choose these particular answers over other answers. So I invented other islands to impose rules on how the villagers are allowed to answer questions.
Now you travel to the next island that is called a straightforward island, where everyone answers your question exactly. You are in the same situation, and ask the same question, with the following result:
A: One.
B: One.
C: Ten.
Can you determine who is a truth-teller and who is a liar?
Once again we wonder about their motivation. This time C told an obvious lie, an answer that is impossible. Why on earth did he say 10? Isn't the goal of lying to deceive and confuse people? There is nothing confusing in the answer "ten."
Now you come to the third island, which is a straightforward inconspicuous island. To answer your question, a liar wouldn't tell you an obvious lie. For this particular situation, the liar has to choose one of the four answers that are theoretically possible: zero, one, two, or three. You are again in the same situation of asking three people how many truth-tellers are among them, and these are the answers:
A: Two.
B: Zero.
C: One.
Can you determine who is a truth-teller and who is a liar?
When you think about it, a truth-teller cannot answer zero to this question. So although zero is a theoretically possible answer, we can deduce that the person who said it is a liar. If liars are trying to confuse a stranger, and they're smart, they shouldn't answer "zero."
The next island is a straightforward inconspicuous smart island. The liars on this island are smart enough not to answer zero. You are in the same situation again and ask the same question with the following outcome:A: Two.
B: Two.
C: One.
Can you determine who is a truth-teller and who is a liar? You shouldn't be able to. There are three possibilities. There are two truth-tellers (A and B), one truth-teller (C), or zero truth-tellers.
Let us assign probabilities to liars' answers. Assume that liars pick their answers randomly from the subset of wrong answers out of the set: one, two, three. If two of these answers are incorrect, they pick a wrong answer with probability one half. If all three of the answers are incorrect, they pick one of them with probability one-third. Suppose the people you meet are picked at random. Suppose that the probability that a random person is a truth-teller is 1/2. Given the answers above, what is more probable: that there are two truth-tellers, one truth-teller, or zero truth-tellers?A movie passes the Bechdel Test if these three statements about it are true:
Surely there should be a movie where two women talk about the Bechdel test. But I digress.
The Bechdel test website rates famous movies. Currently they have rated 4,683 movies and 56% pass the test. More than half of the movies pass the test. There is hope. Right? Actually they have a separate list of the top 250 famous movies. Only 70 movies, or 28%, from this list pass the test.
My son Alexey suggested the obvious reverse Bechdel test, which is more striking than the Bechdel test. A movie doesn't pass the test if it
I can't think of any movie like that. Can you?
* * *
A Roman walks into a bar, holds up two fingers, and says, "Five beers, please."
* * *
To understand what a recursion is, you must first understand recursion.
* * *
A guy is complaining to his mathematician friend:
— I have a problem. I have difficulty waking up in the morning.
— Logically, counting sheep backwards should help.
* * *
— Can I ask you a question?
— You can, but you have already just done that.
— Darn, what about two questions?
— You can, but that was your second question.
* * *
The Internet ethics committee worked hard to generate a list of words that should never be used on the Internet. The problem is, now they can't post it.
* * *
Quantum entanglement of a pair of socks: As soon as one is designated as the left, the other instantly becomes the right.
The last MIT Mystery Hunt was well-organized. It went smoothly—unlike the hunt that my team designed the year before. Sigh. As I do every year, here is the list of 2014 puzzles related to math.
There were also several puzzles requiring decoding or having a CS flavor.
I want to mention one non-mathematical puzzle.
You visit an island of three towns: Trueton, Lieberg and Alterborough. Folks living in Trueton always tell the truth. Those who live in Lieberg, always lie. People from Alterborough alternate strictly between truth and lie. You meet an islander who says:
Two plus two is five. Also, three plus three is six.
Can you determine which town he is from?
It should be easy. He made two statements: the first one is false, the second is true. So he must be from Alterborough.
But what about "also"? How should we interpret this transition? There are many ways to interpret this "also." On one hand it could mean: In addition to the previous statement I am making another statement. On the other hand it could mean: The previous pause shouldn't be considered as the end of the statement; the whole thing should be interpreted as one statement. Besides this person was speaking not writing. Are we sure that the first period was not meant to be a comma or a semi-colon? If we assume that the quote is one statement, then the speaker might be either a liar or an alternator.
Here is a puzzle for you from the same island:
One night a call came into 911: "Fire, help!" The operator couldn't ID the phone number, so he asked, "Where are you calling from?" "Lieberg." Assuming no one had overnight guests from another town, is there an emergency? If so, where should help be sent? And was it a fire?
Now find the ambiguities.
The MIT Mystery Hunt starts on Friday. My old team—Manic Sages— fell apart after last years' hunt. My new team—Death and Mayhem—started sending us daily practice puzzle to prepare for the hunt.
Today's puzzle was written by Paul Hlebowitsh. As usual, the answer is a word or a phrase.
Puzzle 3: Humans
"After a long day taking care of the animals, it's good to unwind by letting the taps flow."
Tap 1: Sierra Nevada Pale Ale
Tap 2: Lagunitas Pale Ale
Tap 3: 21st Amendment "Brew Free! or Die IPA"
Tap 4: Wachusett Blueberry
Tap 5: Harpoon UFO
Alice, Bob, Carol, Danny, Erica, Fred, and Gregario, the seven children of Noah, went to a local bar before the flood to get drinks. Accounts of night vary, and no one remembers exactly what happened, but some facts have become clear:
Who ordered which drinks and in what order?
I stopped losing weight. My Yellow Road plan stopped working. If you recall I draw a line on the weight/time plane which I call my target weight. If I'm more than one pound above my target weight, then I'm in the red zone and must restrict my evening food to apples. The hour at which the evening starts depends on how many pounds I'm above my target weight.
And now, back to my bump.
First I stopped following the plan exactly. I realized that I didn't need to restrict myself to eating only apples in the evening when I am in the red zone. I can use salad or anything light.
One day I found myself in the red zone weighing two pounds over my target weight. I was invited to dinner that evening. I decided to accept the invitation and skip the plan for one day. At the party the food was so good I couldn't resist it. The next day I was four pounds over my target weight.
My Yellow Road plan requires me in this situation to eat only apples from 2:00pm onward, but I knew that applying this restriction after 6:00pm worked for me. I decided not to torture myself and started to restrict my food only from 6:00pm. The weight didn't go down. Even though I went to bed very hungry for two weeks, it didn't work.
After these two weeks, I started to feel hungry all the time and even began dreaming about food. As a result, my food intake increased. Now I am seven pounds over my target weight. I've reached a plateau. For the last two months I've been stuck at 220 pounds.
There is some good news: I now have a partner on this journey. After I started my program, I received an email from Natalia Grinberg from Germany. She offered to join forces. We send each other weekly updates on our progress and cheer each other along. Natalia's path wasn't smooth from the start, so she tried to supplement her diet with Almased, which is very popular in Europe. Because Natalia likes it, I looked into it. While I am afraid of pills and chemical ingredients, Almased seems to be okay. It contains soy, yogurt, honey, and vitamins. I bought one can. It is expensive and tastes awful. I'll experiment with cinnamon or pepper and see if that helps. Will Almased help me get over the Bump?
I collect geeky jokes. I think I've heard most of them. So I was surprised to stumble upon a website with many new ones: 50 People On 'The Most Intellectual Joke I Know'. These are some of them:
* * *
Q: What does the "B" in Benoit B. Mandelbrot stand for?
A: Benoit B. Mandelbrot.
* * *
Entropy isn't what it used to be.
* * *
There are two types of people in the world: those who can extrapolate from incomplete data sets
* * *
This sentence contains exactly threee erors.
* * *
There's a band called 1023MB. They haven't had any gigs yet.
* * *
A logician's wife is having a baby. The doctor immediately hands the newborn to the dad. His wife asks impatiently, "So, is it a boy or a girl"? The logician replies, "Yes."
* * *
The barman says, "We don't serve time travelers in here."
A time traveler walks into a bar.
* * *
The first rule of Tautology club, is the first rule of Tautology club.
* * *
A woman walks in on her husband, a string theorist, in bed with another woman. He shouts, "I can explain everything!"
* * *
What do you get when you cross a joke with a rhetorical question?
I used to love arXiv. I've long thought that it was one of the best things that happened to mathematics. arXiv makes mathematical research available for free and without delay. Moreover, it is highly respected among mathematicians. For example, Grigori Perelman never submitted his proof of the Poincaré conjecture to any journal: he just posted it on arXiv.
When I came back to mathematics, all my math friends explained to me that I should submit my paper to arXiv on the same day that I submit it to a journal. As my trust in arXiv grew, I started submitting to arXiv first, waiting one week for comments, and then submitting to a journal.
Now it seems that arXiv might not love its contributors as much as they used to. arXiv moderators seem to be getting harsher and harsher. Here is my story.
In June of 2013, I submitted my paper "A Line of Sages" to arXiv. This paper is about a new hat puzzle that appeared at the Tournaments of the Towns in March 2013. The puzzle was available online at the Tournaments of the Towns webpage in Russian. After some thought I decided that it is better to cite the Tournament itself inside the body of the paper, rather than to have a proper reference. Online references in general are not stable, and this particular one was in Russian. Very soon this competition will be translated into English and the puzzle will appear in all standard math competition archives.
arXiv rejected my paper. A moderator complained that I didn't have a bibliography. So I created a bibliography with the link to the puzzle. My paper was rejected again saying that the link wasn't stable. Duh. That's the reason why I didn't put it there from the start. I Goggled the puzzle and I still didn't find any other links.
I argued with my moderator that the standards for papers in recreational mathematics are different from the standards for purely research papers. Short recreational notes do not require two pages of history and background, nor a long list of references. A recreational paper doesn't need to have theorems and lemmas.
Meanwhile, the moderator complained that the paper was "not sufficiently motivated to be interesting to the readership."
I got tired of exchanging emails with this moderator and submitted my paper to The Mathematical Intelligencer, where it was immediately accepted. So I dropped my submission to arXiv.
Now my paper is not available for free on arXiv. But anyone can freely buy it from Springer for $39.95.
I started my life wanting to be a mathematician. At some point I had to quit academia in order to feed my children. And so I went to work in industry for ten years. Now that my children have grown, I am trying to get back to academia. So I am the right person to compare the experience of working in the two sectors. Just remember:
Money. The pay is much better in industry. About twice as high as academia.
Time. I almost never had to work overtime while working in industry. That might not be true for programmers and testers. As a designer, I worked at the beginning of the project stage. Programmers and testers are closer to deadlines, so they have more pressure on them. The industrial job was more practical than conceptual, so I didn't think about it at home. My evenings and weekends were free, so I could relax with my children. In academia I work 24/7. There are 20 mathematical papers that I have started and want to finish. This is a never-ending effort because I need those papers to find my next job. Plus, I want to be a creative teacher, so I spend a lot of time preparing for classes. I do not have time to breath.
Respect. When I was working in industry, some of my co-workers would tell me that I was the smartest person they ever met. In any case, I always felt that my intelligence and my skills were greatly appreciated. In academia, I am surrounded by first-class mathematicians who rarely express respect and mostly to those who supersede them in their own fields.
Social Life. Mathematics is a lonely endeavor. Everyone is engrossed in their own thoughts. There is no urge to chat at the coffee machine. In industry we were working in teams. I knew everyone in my group. I was closer to my co-workers when I worked in industry.
Freedom. In both industry and academia there are bosses who tell you what to do. But while building my university career, a big part of my life is devoted to writing papers. It is not a formal part of my job, but it is a part of the academic life style. And in my papers I have my freedom.
Motivation. In academia, one must be self-motivated.
Rejection. The output of an academic job is published papers. Most journals have high rejection rates. For me, it's not a big problem because from time to time I get fantastic reviews and I usually have multiple papers awaiting review. I have enough self-confidence that if my paper is rejected, I don't blink. I revise it and send it to a different journal. But this is a huge problem for my high school students who submit their first paper and get rejected. It is very discouraging.
Perfectionism. In industry I was working on deadlines. The goal was to deliver by the deadline a project that more or less worked. Time was more important than quality. My inner perfectionist suffered. When I write papers, I decide myself when they are ready for publication.
Impact. When I was working at Telcordia I felt that I was doing something useful. For example, we were building a local number portability feature, the mechanism allowing people to take their phone numbers with them when they moved. I wish Verizon had bought our product. Just a couple of months ago I had to change my phone number when I moved five blocks from Belmont to Watertown. Bad Verizon. But I digress. When I was working at Alphatech/BAE Systems, I was designing proofs of concepts for future combat systems. I oppose war and the implementation was sub-standard. I felt I was wasting my time. Now that I am teaching and writing papers, I feel that I am building a better world. My goal is to help people structure their minds and make better decisions.
Fame. All the documents I wrote in industry were secret. The world would never know about them. Plus, industry owns the copyright and takes all the credit. There is no trace of what I have done; there is no way to show off. People in academia are much more visible and famous.
Happiness. I am much happier now. I do what I love.
* * *
I just received a call from the delivery man working for the store where I ordered a new GPS device. He got lost and asked directions to my house.
* * *
I have two problems: I can't count.
* * *
Do you want to double your cash?
Hold your money in front of a mirror.
* * *
- Doctor, I think I have paranoia.
- Why is that?
- Yesterday I left my computer for ten minutes to go to the bathroom, and when I came back all my browser windows were filled with bathroom tissue ads.
* * *
- Why is your disc drive so noisy?
- It is reading a disc.
- Aloud?
* * *
If only DEAD people understand hexadecimal, how many people understand hexadecimal?
I gave the following puzzle from Raymond Smullyan's book What is the Name of this Book? to my AMSA students.
What happens if an irresistible cannonball hits an immovable post?
This puzzle is known as the Irresistible Force Paradox. The standard answer is that the given conditions are contradictory and the two objects cannot exist at the same time.
My AMSA student gave me a much cuter answer: The post falls in love with the cannonball as it is so irresistible.
It has been a while since I wrote my last essay and my readers have started to worry. Sorry for being out of touch, but let me tell you what is going on in my life.
In September I received an offer from MIT that changes my status there. In exchange for a slight increase in pay, I am now conducting recitations (supplemental seminars) in linear algebra In addition to my previous responsibilities.
My readers will know that just a slight increase in money in exchange for significant demands on my time would not appeal to me. But this offer comes with perks. First, my position at MIT changes from an affiliate to a lecturer, which looks so much better on my CV. Second, it includes benefits, the most important of which is medical insurance.
I lived without insurance for three years. On the bright side, lack of insurance made me conscious of my health. I developed many healthy habits. I read a lot about the treatments for colds and other minor problems that I had. On the other hand, it is a bit scary to be without insurance.
Many people are surprised to hear that I didn't have any insurance: Doesn't Massachusetts require medical insurance for everyone?
The Commonwealth levies a fine on those who do not have insurance. But I was in this middle bracket in which my income was too high for a subsidized plan, and too low to be fined. You see, the fine is dependent on one's income and is pro-rated. So I didn't have to pay it at all.
I got my insurance from MIT in October, but ironically my doctor's waiting list is so long, that my first check-up will not be until January.
Anyway, I sort of have four jobs now. I am coaching students for math competitions at the AMSA charter school six hours a week. I am the head mentor at the RSI summer program where I supervise the math research projects of a dozen high school students. I do the same thing for the PRIMES program, in which I have the additional responsibility of mentoring my own students. And as I mentioned, I am also teaching two recitation groups in linear algebra at MIT.
Teaching linear algebra turned out to be more difficult than I expected. I love linear algebra, but I had to learn the parts of it that are related to applications and engineering. Plus, I didn't know linear algebra in English. And my personality as a perfectionist didn't help because to teach linear algebra up to my standards would have taken more time than I really had.
This semester I barely had time to breathe, and I certainly couldn't concentrate on essay pieces. Now that this semester is almost over, my as yet unwritten essays are popping up in my head. It's nice to be back.
Hooray! My weight is down to two digits in kilograms: below 100. This is a big deal for me. I reached the desired number of digits. In pounds it means I weigh less than 220 and I've lost 25 pounds.
My friends have started to notice. The chubbier ones ask me to tell them about my Yellow Road. And I don't actually know what to reply, because the Yellow Road is not a solution. I took many steps before I approached the Yellow Road. The Yellow Road is, I hope, the end of the road.
The idea of the Yellow Road is simple. If I weigh more than I want, I decrease food. All my skinny friends have always lived that way. The problem is that the rest of us do not know how exactly to reduce food intake and then how to sustain that reduction.
So today, I would like to explain to my friends and my readers what I really think helped me to lose weight.
1. I got desperate. I was ready to do whatever it takes. I was prepared not to ever eat again. If I had to extract my calories from the air, I was prepared to do that. I was ready to be hungry and restrain myself for the rest of my life. In short, I was totally motivated.
2. I fought my sugar addiction. I used to crave sugar. I used to think that sugar helps my brain. But once I looked into it, I realized that I might be wrong. I decided to experiment and cut off my carbohydrates intake significantly. That was the most painful thing I had to do. But after a week of withdrawal symptoms, I felt better and stopped craving foods and sugars as much.
3. I wrote my weight down everyday. Having numbers staring me in the face reminded me what I ate the day before. This moment of reflection allowed me to understand what causes the increase or decrease in my weight. Now I know that some foods provoke my appetite: carbohydrates, dairy, mayonnaise. I eat them in small portions, but I do not start my day with them. It's better to have an increased appetite for a couple of hours in the evening, than for the whole day.
4. I had already changed some bad habits. I tried to build new healthy habits before I started my Yellow Road. These alone didn't help me lose weight, but I think they contribute to my weight loss. I still do the following:
I feel that an internal switch was turned off. I don't feel that hungry anymore. My son thinks that I'm in my hibernating state. I wonder if he is right and I will awake one day as hungry as ever.
My son, Alexey Radul, is gainfully unemployed. While looking for a new job he wrote several essays about his programming ideas. I am a proud and happy mother. While I can't understand his code, I understand his cutting-edge essays. Below are links to the four essays he has posted so far. He is also a superb writer. You do not need to take my word for it. Each link is accompanied by the beginning of the essay.
The successful fox must know more than the sum of what the hedgehogs know, for it must know the connections from one thing to another. This fact is key to the design of computer systems for solving certain kinds of problems. Read more
Automatic differentiation may be one of the best scientific computing techniques you’ve never heard of. If you work with computers and real numbers at the same time, I think you stand to benefit from at least a basic understanding of AD, which I hope this article will provide; Read more
The “Sufficiently Clever Compiler” has become something of a trope in the Lisp community: the mythical beast that promises language and interface designers near-unlimited freedom, and leaves their output in a performance lurch by its non-appearance. A few years ago, I was young enough to join a research project to build one of these things. Neglecting a raft of asterisks, footnotes, and caveats, we ended up making something whose essence is pretty impressive: you pay for abstraction boundaries in compile-time resources, but they end up free at runtime. One prototype was just open-sourced recently, so that makes now a good time to talk about it. Read more
The Cleverness of Compilers essay described the name of the hyperaggressive compilation game in broad, philosophical strokes. Here, I would like to walk through the Mandelbrot example in some detail, so that the interested reader may see one particular way to actually accomplish that level of optimization. Read more
I submitted four papers to the arXiv this Spring. Since then I wrote four more papers:
(with Leigh Marie Braswell) Cookie Monster Devours Naccis. History and Overview arXiv: arXiv 1305.4305.
In 2002, Cookie Monster appeared in The Inquisitive Problem Solver. The hungry monster wants to empty a set of jars filled with various numbers of cookies. On each of his moves, he may choose any subset of jars and take the same number of cookies from each of those jars. The Cookie Monster number is the minimum number of moves Cookie Monster must use to empty all of the jars. This number depends on the initial distribution of cookies in the jars. We discuss bounds of the Cookie Monster number and explicitly find the Cookie Monster number for Fibonacci, Tribonacci and other nacci sequences.
A Line of Sages.
A new variation of an old hat puzzle, where sages are standing in line one behind the other.
(Jesse Geneson and Jonathan Tidor) Convex geometric (k+2)-quasiplanar representations of semi-bar k-visibility graphs. Combinatorics arXiv: arXiv 1307.1169.
We examine semi-bar visibility graphs in the plane and on a cylinder in which sightlines can pass through k objects. We show every semi-bar k-visibility graph has a (k+2)-quasiplanar representation in the plane with vertices drawn as points in convex position and edges drawn as segments. We also show that the graphs having cylindrical semi-bar k-visibility representations with semi-bars of different lengths are the same as the (2k+2)-degenerate graphs having edge-maximal (k+2)-quasiplanar representations in the plane with vertices drawn as points in convex position and edges drawn as segments.
(with Leigh Marie Braswell) On the Cookie Monster Problem. History and Overview arXiv: arXiv 1309.5985.
The Cookie Monster Problem supposes that the Cookie Monster wants to empty a set of jars filled with various numbers of cookies. On each of his moves, he may choose any subset of jars and take the same number of cookies from each of those jars. The Cookie Monster number of a set is the minimum number of moves the Cookie Monster must use to empty all of the jars. This number depends on the initial distribution of cookies in the jars. We discuss bounds of the Cookie Monster number and explicitly find the Cookie Monster number for jars containing cookies in the Fibonacci, Tribonacci, n-nacci, and Super-n-nacci sequences. We also construct sequences of k jars such that their Cookie Monster numbers are asymptotically rk, where r is any real number between 0 and 1 inclusive.
This is my toast at the Gelfand's Centennial Conference:
I moved to the US twenty years ago, right after I got my Ph.D in mathematics under the supervision of Israel Gelfand. My first conversation with an American mathematician went like this:
The guy asks me, "What do you do?"
I say, "Mathematics."
"No, I mean what is your field?"
I can't understand what he wants, and repeat "Mathematics."
He says, "No, no, I mean, I do differential geometry. What do you do?"
I do not know how to answer him. My teacher, Israel Gelfand, never mentioned that mathematicians divide mathematics into pieces. So I had to repeat, "My field is mathematics."
I got asked this question many times and I couldn't figure out how to give a satisfactory answer, so I quit academia. Well, I quit it not because of the question, but for many other reasons... But answering the question became so much easier when I worked for industry.
A guy asks me, "What do you do?"
I say, "Battle management."
He says, "What?"
I say, "Battle management. I manage battles, in case there is a war." And this is it, he doesn't ask any more questions … ever.
I always knew that industry was not the right place for me. Five years ago, when my children grew up, I realized that it was time to take some risks. So I resigned from my job, and came back to mathematics. But now I know how to answer the question. When someone asks me, What is your field in mathematics? I say, … brag, "I am a student of Israel Gelfand, I just do mathematics."
I would like to drink to the Unity of Mathematics.
Now that my weight loss is under way, I want to build an exercise program.
I already exercise. I am a member of the MIT ballroom dance team and I go to the gym, where I use the machines, swim, and take gentle yoga and Zumba classes.
Doesn't sound bad, right? But the reality is that I went to Zumba class a total of three times last year. I skipped half the dance classes I signed up for because I was too tired to attend. I had sincerely planned to go to them, but they're held in the evenings, when I'm already drooping.
The yoga situation is the worst. I'm scheduled to go to yoga twice a week, but just as it is time to leave for yoga, I get very hungry and cannot resist having a snack. Then I remember that they do not recommend practicing yoga after eating a meal, and voilà—I have found my excuse. I am all set up to exercise five hours a week, but in reality most weeks I do not exercise at all.
How can I motivate myself? I know that with food, if I have gained weight one day, I eat less the next. So if I've missed my goal of exercising one week, do I add those hours to the next week? That's unlikely to actually help, since the problem is that I'm not meeting my exercise goal, period.
Many people suggested that I punish myself. For example, if I do not meet my goal, I should donate money to a cause I do not support. This feels wrong. If I fail, I'll feel doubly guilty. I'll go broke paying for psychotherapy.
I can try to reward myself. But what should the reward be? Should I reward myself with a piece of tiramisu? Since I am trying to persuade myself that sugar is bad, I shouldn't create a situation that makes sugar desirable. So rewarding with food won't work. Should I buy myself something? If I really want it, I will buy it anyway.
I've been thinking about a plan for a long time. Finally I realized that I should find other people to reward me. I do have a lot of friends, and I came up with an idea of how they could help.
The next time I saw my friend Hillary I asked her if she wanted to sponsor my new exercise plan. She said, "I'm in," without even hearing the plan. Hillary is a true friend. This is what she blindly signed up for.
I decided to push myself to exercise five hours a week. Because of weather and health fluctuations, I pledged to spread 20 hours of exercise over four weeks. I will sent Hillary weekly reports of what I do. This is in itself a huge motivating factor. After the four weeks, we will go to lunch together, which is a great reward for me to look forward to. If I succeed with my plan, she pays for lunch. If I fail, I pay.
Once I saw how enthusiastic Hillary was, I lined up four other friends for the next four-week periods. I hope that after several months of exercise, I will learn to enjoy it. Or at least, I will start feeling the benefits and that itself will be a motivating factor.
Hillary liked my plan so much that she designed a similar exercise plan for herself. Now I am looking forward to two lunches with Hillary.
I was driving on Mass Pike, when the cars in front of me stopped abruptly. I hit the brakes and was lucky to escape the situation without a scratch.
Actually, it wasn't just luck. First of all, I always keep a safe distance from the other cars. Second, if I see the brake lights of the car in front of me, I automatically remove my foot from the gas pedal and hold it over the brake pedal until I know what the situation is.
On a highway, if the car in front of me has its brake lights on, usually that means that the driver is adjusting their speed a little bit. So, most of the time I don't have to do anything. Seeing that the car in front of me has its brake lights on is not a good predictor of what will happen next. Only after I see that the distance between me and the car in front of me is decreasing rapidly, do I know to hit my brakes. That means that brake lights alone are not enough information. Differentiating between insignificant speed adjustments and serious braking requires time and can cost lives.
I have a suggestion. Why not create smart brake lights. The car's computer system can recognize the difference in the strength with which the brakes are hit and the lights themselves can reflect that. They can be brighter or a different color or pulsing, depending on the strength of the pressure.
The drivers behind will notice these things before they will notice the decrease in the distance. This idea could save lives.
I recently posted the following 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 than all the genuine coins, which weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?
The author's solution in Russian is available at his blog. Also, two of my readers, David Reynolds and devjoe, solved it correctly.
Here I want to explain the solution for any number of required weighings.
It is easy to see that for n weighings the information theoretical bound is 5^{n}. Indeed, each weighing divides coins into five groups: four pans and the leftover pile. To distinguish between coins, there can't be two coins in the same pile at every weighing.
Suppose we know the faking potential of every coin, that is, each coin is assigned a value: potentially light or potentially heavy. If a potentially light coin is ever determined to be fake, then it must be lighter than a real coin. The same story holds for potentially heavy coins. How many coins with known potential can we process in n weighings?
If all the coins are potentially light then we can find the fake coin out of 5^{n} coins in n weighings. What if there is a mixture of coins? Can we expect the same answer? How much more complicated could it be? Suppose we have five coins: two of them are potentially light and three are potentially heavy. Then on the first scale we compare one potentially light coin with the other such coin. On the other scale we compare one potentially heavy coin against another potentially heavy coin. The fake coin can be determined in one weighing.
The discussion above shows that there is a hope that any mixture of coins with different potential can be resolved. After each weighing, we want the number of coins that are not determined to be real to be reduced by a factor of 5. If one of the weighings on one scale is unbalanced, the potentially light coins on the lighter pan, plus the potentially heavy coins on the heavier pan would contain the fake coin. We do not want this number to be bigger than one-fifth of the total number of coins we are processing. So we divide coins in pairs with the same potential, and from each pair we put the coins on different pans of the same scale. So in one weighing we can divide the group into five equal groups. If there is an odd number of coins with the same potential, then the extra coin doesn't go on the scales.
The only thing that we is left to check is what happens if the number of coins is small. Namely, we need to check what happens when the number of potentially light coins is odd and the number of potentially heavy coins is odd, and the total number of coins is not more than five. In this case the algorithm requires us to put aside the extra coin in each group, but the put-aside pile can't have more than one coin.
After checking small cases, we see that we can't resolve the problem in one weighing when there are 2 coins of different potential, or when the 4 coins are distributed as 1 and 3.
On the other hand, if we have extra coins that are known to be real, then the above cases can be resolved. Hence, any number of coins with known potential greater than four can be resolved in ⌈log_{5}n⌉ weighings.
Now let's go back to the original problem in which we do not know the coins' potential at the start. After a weighing, if both scales balance, then all the coins on the scale are real and the fake coin is in the leftover pile and we do not know its potential. If a scale doesn't balance then the fake coin is in one of its two pans: the lighter pan has coins that are potentially light and the heavier pan has coins that are potentially heavy.
Let's add an additional assumption to the original problem. Suppose we have an unlimited supply of coins that we know to be real. Let u(n) be the maximum number of coins we can process in n weighings if we do not know their potential.
What would be the first weighing? Both scales might be balanced, meaning that the fake coin is in the leftover pile of coins with unknown potential. So we have to leave out not more than u(n−1) coins. On the other hand, exactly one scale might be unbalanced. In this case, all the coins on this scale will get their potential known. The number of these coins can't be more than 5^{n-1}. But this is an odd number, so we can use one extra real coin to make this number even, in order to put the same number of coins in each pan on this scale.
So u(n) = 2 · 5^{n-1} + u(n−1), and u(1) = 3. This gives the answer of (5^{n}+1)/2. Now we need to go back and remember that we got this bound using an additional assumption that we have an unlimited supply of real coins. Looking closer, we do not need our additional supply of real coins to be unlimited; we just need not more than two real coins. The good news is that we will have these extra real coins after the first weighing. The bad news is that for the first weighing we do not have extra real coins at all. So in the first weighing we should put unknown coins against unknown coins, not more than 5^{n-1} on each scale, and as the number on each scale must be even, the best we can do is put 5^{n-1}−1 coins on each scale.
Thus the answer is (5^{n}−3)/2 for n more than 1.
We can generalize this problem to any number of scales used in parallel. Suppose the number of scales is k. Suppose the number of weighings is more than 1, then the following problems can be solved in n weighings:
The methods I described can be used to answer another common question in the same setting: Find the fake coin and say whether it is heavier or lighter. Let us denote by U(n) the number of coins that can be resolved in n weighings when there is an unlimited supply of extra real coins. Then the recurrence for U(n) is the same as the recurrence for u(n): U(n) = 2·5^{n-1} + U(n−1). The only difference is in the initial conditions: U(1) = k. This means that U(n) = ((2k+1)^{n}−1)/2. If we don't have extra real coins then the answer is: U(n) = ((2k+1)^{n}−1)/2 − k.
When we don't need to say whether the fake coin is heavier or lighter, we can add one extra coin to the mix: the coin that doesn't participate in any weighing and is fake if the scales always balance.
I valued my previous analog scale: I brought it from Europe and it showed my weight in kilograms.
I don't remember why, a year ago, I decided to buy a digital scale: after all, my old scale is still functioning. But my new digital scale became an important instrument in my weight loss program.
Now I understand how my old scale and my lazy nature played tricks on my mind. For example, let's say I decide that as soon as I reach one hundred kilograms (220 pounds), I would do something about it. Some time passes, I step on the scale, and it shows 100 kilograms — and maybe more. Is it more or not quite? If I lean to the right, it looks lower. The scale is not precise. So, I step back and adjust the scale at zero a little bit, to make sure it is not more than zero, but it actually could be slightly less. So I have reached my limit, but the scale's imprecision allows me to pretend that there is a chance I am not over 100 kilograms, and thus do not have to do anything. More time passes, the scale shows 103 kilograms. I realize that I have been deceiving myself, but then it is too late for a fast fix that would lower my weight below 100. So I push the limit, and decide to wait until 105 kilograms.
My digital scale erased any ambiguity. The scale, however, has its own doubts. When I stand on it, the display flips between two numbers for a while. One of the numbers means no dinner tonight. But at the end the scale calls it: it spits out the final number with a beep. I had already decided that the scale is the boss. The flipping is irrelevant; the decision is irrevocable. If it means no dinner, so be it.
My new scale prevents inaction. It also allowed me to design my new plan: my Yellow Road. In this plan, my target weight decreases by 0.1 pounds per day. My behavior depends on my real weight with respect to my target weight. A small difference changes my behavior. So precision is essential.
As you can see in the picture the plan continues to work. I've lost 16 pounds since the start of the plan. Actually, I've lost 18 pounds, if I weigh myself without the camera.
I recently posted the following problem from the 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 acquaintances at this meeting.
- Show that n can be greater than 4.
Here is the proof for the first bullet. Choose a person X. Take a pair of X's acquaintances. These two acquaintances have to share two acquaintances between themselves one of whom is X. In other words, we have described a function from all pairs of X's acquaintances to people who are not X. On the other hand, for every person who is not X, s/he and X share a pair of acquaintances. Hence, there is a bijection between people other than X and all pairs of X's acquaintances. If the number of X's acquaintances is a, and the total number of people is t, then we have shown that (a choose 2) = t−1. As this is true for any X, we see that everyone has the same number of acquaintances. Moreover, this situation can happen only if t−1 is a triangular number.
But wait. There is more work that needs to be done. The smallest triangular number is 1. That means that t might be 2. If there are two people at the meeting, then the condition holds: they have 0 common acquaintances. The next triangular number is 3. So we need to see what would happen if there are four people. In this case, if everyone knows each other, it works. This is why the second bullet asks us to find an example of the situation with more than four people, because four people is too easy.
Let's look at larger triangular numbers. The situation described in the problem might also happen when there are:
The official Olympiad solution suggests the following example for 16 people total. Suppose we put 16 people in a square formation so that everyone knows people in the same row and column. I leave it to the reader to check that every two people share exactly two acquaintances.
Let me prove that there is no solution for a total of seven people. If there were a solution, then each person would have to know four people. My first claim is that the acquaintance graph can't contain a four-clique. Suppose there is a four-clique. Then each person in the clique has to have another acquaintance outside of the clique to make it up to four. In addition, this extra acquaintance can't be shared with anyone in the clique, because the clique contains all the acquaintances that they share. This means we need to have at least four more people.
Next, suppose two people a and b know each other and share an acquaintance c. Any two people in this group of three has to have another shared acquaintance, who is not shared with the third person. That is, there should be another person who is the acquaintance of a and b, a different person who is an acquaintance of a and c, and a third person who is acquainted with b and c. These three extra people are all the acquaintances of a, b, and c. Which means the last person who is not acquainted with a, b, or c, has less than for four acquaintances.
Let's look at a more difficult problem that I offered at the same posting:
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 acquaintances at this meeting.
- Is it possible that n can be greater than 2k?
As in the previous solution, we see that a, the number of acquaintances of a person and t, the total number of people, satisfy the following equation: (a choose k) = (t−1 choose k−1).
For example, if k = 3, the equation becomes (a choose 3) = (t−1 choose 2). This is a question of finding numbers that are both tetrahedral and triangular. They are known and their sequence, A027568, is finite: 0, 1, 10, 120, 1540, 7140. The corresponding number of acquaintances is 3, 5, 10, 22, 36 and the total number of people is 3, 6, 17, 57, 121. The first trivial example involves 3 people who do not know each other. The next example is also simple: it has 6 people and everyone knows everyone else.
What about non-trivial examples? If there are 17 people in the group, then each person has to know 10 people. Does the acquaintance graph exist so that every group of three people share 3 acquaintances?
We see that the problem consists of two different parts. First, we have to solve the equation that equates two binomial coefficients. And second, we need to build the acquaintance graph. Both questions are difficult. We see that for k = 2 we have an infinite number of solutions to the equation with binomial coefficients. For k = 3, that number is finite. What happens with other k? If there are 2k people and they all know each other, then this works. But are there other non-trivial solutions? I am grateful to Henry Cohn for directing me to the works of Singmaster who studied non-trivial repetitions of numbers in Pascal's triangle. In particular, Singmaster showed that the equation (n+1 choose k+1) = (n choose k+2) has infinitely many solutions given by n = F_{2i+2}F_{2i+3}−1 and k = F_{2i}F_{2i+2}−1.
This sequence generates the following non-trivial examples (15 choose 5) = (14 choose 6), (104 choose 39) = (103 choose 40), and so on. That means it might be possible that there is a group of 16 people so that every 6 people share 6 acquaintances. In this situation every person must know everyone else except for one other person. That leads us to the structure of the acquaintance graph: it is a complement to the perfect matching graph. I leave it to my readers to check that the corresponding acquaintance graph doesn't exist. Are there examples of two binomial coefficients that equal each other and that lead to an acquaintance graph that can be built?
Now that I've tackled the solution to this Olympiad problem, I see that I generated more questions than I answered.
Many years ago at Gelfand's seminar in Moscow, USSR, someone pointed out a young girl and told me: "This is Natalia Grinberg. In her year in the math Olympiads, she was the best in the country. She is the next you."
We were never introduced to each other and our paths never crossed until very recently.
Several years ago I became interested in the fate of the girls of the IMO (International Math Olympiad). So, I remembered Natalia and started looking for her. If she was the best in the USSR in her year, she would have been a gold medalist at the IMO. But I couldn't find her in the records! The only Grinberg I found was Darij Grinberg from Germany who went to the IMO three times (2004, 2005, and 2006) and won two silver medals and one gold.
That was clearly not Natalia. I started doubting my memory and forgot about the whole story. Later I met Darij at MIT and someone told me that he was Natalia's son.
I was really excited when I received an email from Natalia commenting on one of my blog posts. We immediately connected, and I asked her about past events.
Natalia participated in the All-Soviet Math Olympiads three times. In 1979 as an 8th grader she won a silver medal, and in 1980 and 1981 she won gold. That indeed was by far the best result in her year. So she was invited to join the IMO team.
That year the IMO was being held in the USA, which made Soviet authorities very nervous. At the very last moment four members of the team did not get permission to travel abroad. Natalia was one of them. The picture below, which Natalia sent to me, was taken during the Soviet training camp before the Olympiad. These four students were not allowed to travel to the IMO: Natalia Grinberg, Taras Malanyuk, Misha Epiktetov, and Lenya Lapshin.
Because of the authorities' paranoia, the Soviet team wasn't full-sized. The team originally contained eight people, but as they rejected four, only six traveled to the USA, including two alternates.
I have written before how at that time the only way for a Jewish student to get to study mathematics at Moscow State University was to get to the IMO. I wrote a story about my friend Sasha Reznikov who trained himself to get to the IMO, but because of some official machinations, still was not accepted at MSU.
Natalia's story surprised me in another way. She didn't get to the IMO, but she was accepted at MSU. It appears that she was accepted at MSU as a member of the IMO team, because that decision was made before her travel documents were rejected.
Natalia became a rare exception to the rule that the only way for a Jewish person to attend MSU was to participate in the IMO. It was a crack in the system. They had to block visas at the last moment, so that people wouldn't have time to make a fuss and do something about it. Natalia slipped through the crack and got to study at the best university in the Soviet Union.
Unfortunately, the world lost another gold IMO girl. Three Soviet team members won gold medals that year. Natalia, being better then all of them, would have also won the gold medal.
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:
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 p_{1}. Suppose the probability of a son of a two-sons mother to be a homosexual is p_{2}. The data shows that p_{2} is greater than p_{1}. What is the consequence? Suppose the number of mothers with one son is m_{1} and the number of mothers with two sons is m_{2}. Then in the whole population the probability of a boy who is the first son to be gay is (p_{1}m_{1}+p_{2}m_{2})/(m_{1}+m_{2}) and the probability of a boy who is the second son to be gay is p_{2}. 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 5^{5} = 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:
This is the promised solution to the puzzle Integers and Sequences that I posted earlier. The puzzle is attached below.
Today I do not want to discuss the underlying math; I just want to discuss the puzzle structure. I'll assume that you solved all the individual clues and got the following lists of numbers.
Since the title mentions sequences, it is a good idea to plug the numbers into the Online Encyclopedia of Integer Sequences. Here is what you will get:
Your first "aha moment" happens when you notice that the sequences are in alphabetical order and each has exactly one number missing. The alphabetical order is a good sign that you are on the right track; it can also narrow down the possible names of the sequences that you haven't yet identified. Alphabetical order means that you have to figure out the correct order for producing the answer.
Did you notice that some groups above are as long as nine integers and some are as short as four? In puzzles, there is nothing random, so the lengths of the groups should mean something. Your second "aha moment" will come when you realize that, together with the missing number, the number of the integers in each group is the same as the number of letters in the name of the sequence. This means you can get a letter by indexing the index of the missing number into the name of the sequence.
So each group of numbers provides a letter. Now we need to identify the remaining sequences and figure out in which order the groups will produce the word that is the answer.
Let's go back and try to identify the remaining sequences. We already know the number of letters in the name of each sequence, as well as the range within the alphabet. The third sequence might represent a challenge as its numbers are small and there might be many sequences that fit the pattern, but let's try. The results are below with the capitalized letter being the one that is needed for the answer.What is going on? There are two sequences that fit the pattern of the third group and the sequence for the sixth group has many names, two of which fit the profile but produce different letters. Now we get to your third "aha moment": you have already seen some of the sequence names before, because they are in the puzzle. This will allow you to disambiguate the names.
Now that we have all the letters, we need the order. Sequences are mentioned inside the puzzle. You were forced to notice that because you needed the names for disambiguation. Maybe there is something else there. On closer examination, all but one of the sequence names are mentioned. Moreover, with one exception the clues for one sequence mention exactly one other sequence. Once you connect the dots, you'll have your last "aha moment:" the way the sequences are mentioned can provide the order. The first letter G will be from the pentagonal sequence, which was not mentioned. The clues for the pentagonal sequence mention the primeval sequence, which will give the second letter R, and so on.
The answer is GRANTER.
Many old-timers criticized the 2013 MIT Mystery Hunt. They are convinced that a puzzle shouldn't have more than one "aha moment." I like my "aha moments."
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Should I eat this piece of cake or not? I will certainly enjoy it very much. What harm will it do? Will this piece increase my weight? Maybe not. The next piece might, but this particular one looks harmless. Even if my weight increases by half a pound, it could be muscle weight. Yes, it probably would be due to muscle weight: I just went out of my house to throw away my garbage and this has to count as exercise.
Do you see the problem? Eating the cake provides an immediate reward, but the punishment is vague and in the far distant future. That is why I got excited when my son Alexey sent me the link to Beeminder, a company that creates an artificial non-vague and not far-in-the-future punishment for eating that piece of cake.
Here is how it works. You give them your target number — in my case my desired weight, but it could be any measurable goal — and the date by which you want to hit it. They draw a yellow path on a weight chart. You must weigh yourself every day. Whenever your weight is above your path, you have to pay real money to the company. Five dollars!
This is a great idea. Suddenly that piece of cake looks threatening. The only problem with using their system is that I have no clue how to lose weight. The company doesn't provide tools to lose weight: it just provides a commitment device. So it is difficult to stick with the weight-loss commitment without having a proven weight-loss plan.
The truth is that my son sent me the link, I laughed, and forgot about it. Besides, if I ever want to pay money for failing in my commitments, I would rather choose the beneficiary myself. Then I realized that I can use the yellow-road idea to try to lose weight while figuring out what works for me. I call my new plan the Adaptive Diet.
Starting from my actual weight on Day One, I drew a line that represents my target weight, assuming a daily decrease of 0.1 pounds. A deviation of one pound from my target weight on my daily weigh-in is what I call my Yellow Zone. When I am in the Yellow, I continue doing what I was doing before: trying to build new, healthier habits.
If I am more than one pound below my target weight, then I have entered what I call the Green Zone. When I am in the Green, I can allow myself to indulge my cravings. However, when I am one pound above my target weight, I call that the dreaded Red Zone. This Zone has different shades of red. If I am between 1 and 2 pounds above my target weight, I have to eat only apples after 8:00pm. If I am 2 to 3 pounds above my target weight, only-apples time starts at 6:00pm. And so on. Every extra pound above my target weight moves the cut-off time by two hours. That means that if I am 7 pounds above my target weight, I would have to eat apples all day long.
The system has to work: I do not like apples.
In skyscraper puzzles you have to put an integer from 1 to n in each cell of a square grid. Integers represent heights of buildings. Every row and column needs to be filled with buildings of different heights and the numbers outside the grid indicate how many buildings you can see from this direction. For example, in the sequence 213645 you can see 3 buildings from the left (2,3,6) and 2 from the right (5,6).
In mathematical terminology we are asked to build a Latin square such that each row is a permutation of length n with a given number of left-to-right and right-to-left-maxima. The following 7 by 7 puzzle is from the Eighth World Puzzle Championship.
Latin squares are notoriously complicated and difficult to understand, so instead of asking about the entire puzzle we discuss the mathematics of a single row. What can you say about a row if you ignore all other info? First of all, let us tell you that the numbers outside have to be between 1 and n. The sum of the left and the right numbers needs to be between 3 and n+1. We leave the proof as an exercise.
Let's continue with the simplest case. Suppose the two numbers are n and 1. In this case, the row is completely defined. There is only one possibility: the buildings should be arranged in the increasing order from the side where we see all of them.
Now we come to the question we are interested in. Given the two outside numbers, how many permutations of the buildings are possible? Suppose the grid size is n and the outside numbers are a and b. Let's denote the total number of permutations by f_{n}(a, b). We will assume that a is on the left and b is on the right.
In a previous example, we showed that f_{n}(n, 1) = 1. And of course we have f_{n}(a, b) = f_{n}(b, a).
Let's discuss a couple of other examples.
First we want to discuss the case when the sum of the border numbers is the smallest — 3. In this case, f_{n}(1, 2) is (n−2)!. Indeed, we need to put the tallest building on the left and the second tallest on the right. After that we can permute the leftover buildings anyway we want.
Secondly we want to discuss the case when the sum of the border numbers is the largest — n+1. In this case f_{n}(a,n+1-a) is (n-1) choose (a-1). Indeed, the position of the tallest building is uniquely defined — it has to take the a-th spot from the left. After that we can pick a set of a-1 buildings that go to the left from the tallest building and the position is uniquely defined by this set.
Before going further let us see what happens if only one of the maxima is given. Let us denote by g_{n}(a) the number of permutations of n buildings so that you can see a buildings from the left. If we put the shortest building on the left then the leftover buildings need to be arrange so that you can see a-1 of them. If the shortest building is not on the left, then it can be in any of the n-1 places and we still need to rearrange the leftover buildings so that we can see a of them. We just proved that the function g_{n}(a) satisfies the recurrence:
Actually g_{n}(a) is a well-known function. The numbers g_{n}(a) are called unsigned Stirling numbers of the first kind (see http://oeis.org/A132393); not only do they count permutations with a given number of left-to-right (or right-to-left) maxima, but they also count permutations with a given number of cycles, and they appear as the coefficients in the product (x + 1)(x + 2)(x + 3)...(x + n), among other places. (Another pair of exercises.)
We are now equipped to calculate f_{n}(1, b). The tallest building must be on the left, and the rest could be arranged so that, in addition to the tallest building, b-1 more buildings are seen from the right. That is f_{n}(1, b) = g_{n-1}(b-1).
Here is the table of non-zero values of f_{n}(1, b).
b=2 | b=3 | b=4 | b=5 | b=6 | b=7 | |
---|---|---|---|---|---|---|
n=2 | 1 | |||||
n=3 | 1 | 1 | ||||
n=4 | 2 | 3 | 1 | |||
n=5 | 6 | 11 | 6 | 1 | ||
n=6 | 24 | 50 | 35 | 10 | 1 | |
n=7 | 120 | 274 | 225 | 85 | 15 | 1 |
Now we have everything we need to consider the general case. In any permutation of length n, the left-to-right maxima consist of n and all left-to-right maxima that lie to its left; similarly, the right-to-left maxima consist of n and all the right-to-left maxima to its right. We can take any permutation counted by f_{n}(a, b) and split it into two parts: if the value n is in position k + 1 for some 0 ≤ k ≤ n-1, the first k values form a permutation with a - 1 left-to-right maxima and the last n - k - 1 values form a permutation with b - 1 right-to-left maxima, and there are no other restrictions. Thus:
Let's have a table for f_{7}(a,b), of which we already calculated the first row:
b=1 | b=2 | b=3 | b=4 | b=5 | b=6 | b=7 | |
---|---|---|---|---|---|---|---|
a=1 | 0 | 120 | 274 | 225 | 85 | 15 | 1 |
a=2 | 120 | 548 | 675 | 340 | 75 | 6 | 0 |
a=3 | 274 | 675 | 510 | 150 | 15 | 0 | 0 |
a=4 | 225 | 340 | 150 | 20 | 0 | 0 | 0 |
a=5 | 85 | 75 | 15 | 0 | 0 | 0 | 0 |
a=6 | 15 | 6 | 0 | 0 | 0 | 0 | 0 |
a=7 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
We see that the first two rows of the puzzle above correspond to the worst case. If we ignore all other constrains there are 675 ways to fill in each of the first two rows. By the way, the sequence of the number of ways to fill in the most difficult row for n from 1 to 10 is: 1, 1, 2, 6, 22, 105, 675, 4872, 40614, 403704. The maximizing pairs (a,b) are (1, 1), (1, 2), (2, 2), (2, 2), (2, 2), (2, 3), (2, 3), (2, 3), (3, 3).
The actual skyscraper puzzles are designed so that they have a unique solution. It is the interplay between rows and columns that allows to reduce the number of overall solutions to one.
I just compared two searches on Google Trends:
The most personal puzzle that I wrote for the 2013 MIT Mystery Hunt was Integers and Sequences based on my Number Gossip database. I named it after the first lecture that I prepared after I decided to return to mathematics. It is still my most popular lecture.
Many of the clues in this puzzle are standard math problems that are very good for math competition training. Other clues are related to sequences and integer properties.
You might wonder why I often ask for the second largest integer with some property. Isn't the largest one more interesting than the second largest? I do think that the largest number is more interesting, but exactly for this reason the largest number is available on my Number Gossip website and therefore is googleable. For example, my Number Gossip properties for 3000 contain the fact that 3000 is the largest palindrome in Roman numerals. This is why in the puzzle I used a slightly different clue, i.e. "the second largest three-letter palindrome in Roman numerals."
It took me many hours to find non-googleable variations of interesting properties for this puzzle. Unfortunately, its non-googleability evaporated as soon as my solution was posted, right after the hunt. In any case some clues in this puzzle are useful for math competition training, and I plan to use them myself in my classes. The puzzle is attached below. I will post the solution in a couple of weeks.
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The ultimate goal of each MIT Mystery Hunt is to find a hidden coin. So it was highly appropriate that our 2013 team created a coin-weighing puzzle (written by Ben Buchwald, Darby Kimball, and Glenn Willen) as a final obstacle to finding the winning coin:
There are nine coins, one real and eight fake. Four of the fake coins weigh the same and are lighter than the real coin. The other four fake coins weigh the same and are heavier than the real coin. Find the real coin in seven weighings on the balance scale.
Actually, it is possible to find the real coin in six weighings. Can you do that?
My weight used to be my most guarded secret. In general, I am a very open person: I'll tell anyone anything about me, unless it involves other people. However, there were two exceptions, both of them numbers, interestingly enough: my age and my weight. The closest I came to revealing my weight was with my sister, because we often discuss our similar health issues. Unfortunately, she knows my age, so the only missing number is my weight. I am so tired of my struggle to lose weight, that I've stopped caring about keeping the number secret. I am ready to tell it to the whole world.
Let me start from the beginning. I grew up in a country and at a time when men liked plump women. I was never thin, and didn't have to worry about my weight like my thin girlfriends did. I'll never forget my high school boyfriend telling me, "Ninety percent of men like fat women, and the other ten percent like very fat women." When in college I weighed 70 kilograms (154 pounds) and I felt fine. I had my first child when I was 23. I gained 20 kilos while I was breastfeeding, reaching 90 kilos (200 pounds). My husband Andrey kept telling me that he liked Rubenesque women. I wasn't even slightly concerned about my weight. When we divorced in 1988, I felt that my world was crushed and I didn't want to go on living. As a result, I lost about 20 pounds.
By 1990 I recovered from my depression, married my next husband, and moved to the US to live with him. The US made me aware of my weight immediately. It didn't help that Andrey remarried a woman who was the opposite of Rubenesque. From this point on, I wanted to lose weight. After my second child was born, I gained 20 kilograms while breastfeeding, just as I had done with the first child. The result was that I weighed about 220 pounds, much more than I wanted.
I started to look around at what capitalist society had to offer. The pharmacy had many products. I tried Slim Fast, which started to kill my appetite immediately. However, I began to get depressed. The depression felt foreign. As a new mother, I had been very happy before using Slim Fast, and there had been no changes in my life other than consuming Slim Fast. I stopped using it and the depression disappeared. To make sure, I did an experiment. I started using Slim Fast again and the depression reappeared within three days. I stopped it and my depression disappeared. I was so desperate to lose weight that I repeated the experiment. But the result was the same. I stopped using it, and never used any slimming supplement since then. But within that whole process, I lost some weight.
I stayed slightly over 200 pounds for several years. The third time (after the divorce and the Slim Fast) that I lost a lot of weight was when I had my heart broken about 15 years ago. Since then I've been slowly gaining weight.
As you can see from my story, I was never able to lose weight when I wanted to. I lost it three times, but I can't and don't want to reproduce those circumstances. I actually do not know how to lose weight. For the past ten years I've been making changes in my eating habits that I hope, cumulatively, would help me lose weight. I do not buy soda or pizza. I significantly cut my consumption of sweets and starches. I eat more fruits and vegetables. I eat half of what I used to eat in a restaurant. I am still gaining weight.
he only thing I haven't tried is to be hungry. I am afraid of being hungry. Also I am scared that if I decide on a plan which might result in my being hungry, I will not be able to stick to it. I don't want to discover that I don't have enough will power. I am scared to be a failure. I hope that by writing and publishing this I'll gain the courage to replace my half-measures with a more drastic plan.
Oh! I forgot to tell you: I weigh 245 pounds.
* * *
Grigori Perelman's theorem: There is no offer you can't refuse.
* * *
A conversation between two Russians:
— Run to the store and fetch a couple bottles of vodka.
— How much is a couple?
— Seven.
* * *
— Is it true that the Windows operating system was copied from a UFO computer that crashed in Roswell?
— All we know for sure is that the UFO that didn't crash had a different operating system.
* * *
I saw our system administrator's shopping list. The first line was tomatoes.zip for ketchup.
I posted the puzzle In the Details two weeks ago. This is the most talked-about puzzle of the 2013 MIT Mystery Hunt. The author Derek Kisman invented this new type of puzzle and it is now called a Fractal Word Search. I anticipate that people will start inventing more puzzles of this type.
Let's discuss the solution. The words in the given list are very non-random: they are related to fractals. How do fractals work in this puzzle? The grid shows many repeating two-by-two blocks. There are exactly 26 different blocks. This suggests that we can replace them by letters and get a grid that is smaller, for it contains one-fourth of the number of letters. How do we choose which letters represent which blocks? We expect to see LEVEL ONE in the first row as well as many other words from the list. This consideration should guide us into the matching between letters and the two-by-two blocks.
The level one grid contains 18 more words from the list. But where are the remaining words? So we have level one, and the initial grid is level two. The substitution rule allows us to replace letters by blocks and move from level one to level two. When we do this again, replacing letters in level two by blocks, we get the level three grid. From there we can continue on to further levels. There are three words from the list on level three and one word on level four. But this is quickly getting out of hand as the size of the grid grows.
Let's step back and think about the next step in the puzzle. Usually in word search puzzles, after you cross out the letters in all the words you find, the remaining letters spell out a message. What would be the analogous procedure in the new setting of the fractal word search? In which of our grids should we cross out letters? I vote for grid number one. First, it is number one, and, second, we can assume that the author is not cruel and put the message into the simplest grid. We can cross-out the letters from words that we find in level one grid. But we also find words in other levels. Which letters in the level one grid should we cross out for the words that we find in other levels? There is a natural way to do this: each letter in a grid came from a letter in the previous level. So we can trace any letter on any level to its parent letter in the level one grid.
We didn't find all the words on the list, but the missing words are buried deep in the fractal and each can have at most three parent letters. I leave to the reader to explain why this is so. Because there are so few extra letters, it is possible to figure out the secret message. This is what my son Sergei and his team Death from Above did. They uncovered the message before finding all the words. The message says: "SUM EACH WORD'S LEVEL. X MARKS SPOT." Oh no! We do need to know each word's level. Or do we? At this point, the extra letters provide locations of the missing words. In addition, if a word on a deep level has three parents, then it has to be a diagonal word passing through a corner of one of the child's squares. So our knowledge of extra letters can help us locate the missing words faster.
Also, the message says that the answer to this puzzle will be on some level in the part of the grid that is a child of X. Luckily, but not surprisingly, there is only one letter X on level one. The child of X might be huge. But we could start looking in the center. Plus, we know from the number of blanks at the end of the puzzle, that the answer is a word of length 8. So Sergei and his team started looking for missing words and the answer in parallel. Then Sergei realized that the answer might be in the shape of X, so they started looking at different levels and found the answer before finding the last word on the list. The answer was hiding in the X shape in the center of the child of X on level 167: HUMPHREY.
H..Y .UE. .RM. H..P
Cambridge Waldo puzzle from the 2013 MIT Mystery Hunt was supposed to be easy. Its goal was to get people out of the building for some fresh air. I made this puzzle jointly with Ben Buchwald, Adam (Pesto) Hesterberg, Yuri Lin, Eric Mannes and Casey McNamara. The puzzle consists of 50 pictures of different locations in Cambridge; one of the above individuals was hiding in each picture. Let me use this opportunity to thank my friends for starring in my puzzle and being inventive while doing it.
The puzzle starts with a group picture of my stars. The caption to the picture gives their names. The fact that they are standing in alphabetical order is a clue.
Out of the 50 pictures, each person appears in exactly ten pictures. If you mark the locations of one person on a map, they look like a letter. For example, below are Ben's locations that form a letter "S". When you put the letters in the alphabetical order by people names you get the answer to the puzzle: SCAMP.
As you can see, you do not need all ten locations to recognize the letter. You might be able to recognize the letter with five locations, or at least significantly reduce possibilities for the letter. Besides, you do not need all the letters to recognize the answer. We thought that this was an easy puzzle.
And, to make it even easier, the order in which we posted the pictures was not completely random. The pictures of one person were in the order one might walk from one location to another. This played two important roles. If you recognize the person but do not recognize the location on the picture, you can make an approximate estimation of the location because it must be on the path between the previous and next locations. If you recognize the location but not the person, you can guess the person by checking whose path it fits better.
It was difficult to hide people, especially when there were no other people around. So we sometimes used props. We only used one prop per person. Here you can see Pesto with his sarongs in plain view. In the other picture (below) he is hiding under a white sarong. Yuri had a bicycle helmet. In one of the pictures, she hid so well that you couldn't see her — but you could see her helmet. Ben had a bear hat. In one of the pictures you can only see a shadow of a person, but this person was clearly wearing a hat with bear ears. Eric didn't have a prop, but my car was eager to make a cameo appearance at the Mystery Hunt, so I hid him in my car in one of the pictures.
As you might guess I made the pictures of different people at different times. So Ben Buchwald was the one who realized that solvers might differentiate people by looking at the data of the picture files. He carefully removed the original time stamps.
Despite our best intentions, our test-solvers decided not to leave their comfy chairs, but rather to use Google-StreetView. We strategically made some of the pictures not Google-StreetViewable, but our test-solvers still didn't leave the comfort of their chairs. They just became more inventive. I do not know all the things they did to solve the puzzle, but I heard about the following methods:
I have yet to understand why this puzzle was difficult for the Mystery Hunt teams.
Here is the solution to the Open Secrets puzzle I published recently. Through my discussion of this solution, you'll also get some insight into how MIT Mystery Hunt puzzles are constructed in general.
I've included the puzzle (below) so that you can follow the solution. The puzzle looks like a bunch of different cipher texts. Even before we started constructing this puzzle, I could easily recognize the second, the seventh, and the last ciphers. The second is the cipher used by Edgar Allan Poe in his story, The Gold-Bug. The seventh cipher is a famous pigpen (Masonic) cipher, and the last is the dancing men cipher from a Sherlock Holmes story. Luckily you do not need to know all the ciphers to solve the puzzle. You can proceed with the ciphers you do know. With some googling and substitution you will translate these three pieces of text into: COLFERR, OAOSIS OF LIFEWATER, and RBOYAL ARCH.
These look like misspelled phrases, each of which has an extra letter. However, there are no typos in good puzzles. Or, more precisely, "typos" are important and often lead to the answer. So now I will retype the deciphered texts with the extra letter in bold: COLFERR, OAOSIS OF LIFEWATER and RBOYAL ARCH. In the first word it is not clear which R should be bold, but we will come back to that later.
At this point you should google the results. You may notice that the "royal arch" leads you to the Masons, who invented the pigpen cipher. From this, you can infer the structure of the ciphers and the connections among them. Indeed, a translation of one cipher refers to another. So you should proceed in trying to figure out what the texts you have already deciphered refer to. Eoin Colfer is the author of the Artemis Fowl series that contains a Gnomish cipher, and Oasis of Lifewater will lead you to Commander Keen video games with their own cipher. When you finish translating all of the ciphers, you get the following list:
The bold letters do not give you any meaningful words. So there is more to this puzzle and you need to keep looking. You will notice that the translations are in alphabetical order. This is a sign of a good puzzle where nothing is random. The alphabetical order means that you need to figure out the meaningful order.
To start, the phrases reference each other, so there is a cyclic order of reference. More importantly, the authors of the puzzle added an extra letter to each phrase. They could have put this letter anywhere in the phrase. As there is nothing random, and the placements are not the same, the index of the bold letter should provide information. If you look closely, you'll see that the bold letters are almost all in different places. If we choose the second R in COLFERR as an extra R, then the bold letter in each text is in a different place. Try to order the phrases so that the bold letters are on the diagonal. You'll see that this order coincides with the reference order, which gives you an extra confirmation that you are on the right track. So, let's order:
Now the extra letters read BOKLORYFH. This is not yet meaningful, but what is this puzzle about? It is about famous substitution ciphers. The first and most famous substitution cipher is the Caesar cipher. So it is a good idea to use the Caesar cipher on the phrase BOKLORYFH. There is another hint in the puzzle that suggests using the Caesar cipher. Namely, there are many ways to clue the dancing men code. It could be Conan Doyle, Sherlock Holmes, and so on. For some reason the authors chose to use the word ELEMENTARY as a hint. Although this is a valid hint, you can't help but wonder why the authors of the puzzle are not consistent with the hints. Again, there is nothing random, and the fact that the clues are under-constrained means there might be a message here. Indeed, the first letters read ROMAN CODE, hinting again at the Caesar shift. So you have to do the shift to arrive at the answer to this puzzle, which is ERNO RUBIK.
The most difficult puzzle I wrote for the MIT Mystery Hunt 2013 was Turnary Reasoning. I can't take credit for the difficulty: I designed the checkers positions; they were expectedly the easiest. Timothy Chow created the chess positions, and Alan Deckelbaum created the MTG positions. As the name of the puzzle suggests, you need to find whose turn it is in each position or, as the flavor text suggests, decide that the position is impossible.
I tried to solve the chess positions myself and was charmed by their beauty. The most difficult one was the first chess puzzle presented below. Find whose turn it is or prove that the position is impossible.
The puzzle titled "50/50" was the most difficult puzzle in MIT Mystery Hunt 2013. It is a puzzle in which information is hidden in the probability distribution of coin flips. I consider it the most difficult puzzle of the hunt because it took the longest time to test-solve and we were not able to solve all four layers of the original puzzle. As a result, one of the layers was removed. I think this puzzle is very important and should be included in statistics books and taught in statistics classes. If I were ever to teach statistics, I would teach this puzzle. By the way, this elaborate monstrosity (meant as a compliment) was designed by Derek Kisman.
I am not sure that the puzzle is working on the MIT server. The puzzle is just a coin flip generator and gives you a bunch of Hs (heads) and Ts (tails). Here is the solution.
When you flip a coin, the first thing to check is the probability of heads. In this puzzle it is fifty percent as expected. Then you might check probabilities of different sequences of length 2 and so on. If you are not lazy, you will reach length 7 and discover something interesting: some strings of length seven are not as probable as expected. The two least probable strings are TTHHTTH and HTHHTTT, with almost the same probability. The two most probable strings are TTHHTTT and HTHHTTH, with the matching probability that is higher than expected. All oddly behaving strings of length 7 can be grouped in chunks of four with the same five flips in the middle. In my example above, the five middle flips are THHTT. Five flips is enough to encode a letter. The probabilities provide the ordering, so you can read a message. In the version I tested it was "TAXINUMBLOCKS." In the current version it is "HARDYNUMBLOCKS." Keep in mind that the message encoded this way has to have all different letters. So some awkwardness is expected. The message hints at number 1729, a famous taxicab number, which is a clue on what to do next in the second step.
What do you do with number 1729? You divide the data in blocks of 1729 and see how the k-th flip in one block correlates with the k-th flip in the next block. As expected, for most of the indices there is no correlation. But some of the indices do have correlation. These indices are close together: not more than 26 flips apart. Which means the differences will spell letters. Also, there is a natural way to find a starting point: the group of indices spans only a third of the block. In the original version the message was: "PLEASEHELPTRAPPEDINCOINFLIPPINGFACTORYJKHEREHAVEAPIECEOFPIE."
Now I want to discuss the original version, because its solution is not available online. Here is Derek's explanation of what happens in the third step:
So, this punny message is another hint. In fact the sequence of coin flips conceals pieces of the binary representation of Pi*e. These pieces are of length 14 (long enough to stand out if you know where to look, but not long enough to show up as significant similarities if you compare different sessions of flips), always followed by a mismatch. They occur every 1729 flips, immediately after the final position of the 1729-block message. The HERE in the message is intended to suggest looking there, but you can probably also find them (with more effort) if you search for matches with Pi*e's digits.
The 14-flip sequences start near the beginning of the binary representation of Pi*e and continue to occur in order. (ie, every 1729 flips, 14 of them will be taken straight from Pi*e.) However, between sequences, either 1, 3, or 5 digits will be skipped. These lengths are a sequence of Morse code (1=dot, 3=dash, 5=letter break) that repeats endlessly, with two letter breaks in a row to indicate the start:
- .... ..- ..- ..- - ..- -..- ..- ..- ..- .... ..- .... - ..- ..- .... ..- ....
Translated, this gives the message "THUUUTUXUUUHUHTUUHUH".
(Aside: I didn't use Pi or e individually, because one of the first things I expect some teams will try is to compare the sequence of flips with those constants!)
As I said before, we didn't solve the third step. So Derek simplified it. He replaced "PIECEOFPIE" by "BINARYPI", and made it the digits of Pi, rather than of Pi*e. We still couldn't solve it. So he changed the message from the second step to hint directly at the fourth step: "PLEASEHELPTRAPPEDINCOINFLIPPINGFACTORYJUSTKIDDINGTHUUUTUXUUUHUHTUUHUH." But the binary Pi was still trapped in the coin flipping factory.
Here is Derek’s explanation of the fourth step:
Almost there! This message looks like some sort of flip sequence, because it has several Ts and Hs in there, but what of the Us and Xs? Well, U just stands for "unknown", ie, we don't care what goes there. And there's only one X, so it seems significant!
The final step is to look for every occurrence of this pattern in the sequence. The flips that go where the X is are the final channel of information. You'll find that they repeat in an unvarying pattern (no noise!) with period 323=17*19. There's only one way to arrange this pattern into a rectangular image with a blank border, and it gives the following image:
................. ...X..XXX.XXX.... ...X..X...X.X.... ...X..XXX.XXX.... ...X..X...XX..... ...X..X...X.X.... ...X.....X....... ...X.....X....... ...X....XXX...... ...X.....X....... ...X............. ...X.....XXX...X. ...X....XXXXX.XX. ..X....XX.XXXXX.. .X......XXXXXX... .X...X.XXXXXXXX.. ..XXX...XXXXX.XX. .........XXX...X. .................
The final answer is the French word for fish, POISSON, a word heavily related to statistics!
The answer POISSON didn't fit in the structure of the Hunt. So Derek was assigned a different answer: MOUNTAIN. He changed the picture and it is now available in the official solution to the puzzle. He adjusted his code for coin flips so that the picture of a mountain is hidden there. But the digits of Pi are still trapped in the flips. They are not needed for the solution, but they are still there.
Derek kindly sent to me his C++ program for the latest version of the puzzle. So if the MIT website can't generate the flips, you can do it yourself. And play with them and study this amazing example of the use of statistics in a one-of-a-kind puzzle.
When the MIT Mystery Hunt was about to end, I asked my son Sergei, who was competing with the team "Death from Above," what his favorite puzzle was. I asked the same question to a random guy from team "Palindrome" whom I ran into in the corridor. Surprisingly, out of 150 puzzles they chose the same one as their favorite. They even used similar words to describe it. Calling it a very difficult and awesome puzzle, they both wondered how it was possible to construct such a puzzle.
The puzzle they were referring to is "In the Details" by Derek Kisman, which you can see below.
TWELEVELTWONSHELMUMUOERAIYRANL QAPIUNPIQAYDPEPIRPRPKVOYESOYOR ELRATFDTELDTTFDTBWNLMUTFONYDWJ PIOYJMHAPIHAJMHAAOORRPJMYDANFC MUOZCGTFBWIRYDHIRAIRTFNCUENCUE RPVQUHJMAOHKANJUOYHKJMZKBNZKBN IRONSHOZGOTFUEELTFOEELUEYDOETF HKYDPEVQDNJMBNPIJMKVPIBNANKVJM BWIYNLTFSHHIELTWGOYDONDTYDHIOE AOESORJMPEJUPIQADNANYDHAANJUKV SHDTYDRPBWUEBWIYTWTWTFYDMUELMU PEHAANAJAOBNAOESQAQAJMANRPPIRP ONTWELBWLMSHELTFUEBWBWLMOZEVHI YDQAPIAOGIPEPIJMBNAOAOGIVQUNJU DTCGUEYDRPEVNCIREVIRTWUEUETWON HAUHBNANAJUNZKHKUNHKQABNBNQAYD IRUERAMUTFELTWONTFOEOEEYDTNLYD HKBNOYRPJMPIQAYDJMKVKVHWHAORAN ELGORPNCTFDTYDSHYDELPKTFOZRACG PIDNAJZKJMHAANPEANPIDFJMVQOYUH DTMUWJOETFYDELMUMUGORAONIRDTCG HARPFCKVJMANPIRPRPDNOYYDHKHAUH
BOUNDARY | HENON | LEVY DRAGON | SCALING |
BROWNIAN | HILBERT | LYAPUNOV | SPACE |
CAUCHY | HURRICANE | MANDELBROT | STRANGE |
CURLICUE | ITERATE | NEURON | TAKAGI |
DE RHAM | JULIA | NURNIE | TECTONICS |
DIMENSION | LEIBNIZ | POWER LAW | T-SQUARE |
ESCAPE | LEVEL ONE | RAUZY | WIENER |
HAUSDORFF | LEVEL TWO | RIVER | YO DAWG |
_ _ _ _ _ _ _ _
The puzzle looks like a word search, but I can tell you up-front: you can't find all the words in the grid. You can only find six words there. So there is something else to this puzzle. I will discuss the solution later. Meanwhile I will ask you very pointed questions:
The easiest of the puzzles I made for the MIT Mystery Hunt 2013 was "Something in Common." I collaborated on this puzzle with Daniel Gulotta. Ironically, it was the most time-consuming of my puzzles to design — well over a hundred hours. I can't tell you why it took me so long without revealing hints about the solution, so I will wait until someone solves it.
I received a lot of critique from my editors for suggesting puzzles that were too easy. When, during the test-solve, I realized that this puzzle was one of the easiest in the hunt, I requested permission to make it harder. It would not actually have been difficult to make it harder: I could have just replaced some specific words with more general ones. Unfortunately, we didn't have time for a new test-solve, so the puzzle stayed as it was. That turned out to be lucky.
This puzzle was in the last round. By the time the last round opened, we knew that the hunt was much more difficult than we had anticipated. I was afraid that people were getting angry with difficult puzzles and so I was very happy that I hadn't changed this puzzle. By the time the teams started submitting answers to it, people were exhausted. Manic Sages increased the speed with which options to buy answers to puzzles were released. I was ecstatic that this puzzle was one of the few puzzles in the last round that was solved, not bought with options. Here is the puzzle:
While everyone was discussing which color to paint the room, the counselor was dreaming about a real chocolate sundae. A walk-in walked into the reception in honor of a regular phenomenon visible every four hours or so. The client just wanted to find his dog in order to provide a permanent shortcut to another part of space; instead, he found something else.
The Federation didn't want to buy a lemon, so their lawyer used a spider to type the password. They discovered that the creatures glow in the dark and the perceptive negotiator won the bid. While they were waiting for the results of the shuttle trip, they turned the lights off to hunt them. The bird-like sounds were coming from under the floor. They used alcohol to save the woman's life, which was a big mistake. She refused to be his conscience, instead she used her new untapped power of light to save the world; she already had a job as a counselor.
Their daughter disappeared two months ago, whereas they needed to keep the wounded survivor steady. The African activist was kidnapped by rebels when she was a teenager. While some people were building a boat on the beach, she escaped from her kidnappers. He said that the reward had been withdrawn to eliminate the financial incentive to lie. They needed to perform a blood transfusion, but they didn't have the proper instruments. By his false statement, he removed over-claimers: people lying to get attention. He needed to write vows. They read surprise at the unexpected labor in the woods in the boy's face. It was clear that the boy had never looked into the backpack.
She described horrors she had seen: rape, torture, massacres carried out by child soldiers, the doctor losing his own blood to help the wounded survivor. There was no deception leakage, but she seemed anxious, so the doctor couldn't go and deliver the baby. They needed to find the other girl before noon, and the crushed leg made blood transfusion impractical. His vows were self-referential. He talked about how he couldn't write the vows. While the doctor played 20 questions one man died and another was born. The girl wasn't answering, but the doctor read her micro-expressions. The writer touched her ear and exposed her lies. At the end the doctor suspected murder, and the book was withdrawn.
She took pictures in an abandoned house and jumped off the roof. Two girls came back to investigate. He was surprised by a basket of cookies and a statue that was in the wrong place. A new drug hit the streets making people see things, like old pictures of a woman who looked exactly like her friend. He started seeing his dead father: when people die in the past, they produce energy in the present.
They were the two most powerful people on the planet. For wedding preparations she got a list of 17 DVDs. His partner started hallucinating about her ex. People do not understand time. Naomi's ability caused all this mess: the box left them behind. The only place where she didn't use gloves was her garden. The dinner guests tried her chicken, and, as a consequence, they were stuck looking at each other forever. The drugged cookies were a gift to help resolve their issues. Because of that, she gave him the shorthand transcripts to complete the cycle.
Did she fall asleep while they were discussing the dead boy's name? The price tag was exorbitant for a witness to identify the defendant. The woman would be everything he wanted and needed. She was a security guard chewing gum. The witness changed her testimony: they went to the woods and were water-rafting, rock-climbing, and arrow-shooting. As she was not the first girl he brought to the woods, the jury gave a verdict of "Not Guilty."
The detective was in court, though it wasn't his case. She needed to stop talking and start running. Shit always rolls downhill. Poisoned water made her see things she was not supposed to remember like a shitty orange couch outside. In spite of the background check the client was a psychopath who was moved from uptown to the fucking low-rises. Shoulder to the wheel. Dead witness, mother-fucker.
A little girl was humming about the dark side. The Wicca group was boring, and the new invention was making whooshing sounds. While a laryngitis epidemic took over the town, the noise-eater got turned on by accident.
The princess's scream would kill the monsters, while the monster machine killed its own remote control. She and her future boyfriend discovered each other's secrets. The dark side is waiting.
The experiment in the gay club was very important to him. He was a semi-cute boy-next-door type and his shuttle trip was supposed to be easy. There is more to a guy than cock size, and he was being pulled. He was injected with something and started to understand everyone. They were a bald priestess and a warrior with a very nice butt. They offered you a discount if you bought the butt and the bulge together.
They were approaching a commerce planet in Pennsylvania. They escaped and were standing on the roof of a hospital. She was sending a transmission, while the older man was running his tongue along the young man's spine. The girl tried to protect him, believing his story, but was arrested for contamination. Fuck. He had a baby, two babies. He distracted the guards with a puzzle ring. And while they were painting his black car with a pink derogatory word, they escaped.
He was drinking before operating on his fiancée. What kind of a father kills people? Tonight was the night he would finally sleep, and drinking was the only way for him to stop his hands from shaking. The limo driver kidnapped him while she watched herself die in the mirror. While being sleep-deprived he conducted a test. Nothing was wrong, except he brought the wrong documents to court. The criminal walked, but, unexpectedly, he couldn't feel his leg and couldn't walk anymore. There was a lot of blood and the body was not filtering iodine. He took a picture of the actor falling into a coma with his blood-spatter camera.
His girlfriend wanted to buy a firm mattress to have sex. He preferred to sleep on a water-bed and told the secret of his bloody hobby to his baby. It was not an infection, it was an allergy, so he changed his plan of where to grab the victim. The case was solved because of the bubbles in the glass. He was killing for his son now. The patient was indeed allergic and crashed his car.
The woman died two weeks ago, but her car accident was yesterday. There was so much blood and, for some reason, ice. Jane Doe wasn't the girl's sister. Melting speed and surface tension could help the calculation. They came to pay their respects but the celebration was postponed. They brought personal effects to drop off. If the couple's daughter was like their mother, then they themselves were like their grandparents. The idea to ask the grandmother resulted in a kiss.
Their planet had been on the verge of a golden age, then everything fell apart. She brought some clothes, a toothbrush, shampoo and some other things to the detention center. "Don't let the history repeat itself," the worshiper of the second in command said. She didn't know what the red string was for. The teen's sister told him stories about another galaxy, and the war and the algorithm that might find the dead man's favorite coffee place and, consequently, solve the case. The kidney operation was in a hotel. They didn't expect to see the body, but there it was. She touched the body to reveal emptiness inside. At the end they gathered in his house for dinner as usual. He had to save Courtney's future body before everything blew up and put the red sticker on his license.
After retirement, he came back to an unsolved murder case; he remembered a kiss in flashbacks. He drove to the lake where the accidental drowning happened. He lured everyone involved to the lake, and, after having sex, he told her that he wanted a real marriage. They wanted their child to be born on the new planet. The doctor was one of the suspects. The couple was already in the cabin, and he doubted that he was the one who she really wanted.
The young man was murdered, and everyone thought that the detective was the real target. The man went to look for sneakers during the boxing match. The nosy lady-writer dropped names to persuade the sheriff to look into the cabin. The admiral challenged the chief to join him in the ring. They didn't find the sneakers, and they shouted this to the skies. The lady solved the murder, as usual, while the boxing match became too personal and people started leaving.
The second "instructioned" puzzle is Portals by Palmer Mebane. It is an insanely beautiful and difficult logic puzzle that consists of known puzzle types interconnected to each other through portals. Here Palmer Mebane explains how portals work:
"Each of the ten puzzles corresponds to a color, seen above the grid where the name of the puzzle is written. The grid contains nine square areas, one each of the other nine colors. These are portals that connect the puzzle to one of the other nine, as denoted by the portal color. Each puzzle's rules define which squares of their solution are "black". On the portal squares, the two puzzles must agree on which squares are black and which are not. For instance, if in the red grid the top left square of the blue portal is black, then in the blue grid the top left square of the red portal must also be black, and vice versa."
On the Portals puzzle page you can find the rules for how each individual game is played and how to shade areas. The puzzle requires a lot of attention. It took us a long time to test-solve it. If you make a mistake in one grid it will propagate and will lead to a contradiction in another grid, so it is difficult to correct mistakes. If you do make a mistake, you are not alone: we kept making mistakes during our test-solve. Because of the difficulty of tracing back to the source of the error, we just started anew, but this time making sure that every step was confirmed by two people. Working together in this way, we were able to finish it.
If you do not care about the extraction and the answer, ignore the letter grid in the middle and enjoy the logic of it.
There were a couple of puzzles during the MIT Mystery Hunt that were not so mysterious. Unlike in traditional hunt puzzles, these puzzles were accompanied by instructions. As a result you can dive in and just enjoy solving the logic part of the puzzle without bothering about the final phase, called the extraction, where you need to produce the answer.
The first puzzle with instructions is Random Walk by Jeremy Sawicki. I greatly enjoyed solving it. In each maze, the goal is to find a path from start to finish, moving horizontally and vertically from one square to the next. The numbers indicate how many squares in each row and column the path passes through. There are nine mazes in the puzzle of increasing difficulty. I am copying here two such mazes: the easiest and the toughest. The colored polyomino shapes are needed for the extraction, so you can ignore them here.
Today I have a special treat for you. Here is the first of several puzzles that I plan to present from the 150 that we used in the MIT Mystery Hunt 2013. Keep in mind that although the puzzles have authors, they were the result of a collaboration of all the team members. In many instances editors, test-solvers and fact-checkers suggested good ways to improve the puzzles.
I wrote the puzzle Open Secrets jointly with Rob Speer. The puzzle was in the opening round, which means it is not too difficult. By agreement the answers to the puzzles are words or phrases. I invite my readers to try this puzzle. I will post the explanation in about two weeks.
I dropped my blog for two months. Some of my readers got worried and wrote to ask if I was okay. Thanks for your concern.
I am okay. I was consumed by the MIT Mystery Hunt. My team, Manic Sages, won the hunt a year ago, and as a punishment — oops, I meant as a reward — we got to write the 2013 hunt, instead of competing in it. I myself ended up writing about ten problems for the hunt. This was in addition to test-solving about 150 problems my whole team prepared for the hunt.
I could only think about the hunt. My mind was full of ideas for the hunt so I was afraid to write in my blog about something that I might later want to use for my problems. Or even worse, I was nervous that my blog posts might be unconsciously revealing hunt secrets. Moreover, I didn't want to advertise the fact that I was working on the hunt, thereby drawing people to my blog to scrutinize my interests as they prepared for the hunt.
So I just disappeared.
I apologize; please forgive me.
The father of my son has four children. My son is my only child. How many children do we have in total together?
"I will win the next International Chopin Piano Competition."
No matter how good I am at positive affirmations, that won't work: I do not play piano.
I tried to read books on positive thinking, but they made me mistrust the genre. The idea that you can achieve anything by positive thinking makes no sense. For example:
Positive thinking might actually be harmful. I can invest tons of time into trying to change my natural eye color by using my thoughts, when instead I could just use my money and buy some colored contact lenses. Or, if I think myself rich, I might start spending more money than I have and end up bankrupt.
However, perhaps I should not have totally dismissed the idea of positive thinking. While it does have logical inconsistencies, such as those in my examples above, maybe there are ways in which positive thinking is helpful.
First, we should treat these beliefs not as a guarantee, but probabilistically. For example, if you think that you can win the piano competition, the judges will feel your confidence, and may give you slightly better marks.
Second, positive thinking can work, if we choose our affirmations correctly. I recently discovered that I am deceiving myself into believing that I am hungry when I'm not. I should be able to reverse that. I should be able to persuade myself that I am not hungry when I am.
I decided to start small. I tried to persuade myself that tiramisu doesn't really taste good. Once that seemed to be working, I got more serious. I bought a couple of CDs with affirmations for weight loss.
Unfortunately, they want me to lie down and relax. I do not have time to lie down. I could listen when I am driving or when I am cleaning my kitchen. Hey, does anyone know some good weight-loss affirmations CDs that do not require relaxation?
One of the 2012 PRIMES projects, suggested by Professor Jacob Fox, was about bounds on the number of halving lines. I worked on this project with Dai Yang.
Suppose there are n points in a general position on a plane, where n is an even number. A line through two given points is called a halving line if it divides the rest of n−2 points in half. The big question is to estimate the maximum number of halving lines.
Let us first resolve the small question: estimating the minimum number of halving lines. Let's take one point from the set and start rotating a line through it. By a continuity argument you can immediately see that there should be a halving line through any point. Hence, the number of halving lines is at least n/2. If the point is on the convex hull of the set of points, then it is easy to see that it has exactly one halving line through it. Consequently, if the points are the vertices of a convex n-gon, then there are exactly n/2 halving lines. Thus, the minimum number of halving lines is n/2.
Finding the maximum number of halving lines is much more difficult. Previous works estimated the upper bound by O(n^{4/3}) and the lower bound by O(ne^{√log n}). I think that Professor Fox was attracted to this project because the bounds are very far from each other, and some recent progress was made by elementary methods.
Improving a long-standing bound is not a good starting point for a high school project. So after looking at the project we decided to change it in order to produce a simpler task. We decided to study the underlying graph of the configuration of points.
Suppose there is a configuration of n points on a plane, and we are interested in its halving lines. We associate a graph to this set of points. A vertex in the configuration corresponds to a vertex in the graph. The graph vertices are connected, if the corresponding vertices in the set have a halving line passing through them. So we decided to answer as many questions about the underlying graph as possible.
For example, how long can the longest path in the underlying graph be? As I mentioned, the points on the convex hull have exactly one halving line through them. Hence, we have at least three points of degree 1, making it impossible for a path to have length n. The picture below shows a configuration of eight points with a path of length seven. We generalized this construction to prove that there exists a configuration with a path of length n−1 for any n.
We also proved that:
After we proved all these theorems, we came back to the upper bound and improved it by a constant factor. Our paper is available at arXiv:1210.4959.
Samuel Hansen has an unusual profession: he is a mathematics podcaster. He interviewed me for his Relatively Prime podcast titled 0,1,2,3,…, where we discussed my Number Gossip project. The podcast also includes interviews with Neil Sloane, Michael Shamos, and Alex Bellos.
My previous interview with Samuel is at acmescience.com. There I discuss both math education and gender in math issues.
When I listened to myself, I found it strange that I seemed to have a British accent on top of my Russian accent. Did you notice that too?
I discovered the following chess puzzle on a Russian blog for puzzle lovers. It is a helpmate-type puzzle. Black cooperates with White in checkmating himself. In this particular puzzle Black starts and helps White to win in one move.
Oops. Something is not quite right. There are not enough pieces on the board. To recover the missing pieces in order to solve the puzzle, you need to retrace your steps. If Black and White go back one move each, they will be able to cooperatively checkmate Black in one move. Find the position one move back and the cooperative checkmate.
I am Russian; I know how to play chess. My father taught me when I was three or four. We played a lot and he would always win. I got frustrated with that and one day, when I was five, I didn't announce my check. On the next move, I grabbed his king and claimed my victory.
He was so angry that he turned red and almost hit me. This frightened me so much that I lost my drive for chess that very moment.
I still understand its beauty and solve a chess problem about once a decade. Look for a cute chess puzzle in my next post.
For my every class I try to prepare a challenge problem to stretch the minds of my students. Here is a problem I took from Adam A. Castello's website:
There is a ceiling a hundred feet above you that extends for- ever, and hanging from it side-by-side are two golden ropes, each a hundred feet long. You have a knife, and would like to steal as much of the golden ropes as you can. You are able to climb ropes, but not survive falls. How much golden rope can you get away with, and how? Assume you have as many hands as you like.
The next problem I heard from my son Sergei:
You are sitting at the equator and you have three planes. You would like to fly around the equator. Each plane is full of gas and each has enough gas to take you half way around. Planes can transfer gas between themselves mid-air. You have friends, so that you can fly more than one plane at once. How do you fly around the equator?
* * *
Right clicking a file with a mouse will allow you to change it, check it for viruses, or revert to the previous version. I wonder where I can buy a mouse that can do the same thing with my husband.
* * *
— Let's have sex.
— Sure, but today I want to be the numerator.
* * *
Attention! We want to check that you are not a robot. Please, undress and turn on a web-camera.
* * *
In a drug store:
— I would like input and output cleaners.
— ???
— Toothpaste and toilet paper.
* * *
I used to recount the multiplication tables to delay my ejaculation. Now, each time I see the multiplication tables I get a hard on.
* * *
— Tonight my parents are away. Let's finally try a forbidden thing.
— Dividing by zero?
* * *
My friend put his mistress in his phone's contact list as 'low battery'.
In what base does 2012! have more trailing zeros: base 15 or 16?
Explain why the result shouldn't be too surprising.
I would like to report on my weight loss progress. Last time I added two new habits, walking my toy dog every day, and drinking more water from the enticing cute bottles I bought.
I named my stuffed dog Liza and I walk with her every day. I didn't expect immediate weight loss due to this new regime, because my first goal was to get out of the house every day, even if only for two seconds. The next step will be to increase walking time to ten minutes.
Drinking a lot of water doesn't work well. I spend too much time looking for bathrooms and panicking that I will not make it. I like the idea of drinking a lot of water, but I am not sure I can hold to it, if you understand what I mean.
Since taking on this challenge, I've gained two habits, but I haven't lost a pound.
Now I'm upping my game. Below is my analysis of why I eat. When I eat, I believe that I am hungry. But looking at this more objectively I think this is not always the case: sometimes there are other reasons. I am listing these other reasons so I can fight them face-to-face. Here we go:
Hmm. That was painful to write. My psychoanalyst taught me that pain means I am on the right track.
Davidson Institute for Talent Development announced their 2012 Winners. Out of 22 students, three were recognized for their math research. All three of them are ours: that is, they participated in our PRIMES and RSI programs:
I already wrote about Xiaoyu's project. Today I want to write about Sitan's project and what I do as the math coordinator for RSI.
RSI students meet with their mentors every day and I meet with students once a week. On the surface I just listen as they describe their projects. In reality, I do many different things. I cheer the students up when they are overwhelmed by the difficulty of their projects. I help them decide whether they need to switch projects. I correct their mistakes. Most projects involve computer help, so I teach them Mathematica. I teach them the intricacies of Latex and Beamer. I explain general mathematical ideas and how their projects are connected to other fields of mathematics. I never do their calculations for them, but sometimes I suggest general ideas. In short, I do whatever needs to be done to help them.
I had a lot of fun working with Sitan. His project was about the rank number of grid graphs. A vertex k-ranking is a labeling of the vertices of a graph with integers from 1 to k so that any path connecting two vertices with the same label passes through a vertex with a greater label. The rank number of a graph is the minimum possible k for which a k-ranking exists for that graph. When Sitan got the project, the ranking numbers were known for grid graphs of sizes 1 by n, 2 by n, and 3 by n. So Sitan started working on the ranking number of the 4 by n graph.
His project was moving unusually fast and my job was to push him to see the big picture. I taught him that the next step, once he finishes 4 by n graphs is not to do 5 by n graphs, as one might think. After the first step, the second step should be bigger. He should use his insight and understanding of 4 by n graphs to try to see what he can do for any grid graphs.
This is exactly what he did. After he finished the calculation of the rank number of the 4 by n graphs, he found a way to improve the known bounds for the ranking number of any grid graph. His paper is available at the arXiv.
I just looked at my notes for my work with Sitan. The last sentence: "Publishable results, a potential winner."
PRIMES-USA: A new MIT program for talented math students from across the country.
I've been working as a math coordinator for RSI, the most competitive summer program for high school juniors. RSI arranges for these select students to do scientific research. I only work with kids who do math, and usually we have a dozen of them. Every student has an individual mentor, usually a graduate student, with whom they meet daily. I supervise all the projects and meet with each high school student about once a week. My job was described as "going for the biggest impact": when the project is in trouble, I jump in to sort it out; when the project is doing well, I push it to further limits.
RSI is a great program: kids enjoy it and we produce interesting research. My biggest concern is that the program is too short. The kids do math for five weeks and they usually approach a good result, but at the end of RSI we generally see just a hint of what they could truly achieve. Kids who continue to work on their own after the program ends are more successful. Unfortunately most of the students stop working at the end of the program just as they are approaching a big theorem.
I discussed this dissatisfying trend with Pavel Etingof and Slava Gerovitch and we decided to do something about it. Pavel and Slava conceived and found funding for a new program called PRIMES that is similar to RSI, but runs for a year. From February through May, PRIMES students meet with their mentors weekly. In fact, we require on the application that the students commit to coming to MIT once a week, thereby limiting us to local students. Theoretically, someone from Detroit with a private jet who can fly to MIT weekly would be welcomed.
Before the first year, we wondered whether the smaller pool of local students would be weaker than national and international RSI students. To our delight, that wasn't the case. In the first year we got fantastic students. One explanation is that PRIMES is much more flexible. We do not mind when our students go to IMO in the summer or to math camps or when they go away on vacation with their parents. As a result, we get students who would never apply to RSI because of their summer schedules. Our PRIMES students have won so many prizes that I do not remember them all. We post our successes on the website.
Our success in PRIMES suggests that there are likely many talented kids in other states who never even apply to RSI because of a scheduling conflict. This led us to try to adapt PRIMES to national needs. So we created a new program called PRIMES-USA that will accept students from across the country. We will work with them through Skype. These students must commit to travel to MIT for a PRIMES conference in May. Because this is our pilot program, we will only accept five students.
Thank you to everyone who helped me to find a host for my Number Gossip website. Some readers and friends even offered me free hosting on their servers. I decided to pay for hosting because I have many specific requirements and that might be a burden on my friends.
On the basis of my readers' recommendations, I chose Dreamhost as my new webhosting provider. I apologize for the interruption in the flow of the gossip. I know that many people use Number Gossip for birthday gift ideas. I can tell you that on my previous birthday, you could have congratulated me on becoming prime and evil.
I visited Raymond Smullyan on my way home from Penn State. We went for lunch at Selena's Diner. What do two mathematicians do during lunch? Exchange magic tricks and jokes, of course. Here is a story Raymond told me:
Raymond: What is the date?
Stranger: I do not know.
Raymond: But you have a newspaper in your pocket!
Stranger: It's no use. It's yesterday's.
Consider central symmetry: squares and circles are centrally symmetric, while trapezoids and triangles are not. But if you have two trapezoids, which of them is more centrally symmetric? Can we assign a number to describe how symmetric a shape is?
Here is what I suggest. Given a shape A, find a centrally symmetric shape B of the largest area that fits inside. Then the measure of central symmetry is the ratio of volumes: B/A. For centrally symmetric figures the ratio is 1, and otherwise it is a positive number less than 1.
The measure of symmetry is positive. But how close to 0 can it be? The picture on the left is a shape that consists of five small disks located at the vertices of a regular pentagon. If the disks are small enough than the largest symmetric subshape consists of two disks. Thus the measure of symmetry for this shape is 2/5. If we replace a pentagon with a regular polygon with a large odd number of sides, we can get very close to 0.
What about convex figures? Kovner's theorem states that every convex shape of area 1 contains a centrally symmetric shape of area at least 2/3. It is equal to 2/3 only if the original shape is a triangle. That means every convex shape is at least 2/3 centrally symmetric. It also means that the triangle is the least centrally symmetric convex figure. By the way, a convex shape can have only one center of symmetry.
After I started writing this I discovered that there are many ways in which people define measures of symmetry. The one I have defined here is called Kovner-Besicovitch measure. The good news is that the triangle is the least symmetric planar convex shape with respect to all of these different measures.
I'm trying to lose weight. Many books explain that dieting doesn't work, that people need to make permanent changes in their lives. This is what I have been doing for several years: changing my habits towards a healthier lifestyle.
This isn't easy. I am a bad cook; I hate shopping; and I never have time. Those are strong limitations on developing new habits. But I've been a good girl and have made some real changes. Unfortunately, my aging metabolism is changing faster than I can adopt new habits. Despite my new and improved lifestyle, I am still gaining weight.
But I believe in my system. I believe that one day I will be over the tipping point and will start losing weight, and it will be permanent. Meanwhile I would like to share with you the great ideas that will work someday.
I have many other ideas that for different reasons haven't yet become habits. So I am thinking about tricks to turn them into habits.
I have many more ideas, but I gotta run now. I need to walk my dog.
I received the book Taking Sudoku Seriously by by Jason Rosenhouse and Laura Taalman for review and put it aside to collect some dust. You see, I have solved too many Sudokus in my life. The idea of solving another one made me barf. Besides, I thought I knew all there is to know about the mathematics of Sudoku.
One day out of politeness or guilt I opened the book — and couldn't stop reading.
The book is written for people who like Sudoku, but hate math. This is so strange. Sudoku is math. People who are good at Sudoku are good at math, or at least they are supposed to be. It seems that math education in the United States is so bad that people who were born to be good at math and to like math, hate it instead. So the goal of the book is to establish a bridge from Sudoku to math. And the book does a superb job of it.
This well-written book moves from puzzles to discussions in such a natural way that math becomes a continuation of puzzles.
Taking Sudoku Seriously covers a lot of fun material: methods to solve Sudoku, how to count the number of different Sudoku puzzles, and how to find the smallest number of clues that are needed for a unique puzzle. The book travels into the neighboring area of Latin and Greco-Latin squares. While discussing all those fun things it covers groups, symmetries, number theory, graph theory (including book thickness) and more.
I am not the target audience for this book, because I do not need convincing that math is fun. The best part for me was the hundred puzzles. Only a portion of them were standard Sudoku puzzles — and I skipped those. The others were either Sudoku with a twist or plain math puzzles.
The puzzles are all very different and I was so excited by them, that I went ahead and solved them, and caught up with reading the text later. And I enjoyed both: reading and solving.
Here is puzzle 91 from the book. Fill in the grid so that every row, column, and block contains 1-9 exactly once. In addition, each worm must contain entries that increase from tail to head. For blue worms you must figure out yourself which end is the head.
It's easy to judge who is the fastest runner or swimmer. Judges do not need to be runners and swimmers themselves. They simply need a stopwatch and a camera.
Other competitions are more difficult to judge. Take for example the Fields medal. The judges need to be mathematicians. Since they can't be experts in all the different areas of mathematics, they have to rely on recommendation letters. The mathematicians who write recommendation letters are biased, because they are interested in promoting their own field. The committee's job is not simple, not the least because it involves a lot of politics. It is easy to award the medal to Grigory Perelman. He solved a high-profile long-standing conjecture. But other cases are not that straightforward.
Imagine a genius mathematician with a new vision. He or she might be so far ahead of everyone else, that the Fields committee would fail to appreciate the new concept. I wish the math community would create a list of mathematicians who deserved the Fields medal, but were passed over. As time goes by, perhaps a new Einstein will emerge on this list.
The reason the Fields committee more or less works is that the judges do not need to be as talented mathematicians as the awardees. They do not need to create mathematics, they need to understand it. And the latter is easier than the former.
A completely different story happens with IQ tests. Someone has to write those tests. There is no reason to think that writers of the IQ tests are anywhere close to the end tail of the IQ distribution. Hence, the IQ tests are not qualified to find the IQ geniuses.
Now might be a good time to complain about the IQ test I took myself. Many years ago I tried an IQ test online through tickle.com. I was so disappointed with my non-perfect score that I never looked at my answers. Recently, while cleaning my apartment, I discovered the printout of the test. I made one mistake in the following question.
Which one of the designs is least like the other four?
The checkmark is the expected answer. They think that the circle is the odd one out because all the other shapes are polygons. The arrow points to my answer. I chose the right triangle because it is the only shape without symmetries. Who says that polygonality is more important than symmetry?
I recently received an invoice from Jumpline, Inc. requesting a payment for hosting www.tanya-khovanova-temp.com. I had never heard of Jumpline before and I didn't have a webpage with that address. So I thought that it was spam.
Because the invoice had my name and address, I decided to call them and check what was going on. It appeared that Jumpline had swallowed Hosting Rails, the company that was hosting my Number Gossip page. Still, I didn't have a clue what the invoice was about.
I asked the representative whether the web address was related to Number Gossip, and he said no. So I canceled the hosting. My work schedule is the busiest in July, so I forgot about the invoice and didn't check my website.
Then I received a letter from Christian, a Number Gossip fan, who told me that the website was down. I called Jumpline again.
It appears that the representative didn't know what he was doing and misled me. The web address www.tanya-khovanova-temp.com was an internal name for my Number Gossip site. They had deleted all the files and were unable to restore my website.
Now I have to decide what to do. I do not want to go back to Jumpline as they are very unprofessional in these ways:
Can anyone suggest a company that can host a website that is written in Ruby on Rails?
The Fibonacci sequence is all about addition, right? Indeed, every element F_{n} of the Fibonacci sequence is the sum of the two previous elements: F_{n} = F_{n-1} + F_{n-2}. Looking closer we see that the Fibonacci sequence grows like a geometric progression φ^{n}, where φ is the golden ratio. In addition, the Fibonacci sequence is a divisibility sequence. Namely, if m divides n, then F_{m} divides F_{n}.
My point: we define the sequence through addition, and then multiplication magically appears by itself. What would happen if we tweak the rule and combine addition and multiplication there?
John Conway did just that: namely, he invented a new sequence, or more precisely a series of sequences depending on the pair of the starting numbers. The sequences are called Conway's subprime Fibonacci sequences. The rule is: the next term is the sum of the two previous terms, and, if the sum is composite, it is divided by its least prime factor.
Let me illustrate what is going on. First we start with two integers. Let's take 1 and 1 as in the Fibonacci sequence. Then the next term is 2, and because it is prime and we do not divide by anything. The next two terms are 3 and 5. After that the sum of two terms is 8, which is now composite and it is divided by 2. So the sequence goes: 1, 1, 2, 3, 5, 4, 3, 7, 5, 6, 11 and so on.
The subprime Fibonacci sequences excite me very much. Not only does adding some multiplication to the rule make sense to me, but also, the sequences are fun to play with. I got so excited that I even coauthored a paper about these sequences titled, not surprisingly, Conway's Subprime Fibonacci Sequences. The paper is written jointly with Richard K. Guy and Julian Salazar, and is available at the arXiv:1207.5099.
We can start a subprime Fibonacci sequence with any two positive numbers. You can see that such a sequence doesn't grow fast, because we divide the terms too often. We present a heuristic argument in the paper that allows us to conjecture that no subprime Fibonacci sequence grows indefinitely, but they all start cycling. The conjecture is not proven and I dare you to try.
Meanwhile, the sequences are a lot of fun and I suggest a couple of exercises for you:
By the way, a trivial cycle is the boring thing that happens if we start a sequence with two identical numbers n bigger than one: n, n, n, n, ….
Have fun.
Kvant is a very popular Russian math and physics journal for high school-children. My favorite page is the one with puzzles directed to younger readers. Here are two puzzles from the latest online issue: 2012 number 3.
The first one, by N. Netrusova, is optimistic about the next year.
An astrologist believes that a year is happy if its digit representation contains four consecutive digits. For example, the next year, 2013, will be happy. When was the previous happy year?
The second problem is by L. Mednikov and A. Shapovalov. It confused me at first. For a moment I thought that the best answer is 241 rubles:
A big candle lasts one hour and costs 60 rubles. A small candle lasts 11 minutes and costs 11 rubles. Can you measure a minute by spending not more than a) 200 rubles, b) 150 rubles?
My American friends often ask me for insights into why Grigory Perelman refused the one million dollar Clay prize for his proof of the Poincaré conjecture. They are right to ask me: my life experience was very similar to Perelman's.
I went to a high school for children gifted in math. I was extremely successful in competitions. I got my gold medal at IMO and went to college without entrance exams. I received my undergraduate and graduate degrees in one of the best math academic centers in Soviet Russia. Perelman traveled a similar path.
Without ever having met Perelman, I can suggest two explanations of why he might reject the money.
First explanation. To have it publicly known that you have suddenly come into money is very dangerous in Russia. Perelman's life expectancy would have dropped immediately after accepting the million dollars. Russians that have tons of money either hide their wealth or build steel doors way before they make their first million. In addition to being a life hazard, money attracts a lot of bother. He would have been chased by all types of acquaintances asking for help or suggesting marriage proposals.
Second explanation. We grew up in a communist culture where money was scorned and math was idolized. The goal of research was research. Proving the conjecture was the prize itself. In his mind, receiving the award money might diminish the value of what he did. I understand this way of thinking, but I am personally too practical to follow such feelings and would accept the prize.
My first explanation has a flaw. Though valid, it doesn't explain why he rejected the Fields medal. So I reached for the book abour Perelman, Perfect Rigor: A Genius and the Mathematical Breakthrough of the Century by Masha Gessen. I like Gessen's explanation of why he rejected the Fields medal:
His objection to the Fields Medal, though never stated as clearly, seemed to have been twofold: first: he no longer considered himself a mathematician and hence could not accept a price intended for the encouragement of midcareer researchers; and second, he wanted no part of ICM, with all the attendant publicity, speeches, ceremony, and king of Spain.
The reasons are specifically related to the medal, so the Clay prize rejection might not be connected to the medal rejection. This argument slightly rehabilitates my first explanation.
I liked the book. It is a tremendous undertaking — writing about a person who doesn't want to talk to anyone. After reading it, I have one more possible explanation of his refusal of the prize.
Perelman is a loner. One of the closest people to him was his math Olympiad coach. The coaches tend to understand the solutions on the spot, mostly because they already know them. If in his mind Perelman expected all mathematicians to be like his coach, then he might have expected a parade in his honor the day after he solved the conjecture. Instead, he got silence and attempts to steal the prize from him.
Can you imagine doing the century's best math work without receiving congratulations for many years? The majority of mathematicians waited for the judgment of the experts, as did Perelman. The experts were busy and much slower than Perelman expected. The conjecture was extremely difficult, and it was a high-profile situation — after all, $1 million was attached to its solution. So the experts were very cautious in their pronouncements.
Finally, instead of congratulating Grigory, they said that the proof seemed to be correct and that they had not yet found any mistakes. If like Perelman, I was certain of my proof, I would have found this a painfully under-whelming conclusion.
Perelman expected to feel proud, but instead he probably felt unappreciated and attacked. Instead of the parade he may have hoped for, he had to wait for a long time, only to face disappointment and frustration. This reminds me of an old joke:
A genie is trapped in a lantern at the bottom of the sea. He vows, "I will give one million dollars to the person who frees me." One thousand years pass. He changes his vow, "I will give any amount of money to the one who frees me." Another thousand years pass. He ups the ante, "I will give any amount of money and two more wishes to the person who frees me." Another thousand years pass. He promises, "I will kill the one who frees me."
Third explanation. Perelman was profoundly disappointed in the math community. Unlike the genie, Perelman didn't want to kill anyone, but he did want to express his disillusionment. Perhaps that is why he rejected a million dollars.
Do you like challenging puzzles? Are you tired of sudoku? Here's your chance to try your hand at puzzles that are designed for world puzzle championships.
I've already done the homework for you — and it turned out to be more complicated than I anticipated. The world puzzle federation has a website, but unfortunately they are lazy or secretive. It is difficult to find puzzles there. A few puzzles are available in the World Puzzle Federation Newsletters.
Since I am stubborn, I spent a lot of time looking for championship puzzles. I found them in books. Here is the list I compiled so far. If you too are interested in high-level puzzles, this ought to make your search a lot easier. The book titles are confusing, so I added a description of what's in them.
One of my favorite puzzle types is Easy as ABC. You have to fill one of A, B, C, and D in each row and column. The letters outside the grid indicate which letter you see first from that direction. Here is one from the 2011 newsletter:
* * *
Engraved on a mathematician's tombstone: "Q.E.D."
* * *
—You act very brave on the Internet. But could you repeat this looking into my eyes?
—Sure. Send me your picture.
* * *
—Your birthday?
—December 26^{th}.
—What year?
—Every year.
* * *
Teacher: "How much do we get if we cut eight into two halves?"
Student: "Two threes, if we cut vertically; and two zeros, if we cut horizontally."
Two girls. One is older and more experienced. The other is younger and more naive. Which of these two girls will the unnamed male narrator choose? What a great plot for a math book.
I am talking about Hiroshi Yuki's book Math Girls. The plot allows the author to discuss math on different levels. Miruka's math is more advanced and mysterious. Tetra's math is simpler and more transparent.
The book starts discussing sequences and patterns. Can you guess the pattern behind the sequence: 1, 2, 3, 4, 6, 9, 8, 12, 18, 27, …? Can you explain how the beginning of this sequence might be very deceptive?
For the answer, you can read the book, which also discusses tons of fun topics: prime numbers, sum of divisors, absolute values, rotations and oscillations, De Moivre's formula, generating functions, arithmetic and geometric means, differential and difference operators, Catalan numbers, infinite series, harmonic numbers, zeta function, Taylor series, partitions, and more.
I usually do not like math fiction, but this is more math than fiction. It's quite superior to most other math books I've read, for it shows the unity of mathematics. It allows the readers to discover connections among different parts of mathematics, and it accomplishes this in a very thrilling way. Frankly, more thrilling than the romantic sections.
The fictional element brings an additional value to the book. The author uses dialogue to discuss points that are usually skipped in regular text books. The two girls give the narrator an opportunity to explore math on different levels: to talk about heavy stuff with Miruka and to provide explanations with Tetra.
I expected to be more interested in the sections dealing with advanced math. But the book is so well-written that the simpler things were a lot of fun, too. For example, I never before noticed that the column notation for n choose k is exactly the same as for a 2d vector with coordinates n and k. And I will never ever shout "zero" because the exclamation makes it "one".
I have a problem with my binocular vision. The muscles that are responsible for moving my eyes outwards are very weak, much weaker than the muscles that move my eyes inwards. When I am very tired, I can't focus on people or things that are far away. I start seeing doubled monsters with extra eyes and noses.
Luckily, instead of looking scary, the monsters look familiar. In fact, they look exactly like Picasso's portraits. I bet Picasso had problems with his eye muscles.
More than ten years ago I went through a process of psychotherapy which, although very painful, was extremely successful. When I tell my friends about this, they are interested in knowing what can be gained through psychotherapy, so here's my story.
I was living in Princeton, NJ, and I was very tired all the time. My primary care doctor told me that I was depressed and needed to do psychotherapy. A friend of mine recommended Dr. Ella Friedman. During my first visit Ella told me that I block my negative emotions. I protested. All my life I truly tried to be honest with myself. She insisted. I had nothing to lose because I had to solve the problem of my constant exhaustion and I had no other potential solutions. Besides, I liked her very much. So I decided to play along and started my search looking for negative emotions.
For some time I tried to convince Ella that if my best friend broke my favorite mug I wouldn't get angry with her. Ella tried to convince me otherwise. She pushed me back in time to the source of my beliefs and feelings. After several months of therapy, I discovered that I had a strong underlying belief that for my mother to love me, I must be a good girl who is always fair. Since my friend who broke the mug didn't do it on purpose, I wasn't allowed to be angry with her. I repressed all my angry feelings.
It took a lot of time for Dr. Friedman to rewire me and persuade me that my negative emotions do not mean that I am a bad girl. My actions define my goodness, not my emotions. I resisted. She had already convinced me that I might have negative emotions, but I didn't want to look at them. The power forcing me to block my emotions was the threat that my mother would withdraw her love if I wasn't a good girl. Dr. Friedman converted me. I started to believe her and continued more vigorously searching for my hidden emotions. Finally one day I collapsed in the shower. I actually felt my blocked emotions flooding me.
Negative emotions protect us. If someone treats you badly you need to be able to recognize it and get away from the danger. Because I didn't see my emotions I stayed in situations, like toxic relationships, that caused me great pain, without realizing it.
My psychotherapy didn't stop then. We started working on how to understand my emotions and how to process them. Now when someone is talking to me, I listen not only with my ears, but also with my gut. Suppose someone tells me, "I am so glad to see you," but I feel a strange tightness in my stomach. I start wondering what the tightness is about, and usually can figure it out. For the first time I was able to hear my gut and it was more illuminating than what I was hearing with my ears. All my life I processed information as text. Now the sentence "I am so glad to see you" has many different meanings.
The therapy changed my life. It feels as if I added a new sense to my palette of senses. I feel as if I was color blind for many years and at last I can see every color. Now that I've learned to recognize my pain, I can do something about it. I am so much happier today than I ever was before. While my friends may not have consciously recognized the big change in me, they have stopped calling me clueless and now often come to me for advice.
Did this solve my problem of tiredness? When Ella Friedman told me that I was no longer depressed, I still felt tired. I started investigating it further. It turns out that the depression was a result of the tiredness, not the other way around. It seems that I have a sleeping disorder and an iron problem.
SEAHOP created a practice puzzle, called "Making Connections," that includes me. It seems I am making connections.
I found a strange piece of paper in an old pile. I believe that it is a visual proof of the following statement:
If ∞ = 1/0, then 0 = 1/∞.
Proof. Assume ∞ = 1/0. Rotate each side of the equation counterclockwise 90 degrees. We get 8 = −10. Subtract 8, getting 0 = −18. Then rotate both parts back: 0 = 1/∞. QED.
I recently posted a short article on plagiarism. Did you notice that not a word of it was mine?
Baron Münchhausen is famous for his tall tales. My co-author Konstantin Knop wants to rehabilitate him and so invents problems where the Baron is proven to be truthful from the start. We already wrote a paper about one such problem. Here is a new problem by Konstantin:
Kostya has a black box, such that if you put in exactly 3 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 5 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than 3 times?
Actually, Konstantin designed a more complicated problem that was given at the Euler Olympiad, 2012 in Russia.
Let n be a fixed integer. Kostya has a black box, such that if you put in exactly 2n+1 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 4n+1 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than n+2 times?
Note that Kostya can't just put 4n+1 coins in the box. The box accepts exactly 2n+1 coins. The problem that I started with is for n = 1. Even such a simple variation was a lot of fun for me to solve. So, have fun.
Once upon a time there was a land where the only antidote to a poison was a stronger poison, which needed to be the next drink after the first poison. In this land, a malevolent dragon challenges the country's wise king to a duel. The king has no choice but to accept.
By bribing the judges, the dragon succeeds in establishing the following rules of the duel: Each dueler brings a full cup. First they must drink half of their opponent's cup and then they must drink half of their own cup.
The dragon wanted these rules because he is able to fly to a volcano, where the strongest poison in the country is located. The king doesn't have the dragon's abilities, so there is no way he can get the strongest poison. The dragon is confident of winning because he will bring the stronger poison.
The only advantage the king has is that the dragon is dumb and straightforward. The king correctly predicts what the dragon will do. How can the king kill the dragon and survive?
There is an array containing all the integers from 1 to n in some order, except that one integer is missing. Suggest an efficient algorithm for finding the missing number.
A friend gave me the problem above as I was driving him from the airport. He had just been at a job interview where they gave him two problems. This one can be solved in linear time and constant space.
But my friend was really excited by the next one:
There is an array containing all the integers from 1 to n in some order, except that one integer is missing and another is duplicated. Suggest an efficient algorithm for finding both numbers.
My friend found an algorithm that also works in linear time and constant space. However, the interviewer didn't know that solution. The interviewer expected an algorithm that works in n log n time.
The company claims that they are looking for the smartest people in the world, and my friend had presented them with an impressive solution to the problem. Despite his excitement, I predicted that they would not hire him. Guess who was right?
I reacted like this because of my own story. Many years ago I was interviewing for a company that also wanted the smartest people in the world. At the interview, the guy gave me a list of problems, but said that he didn't expect me to solve all of them — just a few. The problems were so difficult that he wanted to sit with me and read them together to make sure that I understood them.
The problems were Olympiad style, which is my forte. While we were reading them, I solved half of them. During the next hour I solved the rest. The interviewer was stunned. He told me of an additional problem that he and his colleagues had been trying to solve for a long time and couldn't. He asked me to try. I solved that one as well. Guess what? I wasn't hired. Hence, my reaction to my friend's interview.
The good news: I still remember the problem they couldn't solve:
A car is on a circular road that has several gas stations. The gas stations are running low on gas and the total amount of gas available at the stations and in the car is exactly enough for the car to drive around the road once. Is it true that there is a place on the road where the car can start driving, stopping to refuel at each station, so that the car completes a full circle without running out of gas? Assume that the car's tank is large enough not to present a limitation.
Sid Dhawan was one of our RSI 2011 math students. He was studying interlocking polyominoes under the mentorship of Zachary Abel.
A set of polyominoes is interlocked if no subset can be moved far away from the rest. It was known that polyominoes that are built from four or fewer squares do not interlock. The project of Dhawan and his mentor was to investigate the interlockedness of larger polyominoes. And they totally delivered.
They quickly proved that you can interlock polyominoes with eight or more squares. Then they proved that pentominoes can't interlock. This left them with a gray area: what happens with polyominoes with six or seven squares? After drawing many beautiful pictures, they finally found the structure presented in our accompanying image. The system consists of 12 hexominoes and 5 pentominoes, and it is rigid. You cannot move a thing. That means that hexominoes can be interlocked and thus the gray area was resolved.
You can find the proofs and the details in their paper "Complexity of Interlocking Polyominoes". As you can guess by the title, the paper also discusses complexity. The authors proved that determining interlockedness of a a system that includes hexominoes or larger polyominoes is PSPACE hard.
The Fomenko drawing on the left is from the original Russian edition of Homotopic Topology by Fuks, Fomenko and Gutenmacher. Dmitry Fuchs signed this book for me after my success in the USSR Math Olympiad when I was in the 9th grade. For many years I didn't know what the picture meant and was mystified by it. Now the book has been republished with explanations and is available in English at a non-affordable price. You can find this picture and many other Fomenko drawings in his book called Mathematical Impressions, which is affordable, although the comments accompanying the illustrations are confusing. So I have my own explanation for the meaning of this illustration.
The bracelet is made out of shells. Each shell is a hollow cone whose vertex is glued to a point on the rim of the cone's opening, thus giving each hollow cone its own handle. In a part of another drawing (at left), Fomenko shows how the bracelet is built by an army of tiny slaves. First they build the shells and then they connect them together.
How do they connect the shells to each other? The rim of the next shell is glued to the handle of the previous shell. Let me remind you that a straight line connecting a point on the rim to the vertex of a cone is called a generatrix. Imagine a generatrix that connects a vertex of a cone to the point on the rim to which this vertex is glued. This generatrix becomes a circle in a shell, which I call the handle circle. So the rim of the next shell is glued to the handle circle of the previous shell.
Now consider the fundamental group of a shell. The rim can be contracted to the handle circle. Moreover, the cone itself can be contracted to the handle circle. If we glue several shells together, the result is contractible to the handle circle of the last shell.
Now let's go back to the bracelet. The shells become smaller in both directions and end in two points. The front end point is more interesting topologically than the one in back. Every point other than the front end has a contractible neighborhood, while the front end point does not. Or in scientific terms: The bracelet gives an example of a space with a point at which the space is "1-lc" but with no open neighborhoods on which every (Cech) 1-cycle bounds.
My son Sergei invented the following game a couple of years ago. Two people, Alice and Bob, agree on a number, say, four. Alice takes a clean Rubik's cube and secretly makes four moves. Bob gets the resulting cube and has to rotate it to the initial state in not more than four moves. Bob doesn't need to retrace Alice's moves. He just needs to find a short path back, preferably the shortest one. If he is successful, he gets a point and then it is Alice's turn.
If they are experienced at solving Rubik's cube, they can increase the difficulty and play this game with five or six moves.
By the way, how many moves do you need to solve any position on a Rubik's cube if you know the optimal way? The cube is so complicated that people can't always know the optimal way. They think that God can, so they called the diameter of the set of all possible Rubik's cube positions, God's Number. It was recently proven that God's Number is 20. If Alice and Bob can increase the difficulty level to 20, that would mean that they can find the shortest path back to the initial state from any position of the cube, or, in short, that they would master God's algorithm.
One hundred people play the following game. Their names are written on pieces of paper and put into 100 labeled boxes at random. Each box is labeled with a number from 1 to 100 and one name has been placed inside each box. The boxes are placed on a table in a separate room. The players go into the room one by one and each has to open 99 boxes one after another. After each player finishes and leaves the room, the boxes are closed again. The players are not allowed to communicate with each other in any way, although they have been given one day before the event to discuss their strategies. They only win if every one of the one hundred players avoids opening the box with his or her own name. What is the optimal strategy?
Let me first discuss a simpler version of the game. Each player has to open exactly one box and they win if each one of them finds their name. After each player finishes and leaves the room, the boxes are closed again and the room is re-set.
If all of them decide to open box number 42, they are guaranteed to lose. They can try to open random boxes, then they win with probability (1/100)^{100}. Can they use a joint strategy that is better than random?
Yes, they can. Clearly, two people shouldn't open the same box. So on the day before, if each agrees to open a box with a different assigned number, their probability to win is one over 100!. I leave it to my readers to prove that this is the best strategy.
What is the difference between this problem and the original problem? Isn't choosing the last box the same as choosing the first? Aha! When they open 99 boxes they see the names, so they can use this information as part of their strategy.
I hope that this new version is so intriguing that you will start solving this puzzle right away.
* * *
Decimals have a point.
* * *
During the show "Are You Smarter Than a 5th Grader?" the following question was asked:
What is superfluous in the following list: a carrot, an onion, a potato, a Lexus?
A smart 5th grader answered: a carrot, an onion, and a potato.
* * *
If you buy 3 DVDs for the price of 4, you will get one more as a bonus.
* * *
Only yesterday, today was tomorrow.
* * *
By definition, one divided by zero is undefined.
* * *
Finally artificial intelligence has caught up with humans: when filling out electronic forms, many humans need several tries to prove they are not robots.
* * *
Be back in 5 minutes. If I am late, reread this sms.
* * *
— We'll split the money 50-50.
— I want 70.
— Okay, 70-70!
I recently published my new favorite math problem:
A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic's goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic's assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.
If the psychic is only allowed to look at the backs of the cards, then the amount of transmitted information is 2^{36}, which is the same amount of information as suits for 18 cards. This number of guesses is achievable: the backs of every two cards can clue in the suit of the second card in the pair. This way the psychic can guess the suits of all even-numbered cards in the deck. So the problem is to improve on that. Using the info from the cards that the psychic is permitted to turn over can help too.
The problem is from the book Moscow Mathematical Olympiads, 2000-2005. The book and Russian blog discussions provide many different ideas on how to guess more than half of the deck.
Here is the list of ideas.
Idea 1. Counting cards. If you count cards you will know the suits of the last cards.
Idea 2. Trading. As we discussed before, the psychic can correctly guess the suits of even-numbered cards. By randomly guessing the odd-numbered cards she can correctly guess on average the suits of 4.5 additional cards. Unfortunately, this is not guaranteed. But wait. What if we trade the knowledge of the second card's suit for the majority suit among odd-numbered cards?
Idea 3. Three cards. Suppose we have three cards. Three bits can provide the following knowledge: the majority color, plus the suit of the first and of the second cards in the majority color. Thus, three bits of information will allow the psychic to guess the suits of two cards out of three.
Idea 4. Which card. Suppose the assistant signals the suits of even-numbered cards. With no loss, the psychic can guess the even-numbered card and repeat the same suit for the next card. If this is the plan, the assistant can choose which of the two cards to describe. Which card of the two matches the psychic's guess provides an additional bit of information.
Idea 5. Surprise. Suppose we have a strategy to inform the psychic about some cards. Suppose the assistant deliberately fails on one of the cards. Then the index of this card provides info to the psychic.
I leave it to my readers to use these ideas to find the solution for 19, 23, 24 and maybe even for 26 cards.
Imagine a slice of buttered white bread with a heap of sugar on top. That was my favorite lunch when I was a kid. My mom was working very hard, I was the oldest sister, and this was what I would make for myself almost every day.
Later someone told me that sugar is brain food. I believed that sugar and chocolate helped me do mathematics, so my love for sugar got theoretical support. I finally figured out the source of this love when my first son was born. To teach my son to stop requesting milk at night, my mother pushed me to give him sugar-water instead. At that moment, I realized that I developed my love of sugar with my mother's milk. Or, more precisely, instead of my mother's milk.
Now there is more and more evidence that the love of my life is a mistake. See for example Is Sugar Toxic?. Will I ever be able to break my oldest bad habit, the one I developed before I can remember myself doing it?
This is how my ex-husband Joseph Bernstein used to start his courses in representation theory.
Suppose there is a four-armed dragon on every face of a cube. Each dragon has a bowl of kasha in front of him. Dragons are very greedy, so instead of eating their own kasha they try to steal kasha from their neighbors. Every minute every dragon extends four arms to the neighboring cube's faces and tries to get the kasha from the bowls there. As four arms are fighting for every bowl of kasha, each arm manages to steal one-fourth of what is in the bowl. Thus each dragon steals one-fourth of of the kasha of each of his neighbors, while all of his own kasha is stolen too. Given the initial amounts of kasha in every bowl, what is the asymptotic behavior of the amounts of kasha?
You might ask how this relates to representation theory. First, it relates to linear algebra. We can consider the amounts of kasha as a six-dimensional vector space and the stealing process as a linear operator. As mathematicians, we can easily assume that a negative amount of kasha is allowed.
Now to representation theory. The group of rotations of the cube naturally acts on the 6-dimensional vector space of kashas. And the stealing operator is an intertwining operator of this representation. Now for a spoiler alert: I'm about to finish the solution, so stop here if you want to try it on your own.
The intertwining operator acts as a scalar on irreducible representations of the group. Thus we should decompose our representation into irreducible ones. The group has five irreducible representations with dimensions 1, 1, 2, 3, and 3.
We can decompose the kasha into the following three representations:
We see that asymptotically every dragon will have the same amount of kasha.
Now it is your turn to use this method to solve a similar problem, where there are n dragons sitting on the sides of an n-gon. Each dragon has two arms, and steals half of the kasha from his neighbors. Hey, wait a minute! Why dragons? There are people around the table stealing each other's kasha. But the question is still the same: What is the asymptotic behavior of the amounts of kasha?
Students should use a different strategy for the AIME than for the AMC. So students who are approaching the AIME for the first time need to question the habits they have developed after years of doing multiple choice tests. Here are some suggestions.
Checking. I've noticed that the accuracy level of students who take the AIME for the first time drops significantly. It seems that they are so used to multiple choice questions that they rely on multiple choices as a confirmation that they are right. So when someone solves a problem, they compare their answer to the given choices and if the answer is on the list they assume that the answer must be correct. Their pattern is broken when there are no choices. So they arrive at an answer and since there is no way to check it against choices, they just submit it. Because of this lack of confirmation, checking their answer in other ways becomes more important.
Timing.
At the AMC we have 3 minutes per problem. At the AIME — 12. That means the timing strategies need to be different. Indeed, the AMC is so fast-paced that it is reasonable to save time by not reading a problem twice. If you read it, you either solve it or skip it and go on. The student who is not trying to achieve a perfect score can decide in advance not to read those final, highly-difficult problems.
For the AIME it is not expensive, in relative terms of time, to read all the problems. The student can read the problems and choose the most promising ones to start with, knowing that if there is time they can always come back to other problems.
Guessing. Guessing at the AMC is very profitable if you can exclude three choices out of the given five. Guessing for the AIME is a waste of time because the answers are integers between 000 and 999. So the probability of a random guess is one in a thousand. Actually, this is not quite right, because the problem writers are human and it is much easier to write a problem with an answer of 10 than one with an answer of 731. But the AIME designers are trying very hard to make answers that are randomly distributed. So the probability of a random guess is not one in a thousand, but it is very close. You can improve your chances by an intelligent guess. For example, you might notice that the answer must be divisible by 10. But guessing is still a waste of time. Thinking about a problem for two minutes in order to increase the probability of a correct guess to one in a 100 means that your expected gain is 1/200 points per minute. Which is usually much less than the gain for checking your answers. You can play the guessing game if you have exhausted your other options.
What saddens me is that the students who are not trained in checking use their first guess to make their life choices. But this is a subject for a separate discussion.
I have already written about how American math competition are illogically structured, for the early rounds do not prepare students for the later rounds. The first time mathletes encounter proofs is in the third level, USAMO. How can they prepare for problems with proofs? My suggestion is to look East. All rounds of Russian math Olympiads — from the local to the regional to the national — are structured in the same way: they have a few problems that require proofs. This is similar to the USAMO. At the national All-Russian Olympiad, the difficulty level is the same as USAMO, while the regionals are easier. That makes the problems from the regionals an excellent way to practice for the USAMO. The best regional Olympiad in Russia is the Moscow Olympiad. Here is the problem from the 1995 Moscow Olympiad:
We start with four identical right triangles. In one move we can cut one of the triangles along the altitude perpendicular to the hypotenuse into two triangles. Prove that, after any number of moves, there are two identical triangles among the whole lot.
This style of problems is very different from those you find in the AMC and the AIME. The answer is not a number; rather, the problem requires proofs and inventiveness, and guessing cannot help.
Here is another problem from the 2002 Olympiad. In this particular case, the problem cannot be adapted for multiple choice:
The tangents of a triangle's angles are positive integers. What are possible values for these tangents?
The problems are taken from two books: Moscow Mathematical Olympiads, 1993-1999, and Moscow Mathematical Olympiads, 2000-2005. I love these books and the problems they present from past Moscow Olympiads. The solutions are nicely written and the books often contain alternative solutions, extended discussion, and interesting remarks. In addition, some problems are indexed by topics, which is very useful for teachers like me. But the best thing about these books are the problems themselves. Look at the following gem from 2004, which can be used as a magic trick or an idea for a research paper:
A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic's goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic's assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.
Do you remember how to divide three apples among four people? Make apple sauce, of course. In the following two puzzles you are not allowed to cut apples. Here is an old riddle:
There are four apples in a basket. How can you divide them among four people, so that one apple remains in the basket?
Here is a variation from Konstantin Knop's blog:
There are four apples in a basket. How can you divide them among three people, so that no one has more than the others and one apple remains in the basket?
Imagine a magician who comes on stage and performs the following magic trick:
He asks someone in the audience to think of a two-digit number, then subtract the sum of its digits. He waves his wand and guesses that the result is divisible by nine. Ta-Da!
This is not magic. This is a theorem. To make it magical we need to disguise the theorem.
First, there are many ways to hide the fact that we subtract the sum of the digits. For example, we can ask to subtract the digits one by one, while chatting in between. It is better to start with subtracting the first digit. Indeed, if we start with subtracting the second digit, the audience might notice that the result is divisible by 10 and start suspecting that some math is involved here. You can be more elaborate in how you achieve the subtraction of the sum of digits. For example, subtract twice the first digit, then the second, then add back the original number divided by 10.
Second, we need to disguise that the result is divisible by 9. A nice way to do this is implemented in the online version of this trick. The website matches the resulting number to a gift that is described on the page in pale letters. Paleness of letters is important as it is difficult to see that the same gift reappears in a pattern. In my work with students I use the picture on the left. At the end I tell them, "Ta-Da! the resulting number is blue." Here is the full sized version of the same picture that you can download.
My students are too smart. They see through me and guess what is going on. Then I ask them the real question, "Why do I have some cells with question marks and other symbols?" To give you a hint, I can tell you that the symbols are there for the same reason some blue numbers are not divisible by 9.
We got this problem from Rados Radoicic:
A 7 by 7 board is covered with 38 dominoes such that each covers exactly 2 squares of the board. Prove that it is possible to remove one domino so that the remaining 37 still cover the board.
Let us call a domino covering of an n by n board saturated if the removal of any domino leaves an uncovered square. Let d(n) be the number of dominoes in the largest saturated covering of an n by n board. Rados' problem asks us to prove that d(7) < 38.
Let's begin with smaller boards. First we prove that d(2) = 2. Suppose that 3 dominoes are placed on a 2 × 2 board. Let us rotate the board so that at least two of the dominoes are horizontal. If they coincide, then we can remove one of them. If not, they completely cover the board and we can remove the third one. Similarly, you can check all the cases and show that d(3) = 6.
Now consider a saturated domino covering of an n × n board. We can view the dominoes as vertices of a graph, joining two if they share a cell of the board. No domino can share both cells with other dominoes, or we could remove it. Hence, each domino contains at most one shared cell. This means that all the dominoes in a connected component of the graph must overlap on a single shared cell. Hence, the only possible connected components must have the following shapes:
The largest shape in the picture is the X-pentomino. We can describe the other shapes as fragments of an X-pentomino, where the center and at least one more cell is intact. We call these shapes fragments.
A saturated covering by D dominoes corresponds to a decomposition of the n × n board into F fragments. Note that a fragment with k cells is made from k − 1 dominoes. Summing over the dominoes gives: D = n^{2} − F. Thus, in order to make D as large as possible, we should make F as small as possible. Let f(n) be the minimal number of fragments that are required to cover an n by n board without overlap. Then d(n) = n^{2} − f(n).
Consider the line graph of the n by n board. The vertices of the line graph correspond to cells in the original board and the edges connect vertices corresponding to neighboring cells. Notice that in the line graph our fragments become all star graphs formed by spokes coming out from a single central node. Thus a decomposition of a rectangular board into fragments corresponds to a covering of its line graph by star graphs. Consider an independent set in the line graph. The smallest independent set has the same number of elements as the smallest number of stars that can cover the graph. This number is called a domination number.
Now let's present a theorem connecting domino coverings with X-pentomino coverings.
Theorem. f(n) equals the smallest number of X-pentominoes that can cover an n by n board allowing overlaps and tiles that poke outside, which is the same as the domination number of the corresponding line graph.
The proof of this theorem and the solution to the original puzzle is available in our paper: "Saturated Domino Coverings." The paper also contains other theorems and discussions of other boards, not to mention a lot of pictures.
The practical applications of star graph coverings are well-known and widely discussed. We predict a similar future for saturated domino coverings and its practical applications, two examples of which follow:
First, imagine a party host arranging a plate of cookies. The cookies must cover the whole plate, but to prevent the kids sneaking a bite before the party, the cookies need to be placed so that removal of just one cookie is bound to expose a chink of plate. This means the cookies must form a saturated covering of the plate. Of course the generous host will want to use a maximal saturated covering.
For the second application, beam yourself to an art museum to consider the guards. Each guard sits on a chair in a doorway, from where it is possible to watch a pair of adjacent rooms. All rooms have to be observed. It would be a mistake to have a redundant guard, that is, one who can be removed without compromising any room. Such a guard might feel demotivated and then who knows what might happen. This means that a placement of guards must be a saturated domino covering of the museum. To keep the guards' Union happy, we need to use a maximal saturated covering.
We would welcome your own ideas for applications of saturated coverings.
Several years ago my son Sergei started a new movement: Sergeism. Followers of this philosophy seek to maximize Sergei's happiness. Since Sergei's happiness involves everyone being happy, becoming happy is a consequential goal of his followers.
Let me explain why this might be a perfect religion for many people, not the least myself. My parents didn't teach me to love myself. They taught me to sacrifice myself and put other peoples' interests ahead of my own. After reading tons of books and spending hours in therapy, I've learned to love myself — well, somewhat. But the truth is, I still feel guilty when I pamper myself. Sergeism eliminates this guilt. I can freely invest in my happiness as a committed member of this movement.
I became a Sergeist when I lost all hope of losing weight. I realized that my own health wasn't a strong enough motivation. But I'm always glad to skip a cookie in tribute to Sergeism. If, like me, you put others ahead of yourself and never find the time to exercise or the will to refuse deserts, join me. Become a Sergeist and lose weight for Sergei.
I recently posted an essay Binary Bulls without Cows with the following puzzle:
The test Victor is taking consists of n "true" or "false" questions. In the beginning, Victor doesn't know any answers, but he is allowed to take the same test several times. After completing the test each time, Victor gets his score — that is, the number of his correct answers. Victor uses the opportunity to re-try the test to figure out all the correct answers. We denote by a(n) the smallest numbers of times Victor needs to take the test to guarantee that he can figure out all the answers. Prove that a(30) ≤ 24, and a(8) ≤ 6.
There are two different types of strategies Victor can use to succeed. First, after each attempt he can use each score as feedback to prepare his answers for the next test. Such strategies are called adaptive. The other type of strategy is one that is called non-adaptive, and it is one in which he prepares answers for all the tests in advance, not knowing the intermediate scores.
Without loss of generality we can assume that in the first test, Victor answers "true" for all the questions. I will call this the base test.
I would like to describe my proof that a(30) ≤ 24. The inequality implies that on average five questions are resolved in four tries. Suppose we have already proven that a(5) = 4. From this, let us map out the 24 tests that guarantee that Victor will figure out the 30 correct answers.
As I mentioned earlier, the first test is the base test and Victor answers every question "true." For the second test, he changes the first five answers to "false," thus figuring out how many "true" answers are among the first five questions. This is equivalent to having a base test for the first five questions. We can resolve the first five questions in three more tests and proceed to the next group of five questions. We do not need the base test for the last five questions, because we can figure out the number of "true" answers among the last five from knowing the total score and knowing the answers for the previous groups of five. Thus we showed that a(mn) ≤ m a(n). In particular, a(5) = 4 implies a(30) ≤ 24.
Now I need to prove that a(5) = 4. I started with a leap of faith. I assumed that there is a non-adaptive strategy, that is, that Victor can arrange all four tests in advance. The first test is TTTTT, where I use T for "true" and F for "false." Suppose for the next test I change one of the answers, say the first one. If after that I can figure out the remaining four answers in two tries, then that would mean that a(4) = 3. This would imply that a(28) ≤ 21 and, therefore, a(30) ≤ 23. If this were the case, the problem wouldn't have asked me to prove that a(30) ≤ 24. By this meta reasoning I can conclude that a(4) ≠ 3, which is easy to check anyway. From this I deduced that all the other tests should differ from the base test in more than one answer. Changing one of the answers is equivalent to changing four answers, and changing two answers is equivalent to changing three answers. Hence, we can assume that all the other tests contain exactly two "false" answers. Without loss of generality, the second test is FFTTT.
Suppose for the third test, I choose both of my "false" answers from among the last three questions, for example, TTFFT. This third test gives us the exactly the same information as the test TTTTF, but I already explained that having only one "false" answer is a bad idea. Therefore, my next tests should overlap with my previous non-base tests by exactly one "false" answer. The third test, we can conclude, will be FTFTT. Also, there shouldn't be any group of questions that Victor answers the same for every test. Indeed, if one of the answers in the group is "false" and another is "true," Victor will not figure out which one is which. This uniquely identifies the last test as FTTFT.
So, if the four tests work they should be like this: TTTTT, FFTTT, FTFTT, FTTFT. Let me prove that these four tests indeed allow Victor to figure out all the answers. Summing up the results of the last three tests modulo 2, Victor will get the parity of the number of correct answers for the first four questions. As he knows the total number of correct answers, he can deduce the correct answer for the last question. After that he will know the number of correct answers for the first four questions and for every pair of them. I will leave it to my readers to finish the proof.
Knop and Mednikov in their paper proved the following lemma:
If there is a non-adaptive way to figure out a test with n questions by k tries, then there is a non-adaptive way to figure out a test with 2n + k − 1 questions by 2k tries.
Their proof goes like this. Let's divide all questions into three non-overlapping groups A, B, and C that contain n, n, and k − 1 questions correspondingly. By our assumptions there is a non-adaptive way to figure out the answers for A or B using k tries. Let us denote subsets from A that we change to "false" for k − 1 non-base tests as A_{1}, …, A_{k-1}. Similarly, we denote subsets from B as B_{1}, …, B_{k-1}.
Our first test is the base test that consists of all "true" answers. For the second test we change the answers to A establishing how many "true" answers are in A. In addition we have k − 1 questions of type Sum: we switch answers to questions in A_{i} ∪ B_{i} ∪ C_{i}; and type Diff: we switch answers to (A ∖ A_{i}) ∪ B_{i}. The parity of the sum of "false" answers in A − A_{i} + B_{i} and A_{i} + B_{i} + C_{i} is the same as in A plus C_{i}. But we know A's score from the second test. Hence we can derive C_{i}. After that we have two equations with two unknowns and can derive the scores of A_{i} and B_{i}. From knowing the number of "true" answers in A and C, we can derive the same for B. Knowing A and A_{i} gives all the answers in A. Similarly for B. QED.
This lemma is powerful enough to answer the original puzzle. Indeed, a(2) = 2 implies a(5) ≤ 4, and a(3) = 3 implies a(8) ≤ 6.
The following variation of a Bulls and Cows problem was given at the Fall 2008 Tournament of the Towns:
A test consists of 30 true or false questions. After the test (answering all 30 questions), Victor gets his score: the number of correct answers. Victor doesn't know any answer, but is allowed to take the same test several times. Can Victor work out a strategy that guarantees that he can figure out all the answers after the 29th attempt? after the 24th attempt?
Let's assume that we have a more general problem. There are n questions, and a(n) is the smallest number of times we need to take the test to guarantee that we can figure out the answers. First we can try all combinations of answers. This way we are guaranteed to know all the answers after 2^{n} attempts. The next idea is to start with a baseline test, for example, to say that all the answers are true. Then we change answers one by one to see if the score goes up or down. After changing n − 1 answers we will know the answers to the first n − 1 questions. Plus we know the total number of true answers, so we know the answers to all the questions. We just showed that a(n) ≤ n.
This is not enough to answer the warm-up question in the problem. We need something more subtle.
Let's talk about the second part of the problem. As we know, 24 = 4 ⋅ 6. So to solve the second part, on average, we need to find five correct answers per four tests. Is it true that a(5) ≤ 4? If so, can we use it to show that a(30) ≤ 24?
The following three cases are the most fun to prove: a(5) = 4, a(8) ≤ 6, and a(30) ≤ 24. Try it!
By the way, K. Knop and L. Mednikov wrote a paper (available in Russian) where they proved that a(n) is not more than the smallest number k such that the total number of ones in the binary expansion of numbers from 1 to k is at least n − 1. Which means they proved that a(30) ≤ 16.
On the left is a puzzle from the 2000 Qualifying Test for USA and Canada teams to compete in the world puzzle championship. A set of all 21 dominoes has been placed in a 7 by 6 rectangular tray. The layout is shown with the pips replaced by numbers and domino edges removed. Draw the edges of the dominoes into the diagram to show how they are positioned.
We would like to discuss the mathematical theory behind this puzzle using a toy example below. Only three dominoes: 1-1, 1-2, 2-2 are positioned on the board and the goal is to reconstruct the positioning:
Let's connect adjacent numbers with segments to show potential dominoes and color the segments according to which domino they represent. The 1-1 edge is colored green, the 1-2 — blue, and the 2-2 — red. Now our puzzle has become a graph, where the numbers are vertices, the segments are edges, and the edges are colored. In this new setting, the goal of the puzzle is to find edges of three different colors so that they do not share vertices.
The next picture represents the line graph of the previous graph. Now the colors of the vertices correspond to different potential dominoes. Vertices are connected if the corresponding dominoes share a cell. In the new setting finding dominoes that do not share a cell is equivalent to finding an independent set. The fact that we need to use all possible dominoes means that we want the most colorful independent set.
* * *
— If a black cat crosses in front of you and then crosses back, what does it mean? Is your bad luck doubled or canceled?
— Is this a scalar or a vector cat?
— Huh?
— A scalar cat doubles and a vector cat cancels.
* * *
Unbuttered bread, unable to cause the usual harm, tries to fall on the dirtiest spot.
* * *
Chance is a design carefully planned by someone else.
* * *
Wikipedia: I know everything.
Google: I can find anything.
Facebook: I know everyone.
Internet: You are nothing without me.
Electricity: Shut up, jerks.
* * *
Yesterday I bought pills to increase my IQ. Couldn't open the jar.
* * *
Today I opened my desktop's case and finally understood whither my trash is emptied.
I just discovered a Russian Internet Linguistics Olympiad. Even though most linguistics problems are not translatable, this time we are lucky. My favorite problem from this Olympiad is related to chemical elements — their names in Russian have the same logical structure as in English. Keep in mind, the problem doesn't assume any knowledge of chemistry. Here is the problem:
The formulae for chemical elements and their names are given below in mixed order:
C_{3}H_{8}, C_{4}H_{6}, C_{3}H_{4}, C_{4}H_{8}, C_{7}H_{14}, C_{2}H_{2};
Heptene, Butine, Propane, Butene, Ethine, Propine.
- Match the formulae with their names. Explain your solution.
- Write the names of the elements with the following formulae: C_{2}H_{4}, C_{2}H_{6}, C_{7}H_{12}.
- Write the formulae for the following elements: Propene, Butane.
I have two admirers, Alex and Mike. Alex lives next to my home and Mike lives next to my MIT office. I have a lousy memory, so I invented the following system to guarantee that I see both of my friends and also manage to come to my office from time to time. I have a sign hanging on the inside of my home door that says Office on one side and Alex on the other. When I approach the door, I can see right away where I went last time. So I flip the sign and that tells me where next to go. I have a similar sign inside my office door that tells me to go either to home or to Mike. Every evening I spend with one of my admirers discussing puzzles or having coffee. Late at night I come home to sleep in my own bed. Now let's see what happens if today my home sign shows Office and the office sign shows Mike:
After three days the signs return to their original positions, meaning that the situation is periodic and I will repeat this three-day pattern forever.
Let's get back to reality. I am neither memory-challenged nor addicted to coffee. I invented Alex and Mike to illustrate a rotor-router network. In general my home is called a source: the place where I wake up and start the day. There can only be one source in the network. My admirers are called targets and I can have an infinite number of them. The network needs to be constructed in such a way that I always end up with a friend by the end of the day. There could be many other places that I can visit, other than my office: for example, the library, the gym, opera and so on. These places are other vertices of a network that could be very elaborate. Any place where I go, there is a sign that describes a pattern of where I go from there. The sign is called a rotor.
The patterns at every rotor might be more complicated than a simple sign. Those patterns are called rotor types. My sign is called 12 rotor type as it switches between the first and the second directions at every non-friend place I visit.
The sequence of admirers that I visit is called a hitting sequence and it can be proved that the sequence is eventually periodic. Surprisingly, the stronger result is also true: the hitting sequence is purely periodic.
The simple 12 rotor is universal. That means that given a set of friends and a fancy periodic schedule that designates the order I want to visit them in, I can create a network of my activities where every place has a sign of this type 12 and where I will end up visiting my friends according to my pre-determined periodic schedule.
It is possible to see that not every rotor type is universal. For example, palindromic rotor types generate only palindromic hitting sequences, thus they are not universal. The smallest such example, is rotor type 121. Also, block-repetitive rotor types, like 1122, generate block-repetitive hitting sequences.
It is a difficult and an interesting question to describe universal rotor types. My PRIMES student Xiaoyu He was given a project, suggested by James Propp, to prove or disprove the universality of the 11122 rotor type. This was the smallest rotor type the universality of which was not known. Xiaoyu He proved that 11122 is universal and discovered many other universal rotor types. His calculations support the conjecture that only palindromic or block-repetitive types are not universal. You can find these results and many more in his paper: On the Classification of Universal Rotor-Routers.
My co-author Konstantin Knop wrote a charming book, Weighings and Algorithms: from Puzzles to Problems. The book contains more than one hundred problems. Here are a couple of my favorites that I translated for you:
There is one gold medal, three silver medals and five bronze medals. It is known that one of the medals is fake and weighs less than the corresponding genuine one. Real medals made of the same metal weigh the same and from different metals do not. How can you use a balance scale to find the fake medal in two weighings?
There are 15 coins, out of which not more than seven are fake. All genuine coins weigh the same. Fake coins might not weigh the same, but they differ in weight from genuine coins. Can you find one genuine coin using a balance scale 14 times? Can you do it using fewer weighings?
You might get the impression that the latter problem depends on two parameters. Think about it: It is necessary that the majority of the coins are genuine in order to be able to solve the problem. In fact, the number of weighings depends on just one parameter: the total number of coins. Denote a(n) the optimal number of weighings needed to find a genuine coin out of n coins, where more than half of the coins are genuine. Can you calculate this sequence?
Hint. I can prove that a(n) ≤ A011371(n-1); that is, the optimal number of weighings doesn't exceed n − 1 − (number of ones in the binary expansion of n−1).
We all heard this paradoxical statement:
This statement is false.
Or a variation:
True or False: The correct answer to this question is 'False'.
Recently we received a link to the following puzzle, which is similar to the statement above, but has a cute probabilistic twist:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 60%
- 25%
There are four answers, so you can choose a given answer with probability 25%. But oops, this answer appears twice. Is the correct answer 50%? No, it is not, because there is only one answer 50%. You can see that none of the answers are correct, hence, the answer to the question—the chance to be correct—is 0. Now is the time to introduce our new puzzle:
If you choose an answer to this question at random, what is the chance you will be correct?
- 25%
- 50%
- 0%
- 25%
This fractal was designed by Ross Hilbert and is named "Weathered Steel Weave." You can find many other beautiful pictures in his fractal gallery.
The fractal is based on iterations of the following fractal formula z_{new} = cos(c z_{old}), where the Julia Constant c is equal to −0.364444444444444+0.995555555555556i. To produce the image, you need to start with a complex value of z and iterate it many times using the formula above. The color is chosen based on how close the iteration results are to the border of the unit circle.
I found a new Russian Olympiad for high schools related to universities. I translated my favorite problems from last year's final round. These are the math problems:
8th grade. In a certain family everyone likes their coffee with milk. At breakfast everyone had a full cup of coffee. Given that Alex consumed a quarter of all consumed milk and one sixth of all coffee, how many people are there in the family?
8th grade. How many negative roots does the equation x^{4} − 5x^{3} − 4x^{2} − 7x + 4 = 0 have?
10th grade. Find a real-valued function f(x) that satisfies the following inequalities for any real x and y: f(x) ≤ x and f(x+y) ≤ f(x) + f(y).
I liked the physics problems even more:
8th grade. Winnie-the-Pooh weighs 1 kg. He hangs in the air with density 1.2kg/m^{3} next to a bee hive. He is holding a rope connected to a balloon. Estimate the smallest possible diameter of the balloon, assuming that this happens on Earth.
10th grade. Two containers shaped like vertical cylinders are connected by a pipe underneath them. Their heights are the same and they are on the same level. The cross-sectional area of the right container is twice bigger than the left's. The containers are partially filled with water of room temperature. Someone put ice into both containers: three times more ice into the right one than into the left one. After that, the containers are closed hermetically. How will the water level will change after the ice melts completely:
- The levels will not change.
- The level on the left will be higher than on the right.
- The level on the left will be lower than on the right.
- The answer depends on the initial volume of water in the containers.
Once I talked to my friend Michael Plotkin about IQ tests, which we both do not like. Michael suggested that I run an experiment and send a standard IQ question for children to my highly-educated friends. So I sent a mass email asking:
What's common between an apple and an orange?
I believe that the expected answer is that both are fruits.
Less than half of my friends would have passed the IQ test. They gave four types of answer. The largest group chose the expected answer.
The second group related the answer to language. For example, apples and oranges both start with a vowel and they both have the letters A and E in common.
The third group connected the answer to what was on their minds at the time:
And the last group were people who just tried to impress me:
If my friends with high IQs have given so many different answers, I would expect children to do the same. The variety of answers is so big that no particular one should define IQ. By the way, my own well-educated kids' answers are quoted above — and they didn't go with the standard answer. I'm glad they never had IQ tests as children: I'm sure they would never have passed.
I have an idea for a start-up medical insurance company for Massachusetts. My insurance will have an infinite deductible. That means you pay your own bills. The cost of insurance can be very low, say $100 a year, as I do not need to do anything other than to send you a letter confirming that you have medical insurance. People who otherwise will be fined up to $900 for being uninsured will run in droves to buy my insurance.
I have an even better idea. For an extra fee, I will negotiate with doctors so that you will pay the same amount as medical insurance companies pay to them, which is often three times less than you would pay on your own.
Who am I kidding? I am not a business person, I can't build a company. But I am looking to buy the insurance I just described.
I am just wondering:
What is the largest integer consisting of distinct digits such that, in its English pronunciation, all the words start with the same letter?
I continue to wonder:
What is the largest integer consisting of the same digit such that, in its English pronunciation, all the words start with distinct letters?
When you name your child there are many considerations to take into account. For example, you should always check that your kids' initials don't embarrass them. For example, if the Goldsteins want to name their son Paz, because it means golden in Biblical Hebrew, the middle name shouldn't be Isaak, or anything starting with I.
Contemporary culture adds another consideration: how easy would it be to find your child on the Internet? I personally find it extremely convenient to have a rare name, because my fans can find my webpage and blog just by googling me. Parents need to decide whether they want their children to be on the first page of the search engine or hidden very far away when someone googles them.
When I named my son Sergei, I knew that there was another mathematician named Sergei Bernstein. But I didn't think about the Internet. As a result, I confused the world: is my son more than a hundred years old or did Sergei Natanovich Bernstein compete at Putnam?
I decided to see the film The Oxford Murders
At the core of the movie are sequences of numbers and symbols. When the characters started a discussion about how to continue a sequence, I immediately tensed up. Why? Because when people ask what the next element in the sequence is, I get ready to confront them, by explaining that there are many ways to continue a sequence. For example, the sequence — 1, 2, 4 — could be powers of two, or could be Tribonacci numbers, or any of 10,000 sequences that the Online Encyclopedia of Integer Sequences spills out if you plug in 1, 2, 4. That is, if we do not count the infinity of sequences that are not in the Encyclopedia.
To my surprise and relief, the logic Professor, one of the main characters in the movie, explained that there is no unique way to continue a sequence. From that moment on, I relaxed and fell in love with the movie.
The movie is a detective story with a lot of twists and turns. The crimes are related to symbols. The first two symbols are in the picture below. Can you guess the next symbol?
I cannot. There is an irony in the film at this point, because the Professor and the student need to guess the sequence in order to solve the crimes. But the Professor has already explained that there is no unique way to continue. So illogical for a movie about logic.
And what's worse, the sequence of symbols they finally discover doesn't make sense. I guess I fell in love with this movie too quickly.
* * *
— I've noticed that fools are always sure of themselves, while clever people are doubtful.
— No doubt.
* * *
— What happened to your girlfriend, that really cute math student?
— She's no longer my girlfriend. I caught her cheating on me.
— I don't believe that she cheated on you!
— Well, a couple of nights ago I called her, and she told me that she was in bed wrestling with three unknowns.
* * *
A programmer calls the library:
— Can I talk to Kate please?
— She's in the archive.
— Can you unzip her?
* * *
To protect the population from airplane disasters, Congress has ratified an addendum to the law of gravity.
* * * (invented by David Bernstein)
Energy conservation: it's not just a good idea; it's the law.
* * *
— Your computer is such a mess.
— It got a nasty virus.
— And it poured coffee on your keyboard?
* * *
After little Tom learned to count, his father had to start dividing dumplings evenly.
* * *
In spite of the crisis, inflation, and erratic fluctuations of the market, Russian mathematicians promised the president to keep number Pi between 3 and 4 until at least the end of the year.
* * *
A logician rides an elevator. The door opens and someone asks:
— Are you going up or down?
— Yes.
My webpage and my blog generate a lot of emails. I love receiving most of the emails, but if I reply to them, I won't have time to work on my blog. My favorite type of message is one that is full of compliments, with a note that the writer doesn't expect a reply.
I am grateful to people who send me things I requested, like pictures of Russian plates, or some interesting number properties. I apologize that it takes me so long to reply.
The emails that I don't enjoy reading contain amazing elementary proofs of Fermat's last theorem, or any other theorem on the Millennium list, for that matter. I also do not like when my readers ask me for help with their homework.
Like most people, I'm already dealing with spammers who want to enlarge the body parts I do not have or to slim the ones I do have. However, if you do need to send me millions of dollars that I won in your lottery, there is no reason to waste time on email exchanges: you can process them through my "donate" button.
You are welcome to contact me, but ….
I already gave an example of the kinds of problems that were given to Jewish people at the oral entrance exam to the math department of Moscow State University. In fact, I have a whole page with a collection of such problems, called Jewish problems or Coffins. That page was one of the first pages I created when I started my website more than ten years ago.
When my son Alexey was in high school, I asked him to help me type these problems into a file and to recover their solutions from my more than laconic notes, and solve the problems that I didn't have notes for. He did the job, but the file was lying dormant on my computer. Recently I resurrected the file and we prepared some of the solutions for a publication.
The problems that were given during these exams were very different in flavor: some were intentionally ambiguous questions, some were just plain hard, some had impossible premises. In our joint paper "Jewish Problems" we presented problems with a special flavor. These are problems that have a short and "simple" solution, that is nonetheless very difficult to find. This way the math department of MSU was better protected from appeals and complaints.
Try the following problem from our paper:
Find all real functions of real variable F(x) such that for any x and y the following inequality holds: F(x) − F(y) ≤ (x − y)^{2}.
I will give a talk on the subject for UMA at MIT on October 18, at 5pm.
What's "plagiarism"? It's when you take someone else's work and claim it's your own. It's basically STEALING.
Ideas improve. The meaning of words participates in the improvement. Plagiarism is necessary. Progress implies it. It embraces an author's phrase, makes use of his expressions, erases a false idea, and replaces it with the right idea.
Perhaps the Russians have done the right thing, after all, in abolishing copyright. It is well known that conscious and unconscious appropriation, borrowing, adapting, plagiarizing, and plain stealing are variously, and always have been, part and parcel of the process of artistic creation. The attempt to make sense out of copyright reaches its limit in folk song. For here is the illustration par excellence of the law of Plagiarism. The folk song is, by definition and, as far as we can tell, by reality, entirely a product of plagiarism.
If you copy from one author, it's plagiarism. If you copy from two, it's research.
In my essays The Oral Exam and A Math Exam's Hidden Agenda, I gave some examples of math problems that were used during the entrance exams to Moscow State University. The problems were designed to prevent Jewish and other "undesirable" students from studying at the University. My readers might have supposed that an occasional bright student could, by solving all the problems, get in. Here is the story of my dear friend Mikhail (Misha) Lyubich; it shows that being extremely bright was not enough.
Misha passed the first three exams and was facing his last exam: oral physics. He answered all the questions. None of his answers were accepted: all of them were declared wrong. Misha insisted that he was right and requested that the examiners explain themselves. Every time their reply was the same:
This is not a consultation, it's an exam.
Misha failed the exam. The solution to the last problem was a simple picture: a document that seemed to be impossible to deny, so Misha decided that he had grounds for an appeal. The person in charge denied the appeal. When Misha requested an explanation, can you guess the answer?
This is not a consultation, it's an appeal.
Misha ended up studying at Kharkov State University. Now he is a professor at Stony Brook and the director of the Institute for Mathematical Sciences at Stony Brook.
You know that the negation of a true statement is a false statement, and the negation of a false statement is a true statement. You also know that you can negate a sentence by preceding it with "It is not true that …."
Now look at the following statement and its negation, invented by David Bernstein. Which one is true?
How about this pair?
My son Alexey taught me to always plug unused power strips into themselves, so that we can call them "The Rings of Power." These are my Borromean Rings of Power:
Let's call a projection of a body L onto a hyperplane a shadow. Here is a mathematical way to hide behind. An object K can hide behind an object L if in any direction the shadow of K can be moved by a translation to be inside the corresponding shadow of L. If K can hide inside L, then obviously K can hide behind L. Dan Klain drew my interest to the following questions. Is the converse true? If K can hide behind L can it hide inside L? If not, then if K can hide behind L, does it follow that the volume of K is smaller than that of L?
We can answer both questions for 2D bodies by using objects with constant width. Objects with constant width are ones that have the same segment as their shadow in every direction. The two most famous examples are a circle and a Reuleaux triangle:
Let's consider a circle and a Reuleaux triangle of the same width. They can hide behind each other. Barbier's Theorem states that all objects of the same constant width have the same perimeter. We all know that given a fixed perimeter, the circle has the largest area. Thus, the circle can hide behind the Reuleaux triangle which has smaller area and, consequently, the circle can't hide inside the Reuleaux triangle. By the way, the Reuleaux triangle has the smallest area of all the objects with the given constant width.
To digress. You might have heard the most famous Microsoft interview question: Why are manhole covers round? Presumably because round manhole covers can't fall into slightly smaller round holes. The same property is true for manhole covers of any shape of constant width. On the picture below (Flickr original) you can see Reuleaux-triangle-shaped covers.
Let's move the dimensions up. Dan's questions become both more difficult and more interesting, because the shadows are not as simple as segments any more.
Before continuing, I need to introduce the concept of "Minkowski sums." Suppose we have two convex bodies in space. Let's designate the origin. Then a body can be represented as a set of vectors from the origin to the points in the body. The Minkowski sum of two bodies are all possible sums of two vectors corresponding to the first body and the second body.
Another way to picture the Minkowski sum is like this: Choose a point in the second body. Then move the second body around by translations so that the chosen point covers the first body. Then the area swept by the second body is the Minkowski sum of both of them.
Suppose we have two convex bodies K and L. Their Minkowski interpolation is the body tK + (1-t)L, where 0 ≤ t ≤ 1 is a scaling coefficient. The picture below made by Christina Chen illustrates the Minkowski interpolation of a triangle and an inverted triangle.
If two bodies can hide behind L, then their Minkowski interpolation can hide behind L for any value of parameter t. In particular if K can hide behind L, then the Minkowski interpolation tK + (1-t)L can hide behind L, for any t.
In my paper co-authored with Christina Chen and Daniel Klain "Volume bounds for shadow covering", we found the following connection between hiding inside and volumes. If L is a simplex, and K can hide behind it, but can't hide inside L, then there exists t such that the Minkowski interpolation tK + (1-t)L has a larger volume than the volume of L.
In the paper we conjecture that the largest volume ratio V(K)/V(L) for a body K that can hide behind another body L is achieved if L is a simplex and K is a Minkowski interpolation of L and an inverted simplex. The 3D object that can hide behind a tetrahedron and has 16% more volume than the tetrahedron was found by Christina Chen. See her picture below.
The main result of the paper is a universal constant bound: if K can hide behind L, then V(K) ≤ 2.942 V(L), independent of the dimension of the ambient space.
Question 1. Holodeck. After a long and difficult assignment on an uninhabited planet, Commander Riker went to Holodeck III to unwind. While there he ate three cheeseburgers generated by the holodeck program. Is Commander Riker hungry after he ends the program?
Question 2. Relativity. We know that speed in space is relative, there is no absolute speed. What does Captain Picard mean when he orders a "full stop"?
Question 3. The Replicator. Captain Picard approached a replicator and requested: "Tea, Earl Grey. Hot." The replicator immediately created a glass with hot Earl Grey tea. How much energy would the Enterprise have saved in seven years if they used a dish-washing machine, rather than creating glasses from atoms each time and dissolving them afterwards?
Question 4. Contractions. Commander Data hasn't mastered contractions in English speech. In what year do you think the first program was written to convert formal English into English with contractions?
Question 5. Data. Commander Data is fully functional and absolutely superior to a vibrator. Given that there are more than a thousand people on board the Enterprise, estimate how many times a year on average Data will receive sexual requests.
The next two questions are related to particular episodes.
Question 6. "Up The Long Ladder". Mariposans reproduce by cloning. Why do all the identical sets of clones appear to be the same age? Does it mean that upon the reproduction the clone is the age of the host? If so, they all should be 300 years old.
Mariposans steal sample DNA from Commander Riker and Dr. Pulaski. If Riker and Pulaski didn't destroy their maturing clones what age would those clones be? Would they know how much two plus two is when they awaken? If clones awaken as adults, what is their life span?
Question 7. "Force of Nature". Serova sacrifices herself to save her world from the effects of warp drive, but in doing so, she herself creates the rift that will destroy her world. Explain the logic.
* * *
Logic: if an empty yogurt container is in the sink, a spoon is in the garbage can.
* * *
Logically, a wireless mouse should be called a hamster.
* * *
— I started a new life today.
— You quit smoking and drinking?
— No, I changed my email and Facebook accounts.
* * *
— The reviewer has rejected your paper submitted to our math journal because it doesn't contain any theorems or fomulae or even numbers.
— Wait a minute. Your reviewer is mistaken. There are page numbers on every page.
* * *
A kyboard for sal: only on ky dosn't work.
* * *
My computer always beats me in chess. In revenge, I always beat it in a boxing match.
* * *
— Were your parents married when you were born?
— 50%.
— 50%?
— Yes, my father was married and my mother was not.
* * *
Two programmers are talking:
— I can't turn on my oven.
— What's the error message?
Have you ever been punished for being too good at spider solitaire? I mean, have you ever been stuck because you collected too many suits? Many versions of the game don't allow you to deal from the deck if you have empty columns, nor do they allow you to get back a completed suit. If the number of cards left on the table in the middle of the game is less than ten — the number of columns — you are stuck. I always wondered what the probability is of being stuck. This probability is difficult to calculate because it depends on your strategy. So I invented a boring version of spider solitaire for the sake of creating a math problem. Here it goes:
You start with two full decks of 104 cards. Initially you take 54 cards. At each turn you take all full suits out of your hand. If you have less than ten cards left in your hand, you are stuck. If not, take ten more cards from the leftover deck and continue. What is the probability that you can be stuck during this game?
Let us simplify the game even more by playing the easy level of the boring spider solitaire in which you have only spades. So you have a total of eight full suits of spades. I leave it to my readers to calculate the total probability of being stuck. Here I would like to estimate the easiest case: the probability of being stuck before the last deal.
There are ten cards left in the deck. For you to be stuck, they all should have a different value. The total number of ways to choose ten cards is 104 choose 10. To calculate the number of ways in which these ten cards have different values we need to choose these ten values in 13 choose 10 ways, then multiply by the number of ways each card of a given value can be taken from the deck: 8^{10}. The probability is about 0.0117655.
I will leave it to my readers to calculate the probability of being stuck before the last deal at the medium level: when you play two suits, hearts and spades.
No, I will not tell you how many times I played spider solitaire.
Is there a way to put a sequence of numbers to music? The system that comes immediately to mind is to match a number to a particular pitch. The difference between any two neighboring integers is the same, so it is logical to assume that the same tone interval should correspond to the same difference in integers. After we decide which tone interval corresponds to the difference of 1, we need to find our starting point. That is, we need to choose the pitch that corresponds to the number 1. After that, all numbers can be automatically matched to pitches.
After we know the pitches for our numbers, to make it into music we need to decide on the time interval between the notes. The music should be uniquely defined by the sequence, hence the only logical way would be to have a fixed time interval between two consecutive notes.
We see that there are several parameters here: the starting point, the pitch difference corresponding to 1, and the time interval between notes. The Online Encyclopedia of Integer Sequences offers the conversion to music for any sequence. It gives you freedom to set the parameters yourself. The sequences do not sound melodic because mathematical sequences will not necessarily follow rules that comply with a nice melody. Moreover, there are no interesting rhythms because the time interval between the notes is always the same.
One day I received an email from a stranger named Michael Blake. He sent me a link to his video on YouTube called "What Pi Sounds Like." He converted the digits of Pi to music. My stomach hurt while I was listening to his music. My stomach hurts now while I am writing this. He just numbered white keys on the piano from 1 to 9 starting from C. Then he played the digits of Pi. Clearly, Michael is not a mathematician, as he does not seem to know what to do with 0. Luckily for him the first 32 digits of Pi do not contain zero, so Michael played the first several digits over and over. My stomach hurts because he lost the basic math property of digits: the difference between the neighboring digits is the same. In his interpretation the digits that differ by one can have a tone interval of minor or major second in a random order corresponding to his random starting point.
I am not writing this to trash Michael. He is a free man in a free country and can do whatever he wants with the digits of Pi. Oops, I am sorry, he can't do whatever he wants. Michael's video was removed from YouTube due to an odd copyright infringement claim by Lars Erickson, who wrote a symphony using the digits of Pi.
Luckily for my readers Michael's video appears in some other places, for example at the New Scientist channel. As Michael didn't follow the symmetry of numbers and instead replaced the math rules with some music rules, his interpretation of Pi is one of the most melodic I've heard. The more randomly and non-mathematically you interpret digits, the more freedom you have to make a nice piece of music. I will say, however, that Michael's video is nicely done, and I am glad that musicians are promoting Pi.
Other musicians do other strange things. For example, Steven Rochen composed a violin solo based on the digits of Pi. Unlike Michael, he used the same tone interval for progressing from one number to the next, like a mathematician would do. He started with A representing 1 and each subsequent number corresponded to an increase of half a tone. That is, A# is 2 and so on. Like Michael Blake he didn't know what to do with 0 and used it for rest. In addition, when he encountered 10, 11, and 12 as part of the decimal expansion he didn't use them as two digits, but combined them, and used them for F#, G, G# respectively. To him this was the way to cover all possible notes within one octave, but for me, it unfortunately caused another twinge in my stomach.
In August I visited my son Alexey Radul, who currently works at the Hamilton Institute in Maynooth, Ireland. One of the greatest Irish attractions, Broom Bridge, is located there. It's a bridge over the railroad that connects Maynooth and Dublin. One day in 1843, while walking over the bridge, Sir William Rowan Hamilton had a revelation. He understood how the formulae for quaternions should be written. He scratched them into a stone of the bridge. Now the bridge has a plaque commemorating this event. The plaque contains his formulae. I don't remember ever seeing a plaque with math, so naturally I rushed off to make my pilgrimage to Broom Bridge.
Quaternions have very pronounced sentimental value for me, since my first research was related to them. Let's consider a simple graph. We can construct an algebra associated with this graph in the following way. For each vertex we have a generator of the algebra. In addition we have some relations. Each generator squared is equal to −1. If two vertices are connected the corresponding generators anti-commute, and they commute otherwise. The simplest non-commutative algebra associated with a graph corresponds to a graph with two vertices and one edge. If we call the generators i and j, then the we get the relations: i^{2} = j^{2} = −1, and ij = −ji. I we denote ij as k, the algebra as a vector space has dimension 4 and a basis: 1, i, j, k. These are exactly the quaternions. In my undergraduate research I studied such algebras related to Dynkin diagrams. Thirty years later I came back to them in my paper Clifford Algebras and Graphs. But I digress.
I was walking on the bridge hoping that like Hamilton I would come up with a new formula. Instead, I was looking around wondering why the Broombridge Station didn't have a ticket office. I already had my ticket, but I was curious how other people would get theirs. I asked a girl standing on the platform where to buy tickets. She said that there is no way to buy tickets there, so she sometimes rides without a ticket. The fine for not having tickets is very high in Ireland, so I expressed my surprised. She told me that she just says that she is from the town of Broombridge if she is asked to present her ticket.
Being a Russian I started scheming: obviously people can save money by buying tickets to Broombridge and continuing without a ticket wherever they need to go. If the tickets are checked, they can claim that they are traveling from Broombridge. Clearly Ireland hasn't been blessed with very many Russians visitors.
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and is offered a bet. She may pay $550 in which case she will get $1000 if the coin was tails. If the coin was tails, she is put back to sleep with her memory erased, and awakened on Tuesday and given the same bet again. She knows the protocol. Should she take the bet?
As we discussed in our first essay about Sleeping Beauty, she should take the bet. Indeed, if the coin was heads her loss is $550. But if the coin was tails her gain is $900.
To tell you the truth, when Beauty is offered the bet, she dreams: "It would be nice to know the day of the week. If it were Tuesday, then the coin must have been tails and I would gladly take the winning bet."
In our next variation of the riddle her dream comes true.
Every time she is awakened she is offered to buy the knowledge of the day of the week. How much should she be willing to pay to know the day of the week?
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning and instead of the expected questions she is offered a bet. She may pay $600 in which case she will get $1000 if the coin was tails. Should she take the bet?
We know that tripling the triangular number 1 yields the triangular number 3. The figure shows how we can use this fact to conclude that tripling the triangular number 15 yields the triangular number 45.
Using this new fact, can you modify the figure to find even larger examples of tripling triangles?
This post is inspired by the following problem:
Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning: What is her probability that the coin was heads?
Some people argue: asleep or awake, the probability of a fair coin being heads is one half, so her probability should be one half.
Other people, including us, argue that those people didn't study conditional probability. On the information of the setup to the problem and the information of having awakened, the three situations "Coin was heads and it is Monday", "Coin was tails and it is Monday", and "Coin was tails and it is Tuesday" are symmetric and therefore equiprobable; thus the probability that the coin was tails is, on this information, two thirds.
So who is right? We are, of course. A good way to visualize probability judgements is to turn them into bets. Suppose each time Beauty wakes up she is offered the following bet: She pays $600 and gets $1000 if the coin was tails. Should she take it? If her probability of the coin being tails were one half, then obviously not; if her probability of the coin being tails were two thirds, obviously yes. So which is it? Consider the situation from her perspective as of Sunday. She can either always take this bet or always refuse it. If she always refuses, she gets nothing. If she always accepts: If the coin turns up heads, she will be asked the question once and will lose $600. If the coin turns up tails, she will be asked the question twice and will gain $800. So on average she will win, so she should take the bet. By this thought experiment, her probability of tails is clearly not one half.
To make matters more interesting, let's try another bet. Suppose she is given the above bet just once, in advance, on Sunday. She pays $600, and she gets paid $1000 on Wednesday if the coin was tails. This has nothing to do with sleeping and awakening. If she takes the bet she loses $600 with probability one half and gains $400 otherwise. So she shouldn't take the bet. Her probability on Sunday that the coin will come up heads is, of course, one half. The point is that just as these two bets are different bets, the sets of information Beauty has on Sunday vs at awakening are different, and lead to different conclusions. On Sunday she knows that the next time she wakes up it will be Monday, but when she then wakes up, she doesn't know that it's Monday.
Parting thought: The phenomenon of predictably losing information leads to the phenomenon of predictably changing one's assessments. Suppose for some reason she decided to take that unprofitable bet on Sunday. When she wakes up during the experiment, should she feel happy or sad? From her perspective during the experiment, the odds of gaining $400 vs losing $600 are two to one, so she should be happy. Given that she knows on Sunday how she will (with complete certainty!) feel about this bet on Monday, should she take it, even given her Sunday self's assessment that it's a bad bet?
If you buy one Mega Millions ticket, your probability of hitting the jackpot is one in 175,000,000. For all practical purposes it is zero. When I give my talk on lotteries, there is always someone in the audience who would argue that "but someone is winning and so can I." The fact that someone is winning depends on the number of people buying tickets. It is difficult to visualize the large number of people buying tickets and the miniscule odds of winning. For example, the probability of you dying from an impact with a meteorite is larger than the odds of winning the jackpot.
I receive a lot of emails from strangers asking me to advertise their websites on my blog. I always check out their websites and I often find them either unrelated to math or boring. That is why I was pleasantly surprised when I was asked to write about a useful website: Understanding Big Numbers. In each post Liam Gray takes a big number and puts it into some perspective. For example, he estimates Mark Zuckerberg's Hourly Wage by dividing Mark's estimated wealth in 2011 by the number of hours Mark might have worked on Facebook. Facebook has existed for 7 years and, assuming 10 hours of work a day every day, we get 25,000 work hours. That is more than half a million dollars an hour.
Imagine someone calls Mark Zuckerberg and asks to talk to him for a minute. Mark wouldn't be out of line to request nine thousand dollars for that. Lucky am I, that I do not need to talk to Mark Zuckerberg.
Here is a game that John Conway popularizes. It is called "Finchley Central," which is a station of the London Underground. The game goes as follows. Alice and Bob take turns naming London Underground stations, in any order. The first person to say "Finchley Central" wins.
Alice, who starts, can just name the station. But then Bob will give her a look. It is not fun to win a game on the first turn. To avoid appearing rude, Alice will not start with "Finchley Central." It would be impolite of Bob to take advantage of Alice's generosity, so he also won't say "Finchley Central." The game might continue like this for a while.
The game has a hidden agenda: winning it after 10 turns will supply many more bragging rights than winning it right away would. We can make this hidden agenda explicit by assigning a value to the honor of continuing the game. For example, suppose every time Alice (or Bob) says a station, she puts one dollar into the pile. The person who says "Finchley Central" first takes all the money from the pile. The implicit goal of the game becomes explicit: you want to say "Finchley Central" right before your opponent says it.
By the way, Finchley Central is not actually a particularly central station — it is the station between Finchley East and Finchley West, serving the relatively small place called Finchley; and is not even under ground. It has the distinction of being one of the oldest still-standing pieces of London Underground physical plant, because plans to rebuild it were interrupted on account of World War II and never resumed. It also has the distinction of having served the home of the guy (an employee of the Underground system) who had the brilliant idea that since the Underground was, indeed, mostly under ground, the right way to map it was topologically, rather than geographically.
Here is another way to model the game. Alice writes an odd number on a piece of paper, and Bob writes an even number. When they compare, the person who wrote a smaller number wins that number of dollars. This version loses the psychological aspect. When you take turns, it is to your advantage to read the non-verbal signs of your opponent to see when s/he is getting ready to drop the bomb.
People play this game in real life. Here are Alice and Bob looking at the last piece of a mouth-watering Tiramisu:
At this point Alice wins with some extra brownie points for being polite.
We can model the honor points differently. We can say you will be the most proud of the game if you name the station write before you opponent is about to do so. Then the model is: everyone writes down their next move; if your move is Finchley Central when your opponent's next move was going to be Finchley Central, then you win.
Here we suggest another game that we call "Reverse Finchley Central." Alice and Bob name London Underground stations in turns and the person who names "Finchley Central" first loses. This game can continue until all the stations are exhausted, if the players are forbidden to repeat them, or it can continue indefinitely otherwise. But this is quite tiresome. The hidden agenda would be to not waste too much time. Clearly the person who values time less will win.
But let us model this game. We want to fix the value of winning. Let us set aside ten dollars for the winner. On their turn, each player puts one dollar into the pile, and as soon as one of the players says "Finchley Central," the other one wins and takes the ten dollars. The pile goes to charity. Alternatively, Alice and Bob can each write a number. The person with the larger number wins the prize, while both have to pay the smaller number to charity.
We play this game with our parents. They nag us to do the dishes. We resist. Then they give up and do the dishes themselves. They lose, but we all pay with our nerves for nagging or being nagged at. Later our parents get their revenge when we have children of our own.
At the 2011 IMO, Lisa Sauermann received yet another gold medal. Now she tops the Hall of Fame of the IMO with four gold medals and one silver medal.
In addition, in 2011 she achieved the absolute best individual result and was the only person with a perfect score. In previous years, there were several girls who tied for first place, but she is the first girl ever to have an absolute rank of 1.
I told you so. In my 2009 essay Is There Hope for a Female Fields Medalist?, I predicted that a girl will soon become an absolute champion of the IMO.
In that essay I draw a parallel between the absolute champion of IMO and a Fields medalist. Indeed, we get one of each per year. Lisa Sauermann is the best math problem solver in her year. Will she grow up to receive a Fields medal? I am not so sure: the medal is still unfriendly to women. Lisa Sauermann is the best math problem solver ever. Will she grow up to be the best mathematician of our century? I wonder.
My e-friend and coauthor, Konstantin Knop, designed the following problem for the 2011 All-Russia Olympiad:
Some cells of a 100 by 100 board have one chip placed on them. We call a cell pretty if it has an even number of neighboring cells with chips. Neighbors are the cells that share a side. Is it possible for exactly one cell to be pretty?
The problem is not easy. Only one person at the Olympiad received full credit for it.
From time to time my female colleagues share stories with me of great unfairness or horrible sexual harassment in the world of mathematics. I can't reciprocate — certainly not on that level.
I do not have any horror stories to tell. Generally I am treated with great respect, at least to my face. In fact, some men have told me that I am the smartest person they ever met.
The stories I want to share are not about harassment. No single incident is a big deal. But when these things happened time after time after time, I realized: this is gender bias.
First story. A guy told me, "Your proof is unbelievably amazing."
What can I say? It is just a compliment. Though I am not sure why the word "unbelievable" was included. Is it difficult to believe that I can produce an amazing proof? I encounter surprise too often to my taste.
Second story. Another guy tells me after I explain a solution to a math problem, "I didn't realize it was so simple."
Actually it wasn't simple. When I explained the solution, it may have seemed simple, but that was because I was able to explain it to him with such clarity. People tend to downgrade their opinion of the problem, rather than upgrade their opinion of my ability. It actually affects my reputation as a mathematician.
Third story. Another guy said to me (and I quote!), "I am so dumb. I tried for a week to write the program that computes these numbers and you did it in one hour. I feel so dumb. I didn't expect myself to be so dumb. Why am I so dumb?"
After the fourth "dumb", I started wondering what it was all about. Many guys try to compete with me. And they hate losing to a woman. It creates a strong motivation for them to discard my brilliance and to explain away my speed, even if they have to claim temporary dumbness.
Fourth story. Someone asked me, "What is the source of the solutions and math ideas in your blog? Can you refer me to the literature?"
I do invest extra effort in citing the sources of the math puzzles I discuss. Everything else — the solutions, the ideas, new definitions, new sequences — I invent myself. I have even started inventing math puzzles. This is my blog. I thought of it myself, I wrote it myself. Has anyone ever asked Terence Tao where he takes the solutions for his blog from?
Unfortunately, this attitude damages my career. When people think that my ideas come from someone else, they do not cite me.
But all these stories however minor happen all the time, not only to me but to all my female colleagues. Gender bias is real. Next time someone tells me how unbelievably amazing my proof is, I will explode.
Recently I stumbled on a cute xkcd comic with the hidden message:
Wikipedia trivia: if you take any article, click on the first link in the article text not in parentheses or italics, and then repeat, you will eventually end up at "Philosophy".
Naturally, I started to experiment. The first thing I tried was mathematics. Here is the path: Mathematics — Quantity — Property — Modern philosophy — Philosophy.
Then I tried physics, which led me to mathematics: Physics — Natural science — Science — Knowledge — Fact — Information — Sequence — Mathematics.
Then I tried Pierre de Fermat, who for some strange reason led to physics first: Pierre de Fermat — French — France — Unitary state — Sovereign state — State — Social sciences — List of academic disciplines — Academia — Community — Living — Life — Objects — Physics.
The natural question is: what about philosophy? Yes, philosophy goes in a cycle: Philosophy — Reason — Rationality — philosophy.
The original comic talks about spark plugs. So I tried that and arrived at physics: Spark plug — Cylinder head — Internal combustion engine — Engine — Machine — Machine (mechanical) — Mechanical system — Power — Physics.
Then I tried to get far away from philosophy and attempted sex, unsuccessfully: Sex — Biology — Natural science. Then I tried dance: Dance — Art — Sense — Physiology — Science.
It is interesting to see how many steps it takes to get to philosophy. Here is the table for the words I tried:
Word | # Steps |
---|---|
Mathematics | 4 |
Physics | 11 |
Pierre de Fermat | 24 |
Spark plug | 19 |
Sex | 12 |
Dance | 13 |
Mathematics wins. It thoroughly beats all the other words I tried. For now. Fans of sex might be disappointed by these results and tomorrow they might change the wiki essay about sex to start as:
Modern philosophy considers sex …
Not personally. Someone hacked into my website.
I would like to thank my readers Qiaochu Yuan, Mark Rudkin, "ano" and Paul who alerted me to the problem. Viewers who were using the Google Chrome browser and who tried to visit my website got this message: "This site contains content from howmanyoffers.com, a site known to distribute malware."
It took me some time to figure out what was going on. It appears that on June 19 someone from 89-76-135-50.dynamic.chello.pl hacked into my hosting account and added a script to all my html files and to my blog header. It seems that the script was dormant and wasn't yet doing bad things.
As soon as I grasped what was going on, I replaced all the affected files.
I have had my website for many years without changing my hosting password. Unfortunately, passwords, not dissimilar to humans, have this annoying tendency to become weaker with age. I wasn't paying attention to the declining strength of my password and so I was punished.
Now I have fixed the website and my new password is: qwP35q2054uWiedfj052!@#$%.
Just kidding.
I recently had the following chat with a particular calculator:
It seems odd to me that putting a few more e's down the bottom should result in it thinking there were the same number of extra 10s at the bottom. In fact, I've never seen a calculator answer in this form at all. I'm especially intrigued that the final power of ten seems to be the same in all three cases, so it can't even just be estimating. Do you have any thoughts on what screwy counting could be behind these particular answers?
May the Mass times the Acceleration be with you!
John Conway taught me how to tell time at night. But first I need to explain the notions of the "time in the sky" and the "time in the year."
The clock in the sky. Look at Polaris and treat it as the center of a clock. The up direction corresponds to 12:00. Now we need to find a hand. If you find Polaris the way I do, first you locate the Big Dipper. Then you draw a line through the two stars that are furthest away from the Big Dipper's handle. The line passes through Polaris and is your "hour" hand. Now you can read the time in the sky.
The hand of the clock in the sky makes a full rotation in approximately 24 hours. So if you stare at the sky for a long time, you can calculate the time you spent staring. Keep in mind that the hand in the sky clock is twice as slow as the hour hand, and it turns counter-clockwise. So to figure out how long you're looking into the sky, take the sky-time when you start staring, subtract the sky-time when you stop staring and multiply the result by 2.
To calculate the absolute time, we need to adjust for the day in the year.
The clock in the year. A year has twelve months and a clock has twelve hours. How convenient. You can treat each month as one hour. In addition as a month has about 30 days and an hour has exactly 60 minutes, we should count a day as two minutes. Thus, January 25 is 1:50.
Fact: on March 7^{th} at midnight the clock in the sky shows 12:00. March 7^{th} corresponds to 3:15. So to calculate the solar time you need to add up the time in the sky and the time in the year and multiply it by 2. Then subtracting the result from 6:30, which is twice 3:15, you get the solar time.
You are almost ready. You might need to adjust for daylight savings time or for peculiarities of your time zone.
This time formula is not very precise. But if you are looking into the sky and you do not have your watch or cell phone with you, you probably do not need to know the time precisely.
In my life as a female mathematician I have quite often encountered a mathematician's wife who, despite not knowing me, already hated me. It was clear that it had nothing to do with me personally, so being clueless and naive, I assumed that most men were cheaters and that their wives were extremely insecure and jealous.
Then one day one of the wives decided to be frank about her feelings. It wasn't about cheating, she told me. It was that she felt distant from her husband. He lived in a world of mathematics from which she was excluded. I on the other hand shared this world with him.
It was very sad. It meant that I incurred their jealousy, not because of my sins, but because I am a female mathematician.
Let me tell you another story that helped me realize how all-encompassing this world of mathematics can be for some people. Once I had a very close friend who we will call Jack. I do not want to name him as he is a famous mathematician. Jack told me that the strongest emotions he feels are related to mathematics. He can only feel close to someone if he can share a mathematical discussion with them.
Now I understand the wives better. Husbands like Jack invest so much more in their math world and their colleagues than they do in their home life, that it is not surprising the wives are jealous. Because women mathematicians are scarce, when I appear in their husbands' world, it adds another layer of worry.
Another thing that Jack told me is that he gets such a euphoric feeling when he discovers a new math idea that it is better than any orgasm. Of course, this statement made me question the quality of Jack's orgasms, but in any case, for some mathematicians math is an aphrodisiac.
If math is an aphrodisiac, then tattooing a formula on the lover's body may well enhance the orgasm. I just remembered the movie by Ed Frenkel. But I digress.
If math is an aphrodisiac, then I understand jealous wives even better. Without sex I can give their husbands pleasure they can't.
* * *
I am taking my dog to tweet. He'll check other dog's posts at every pole and will leave his comments.
* * *
Not many people know that 1000 chameleons is a chabillion.
* * *
The Internet paradox: it connects people who are far apart, and disconnects those who are close.
* * *
We bought a cell phone for our TV set. We attached it to the remote control, so that we can call our TV when the remote is lost.
* * *
Mary's mom failed arithmetic. Actually, that is why Mary was born.
* * *
Your call is very important to us. Please, hold. And in the meantime, to protect your health, our customer care team encourages you to drink a glass of water at least every two hours.
* * *
Who is your favorite computer game character?
The stick from Tetris.
* * *
Our new boss invited everyone to bring their keyboards to his office. He kept the employees who had worn letters and laid off the ones with worn arrows.
* * *
My son will be a hacker. He started his career before he was born: he found a flaw in the condom.
I recently stumbled upon some notes (in Russian) of a public lecture given by Vladimir Arnold in 2006. In this lecture Arnold defines a notion of complexity for finite binary strings.
Consider a set of binary strings of length n. Let us first define the Ducci map acting on this set. The result of this operator acting on a string a_{1}a_{2}…a_{n} is a string of length n such that its i-th character is |a_{i} − a_{(i+1)}| for i < n, and the n-th character is |a_{n} − a_{1}|. We can view this as a difference operator in the field F_{2}, and we consider strings wrapped around. Or we can say that strings are periodic and infinite in both directions.
Let's consider as an example the action of the Ducci map on strings of length 6. Since the Ducci map respects cyclic permutation as well as reflection, I will only check strings up to cyclic permutation and reflection. If I denote the Ducci map as D, then the Ducci operator is determined by its action on the following 13 strings, which represent all 64 strings up to cyclic permutation and reflection: D(000000) = 000000, D(000001) = 000011, D(000011) = 000101, D(000101) = 001111, D(000111) = 001001, D(001001) = 011011, D(001011) = 011101, D(001111) = 010001, D(010101) = 111111, D(010111) = 111101, D(011011) = 101101, D(011111) = 100001, D(111111) = 000000.
Now suppose we take a string and apply the Ducci map several times. Because of the pigeonhole principle, this procedure is eventually periodic. On strings of length 6, there are 4 cycles. One cycle of length 1 consists of the string 000000. One cycle of length 3 consists of the strings 011011, 101101 and 110110. Finally, there are two cycles of length 6: the first one is 000101, 001111, 010001, 110011, 010100, 111100, and the second one is shifted by one character.
We can represent the strings as vertices and the Ducci map as a collection of directed edges between vertices. All 64 vertices corresponding to strings of length 6 generate a graph with 4 connected components, each of which contains a unique cycle.
The Ducci map is similar to a differential operator. Hence, sequences that end up at the point 000000 are similar to polynomials. Arnold decided that polynomials should have lower complexity than other functions. I do not completely agree with that decision; I don't have a good explanation for it. In any case, he proposes the following notion of complexity for such strings.
Strings that end up at cycles of longer length should be considered more complex than strings that end up at cycles with shorter length. Within the connected component, the strings that are further away from the cycle should have greater complexity. Thus the string 000000 has the lowest complexity, followed by the string 111111, as D(111111) = 000000. Next in increasing complexity are the strings 010101 and 101010. At this point the strings that represent polynomials are exhausted and the next more complex strings would be the three strings that form a cycle of length three: 011011, 101101 and 110110. If we assign 000000 a complexity of 1, then we can assign a number representing complexity to any other string. For example, the string 111111 would have complexity 2, and strings 010101 and 101010 would have complexity 3.
I am not completely satisfied with Arnold's notion of complexity. First, as I mentioned before, I think that some high-degree polynomials are so much uglier than other functions that there is no reason to consider them having lower complexity. Second, I want to give a definition of complexity for periodic strings. There is a slight difference between periodic strings and finite strings that are wrapped around. Indeed, the string 110 of length 3 and the string 110110 of length 6 correspond to the same periodic string, but as finite strings it might make sense to think of string 110110 as more complex than string 110. As I want to define complexity for periodic strings, I want the complexity of the periodic strings corresponding to 110 and 110110 to be the same. So this is my definition of complexity for periodic strings: let's call the complexity of the string the number of edges we need to traverse in the Ducci graph until we get to a string we saw before. For example, let us start with string 011010. Arrows represent the Ducci map: 011010 → 101110 → 110011 → 010100 → 111100 → 000101 → 001111 → 010001 → 110011. We saw 110011 before, so the number of edges, and thus the complexity, is 8.
The table below describes the complexity of the binary strings of length 6. The first column shows one string in a class up to a rotation or reflections. The second column shows the number of strings in a class. The next column provides the Ducci map of the given string, followed by the length of the cycle. The last two columns show Arnold's complexity and my complexity.
String s | # of Strings | D(s) | Length of the end cycle | Arnold's complexity | My complexity |
---|---|---|---|---|---|
000000 | 1 | 000000 | 1 | 1 | 1 |
000001 | 6 | 000011 | 6 | 9 | 8 |
000011 | 6 | 000101 | 6 | 8 | 7 |
000101 | 6 | 001111 | 6 | 7 | 6 |
000111 | 6 | 001001 | 3 | 6 | 5 |
001001 | 3 | 011011 | 3 | 5 | 4 |
001011 | 12 | 011101 | 6 | 9 | 8 |
001111 | 6 | 010001 | 6 | 7 | 6 |
010101 | 2 | 111111 | 1 | 3 | 3 |
010111 | 6 | 111001 | 6 | 8 | 7 |
011011 | 3 | 101101 | 3 | 4 | 3 |
011111 | 6 | 100001 | 6 | 9 | 8 |
111111 | 1 | 000000 | 1 | 2 | 2 |
As you can see, for examples of length six my complexity doesn't differ much from Arnold's complexity, but for longer strings the difference will be more significant. Also, I am pleased to see that the sequence 011010, the one that I called The Random Sequence in one of my previous essays, has the highest complexity.
I know that my definition of complexity is only for periodic sequences. For example, the binary expansion of pi will have a very high complexity, though it can be represented by one Greek letter. But for periodic strings it always gives a number that can be used as a measure of complexity.
I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here's his version:
The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?
When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:
A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.
* * * A Generic Limerick (submitted by Michael Chepovetsky)
There once was an X from place B,
Who satisfied predicate P,
The X did thing A,
In a specified way,
Resulting in circumstance C.
* * *
I just learned that 4,416,237 people got married in the US in 2010. Not to nitpick, but shouldn't it be an even number?
* * *
We are happy to announce that 100% of Russian citizens are computer-savvy and use the Internet on a regular basis (according to a recent Internet survey).
* * *
Two math teachers had a fight. It seems they couldn't divide something.
* * *
Do you know that if you start counting seconds, once you reach 31,556,926 you discover that you have wasted a whole year?
* * *
What I need after a visit to the hairdresser is a "Save" button.
* * *
— Hello! Is this a fax machine?
— Yes.
* * *
— I am not fat at all! My girlfriend tells me that I have a perfect figure.
— Your girlfriend is a mathematician. For her a perfect figure is a sphere.
* * *
A: Hi, how are you?
B: +
A: Will you come to classes today?
B: -
A: You will be kicked out!
B: =
A: Are you using your calculator to chat?
I wonder what the largest number is that can be represented with one character. Probably 9. How about two characters? Is it 9^{9}? What about three or four?
I guess I should define a character. Let's have two separate cases. In the first one you can only use keyboard characters. In the second one you can use any Unicode characters.
I'm awaiting your answers to this.
The last time I talked to John H. Conway, he taught me to walk up the stairs. It's not that I didn't know how to do that, but he reminded me that a nerd's goal in climbing the steps is to establish the number of steps at the end of the flight. Since it is boring to just count the stairs, we're lucky to have John's fun system.
His invention is simple. Your steps should be in a cycle: short, long, long. Long in this case means a double step. Thus, you will cover five stairs in one short-long-long cycle. In addition, you should always start the first cycle on the same foot. Suppose you start on the left foot, then after two cycles you are back on the left foot, having covered ten stairs. While you are walking the stairs in this way, it is clear where you are in the cycle. By the end of the staircase, you will know the number of stairs modulo ten. Usually there are not a lot of stairs in a staircase, so you can easily estimate the total if you know the last digit of that number.
I guess I am not a true nerd. I have lived in my apartment for eight years and have never bothered to count the number of steps. That is, until now. Having climbed my staircase using John's method, I now know that the ominous total is 13. Oh dear.
The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year's Olympiad:
Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?
Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, "My age is equal to the sum of the digits of my birth year." "Mine too," replied Paul. By how many years is Paul younger than Peter?
Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?
Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?
Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.
Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.
A mathematician is someone who pauses when asked "How much is two and two?"
Indeed, the answer might be:
I already wrote about the research of my friend Olga Amosova who studied the sickle-cell anemia mutation. She and her colleagues needed to store short fragments of hemoglobin genes for their experiments. All the fragments were identical. They noticed that with time the fragments always broke down in the same place. It was a mystery. When good scientists stumble on a mystery, they start digging.
They found that one of the nucleotides rips off the DNA fragment at the site of the Sickle-cell mutation. That place on the DNA becomes fragile and later breaks down. These sites need to be repaired. The repair is very error-prone and often leads to a mutation.
When DNA strands are left unattended, they want to pair up. There are four types of nucleotides: A, C, G and T. So mathematically the fragment of DNA is a string in the alphabet A, C, G, T. These nucleotides are matched to each other. When two DNA strands pair up, A on one strand always matches T and C matches G. So it is logical that if there are two complementary DNA pieces on the same fragment, they will find each other and pair up. They form a hydrogen bond. For example, a piece AACGT matches perfectly another piece TTGCA. Suppose a substring of DNA consists of a piece AACGT and somewhere later the reverse of the match: ACGTT. Such a string is called an inverted repeat. The DNA fragment I mentioned contains a string AACGT****ACGTT. Two pieces AACGT and ACGTT are complementary and not too far from each other in space. So it is easy for them to find each other and to bond to form a so-called stem-loop or a hairpin structure. The site of Sickle-cell mutation falls into the loop.
Olga and her colleagues discovered that for some particular loops the orientation in space becomes awkward and one of the nucleotides rips off. Such a rip off is called depurination. In further investigation, Olga found examples of when depurination happens. The first sequence of the pair that will bond later has to have at least five nucleotides and has to end in T. Correspondingly the second part in the pair has to begin with A. In the middle there needs to be four nucleotides GTGG. The first G flies away. Enzymes rush like a first aid squad to repair it and introduce mistakes that lead to mutation and diseases like cancer.
DNA was thought to be simply a passive information storage system, not capable of any action. Now we see that DNA is capable of action. DNA can damage itself. Damage provokes a mutation. For all practical purposes it is self-mutilation. Olga and her colleagues scanned the human genome for other sequences that are capable of self-mutilation. They found that such sequences are overwhelmingly present. They are present in much higher numbers than would be expected statistically. The pieces that are capable of damaging themselves occur 40 times more often than would occur if the nucleotides were distributed randomly. They are especially overrepresented in genes linked to cancer.
Self-damaging shouldn't happen in normal situations. It can be provoked by the environment, for example, the chemistry of the cell. That means, that our cancers are not only in our genes but also in our life-style. There was, for example, a suggestion in a recent NY Times article, Is Sugar Toxic?, that too much sugar in a diet might provoke cancer. If the rate of mutation depends on the environment, we can influence it and prolong our lives.
It is not clear why the ability to self-mutilate survives in the evolutionary process. It is quite possible that if something very bad happens to our planet, we need our genes to be able to mutate very fast in order to adjust to the environment so that humans can survive.
Though I never tried to donate my sperm to a sperm bank, because of my inability to produce it, I know that sperm banks look for people who have ancestors who lived for a very long time. Such sperm is in bigger demand as everyone wants their children to live longer. I wonder if this tendency is a mistake. Global warming is upon us. People with longevity genes might not be flexible enough for their children to survive the changing of the Earth.
As you might have guessed from the title, this essay is about domino tilings.
Suppose a subset of a square grid has area N, and the number of possible domino tilings is T. Let's imagine that each cell is contributing a factor of x tilings to the total independently of the others. Then we get that x^{N} = T. This mental exercise suggests a definition: we call the nth root of T the degree of freedom per square for a given region.
Let's consider a 1 by 2k rectangle. There is exactly one way to tile it with dominoes. So the degree of freedom per square of such a rectangle is 1. Now consider a 2 by k rectangle. It has the same area as before, and we know that there should be more than one tiling. Hence, we expect the degree of freedom to be larger than the one in the previous example. The number of tilings of a 2 by k rectangle is F_{k-2}, where F_{k} is k^{th} Fibonacci number. So the degree of freedom for large k will be approximately the square root of the golden ratio, which is about 1.272.
You might expect that squares should give larger degrees of freedom than rectangles of the same area. The degree of freedom for a large square is about 1.3385. You can find more information in the beautiful paper Tilings by Federico Ardila and Richard P. Stanley.
Let's move from rectangles to Aztec diamonds. They are almost like squares but the side of the diamond is aligned with diagonals of the dominoes rather than with their sides. See the sample diamonds in the picture above, which Richard Stanley kindly sent to me for this essay.
It is easier to calculate the degree of freedom for Aztec diamonds than for regular squares. The degree is the fourth root of 2, or 1.1892…. In the picture below created by James Propp's tiling group you can see a random tiling of a large Aztec diamond.
Look at its colors: horizontal dominoes are yellow and blue; vertical ones are red and aquamarine. You might wonder what rule decides which of the horizontal dominoes are yellow and which are blue. I will not tell you the rule; I will just hint that it is simple.
Back to freedom. As you can see from the picture, freedom is highly non-uniform and depends on where you live. Freedom is concentrated inside a circle called the arctic circle, perhaps because the areas outside it are frozen for lack of freedom.
Now I would like to expand the notion of freedom to give each cell its own freedom. For a large Aztec diamond, I will approximate freedom with a function that is one outside the arctic circle and is uniform inside. The Aztec diamond AZ(n) consists of 2n(n+1) squares, shaped like a square with side-length n√2. So the area of the circle is πn^{2}/2. Hence we can calculate the freedom inside the circle as the π^{th} root of 2, which is about 1.247. This number is still much less than the degree of freedom of a cell in a large square.
Jorge Tierno sent me a link to the following puzzle:
There is a certain country where everybody wants to have a son. Therefore, each couple keeps having children until they have a boy, then they stop. What fraction of the children are female?
If we assume that a boy is born with probability 1/2 and children do not die, then every birth will produce a boy with the same probability as a girl, so girls will comprise half of all children.
Now, I wonder why everyone would want a boy? Y-chromosomes are much shorter than X-chromosomes. If a man wants to pass his genes to the next generation, a daughter should be preferable as she keeps more genes from the father. I am a mother of two boys, so my granddaughters will have my X-chromosome while my grandsons will have my ex-husband's Y-chromosome, so to keep my genes in the pool I should be more interested in granddaughters.
But I digress. I started writing this essay because in the original puzzle link the answer was different from mine. Here is how the other argument goes:
Half of all families have zero girls, a quarter have 1/2 girls, 1/8 have 2/3 girls, and so on. If we sum this up the expected ratio of girls to boys is (1/2)0 + (1/4)(1/2) + (1/8)(2/3) + (1/16)(3/4) + ... which adds to 1 − ln 2, which is about 30%.
What's wrong with this solution?
I found this cute problem in the Russian book Sharygin Geometry Olympiad by Zaslavsky, Protasov and Sharygin.
Find numbers p and q that satisfy the equation: x^{2} + px + q = 0.
The book asks you to find a mistake in the following solution:
By Viète's formulae we get a system of equations p + q = − p, pq = q. Solving the system we get two solutions: p = q = 0 and p = 1, q = −2.
What is wrong with this solution?
I wrote how the written entrance exam was used to keep Jewish students from studying at Moscow State University, but the real brutality happened at the oral exam. Undesirable students were given very difficult problems. Here is a sample "Jewish" problem:
Solve the following equation for real y:
Here is how my compatriots who studied algebra in Soviet high schools would have approached this problem. First, cube it and get a 9th degree equation. Then, try to use the Rational Root Theorem and find that y = 1 is a root. Factoring out y − 1 gives an 8th degree equation too messy to deal with.
The most advanced students would have checked if the polynomial in question had multiple roots by GCDing it with its derivative, but in vain.
We didn't study any other methods. So the students given that problem would have failed it and the exam.
Unfortunately, this problem is impossible to appeal, because it has an elementary solution that any applicant could have understood. It goes like this:
Let us introduce a new variable: x = (y^{3} + 1)/2. Now we need to solve a system of equations:
This system has a symmetry which we can exploit. The graphs of the functions x = (y^{3} + 1)/2 and y = (x^{3} + 1)/2 are reflections of each other across the line x = y. As both functions are increasing, the solution to the system of equations should lie on the line x = y. Hence, we need to solve the cubic y = (y^{3} + 1)/2, one of whose roots we already know.
Now I offer you another problem without telling you the solution:
Four points on a plane used to belong to four different sides of a square. Reconstruct the square by compass and straightedge.
Recently I asked my readers to look at the 1976 written math exam that was given to applicants wishing to study at the math department of Moscow State University. Now it's time to reveal the hidden agenda. My readers noticed that problems 1, 2, and 3 were relatively simple, problem 4 was very hard, and problem 5 was extremely hard. It seems unfair and strange that problems of such different difficulty were worth the same. It is also suspicious that the difficult problems had no opportunity for partial credit. As a result of these characteristics of the exam, almost every applicant would get 3 points, the lowest passing score. The same situation persisted for many years in a row. Why would the best place to study math in Soviet Russia not differentiate the math abilities of its applicants?
In those years the math department of Moscow State University was infamous for its antisemitism and its efforts to exclude all Jewish students from the University. The strange structure of the exam accomplished three objectives toward that goal.
1. Protect the fast track. There was a fast track for students with a gold medal from their high school who got 5 points on the written exam. The structure of the exam guaranteed that very few students could solve all 5 problems. If by chance a Jewish student solved all 5 problems, it was not much work to find some minor stylistic mistake and not count the solution.
2. Avoid raising suspicion at the next exams. The second math entrance exam was oral. At such an exam different students would talk one-on-one with professors and would have to answer different questions. It was much easier to arrange difficult questions for undesirable students and fail all the Jewish students during the oral exam than during the written exam. But if many students with perfect scores on the written exam had failed the oral exam, it might have raised a lot of questions.
3. Protect appeals. Despite these gigantic efforts, there were cases when Jewish students with a failing score of 2 points were able to appeal and earn the minimum passing score of 3. If undesirable students managed to appeal all the exams, they would only get a half-passing grade at the end and would not be accepted because the department was allowed to choose from the many students that the exams guaranteed would have half-passing scores.
I have only heard about one faculty member who tried to publicly fight the written exam system. It was Vladimir Arnold, and I will tell the story some other time.
The following problem appeared at the Gillis Math Olympiad organized by the Weizmann Institute:
A foreign government consists of 12 ministers. Each minister has 5 friends and 6 enemies amongst the ministers. Each committee needs 3 ministers. A committee is considered legitimate if all of its members are friends or all of its members are enemies. How many legitimate committees can be formed?
Surprisingly, this problem implies that the answer doesn't depend on how exactly enemies and friends are distributed. This meta thought lets us calculate the answer by choosing an example. Imagine that the government is divided into two factions of six people. Within a faction people are friends, but members of two different factions dislike each other. Legitimate committees can only be formed by choosing all three members from the same faction. The answer is 40.
We would like to show that actually the answer to the problem doesn't depend on the particular configuration of friendships and enmities. For this, we will count illegitimate committees. Every illegitimate committee has exactly two people that have one enemy and one friend in the committee. Let's count all the committees from the point of view of these "mixed" people. Each person participates in exactly 5*6 committees as a mixed person. Multiply by 12 (the number of people), divide by 2 (each committee is counted twice) and you get the total 180. This gives an answer of 40 for the number of legitimate committees without using a particular example.
What interests us is the fact that the number of illegitimate, as well as legitimate, committees is completely defined by the degree distribution of friends. For any set of people and who are either friends or enemies with each other, the number of illegitimate committees can be calculated from the degree distribution of friends in the same way as we did above.
Any graph can be thought of as representing friendships of people, where edges connect friends. This cute puzzle tells us that the sum of the number of 3-cliques and 3-anti-cliques depends only on the degree distribution of the graph.
As a non mathematical comment, the above rule for legitimate committees is not a bad idea. In such a committee there is no reason for two people to gang up on the third one. Besides, if at some point in time all pairs of friends switch to enemies and vice versa, the committees will still be legitimate.
In 1976 I was about to become a student in the math department at Moscow State University. As an IMO team member I was accepted without entrance exams, but all of my other classmates had to take the exams. There were four exams: written math, oral math, physics, and an essay.
The written math exam was the first, and here are the problems. I want my non-Russian readers to see if they notice anything peculiar about this exam. Can you explain what is peculiar, and what might be the hidden agenda?
Problem 1. Solve the equation
Problem 2. Solve the inequality
Problem 3. Consider a right triangle ABC with right angle C. Angle B is 30° and leg CA is equal to 1. Let D be the midpoint of the hypotenuse AB, so that CD is a median. Choose F on the segment BC so that the angle between the hypotenuse and the line DF is 15°. Find the area of CDF. Calculate its numeric value with 0.001 precision.
Problem 4. Three balls, two of which are the same size, are tangent to the plane P, as well as to each other. In addition, the base of a circular cone lies on the plane P, and its axis is perpendicular to the plane. All three balls touch the cone from the outside. Find the angle between a generatrix of the cone and the plane P, given that the triangle formed by the points of tangency of the balls and the plane has one angle equal to 150°.
Problem 5. Let r < s < t be real numbers. If you set y equal to any of the numbers r, s or t in the equation x^{2} − (9 − y)x + y^{2} − 9y + 15 = 0, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that −1 < r < 1.
Let me describe some background to this exam. Applicants who solve fewer than two problems fail the exam and are immediately rejected. People who solve two or three problems are given 3 points. Four problems earn 4 points, and five problems earn 5 points.
If you still do not see the hidden agenda, here is another clue. People who get 5 points on the first exam and, in addition, have a gold medal from their high school (that means all As) are admitted right after the first exam. For the others, if they do not fail any of the exams, points are summed up with their GPAs to compute their scores. The so-called half-passing score is then calculated. Scores strictly higher than the half-passing score qualify applicants for admission. However, there are too many applicants for the available openings with at least the half-passing score. As a result only some people with exactly the half-passing score are accepted, at the discretion of the department.
Now my readers have enough information to figure out the hidden agenda behind that particular exam.
We worked for several years with RSI where we supervised summer math research projects by high school students. Now, we've started an additional program at MIT's math department called PRIMES, where local high school students do math research during the academic year. In this essay we would like to discuss what makes a good math research project for a high school student.
A doable project. The project should not be believed to be extremely difficult to yield at least results. It is very discouraging for an aspiring mathematician not to produce anything during their first project.
An accessible beginning. The student should be able to start doing something original soon after the start of the project. After all, they don't come to us for coursework, but for research.
Flexibility. It is extremely important to offer them a project that is adjustable; it should go in many directions with many different potential kinds of results. Since we do not know the strength of incoming students in advance, it is useful to have in mind both easier and harder versions of the project.
Motivation. It is important for the project to be well motivated, which means related to other things that have been studied and known to be interesting, to research of other people, etc. Students get more excited when they see that other people are excited about their results.
A computer component. This is not a must for a good project. But modern mathematics involves a lot of computation and young students are better at it than many older professors. Such a project gives young students the opportunity to tackle something more senior people are interested in but might not have enough computer skills to solve. In addition, through computer experiments students get exposed to abstract notions (groups, rings, Lie algebras, representations, etc.) in a more "hands-on" way than when taking standard courses, and as a result are less scared of them.
A learning component. It is always good when a project exposes students to more advanced notions.
The student should like their project. This is very difficult to accomplish when projects are chosen in advance before we meet the students. However, we try to match them to great projects by using the descriptions they give of their interests on their applications. It goes without saying that mentors should like their project too.
Having stated the desired properties of a good project, let us move on to giving examples: bad projects and good projects. We start with a bad one:
Prove that the largest power of 2 that doesn't contain 0 is 2^{86}.
The project satisfies only one requirement: it contains a computer component. Otherwise, it doesn't have an accessible beginning. It is not very flexible: if the student succeeds, the long-standing conjecture will be proven; if s/he doesn't, there is not much value in intermediate results. The question is not very interesting. The only motivation is that it has been open for a long time. Also, there is not much to learn. Though, almost any theoretical question can be made flexible. We can start with the question above and change its direction to make it more promising and enticing.
Another bad example is a project where the research happens after the programs are written. This is bad because it is difficult to estimate the programming abilities of incoming students. It doesn't have an accessible beginning and there is no flexibility until the programming part is finished. If the student can't finish the programming quickly, s/he will not have time to look at the results and produce conjectures. For example, almost any project in studying social networks may fall into this category:
Study an acquaintance graph for some epic movies or fiction, for example Star Wars or The Lord of the Rings. In this graph people are vertices and two people are connected by an edge if they know each other. The project is to compare properties of such graphs to known properties of other social networks.
Though the networks in movies are much smaller than other networks that people study, the amount of programming might be substantial. This project can be a good project for a person with a flexible time frame or a person who is sure in advance that there will be enough time for him/her to look at the data.
Now on to an example of a good project. Lynnelle Ye and her mentor, Tirasan Khandhawit, chose to analyze the game of Chomp on graphs during RSI 2009.
Given a graph, on each turn a player can remove an edge or a vertex together with all adjacent edges. The player who doesn't have a move loses. This game was previously solved for complete graphs and forest graphs, so the project was to analyze the game for other types of graphs.
It is clear how to analyze the game for any particular new graph. So that could be a starting point providing an accessible beginning. After that the next step could be to analyze other interesting sets of graphs. The flexibility is guaranteed by the fact that there are many sets of graphs that can be used. In addition, the project entails learning some graph theory and game theory. And the project has a computational component.
Lynnelle Ye successfully implemented this project and provided a complete analysis of complete n-partite graphs for arbitrary n and all bipartite graphs. She also gave partial results for odd-cycle pseudotrees. The paper is available at the arxiv. Not surprisingly, Lynelle got fourth place in the Intel Science Talent Search and second place in the Siemens Competition.
Suppose that we choose all families with two children, such that one of them is a son named Luigi. Given that the probability of a boy to be named Luigi is p, what is the probability that the other child is a son?
Here is a potential "solution." Luigi is a younger brother's name in one of the most popular video games: Super Mario Bros. Probably the parents loved the game and decided to name their first son Mario and the second Luigi. Hence, if one of the children is named Luigi, then he must be a younger son. The second child is certainly an older son named Mario. So, the answer is 1.
The solution above is not mathematical, but it reflects the fact that children's names are highly correlated with each other.
Let's try some mathematical models that describe how the parents might name their children and see what happens. It is common to assume that the names of siblings are chosen independently. In this case the first son (as well as the second son) will be named Luigi with probability p. Therefore, the answer to the puzzle above is (2-p)/(4-p).
The problem with this model is that there is a noticeable probability that the family has two sons, both named Luigi.
As parents usually want to give different names to their children, many researchers suggest the following naming model to avoid naming two children in the same family with the same name. A potential family picks a child's name at random from a distribution list. Children are named independently of each other. Families in which two children are named the same are crossed out from the list of families.
There is a problem with this approach. When we cross out families we may disturb the balance in the family gender distributions. If we assume that boys' and girls' names are different then we will only cross out families with children of the same gender. Thus, the ratio of different-gender families to same-gender families will stop being 1/1. Moreover, it could happen that the number of boy-boy families will differ from the number of girl-girl families.
There are several ways to adjust the model. Suppose there is a probability distribution of names that is used for the first son. If another son is born, the name of the first son is crossed out from the distribution and following that we proportionately adjust the probabilities of all other names for this family. In this model the probability of naming the first son by some name and the second son by the same name changes. For example, the most popular name's probability decreases with consecutive sons, while the least popular name's probability increases.
I like this model, because I think that it reflects real life.
Here is another model, suggested by my son Alexey. Parents give names to their children independently of each other from a given distribution list. If they give the same name to both children the family is crossed-out and replaced with another family with children of the same genders. The advantage of this model is that the first child and the second child are named independently from each other with the same probability distribution. The disadvantage is that the probability distribution of names in the resulting set of families will be different from the probability distribution of names in the original preference list.
I would like my readers to comment on the models and how they change the answer to the original problem.
I am reading the book Eat to Live by Joel Fuhrman. It contains a formula that as a math formula doesn't make any sense. But as an idea, it felt like a revelation. Here it is:
HEALTH = NUTRIENTS/CALORIES
The idea is to choose foods that contain more nutrients per calorie. The formula doesn't make sense for many reasons. Taken to its logical conclusion, the best foods would be vitamins and tea. The formula doesn't provide bounds: it just emphasizes that your calories should be nutritious. However, too few calories — nutritious or not — and you will die. And too many calories — even super nutritious — are still too many calories. In addition the formula doesn't explain how to balance different types of nutrients.
Let's see why it was a revelation. I often crave bananas. I assumed that I need bananas for some reason and my body tells me that. Suppose I really need potassium. As a result I eat a banana, which contains 800 milligrams of potassium and adds 200 calories as a bonus. If I ate spinach instead, I would get the same amount of potassium at a price of only 35 calories.
The book suggests that if I start eating foods that are high in nutrients, I will satisfy my need for particular nutrients, and my cravings will subside. As a result I will not want to eat that much. If I start my day eating spinach, that might eliminate my banana desire.
I've been following an intuitive eating diet. I am trying to listen to my body hoping that my body will tell me what is better for it. It seems that my body sends me signals that are not precise enough. It's not that my body isn't communicating with me, but it is telling me "potassium" and all I hear is "bananas." What I need to do is use my brain to help me decipher what my body really, really wants to tell me.
As Dr. Fuhrman puts it, we are a nation of overfed and malnourished people. But Fuhrman's weight loss plan is too complicated and time-consuming for me, so I designed my own plan based on his ideas:
I will start every meal with vegetables, as they are the most nutritious. I hope that vegetables will provide the nutrients I need. That in turn will make me less hungry by the next meal, at which time I'll take in fewer calories. I will report to my readers whether or not my plan works. I'm off to shop for spinach. Will I ever love it as much as bananas?
The Cookie Monster is a peculiar creature that appeared in The Inquisitive Problem Solver (Vaderlind, Guy & Larson, MAA, P34). Presented with a set of cookie jars, the Cookie Monster will try to empty all the jars with the least number of moves, where a move is to select any subset of the jars and eat the same number of cookies from each jar in the subset.
Even an untalented Cookie Monster would be able to empty n jars in n moves: to fulfill this strategy the Monster can devour all the cookies of one jar at a time. If the Monster is lucky and some jars have the same number of cookies, the Monster can apply the same eating process to all these identical jars. For example, if all the jars have the same number of cookies, the Monster can gulp down all of them in one swoop.
Now, let us limit our discussion to only cases of n non-empty jars that contain distinct numbers of cookies. If indeed all the numbers are distinct, can the Monster finish eating faster than in n moves?
The answer depends on the actual number of cookies in each jar. For example, if the number of cookies in jars are different powers of 2, then even the most talented Monsters can't finish faster than in n steps. Indeed, suppose the largest jar contains 2^{N} cookies. That would be more than the total number of cookies in all the other jars together. Therefore, any strategy has to include a step in which the Monster only takes cookies from the largest jar. The Monster will not jeopardize the strategy if it takes all the cookies from the largest jar in the first move. Applying the induction process, we see that we need at least n steps.
On the other hand, sometimes the Monster can finish the jars faster. If 2^{k}−1 jars contain respectively 1, 2, 3, …, 2^{k}−1 cookies, the Cookie Monster can empty them all in k steps. Here is how. For its first move, the Monster eats 2^{k-1} cookies from each of the jars containing 2^{k-1} cookies or more. What remains are 2^{k-1}−1 pairs of identical non-empty jars containing respectively 1, 2, 3, …, 2^{k-1}−1 cookies. The Monster can then continue eating cookies in a similar fashion, finishing in k steps. For instance, for k=3 the sequences of non-empty jars are: 1,2,3,4,5,6,7 → 1,1,2,2,3,3 → 1,1,1,1 → all empty.
Now we would like to prove a theorem that shows that the example above is the lowest limit of moves even for the most gifted Cookie Monsters.
Theorem. If n non-empty jars contain distinct numbers of cookies, the Cookie Monster will need at least ⌈log_{2}(n+1)⌉ steps to empty them all.
Proof. Suppose that n jars contain distinct numbers of cookies and let f(n) be the number of distinct non-empty jars after the first move of the Cookie Monster. We claim that n ≤ 2f(n)+1. Indeed, after the first move, there will be at least n − 1 non-empty jars, but there cannot be three identical non-empty jars. That means, the number of jars plus 1 can't decrease faster than twice each time.
Now here is something our readers can play with. Suppose a sequence of numbers represents the number of cookies in the jars. Which sequences are interesting, that is, which can provide interesting solutions for the Cookie Monster problem?
I read an interesting article on the paradoxes involved in allocating seats for the Congress. The problem arises because of two rules: one congressperson has one vote, and the number of congresspeople per state should be proportional to the population of said state.
These two rules contradict each other, because it is unrealistic to expect to be able to equally divide the populations of different states. Therefore, two different congresspeople from two different states may represent different sizes of population.
Let me explain how seats are divided by using as an example a country with three states: New Nevada (NN), Massecticut (MC) and Califivenia (C5). Suppose the total number of congresspeople is ten. Also suppose the population distribution is such that the states should have the following number of congresspeople: NN — 3.33, MC — 3.34 and C5 — 3.33. As you know states generally do not send a third of a congressperson, so the situation is resolved using the Hamilton method. First, each state gets an integer portion of the seats. In my example, each state gets three seats. Next, if there are seats left they are allocated to states with the largest remainders. In my example, the remainders are 0.33, 0.34 and 0.33. As Massecticut has the largest reminder it gets the last seat.
This is not fair, because now each NN seat represents a larger population portion than each MC seat. Not only is this not fair, but it can also create some strange situations. Suppose there have been population changes for the next redistricting: NN — 3.0, MC — 3.4 and C5 — 3.6. In this case, NN and MC each get 3 seats, while C5 gets the extra seat for a total of 4. Even though MC tried very hard and succeeded in raising their portion of the population, they still lost a seat.
Is there any fair way to allocate seats? George Szpiro in his article suggests adding fractional congresspersons to the House of Representatives. So one state might have three representatives, but one of those has only a quarter of a vote. Thus, the state's voting power becomes 2 1/4.
We can take this idea further. We can use the Hamilton method to decide the number of representatives per state, but give each congressperson a fractional voting power, so the voting power of each state exactly matches the population. This way we lose one of the rules that each congressperson has the same vote. But representation will be exact. In my first example, NN got three seats, when they really needed 3.33. So each congressperson from New Nevada will have 1.11 votes. On the other hand MC got four seats, when they needed 3.34. So each MC representative gets 0.835 votes.
Continuing with this idea, we do not need congresspeople from the same state to have the same power. We can give proportional voting power to a congressperson depending on the population in his/her district.
Or we can go all the way with this idea and lose the districts altogether, so that every congressperson's voting power will be exactly proportionate to the number of citizens who voted for him/her. This way the voting power will reflect the popularity — rather than the size of the district — of each congressperson.
Lionel Levine invented a new hat puzzle.
The sultan decides to torture his hundred wise men again. He has an unlimited supply of red and blue hats. Tomorrow he will pile an infinite, randomly-colored sequence of hats on each wise man's head. Each wise man will be able to see the colors of everyone else's hats, but will not be able to see the colors of his own hats. The wise men are not allowed to pass any information to each other.
At the sultan's signal each has to write a natural number. The sultan will then check the color of the hat that corresponds to that number in the pile of hats. For example, if the wise man writes down "four," the sultan will check the color of the fourth hat in that man's pile. If any of the numbers correspond to a red hat, all the wise men will have their heads chopped off along with their hats. The numbers must correspond to blue hats. What should be their strategy to maximize their chance of survival?
Suppose each wise man writes "one." The first hat in each pile is blue with a probability of one-half. Hence, they will survive as a group with a probability of 1 over 2^{100}. Wise men are so wise that they can do much better than that. Can you figure it out?
Inspired by Lionel, I decided to suggest the following variation:
This time the sultan puts two hats randomly on each wise man's head. Each wise man will see the colors of other people's hats, but not the colors of his own. The men are not allowed to pass any info to each other. At the sultan's signal each has to write the number of blue hats on his head. If they are all correct, all of them survive. If at least one of them is wrong, all of them die. What should be their strategy to maximize their chance of survival?
Suppose there is only one wise man. It is clear that he should write that he has exactly one blue hat. He survives with the probability of one-half. Suppose now that there are two wise men. Each of them can write "one." With this strategy, they will survive with a probability of 1/4. Can they do better than that? What can you suggest if, instead of two, there is any number of wise men?
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Today I saw an ad — "A printer for sale" — handwritten. Hmm.
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What do you call a motherboard on your spouse's computer?
The motherboard-in-law.
In a puzzle book by Mari Berrondo (in Russian), I found the following logic problem:
Alfred, Bertran and Charles are asked about their profession. One of them always lies; another one always tells the truth; and the third one [who I will refer to as a "half-liar"] sometimes lies and sometime tells the truth. Here are their answers:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The solution in the book is the following. As Alfred is right about what Bertran would say, Alfred can't be a liar. If Alfred is a half-liar then the other two people would give the opposite statements, since one will be a truth-teller and the other a liar. But they both say that Alfred is a piano-tuner, therefore Alfred must be a truth-teller. Hence, Alfred's statement about everyone's profession must be the truth. Now we know that Charles is an insurance agent. As Charles confirms that, thus telling the truth in this instance, we recognize that he must be a half-liar. The answer to the problem is that the half-liar is an insurance agent.
But I have a problem with this problem. You see, a liar can say many things. He can say that he is a conductor, a mathematician, a beekeeper or whatever. So there is no way of knowing what a person who decides to lie can say. Let's just analyze the statement by Alfred: "Concerning Bertran, if you ask him, he will tell you that he is a painter."
If Alfred tells the truth about what Bertran would say, he needs to know for sure that Bertran will say that he is a painter. Hence, Bertran must be a truth-teller and a painter. If Alfred lies, he needs to be sure that Bertran won't say that he is a painter. So Bertran must be either a truth-teller and not a painter, or a liar and a painter. Bertran can't be a half-liar, because a half-liar can say that he is a painter as well as he can say something else, no matter what his real profession.
There is one interesting aspect of this that many people overlook. There are different types of people who are half-liars. In some books half-liars are introduced as people who, before making a statement, flip a coin to decide whether to lie or to tell the truth. Such a person needs to know in advance exactly what other people are saying, in order to construct a statement about what those people might say that corresponds to the coin flip. On the other hand, other types of half-liars exist. One half-liar can say something and then see later whether it is true. If Alfred is a half-liar who doesn't care in advance about the truth of his statement, he can say that Bertran will claim that he is a painter.
I leave it to my readers to finish my analysis and see that the problem doesn't have a solution. To end my essay on a positive note, I decided to slightly change the problem, so that there is no contradiction. In the same setting:
Bertran: I am a painter, Alfred is a piano-tuner, Charles is a decorator.
Alfred: I am a doctor, Charles is an insurance agent. Concerning Bertran, if you ask him, he will tell you that he is not a painter.
Charles: Alfred is a piano-tuner. Bertran is a decorator, and I am an insurance agent.
What is the profession of the half-liar?
The data from annual surveys carried out by the American Mathematical Society shows the same picture year after year: the percentage of females in different categories decreases as the category level rises. For example, here is the data for 2006:
Category | Percentage of Women |
---|---|
Graduating Math Majors | 41 |
PhDs Granted | 32 |
Fresh PhD hires in academic jobs | 27 |
Full-time Faculty | 27 |
Full-time tenured or tenure-track faculty | 12 |
The high percentage of female math majors means that a lot of women do like mathematics. Why aren't women becoming professors of mathematics? In the picture to the left, little Sanya fearlessly took her first integral. I hope, even as an adult, she will never be afraid of integrals.
I am one of the organizers of the Women and Mathematics Program at the Institute for Advanced Study at Princeton In 2009 we had a special seminar devoted to discussing this issue. Here is the report of our discussion based on the notes that Rajaa Al Talli took during the meeting.
Many of us felt, for the following three reasons, that the data doesn't represent the full picture.
First, the different stages correspond to women of different ages; thus, the number of tenured faculty should be compared, not to the number of current math majors, but rather to women who majored in math many years ago. The percentage of female PhDs in mathematics has been increasing steadily for the past several years. As a result, we expect an eventual increase in the number of full-time female faculty.
Second, international women mathematicians might be having a great impact on the numbers. Let's examine a hypothetical situation. If many female professors come to the US after completing their studies in other countries, it would be logical to assume that they would raise the numbers. But since the numbers are falling, we might be losing more females than we think. Or, it could be the opposite: international graduate students complete a PhD in mathematics in the USA and then go back to their own countries. In this case we would be losing fewer females to professorships than the numbers seem to suggest. Unfortunately, we can't really say which case is true as we do not know the data on international students and professors.
Third, many women who major in mathematics also have second majors. For example, the women who have a second major in education probably plan to become teachers instead of pursuing an academic career. It would be interesting to find the data comparing women who never meant to have careers in science with those women who left because they were discouraged. If we are losing women from the sciences because they decide not to pursue scientific careers, then at least that is their choice.
It is also worth studying why so few women are interested in careers in mathematics in the first place. Changing our culture or applying peer pressure in a different direction might change the ambitions of a lot of people.
We discussed why the data in the table doesn't represent the full picture. On the other hand, there are many reasons why women who can do mathematics and want to do mathematics might be discouraged from pursuing an academic career:
Our group proposed many solutions to help retain women in mathematics:
At the end of our meeting, everyone accepted Ingrid Daubechies' proposal that we do the following:
Each woman in mathematics should take as her responsibility the improvement of the mathematical environment in which she works. If every woman helps change what's going on in her university or the school where she teaches, that will help solve the problem on the larger scale.
The book How to Drive Your Man Wild in Bed by Graham Masterton has a chapter on how to choose a lover. It highlights red flags for men who need to be approached with caution. There is a whole list of potentially bad signs, including neglecting to shower in the previous week and talking only about himself.
The list of bad features also includes professions to avoid. Can you guess the first profession on the list? OK, I think you should be able to meta-guess given the fact that I am writing about it. Indeed, the list on page 64 starts:
Avoid, on the whole, mathematicians…
I am an expert on NOT avoiding mathematicians: in fact, I've married three of them and dated x number of them. That isn't necessarily because I like mathematicians so much; I just do not meet anyone else.
When I was a student I had a theory that mathematicians are different from physicists. My theory was based on two conferences on mathematical physics I attended in a row. The first one was targeted for mathematicians and the second for physicists. The first one was very quiet, and the second one was all boozing and partying. So I decided that mathematicians are introverts and physicists are extroverts. I was sure then that my second husband chose a wrong field, because he liked booze and parties.
By now, years later, I've met many more mathematicians, and I have to tell you that they are varied. It is impossible and unfair to describe mathematicians as a type. One mathematician even became the star of an erotic movie. I write this essay for girls who are interested in dating mathematicians. I am not talking about math majors here, I am talking about mathematicians who do serious research. Do I have a word of advice?
I do have several words of caution. While they don't apply to all mathematicians, it's worth keeping them in mind.
First, there are many mathematicians who, like my first husband, are very devoted to mathematics. I admire that devotion, but it means that they plan to do mathematics on Saturday nights and prefer to spend vacation at their desks. If they can only fit in one music concert per year, it is not enough for me. Of course, this applies to anyone who is obsessed by his work.
Second, there are mathematicians who believe that they are very smart. Smarter than many other people. They expand their credibility in math to other fields. They start going into biology, politics and relationships with the charisma of an expert, when in fact they do not have a clue what they are talking about.
Third, there are mathematicians who enjoy their math world so much that they do not see much else around them. The jokes are made about this type of mathematician:
What is the difference between an extroverted mathematician and an introverted one? The extroverted one looks at your shoes, rather than at his own shoes.
Yes, I have met a lot of mathematicians like that. Do you think that their wives complain that their husbands do not notice their new haircuts? No. Such triviality is not worth mentioning. Their wives complain that their husbands didn't notice that the furniture was repossessed or that their old cat died and was replaced by a dog. My third husband was like that. At some point in my marriage I discovered that he didn't know the color of my eyes. He didn't know the color of his eyes either. He wasn't color-blind: he was just indifferent. I asked him as a personal favor to learn the color of my eyes by heart and he did. My friend Irene even suggested creating a support group for the wives of such mathematicians.
While you need to watch out for those traits, there are also things I like about mathematicians. Many mathematicians are indeed very smart. That means it is interesting to talk to them. Also, I like when people are driven by something, for it shows a capacity for passion.
Mathematicians are often open and direct. Many mathematicians, like me, have trouble making false statements. I stopped playing —Mafia— because of that. I prefer people who say what they think and do not hold back.
There is a certain innocence among some mathematicians, and that reminds me of the words of the Mozart character in Pushkin's poetic drama, Mozart and Salieri: —And genius and villainy are two things incompatible, aren't they?— I feel this relates to mathematicians as well. Many mathematicians are so busy understanding mathematics, they are not interested in plotting and playing games.
Would I ever date a mathematician again? Yes, I would.
33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?
The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.
Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.
First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.
As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.
Let's look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.
Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.
We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.
Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.
If we continue this, we'll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.
We get two interesting new sequences out of this. The sequence of the fastest rider's speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.
The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.
I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.
One day we may all face the necessity of hiring a lawyer. If the case is tricky the lawyer must be smart and inventive. I am collecting puzzles to give to a potential lawyer during an interview. The following puzzle is one of them. It was given at the second Euler Olympiad in Russia:
At a local Toyota dealership, you are allowed to exchange brand new cars. You can exchange three Camrys for one Prius and one Avalon, and three Priuses for two Camrys and one Avalon. Assuming an unlimited supply of cars at the dealership, can collector Vasya exchange 700 Camrys for 400 Avalons?
The beauty of this puzzle is that the answer I may find acceptable from a mathematician is not the same as I want from my future lawyer.
Have I intrigued you? Get to work and send me the solutions.
I am sitting in front of my computer and scheming, or, more precisely, scamming. I am inventing scams as a way of raising awareness of how probability theory can be used for deception.
My first scam is my lottery project. Suppose I create and run a private lottery. I will award minor payments to some participants, while promising a grand prize of one hundred million dollars. However, there will be a very small probability that anyone will win the big payout. My plan is to live lavishly on my proceeds, hoping no one ever wins the big ticket.
The beauty of this scheme is that nobody will complain until someone scores the top prize. After all, everyone has been receiving what I promised, and no one realizes my fraud. If nobody wins the big award until I retire, I will have built my life style on deception without having been caught.
Suppose someone wins the hundred million dollars. Oops. I am in big trouble. On the other hand, maybe I can avoid jail time. I could tell the winner that the money is gone and if s/he complains to the police, I will declare bankruptcy and we will all lose. Alternatively, I can suggest a settlement in exchange for silence. For example, we could share future proceeds. Probability theory will help me run this lottery with only a small chance of being exposed.
But even a small chance of failure will cause me too much stress, so I have come up with an idea for another scam. I will write some complicated mathematical formulas with which to persuade everyone that global warming will necessarily produce earthquakes in Boston in the near future. Then I'll open an insurance company and insure everyone against earthquakes. As I really do not expect earthquakes in my lifetime, I can spend the money. I'll just need to keep everyone scared about earthquakes. This time I can be sure that I won't be caught as no one will have a reason to complain. The only danger is that someone will check my formulas and prove that I used mathematics to lie.
Perhaps I need a scam that covers up the lie better. Instead of inventing an impossible catastrophe, I need to insure against a real but rare event. Think Katrina. I collect the money and put aside money for payouts and pocket the rest. But I actually tweak my formulas and put aside less than I should, boosting my bank account. I will be wealthy for many years, until this event happens. I might die rich but if this catastrophe happens while I'm still alive, I'll declare bankruptcy.
Though I was lying to everyone, I might be able to avoid jail time. I might be able to prove that it was an honest mistake. Mathematical models include some subjective parameters; besides, everyone believes that nature is unpredictable. Who would ever know that I rigged my formulas in my favor? I can claim that the theory ended up being more optimistic than reality is. Who could punish me for optimism?
Maybe I can be accused of lying if someone proves that I knew that the optimistic model doesn't quite match the reality. But it is very difficult for the courts to punish a person for a math mistake.
When I started writing this essay, I wanted to write about the financial crisis of 2008. I ended up inventing scams. In a way, I did write about the financial crisis. My scams are simplified versions of what banks and hedge funds did to us. Will we ever see someone punished?
Most movies related to mathematics irritate me because of simplifications. I especially do not like when a movie pretends to be intelligent and then dumbs it down. I recently watched the Spanish movie Fermat's Room, which, as you may guess, annoyed me several times. In spite of that I enjoyed it very much.
The movie opens with people receiving invitations to attend a meeting for geniuses. To qualify for the meeting they need to solve a puzzle. Within ten days, they must guess the order underlying the following sequence: 5, 4, 2, 9, 8, 6, 7, 3, 1. Right away, at the start of the movie, I was already annoyed because of the simplicity of the question. You do not have to be a genius to figure out the order, not to mention how easy it would be to plug this sequence into the Online Encyclopedia of Integer Sequences to find the order in five minutes.
The participants were asked to hide their real names, which felt very strange to me. All famous puzzle solvers compete in puzzle championships and mystery hunts and consequently know each other.
The meeting presumably targets the brightest minds and promises to provide "the greatest enigma." During the meeting they are given seven puzzles to solve. All of them are from children's books and the so-called "greatest enigma" could easily be solved by kids. Though I have to admit that these were among the cutest puzzles I know. For example:
There are three boxes: one with mint sweets, the second with aniseed sweets, and the last with a mixture of the two. The boxes are labeled, but all the labels are wrong. What is the minimum number of sweets you need to taste to correctly re-label all the boxes?
Another of the film's puzzles includes a light bulb in a room and three switches outside, where you have to correctly find the switch that corresponds to the bulb, but you can only enter the room once. In another puzzle you need to get out of prison by deciding which of two doors leads to freedom. You are allowed to ask exactly one question to one of the two guards, one of whom is a truth-teller and the other is a liar.
The other four puzzles are similar to these three I have just described. To mathematicians they are not the greatest enigmas. They are nice material for a children's math club. For non-mathematicians, they may be fascinating. Certainly it's a good thing that such tasteful puzzles are being promoted to a large audience. But they just look ridiculous as "the greatest enigmas."
So what is it about this film that I so enjoyed?
The intensity of the movie comes from the fact that the people are trapped in a room that starts shrinking when they take more than one minute to solve a puzzle.
I well remember another shrinking room from Star Wars: A New Hope. When Princess Leia leads her rescuers to a room, it turns out to be a garbage compactor. The bad guys activate the compactor and two opposite walls start moving in. In contrast, Fermat's room is shrinking in a much more sophisticated way: all four walls are closing in. Each of the walls in the rectangular room is being pressured by an industrial-strength press. The walls in the corners do not crumble, but rather one wall glides along another. I was more puzzled by this shrinking room than I was by the math puzzles. I recommend that you try to figure out how this can be done before seeing the movie or its poster.
However, the best puzzle in the movie is the plot itself. Though I knew all the individual puzzles, what happened in between grabbed me and I couldn't wait to see what would happen next. I saw the movie twice. After the first time, I decided to write this review, so I needed to check it again. I enjoyed it the second time even better than the first time. The second time, I saw how nicely the plot twists were built.
Maybe I shouldn't complain about the simplicity and the familiarity of the puzzles. If they were serious new puzzles I would have started solving them instead of enjoying the movie. The film's weakness might be its strength.
I collect hats puzzles. A puzzle about hats that I hadn't heard before appeared on the Konstantin Knop's blog (in Russian):
The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat — for both of which he has an inexhaustible supply — on every wizard's head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan's signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide on a strategy to maximize the number of survivors. Suggest a strategy for them.
I'll start the discussion with a rather simple idea: Each wizard writes down a color randomly. In this case the expected number of survivors is 50. Actually, if each wizard writes "red", then the expected number of survivors is 50, too. Can you find a better strategy, with a greater expected number of survivors or prove that such a strategy doesn't exist?
As a bonus question, can you suggest a strategy that guarantees 50 survivors?
Now that you've solved that issue, here's my own variation of the problem.
The wizards are all very good friends with each other. They decide that executions are very sad events and they do not wish to witness their friends' deaths. They would rather die themselves. They realize that they will only be happy if all of them survive together. Suggest a strategy that maximizes the probability of them being happy, that is, the probability that all of them will survive.
The following two problems appeared together in Martin Gardner's Scientific American column in 1959.
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
Many people, including me and Martin Gardner, wrote a lot about Mr. Smith. In his original column Martin Gardner argued that the answer to the first problem is 1/3. Later he wrote a column titled "Probability and Ambiguity," where he corrected himself about Mr. Smith.
… the answer depends on the procedure by which the information "at least one is a boy" is obtained.
This time I would like to ignore Mr. Smith, as I wrote a whole paper about him that is now under consideration for publication at the College Mathematics Journal. I would rather get back to Mr. Jones.
Mr. Jones failed to stir a controversy from the start and was forgotten. Olivier Leguay asked me about Mr. Jones in a private email, reminding me that the answer to the problem about his children also depends on the procedure.
One of the reasons Mr. Jones was forgotten is that for many natural procedures the answer is 1/2. For example, the following procedures will produce an answer of 1/2:
There are many other procedures that lead to the answer 1/2. However, there are many procedures that lead to other answers.
Suppose I know Mr. Jones, and also know that he has two children. I meet Mr. Jones at a mall, and he tells me that he is buying a gift for his older daughter. Most probably I would assume that the other child is a daughter, too. In my experience, people who have a son and a daughter would say that they are buying a gift for "my daughter." Only people with two daughters would bother to specify that they are buying a gift for "my older daughter."
In some sense I didn't forget about Mr. Jones. I wrote about him implicitly in my essay Two Coins Puzzle. His name was Carl and he had two coins instead of two children.
My blog is getting more famous. Now I don't need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov's Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:
Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?
Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:
Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?
Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.
86 is conjectured to be the largest power of 2 not containing a zero. This simply stated conjecture has proven itself to be proof-resistant. Let us see why.
What is the probability that the nth power of two will not have any zeroes? The first and the last digits are non-zeroes; suppose that other digits become zeroes randomly and independently of each other. This supposition allows us to estimate the probability of 2^{n} not having zeroes as (9/10)^{k-2}, where k is the number of digits of 2^{n}. The number of digits can be estimated as n log_{10}2. Thus, the probability is about cx^{n}, where c = (10/9)^{2} ≈ 1.2 and x = (9/10)^{log102} ≈ 0.97. The expected number of powers of 2 without zeroes starting from the power N is cx^{N}/(1-x) ≈ 40 ⋅ 0.97^{N}.
Let us look at A007377, the sequence of numbers such that their powers of 2 do not contain zeros: 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86. Our estimates predicts 32 members of this sequence starting from 6. In fact, the sequence has 30 conjectured members. Similarly, our estimate predicts 2.5 members starting from 86. It is easy to check that the sequence doesn't contain any more numbers below 200 and our estimate predicts 0.07 members after 200. As we continue checking larger numbers and see that they do not belong to the sequence, the probability that the sequence contains more elements vanishes. With time we check more numbers and become more convinced that the conjecture is true. Currently, it has been checked up to the power 4.6 ⋅ 10^{7}. The probability of finding something after that is about 1.764342396 ⋅10^{-633620}.
Let us try to approach the conjecture from another angle. Let us check the last K digits of powers of two. As the number of possibilities is finite, these last digits eventually will start cycling. If we can show that all the elements inside the period contain zeroes, then we need to check the finite number of powers of two until this period starts. If we can find such K, we can prove the conjecture.
Let us look at the last two digits of powers of two. The sequence starts as: 01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04. As we would anticipate, it starts cycling. The cycle length is 20, and 90% of numbers in the cycle don't have zeroes.
Now let's continue to the last three digits. The period length is 100, and 19 of them either start with zero or contain zero. The percentage of elements in the cycle that do not contain zero is 81%.
The cycle length for the last n digits is known. It is 4 ⋅ 5^{n-1}. In particular the cycle length grows by 5 every time. The number of zero-free elements in these cycles form a sequence A181610: 4, 18, 81, 364, 1638, 7371, 33170. If we continue with our supposition that the digits are random, and study the new digits that appear when we move from the cycle of the last n digits to the next cycle of the last n+1 digits, we can expect that 9/10 of those digits will be non-zero. Indeed, if we check the ratio of how many numbers do not contain zero in the next cycle compared to the previous cycle, we get: 4.5, 4.5, 4.49383, 4.5, 4.5, 4.50007. All of these numbers are quite close to our estimation of 4.5. If this trend continues the portion of the numbers in the cycle that don't have zeroes tends to zero; however, the total of such numbers grows exponentially. We can even estimate that the expected growth is 4 ⋅ 4.5^{n-1}. From this estimation, we can derive the conjecture:
Conjecture. For any number N, there exists a power of two such that its last N digits are zero-free.
Indeed, the last N digits of powers of two cycle, and there are an increasing number of members inside that cycle that do not contain zeroes. The corresponding powers of two don't have zeroes among N rightmost digits.
So, how do we combine the two results? First, the expected probability of finding the power of two larger than 86 that doesn't contain zero is minuscule. And second, we most certainly can find a power of two that has as many zeroless digits at the end as we want.
To combine the two results, let us look at the sequences A031140 and A031141. We can deduce from them that for the power 103233492954 the first zero from the right occupies the 250th spot. The total number of digits of that power is 31076377936. So 250 is a tiny portion of the digits.
As time goes by we grow more and more convinced that 86 is the largest power of two without zeroes, but it is not at all clear how we can prove the conjecture or whether it can be proven at all.
My son, Sergei, suggested that I claim that I have a proof of this conjecture, but do not have enough space in the margin to fit my proof in. The probability that I will ever be shamed and disproven is lower than the probability of me winning a billion dollars in the lottery. Though then, if I do win the big bucks, I will still care about being shamed and disproven.
Janet Mertz encouraged me to find IMO girls and compare their careers to that of their teammates. I had always wanted to learn more about the legendary Lida Goncharova — who in 1962 was the first girl to win an IMO gold medal. So I located her, and after an interview, wrote about her. Only 14 years later, in 1976, did the next girl get a gold medal. That was me. I was ranked overall second and had 39 points out of 40.
As I did in the article about Lida, I would like to compare my math career to that of my teammates.
I got my PhD in 1988 and moved to the US in 1990. My postdoc at MIT in 1993 was followed by a postdoc at Bar-Ilan University. In 1996 I got a non-paying visiting position at Princeton University. In 1998 I gave up academia and moved to industry, accepting an offer from Bellcore. There were many reasons for that change: family, financial, geographical, medical and so on.
On the practical level, I had had two children and raising them was my first priority. But there was also a psychological element to this change: my low self-esteem. I believed that I wasn't good enough and wouldn't stand a chance of finding a job in academia. Looking back, I have no regrets about putting my kids first, but I do regret that I wasn't confident enough in my abilities to persist.
I continued working in industry until I resigned in January 2008, due to my feeling that I wasn't doing what I was meant to do: mathematics. Besides, my children were grown, giving me the freedom to leave a job I did not like and return to the work I love. Now I am a struggling freelance mathematician affiliated with MIT. Although my math blog is quite popular and I have been publishing research papers, I am not sure that I will ever be able to find an academic job because of my non-traditional curriculum vitae.
The year 1976 was very successful for the Soviet team. Out of nine gold medals our team took four. My result was the best for our team with 39 points followed by Sergey Finashin and Alexander Goncharov with 37 points and by Nikita Netsvetaev with 34 points.
Alexander Goncharov became a full professor at Brown University in 1999 and now is a full professor at Yale University. His research is in Arithmetic Algebraic Geometry, Teichmuller Theory and Integral Geometry. He has received multiple awards including the 1992 European Math Society prize. Sergey Finashin is very active in the fields of Low Dimensional Topology and Topology of Real Algebraic Varieties. He became a full professor at Middle East Technical University in Ankara, Turkey in 1998. Nikita Netsvetaev is an expert in Differential Topology. He is a professor at Saint Petersburg State University and the Head of the High Geometry Department.
Comparing my story to that of Lida, I already see a pattern emerging. Now I'm curious to hear the stories of other gold-winning women. I believe that the next gold girl, in 1984, was Karin Gröger from the German Democratic Republic. I haven't yet managed to find her, so can my readers help?
I wrote a series of essays about AMC competitions:
This essay is next in the series. Although it is not strictly about AMC, it should be useful during any test when you need to check your answers. There are several important rules which are helpful.
Rule 0. Checking is important. If wrong answers are punished, then correcting a mistake brings more points than solving a new problem. In addition, problems that were solved are often easier than problems yet to be solved, so finding a mistake might be faster than solving a new problem.
Rule 1. Your checking methods must be fast. The tests are generally timed. This means that in order to check your answers, you need to sacrifice your work on the next problem.
Rule 2. Customize how you check according to your strengths and weaknesses. For example, if you tend to jump to conclusions about what the question is going to be, and as a result answer your anticipated question instead of the one that is actually on the test, then when you are checking you should start reading the problem from the question. Or, if you usually make mistakes in geometry problems, you should allocate more time to geometry problems when you are checking. If you never make mistakes in arithmetic problems then you do not need to check those.
Rule 3. Mark problems that might need checking. If you do not have enough time to check all the problems, check only those you are not sure about.
Rule 4. Do not repeat your solution when you check. While solving the problem your brain often creates a pathway from start to finish. If on this pathway your brain decided to believe that two plus two is five, very often during checking, your brain will make the same mistake again. Because of that it is crucial to use other methods for checking than repeating your reasoning. In case you can't find a way to check your answers using a different method and have to repeat your reasoning, you should repeat it in a different order.
This rule is so important, that I am providing some methods to change your brain pathway when you are checking your answers.
Plug in. Plugging in the answer you found is much faster than finding it. Use this method whenever possible. It is perfect for problems like this one below from 2004 AMC10-A:
What is the value of x if |x – 1| = |x – 2|?
Plug in an intermediate result. Sometimes you can't plug in the answer, but you can plug in an intermediate result. In the following problem from 2004 AMC10-B you can plug in the number of nickels and dimes:
Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?
Calculate something else related to your answer. For example a negation. Here is a problem from 2004 AMC10-B:
How many two-digit positive integers have at least one 7 as a digit?
If you calculated the answer directly, to check it you may want to calculate the number of two-digit positive integers that do not contain 7.
Create an example. Sometimes you solve a problem by reasoning, but to check it you might create a particular example. Here is a problem from 2001 AMC10:
Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N?
If we denote the tens digit by a and the units digit by b, then N = 10a + b, P(N) = a*b, and S(N) = a + b. We get an equation a(b+1) = 10a, from which the answer is 9. To check the answer we do not need to repeat the reasoning. It is enough to check that 19 is the sum of the product of its digits plus the digits.
Here is another problem from 2001 AMC10:
Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?
The list of choices is: 6, 14, 21, 28, 42. Your solution might go like this: the product of three consecutive numbers is divisible by 6. Hence, n is divisible by 42. So, the answer must be 28. To check you might consider a product of three consecutive numbers: 5*6*7=210 and see that it is not divisible by 4, hence it is not divisible by 28.
Rule 5. Embrace the partial check. It is very important to check your answers fast. Sometimes you can gain speed if you do not check the problem completely, but check it partially. For example, you can check that your answer is one of the two correct answers. There are many methods for partial checking.
Try an example. Sometimes an example doesn't guarantee that your choice is correct, but it increases your confidence in your answer. Here is another problem from 2001 AMC10:
The sum of two numbers is S. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
The choices are: 2S + 3, 3S + 2, 3S + 6, 2S + 6, 2S + 12. You can reason that increasing each summand by 3, increases the sum by 6. After that doubling each summand increases the resulting sum twice, so the answer is 2S + 12. To check the answer you can use an example. Usually an example doesn't guarantee the confirmation of your answer, but it might help you eliminate some of the wrong answers. For example, if you choose zero and zero as your initial two numbers, then S = 0, and your transformation brings the result to 12, which confirms your answer 2S + 12. In this particular case, a very easy specific example excluded all the wrong answers.
Divisibility. Sometimes it is faster to calculate the remainder of the answer by some number.
For example, look at the following problem from 2003 AMC10:
What is the units digit of 13^{2003}?
The choices are 1, 3, 7, 8, 9. We can immediately say that the answer must be an odd number.
Approximation check. One important example of a partial check is an approximation check. By estimating an approximate answer you might exclude most of the wrong answers. Consider this problem from 2001 AMC12:
How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
The divisibilities by 3, 4 or 5 shouldn't correlate with each other. Approximately one third of those number are multiples of 3 and one quarter are multiples of 4. Let's say that one twelfth are multiples of both 3 and 4. Hence, we estimate the portion of numbers that are multiples of 3 or 4 as 1/3 + 1/4 – 1/12 = 1/2. We have about 1,000 such numbers. The number of numbers that are, in addition, not divisible by 5, are less than that. So out of the given choice of (A) 768, (B) 801, (C) 934, (D) 1067, (E) 1167, we can immediately confirm that the answer is among the first three.
The methods above can be useful even if you do not have multiple choices. But if you do…
Rule 6. Use given choices as extra information. In the previous examples you saw how to use a partial check to exclude some of the choices. Here is a specific example from 2006 AMC10-A of how to exclude choices:
What non-zero real value for x satisfies (7x)^{14} = (14x)^{7}?
The choices are: 1/7, 2/7, 1, 7, 14. If you solved the problem directly, to check it you can reason why other choices do not work. In this particular case it can be done very fast. 1/7 doesn't work because the left part of the equation becomes 1 when the right is clearly not. 1 and 7 do not work because the left part is odd and the right is even; 14 doesn't work because the left is clearly bigger than the right.
Rule 7. Use meta considerations. If you get into the mind of the designers you can better anticipate when you should check more thoroughly. Consider this problem from 2006 AMC10-A:
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
The most common mistake would be to assume that 12:59 supplies the largest sum, which is 17. But look at the choices: 17, 19, 21, 22, 23. When the designers are asking to find the largest number with some property, they assume that some students will make a mistake and chose a smaller number over a larger one. That means the designers would include this potential mistake among the choices. So the answer is extremely unlikely to be the smallest number on the list of choices. Thus, if you think the answer is 17, understanding how these problems are constructed should alert you to thoroughly check your answer. Indeed, the correct answer is 23 which corresponds to 9:59. Not surprisingly, it is the largest on the list of choices.
AMC 10/12 is coming on February 8 and HMMT on February 12. Happy checking.
I've been celebrating my 29th birthday for many years. Once, when I was actually 45 and wanted to have a big party, I invited everyone to the 5th anniversary of my 29th birthday.
Last week my son turned 29 and I realized that it is time to drop this beautiful, prime, evil, deficient, lazy-caterer number, that in addition is the largest power of two to have all different digits. No more celebrating 29.
For my next age, I picked 42. Not because it is the smallest abundant odious number, but rather because it is the answer to life, the universe and everything.
Thank you everyone who congratulated me on my birthday two days ago. For your information, from now on I am 42.
As I did for 2010 and for previous years, here are math-related puzzles from the MIT Mystery Hunt 2011.
Two more puzzles deserve a special mention for their nerdiness. My teammates loved them.
Tired of the same old sudoku? Here's an opportunity to try many variations of it. Thomas Snyder and Wei-Hwa Huang wrote a book called Mutant Sudoku. The authors are both Sudoku champions. I like the book because the authors are trying to bring everyone up to their level, rather than dumbing down their puzzles. So the book is not at all boring as are most Sudoku books.
The book contains about 180 fun puzzles. Look at the variety:
Wei-Hwa Huang kindly sent me this sample Thermo Sudoku puzzle from the book to use on my blog. The grey areas represent thermometers. Every particular thermometer has to have numbers in increasing order (not necessarily consecutive) starting from the bulb.
The second book by the same authors Sudoku Masterpieces: Elegant Challenges for Sudoku Lovers, is itself a masterpiece. With about 100 puzzles, there are fewer than in the first book, but there are more types of puzzles. As a consequence, you'll have less practice for each particular type, but more variety. In addition, as you can see from the cover, the second book is elegantly designed.
I bought both books and immediately started scribbling in the first one. My bad handwriting would seem so out of place in the beautiful second book that I have not even started working in it yet. Maybe I will give it as a gift to someone with better penmanship.
Two days ago I threw at my readers the following problem:
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
The purpose of throwing this problem was to discuss the nature of the implicit assumptions that we are asked to make when solving math problems, and the implicit assumptions we teach our children to make when we teach them to solve math problems. This is especially important for problems like this, that are phrased in terms of a situation in the real world. The real world is too complex to model all of; the great power of mathematics is that sufficiently idealized situations are predictable. But which idealizations are appropriate? How does one choose? How does one teach youngsters what to choose?
Before I get to the actual discussion, however, I want to re-throw this problem at my readers, in an effort to highlight what originally jumped out at me as being wrong with it.
Neglecting the effects of altitude, differential wind, acceleration, relativity, measurement error, finite size and non-superimposability of the planes, and the Earth's deviations from perfect sphericity,
- Find how much time it takes them to become 2000 miles apart, assuming that the planes are starting from Boston and the distance is measured as
- a straight line in 3-space.
- the shortest surface distance.
- How far from the closest pole may the starting point be located, so that the answer to the problem is "never"? Solve separately for
- the 3D distance.
- the shortest surface distance.
- What portion of the Earth's surface do the "never"-locations of the previous question occupy?
- under the 3D distance?
- under the shortest surface distance?
Hint: The easiest question is 2b.
My Jewish ex-father-in-law, Naum Bernstein, is 96 years old and is full of life. He has a joke for every situation. In the last decade he wrote several volumes of memoirs in Russian. One of the books was a collection of his favorite jokes and his explanations of them. I decided to retell some of the jokes from his selection.
An arithmetic teacher calls the student Rabinovich to the blackboard. "It is known that from 1 kilogram of sour cream you can make 200 grams of butter. Imagine, Rabinovich, that your father bought 2 kilograms of sour cream. How much butter can he make?"
"Five hundred grams," Rabinovich replies.
The teacher frowns, "Rabinovich, you do not know arithmetic!"
Rabinovich answers, "Sir, you do not know my father."
An astronomy teacher explains that in the future the Earth will lose its heat energy, continents will collide, and solar radiation will increase. In six billion years life will be extinct. A student looking really scared raises his hand and asks, "In how many years will life become extinct?"
"In about six billion years," the teacher repeats.
"Whew," says the student, "you got me so scared. I thought you said six million."
Two professors are chatting while watching a soccer game. The first one says, "They say that soccer players have their brains in their legs. So their heads are really empty."
"Not quite," the second professor replies. "The player on the right passed my exam yesterday."
The first professor expresses interest, so the second one elaborates. "As a rule, I ask two questions. If the student gives a correct answer to one of them, he passes."
"So, what did you ask that guy?"
"My first question was 'What color are red blood cells?' He answered 'Yellow.' That was an incorrect answer. The second question was 'How is sulfuric acid produced?' To this he replied, 'I do not know,' which was absolutely true, so he passed."
A Russian literature teacher asks a pupil, "Who wrote Eugene Onegin?" The pupil gets scared that he is being blamed for something and replies, "No, not me! I swear I didn't write it!" Everyone laughs. The teacher decides that the pupil disrupted the class on purpose and asks for his father to come by.
The father arrives and after the teacher explains what happened, the father says, "Maybe he is not guilty; maybe he really didn't write it. I doubt that he is capable of writing anything."
The teacher is stunned and later tells the whole story in the teachers' lounge to her colleagues. An astronomy teacher comes home and retells the story to her husband who works for the KGB. The husband comments, "Do not worry, we are on it. Three people already confessed to writing it."
A hardcore anti-Semite was dying. As he got weaker he made a last request. He wanted to convert to Judaism. Everyone was extremely surprised, but decided not to interfere. After the conversion, his wife summoned the courage to ask him what was going on. "Do you think you were mistaken, hating Jews all your life?"
"No," he replied happily, "But now with my death, the world will get rid of one more Jew."
An old Jew comes to a Rabbi and asks if he can shave his beard off, because his children think that he is old-fashioned. The Rabbi tells him that by Jewish law he is not allowed to shave. The old man turns to go home when he realizes that the Rabbi himself doesn't have a beard. He stops and asks, "Dear Rabbi, you just forbade me to shave my beard, but how come you are clean-shaven yourself?"
The Rabbi replies, "I didn't ask anyone's permission."
When Rabinovich came to a bureaucrat with a request, the bureaucrat replied, "Come back tomorrow." Rabinovich returned the next day and received the same reply. Rabinovich was very persistent and returned day after day.
Finally, the bureaucrat lost his patience and attacked Rabinovich, "This is outrageous! Don't you understand simple language? I keep telling you to come tomorrow and you keep coming today."
The communist committee of a supermarket in the USSR received a lot of complaints about the rudeness of their salespeople. The committee decided to improve the quality of service and provided special training in which salespeople were taught politeness. The training emphasized what to do in case a particular item was unavailable. The salespeople were supposed to politely explain that the item is temporarily unavailable and to offer a substitute.
The next thing one of their salespeople said to a customer was "I am very sorry, we are temporarily out of toilet paper. May I offer some sandpaper?"
There is panic in an apartment on the 13th floor. The wife recognizes the sound of her husband's approach, even though he was supposed to be on a business trip. The lover asks, "What should I do, honey?"
"What do people do in such cases? Jump out the window!"
"But we are on the 13th floor!"
"This is no time for superstition!"
A young man had smelly feet, plus he always forgot to change his socks. His girlfriend got tired of it and asked him to promise that he would always change his socks before coming to see her.
Next visit the young man smelled as bad as ever. Outraged, the girl said, "But you promised to change your socks!"
The young man answered, "I did as I promised."
"I don't believe you, you smell awful."
"I was sure you wouldn't believe me. Good thing I brought my dirty socks with me as proof."
At the 50th anniversary of a very happy couple, someone asked the husband for their secret. He said that right before the wedding they agreed that the husband would decide all the crucial and very important things, and the wife would be responsible for all minor decisions. "For example," he continued, "yesterday I decided that the US should withdraw their troops from Iraq, and my wife decided where to buy our vacation house."
Two long-time girlfriends meet after several years without being in touch. "How are your children?" asks one of them.
The other replies, "My daughter is fine, she married a nice young man, who is providing for her. He also helps her with chores and even brings her coffee in bed every morning."
"What about your son?"
"It's a disaster. I don't know what to do. He married a really lazy woman. Even though she's not working, she wants him to help her with the chores. Can you imagine that? She even dared to ask him to bring her coffee in bed every morning."
Two friends are walking along a street very late at night. Robbers attack them with guns, demanding their wallets. One of the friends asks the robbers, "Can you give me 30 seconds?" The robbers agree. He takes out $100 from his wallet and gives it to his friend, "Remember I owed you $100? I am paying back my debt in front of witnesses."
Life is a struggle. Before lunch with hunger, after lunch with sleepiness.
A mother says to her son, "Please, close the window — it's cold outside."
The son replies, "Do you think it will get warmer outside if I close the window?"
How are pessimists and optimists different from normal people?
A pessimist uses both a belt and suspenders, an optimist uses neither.
In a cemetery there is a beautiful monument with a picture of a bald, wrinkled old man. He is smiling, showing his perfect white teeth. His epitaph says:
Here lies Mr. X, who lived more than 100 years, lost his hair, became all wrinkled, but kept his perfect teeth. That is because he always used our company's toothpaste.
A nearby monument has a picture of an old toothless woman with beautiful, voluminous hair. The inscription explains which brand of shampoo she used.
Many other tombstones with ads are scattered throughout the cemetery. But in the middle there is a huge mausoleum with an inscription reading:
No one is buried here and no one ever will be, because his or her parents used condoms made by our company.
A Russian man marries an American woman. After a while he writes a letter home.
My wife must be very dirty. She showers every day.
Rabinovich was asked why he didn't attend the last committee meeting. He replied, "If I knew it was the last, I would certainly have come."
I stumbled upon the following problem in Mathematics Teacher v.73 (September 1980):
A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?
Ooh, boy!
Question for my readers: explain my reaction.
I have a leased Toyota Corolla, and I am happily enrolled in AutoCheck payments with Toyota's Financial Services. So I do not even look at my bills. Once I opened my bill and noticed that the requested payment was twice as high as I expected. I looked closer and the bill had a car tax included in it. I looked even closer and read that:
Your Current Payment Due will be automatically withdrawn from your checking or savings account on the above Payment Due Date or the next banking day.
I decided that everything was taken care of and continued my relaxed life. After several months I checked my bill again, and the car tax was still there. After more careful study of my bill I discovered that Toyota's "Current Payment Due" doesn't include my car tax. Obviously they assume that their definition of "Current Payment Due" is crystal clear to everyone.
I got worried about this delayed car tax payment and went online to pay it. I tried to make this payment, but Toyota's website rejected it. The website informed me that because I am enrolled in AutoCheck, I am not allowed to make separate online payments. I couldn't believe it: to do so, I would have to de-enroll first!
So I just wrote a check.
In one day my feelings for my Toyota Corolla were turned around. If their financial system is designed so stupidly, what can we say about their car designs? Suddenly the sound of my brakes and the squeak in my steering wheel worry me.
A year ago I posted a chessboard puzzle. Recently I stumbled on a September 2008 issue of "Math Horizons" where it was presented as a magic trick.
When the magician leaves the room, the trickees lay out eight coins in a row deciding which side is turned up according to their whim. They also think of a number between 1 and 8 inclusive. The magician's assistant then flips exactly one of the coins, before inviting the magician back in. The magician looks at the coins and guesses the number that the trickees thought of.
The magician's strategy can be derived from the solution to the chessboard puzzle. The assistant numbers the coins from zero to seven from left to right. Then s/he flips the coin so that the parity addition (XORing) of all the numbers corresponding to heads is the number that the magician needs to guess. For this trick to work, the number of coins needs to be a power of 2.
Andrey Zelevinsky posted (in Russian) a cool variation of this trick with two decks of cards.
The magician has two identical card decks and he is out of the room for now. A random person from the audience thinks of a card. Next, the audience chooses several cards from the first deck. Then the assistant adds one card from the second deck to the set of chosen cards, lays them on a table, and then invites the magician back. The magician looks at the cards on the table and guesses the card that was thought of.
Unlike in the coin trick above, the number of cards in the deck doesn't need to be a power of 2. This flexibility is due to the fact that the magician has two decks of cards, as opposed to one set of coins. Having the second deck is equivalent to the assistant in the coin trick being allowed to flip one or ZERO coins.
Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.
An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?
The other one is from 1995.
There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles' orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?
Perfidy is to parity as odious is to odd and evil is to even. As a reminder, odious numbers are numbers with an odd number of ones in their binary expansions. From here you can extrapolate what the evil numbers are and the fact that the perfidy of an integer is the parity of the number of ones in its binary expansion. We live in a terrible world: all numbers are perfidious.
So why are we writing about the perfidy of negative numbers? One would expect it to be a natural extension of the perfidy of positive numbers, but it turns out that the naive way of defining it doesn't work at all. Is there hope? Could negative numbers be innocent of evil and free of odiousness? Is zero an impenetrable bulwark against perfidy? Not quite, but something interesting does happen to evil as it tries to cross zero. Read on.
To define perfidy for negative numbers, let us study how perfidy behaves for positive numbers. It is easiest to think about the perfidies of power-of-two-sized chunks of non-negative integers at a time. Let us denote by T_{n} the string of perfidies of the integers from 0 to 2^{n}−1, with evil being zero and odious being 1. So T_{0} = 0, T_{1} = 01, T_{2} = 0110, T_{3} = 01101001, …. The recurrence relation governing the T_{n} is T_{n+1} = T_{n}T_{n}, where T is the bitwise negation of the string T, and juxtaposition is concatenation. The limit of this as n tends to infinity is the (infinite) sequence of perfidies of non-negative integers. This sequence is called the Thue-Morse sequence: 0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,….
So defining the perfidy of negative numbers is equivalent to extending the Thue-Morse sequence to the left. If we are to define "the" perfidy of negative numbers, that definition should preserve most of the properties of the Thue-Morse sequence after extension.
So, let's see. We asked around, and most people said that the binary expansion of a negative integer should be the binary expansion of its absolute value, but with a minus sign. Defining perfidy as parity of number of ones in this binary expansion corresponds to the following extended Thue-Morse sequence in which we mark values corresponding to negative indices with bold font: … 0, 1, 1, 0, 1, 1, 0, ….
One of the major properties of the Thue-Morse sequence is its fractal property: if you replace every zero of the Thue-Morse sequence by 0,1 and every one by 1,0, you will get the Thue-Morse sequence back. Clearly, our new extended sequence doesn't have this property.
Another set of properties for the Thue-Morse sequence, called avoidance properties, is a long list of patterns that the sequence avoids. For example, the Thue-Morse sequence doesn't contain any overlapping squares — patterns axaxa, where a is a character and x is a word. But you can see above, our first extension contains it. So this definition is wrong, not just once but twice (and two wrongs only make a right under very unusual circumstances). Perfidy is stymied by the cross-over from zero to minus one. Are negative numbers protected from the ravages of evil? (and odiousness?)
Unfortunately, there are many people, for example John Conway, who inadvertently extend the reach of perfidy by arguing that the binary expansion of a negative integer should be different. Indulge in a flight of fancy and imagine the binary expansion that consists of infinitely many ones to the left: …1111. What happens when you add 1 to it? The carry gets pushed infinitely far away, and you get …000000 — zero. So it is quite reasonable to let …1111 be the binary expansion of −1. Similarly, the string …1110 represents −2, …1101 represents −3, etc. Continuing this we see that the binary expansion of a negative integer −n is the bitwise negation of the binary expansion of n − 1 (including the leading zeros). This is called the Two's complement representation.
Why is two's complement a reasonable representation? Suppose you were trying to invent a binary notation for negative numbers, but you wanted to pursue uniformity by not using a minus sign. The problem is that the standard definition of the binary representation allows you to represent only positive numbers. But you can solve this problem with modular arithmetic: modulo any fixed N, every negative number is equivalent to some positive number (by adding enough multiples of N), so you can just represent it by representing that positive number. If you choose N to be a power of two, modding out by it is just truncation of the binary representation. If you let those powers of two tend to infinity, you get the two's complement representation described above.
Aside: When you are building a computer, uniformity is money, because special cases cost special transistors. The two's complement idea lets one build arithmetic units that just operate on positive numbers with some number of bits (effectively doing arithmetic modulo 2^{k}), and leave the question of negativeness to the choice of representatives of those equivalence classes.
If we take two's complement as the binary expansion of negative numbers, how will we define the perfidy? Is the number of ones in the infinite string …1111 corresponding to −1 even or odd?
We can't answer that question, but we know for every binary expansion of negative numbers the parity of the number of zeroes. Thus we can divide all negative integers in two classes with different perfidy. We just do not know which one is which.
Let us consider two cases. In the first case we consider a negative number odious if the number of zeroes in its binary expansion is odd. The corresponding extended Thue-Morse sequence is: … 0, 1, 1, 0, 0, 1, 1, 0, …. The negative half is the reflection of the classical Thue-Morse sequence. In the second case we consider a negative number odious if the number of zeroes in its binary expansion is even. The corresponding extended Thue-Morse sequence is: … 1, 0, 0, 1, 0, 1, 1, 0, …. The negative half is the bitwise negation of the reflection of the classical Thue-Morse sequence.
Can we say that one of the sequences is better than the other? Both of them respect the fractal property of the classical Thue-Morse sequence. Let us look at the avoidance properties. The avoidance properties are symmetric with respect to switching zeroes with ones and with respect to changing the direction of the sequence. Hence, the negation, the reflection, and the reflection of the negation of the Thue-Morse sequence will continue to respect these properties.
Thus, we only need to check the avoidance properties of the finite subsequences that span both negative and non-negative indices. We claim that for both definitions of perfidy, any finite middle subsequence of the extended Thue-Morse sequence occurs as a subsequence in the classical Thue-Morse sequence. So any avoidance properties that are true for the Thue-Morse sequence will also be true for both extensions.
Indeed, it is easy to show that the strings T_{2n} defined above are palindromes. So for the first definition of perfidy the string in the middle will be a substring of T_{2n}T_{2n} for some large n, and for the second definition a substring of T_{2n}T_{2n}. But the classical Thue-Morse sequence contains the subsequence T_{2n}T_{2n}T_{2n}T_{2n}T_{2n}T_{2n}T_{2n}T_{2n}. So whichever way we extend the Thue-Morse sequence to the left any finite middle part will always be a repetition of a piece in the classical Thue-Morse sequence. Thus, all the avoidance properties will hold.
We see that there are two logical ways to define perfidy for negative integers. There are two clear groups of numbers with the same perfidy, but which is called evil and which odious is interchangeable. So evil doesn't stop at zero after all, but at least it gets an identity crisis.
Last revised October 2013