Hat Puzzle: Create a Distribution

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:

  1. 50 wizards will be guaranteed to survive.
  2. 100 wizards will survive with probability 0.5.
  3. 100 wizards will survive with probability 0.25 and 50 wizards will survive with probability 0.5.
  4. 75 wizards will survive with probability 1/2, and 25 wizards survive with probability 1/2.
  5. 75 wizards will survive with probability 2/3.
  6. The wizards will survive according to a given distribution. For which distributions is it possible?

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.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Linear Algebra on a Mission Impossible

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

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

MI3 Skyline

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

MI3 Rope

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Nim Automaton

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.

U-W 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 Ni. Then the P-positions that do not belong to generation i and earlier and from which all moves belong to Ni 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 P1 from generation i + 1 is a P-position P2 in generation i that can be reached from P1 in the game.

The parent-child relationship in the game of Nim is especially easy to explain. A P-position P1 is a parent of a P-position P2 if P1 differs from P2 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.

Nim Automaton

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 P1 is a parent of P2 if there exists an optimal game such that P1 is achieved from P2 in exactly two moves in a game which takes the longest number of possible moves.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Kolmogorov Student Olympiad in Probability

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.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

IQ Migration

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?
Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Math Kangaroo’s Logic Puzzle

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.

  • 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.

How many of the children were telling the truth?
A) 0   B) 1   C) 2   D) 3   E) Impossible to determine

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Express 6

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 22 + 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.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

The Virtue of Laziness

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

One of his rules:

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

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

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

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

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

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

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

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

Beer Jokes and Hat Puzzles

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.

Red Hat Red Hat Red Hat Red Hat Red Hat

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.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

How Well Do You Know Your Dice?

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.

Dice

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail