Tanya Khovanova's Math Blog

The Symmetries of Things by John H. Conway, Heidi Burgiel, and Chaim Goodman-Strauss is out. It is a beautiful edition with great pictures.

The first chapter is very nicely written and might be recommended to high school and undergraduate students. It covers symmetries of finite and infinite 2D objects.

The second chapter adds color to the theory. For beautiful colorful pictures with symmetry, there are two symmetry groups: the group that preserves the picture while ignoring its coloring and the group that preserves the picture while respecting its coloring. The latter group is a subgroup of a former group. This second chapter discusses all possible ways to symmetrically color a symmetric 2D picture. The chapter then continues with a discussion of group theory. This chapter is much more difficult to read than the first chapter, as it uses a lot of notations. The pictures are still beautiful, though.

The third chapter is even more difficult to read and the notations become even heavier. This chapter discusses hyperbolic groups and symmetries of objects in the hyperbolic space. Then the chapter moves into 3D and 4D. I guess that some parts of the second and the third chapters are not meant for light reading; they should be considered more as reference materials.

Here are pictures of a musnub cube (multiplied snub cube), built by John H. Conway. It is an infinite Archimedean polyhedron. The description of it is on page 338 and the diagram is on page 339 of the book. This object was glued together from stars and squares. Each corner of the square is glued to a point of one star and to an inside corner of another star. Mathematically, a star is not a regular polygon. If you look at stars with your mathematical eye, each star in the picture is not just a star, but rather the union of a regular hexagon with 6 regular triangles. That means the list of the face sizes around each vertex of a musnub cube officially is represented as 6.3.4.3.3.

The second picture shows a finished musnub cube. You can’t really finish building an infinite structure. Right? It is finished in the sense that John Conway finished doing what he was planning to do: to construct a part of a musnub cube inside a regular triangular pyramid.

Did I mention that I like the pictures in The Symmetries of Things?

I am now a proud owner of a Zipcard. My old car died a horrible, screaming death recently and I decided to try to go carless as an experiment, hoping to save some money or to lose weight.

Zipcar is the modern way to rent a car. You reserve a car by the hour or by the day through the web, arrive at the site, swipe your card and drive away.

My biggest problem with being a Zipster is that the closest Zipcar location to my home is about one mile away. On second thought, I am losing weight.

But for other Zipsters, the biggest problem is that you have to return the car at the exact location where you picked it up. Obviously, if you allow renters to return their car to a different location the Zipcar company might run out of paid parking spaces in a particular location or, even worse, the cars might migrate to certain places, leaving other locations without any car.

I would like to propose an algorithm that will add some flexibility to where you can return a car, without overwhelming the system.

Here’s how it would work. Suppose we have three cars currently assigned to the Mt Auburn/Homer Ave location. I suggest we name three as a desirable number, but actually allow from three to four cars to be assigned to this location at any particular time.

Now suppose I want to pick up a car at the Mt Auburn/Homer Ave location and to return it to the Somerville Ave/Beacon location. If the number of currently assigned cars to Mt Auburn/Homer Ave location is three (at the lower limit), the Zipcar reservation webpage tells me, “Sorry, you can’t use this location unless you return your car back here,” and shows me the closest location with extra cars. The same goes if the number of assigned cars at the Somerville Ave/Beacon location is at its upper limit. If my starting point has more cars assigned to it than its lowest limit and my destination point has fewer cars assigned to it than its upper limit, then I am allowed to take a car from my starting location and return it to my destination. Zipcar can throw in some financial incentives. If my choice disrupts the most desirable balance of car assignments, I have to pay a fee. If my choice restores the balance, I get a bonus discount.

It would be so cool if zipcars were flexible. I understand that the average cost to the company of parking each car might go up with my flexible algorithm as Zipcar will need more parking spaces than cars. But Zipcar can start implementing this flexibility with a small number of flexible locations. It would be a great feature.

Hey, Zipcar algorithm designers! Can I get a bonus if you implement my algorithm? I can also design a financial incentives formula for you.

I updated my funny pictures page with five new pictures. My new favorites are “Cape Cod Levitation Society” on the left and “Grilled Violators” on the right.

Here I repeat the Conway’s Wizards Puzzle from a previous posting:

Last night I sat behind two wizards on a bus, and overheard the following:

— A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.”
— B: “How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?”
— A: “No.”
— B: “Aha! AT LAST I know how old you are!”

Now what was the number of the bus?

It is obvious that the first wizard has more than two children. If he had one child then his/her age would be the number of the bus and it would be the same as the father’s age. While it is unrealistic, in mathematics many strange things can happen. The important part is that if the wizard A had one child he couldn’t have said ‘No’. The same is true for two children: their age distribution is uniquely defined by the sum and the product of their ages.

Here is a generalization of this puzzle:

Last night I sat behind two wizards on a bus, and overheard the following:

— Wizard A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age. Also, the sum of the squares of their ages is the number of dinosaurs in my collection.”
— Wizard B: “How interesting! Perhaps if you told me your age, the number of your children, and the number of dinosaurs, I could work out your children’s individual ages. ”
— Wizard A: “No.”
— Wizard B: “Aha! AT LAST I know how old you are!”

Now what was the number of the bus?

As usual with generalizations, they are drifting far from real life. For this puzzle, you have to open up your mind. In Conway’s original puzzle you do not need to assume that the wizard’s age is in a particular range, but once you solve it, you see that his age makes sense. In this generalized puzzle, you should assume that wizards can live thousands of years, and keep their libido that whole time. Wizards might spend so much of their youth thinking, that they postpone starting their families for a long time. The wizards’ wives are also generalized. They can produce children in great quantities and deliver multiple children at the same time in numbers exceeding the current world record.

Another difference with the original puzzle is that you can’t solve this one without a computer.

You can continue to the next step of generalization and create another puzzle by adding the next symmetric polynomial on the ages of the children, for example, the sum of cubes. In this case, I do not know if the puzzle works: that is, if there is an “AHA” moment there. I invite you, mighty geeks, to try it. Please, send me the answer.

In case you are wondering why the wizard is collecting dinosaurs, I need to point out to you that John H. Conway is a superb puzzle inventor. His puzzle includes a notation suggestion: a for the wizard’s age, b for the bus, c for the number of children. Hence, the dinosaurs.

Everyone agrees that you snore because some muscles at the back of your mouth are too relaxed when you sleep. Back in Russia, I heard about exercises for snorers that strengthen those muscles, but I never heard about them here. That is, until now.

Even medicine in this country is oriented towards making money. Simple exercises do not make much money. Here they offer you many devices and pills and even operations to relieve snoring, but not much in the way of exercise. Recently, though, I heard a very simple idea — sing before going to bed. Singing will help tone your throat muscles.

Some people are making money off of singing, too: check out the Singing for Snorers CD.

I was waiting for a train in Newark. On the platform, there was an LCD screen that flashed advertising. The ad I was staring at was titled, “Reasons to take NJ Transit to Prudential Center, #15.” I was impressed that the NJ Transit sales people were working so hard to invent that many reasons.

I waited for my train for half an hour. It turned out that the NJ Transit advertising people were not working hard after all. The screen was flipping between four reasons, numbered 3, 6, 12 and 15. This is a case of false advertising. You look at reason number 15 and think that there must be a lot of reasons. They fool with your head. Cheaters.

I hope you noticed that I did the same thing with this posting — purely in order to illustrate my point.

Here is my simplified version of Conway’s wizards puzzle.

Last night when I was coming home from my writing class with Sue Katz, I sat behind two wizards on the bus, and overheard the following:

— Wizard A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is the amount of dollars I have in my pocket.”

At this point I interrupted the wizard. “Excuse me, professor, I overheard your conversation and can’t resist asking you a question. Usually when a father says ‘my children’ everyone assumes that he has at least two children. Can I assume that?”

— Wizard A: “No. I stated my assumptions up front. A positive integral number of children means one or more.”

I started thinking. If I were to explain this to a non-mathematician who assumes that ‘my children’ means more than one child, I would need to change the wizard’s statement into the following:

“I have at least one child. The ages of my one-or-more children are all positive integers. The sum of the ages of my children or the age of my only child is the number of this bus. The product of the ages of my children or my only child’s age is the amount of dollars I have in my pocket.”

Hmm. I like that mathematicians use ‘my children’ to indicate any number of children. Makes puzzles faster to type.

Anyway, the wizards continued their discussion:

— Wizard B: “How interesting! Perhaps if you told me the number of your children, I could work out their individual ages”
— Wizard A: “No.”
— Wizard B: “Aha! AT LAST I know how many children you have!”

If I were John Conway, I would have asked you next, “What is the number of the bus?” As I am not John Conway, I’ll ask you, “Why do we presume that Wizard A hasn’t cheated on his wife?”

The answer is that all wizards are notorious for making precise statements. If he cheats a lot, he would have started the conversation with, “The number of children I know about is a positive integer.” Or maybe, more discreetly, “My wife and I have a positive integral number of children.”

If you have already figured out the number of the bus, the bonus question is, “Why did I change the ‘age of the first wizard’ in Conway’s original puzzle into the ‘amount of dollars’ in my puzzle?”

When I left the bus, I started wondering why on earth anyone would ever want to sum up the ages of their children. And I remembered that I once did it myself. I was trying to persuade my sister to apply for U.S. citizenship. My argument was that by moving here the life expectancy of her children would increase by 30 years. Indeed, she has two sons and the male life expectancy in Russia and the U.S. has an astonishing 15-year difference. I have to admit that my argument is not very clean, as we do not know the causes for this difference and, besides, the data is for life expectancy at birth and it changes while our kids age. My sister dismissed my argument, saying that the low male life expectancy in Russia is due to alcoholism and that her family is not in the high-risk group.

So, there could be a reason to sum up the ages of your children, but why would anyone ever want to multiply the ages of their children? In any case, if the first wizard continues to keep an amount of dollars equaling the product of the ages of his children in his pocket, his pocket will do better than mutual funds for the next several years.

I know I’m a computer addict, because:

• When I just wake up, before opening my eyes, my first thought is not that I need to run to the bathroom, but rather, “I need to check my email.”
• My computer desk is the farthest surface suitable for eating from my kitchen, still I have most of my meals there.
• To take a break from my laptop I play evil level websudoku. To take a break from sudoku, I check my email.
• I gave a name to my laptop — Richard. Now when I travel I never feel lonely — I have my own Dick with me all the time.

If you’re reading this, you might be as bad as I am. Please finish this sentence and add it to comments below: “I know I’m a computer addict, because …”

 This puzzle was given to me by John H. Conway, and he heard it from someone else:

Find a ten-digit number with all distinct digits such that the string formed by the first k digits is divisible by k for any k ≤ 10.

Surprisingly, there is a unique solution to this puzzle. Can you find this very special ten-digit number?

For the contrast, consider ten-digit numbers with all distinct digits such that the string formed by the last k digits is divisible by k for any k ≤ 10. These numbers are not so special: there are 202 of them. My puzzle is: find the smallest not-so-special number.

John Conway sent me a puzzle about wizards, which he invented in the sixties. Here it is:

Last night I sat behind two wizards on a bus, and overheard the following:

— A: “I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.”
— B: “How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?”
— A: “No.”
— B: “Aha! AT LAST I know how old you are!”

Now what was the number of the bus?