Archive for the ‘Game Theory’ Category.

Rubik’s Cube Game

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.


Rock, Paper, Scissor

rpsSergei Bernstein and Nathan Benjamin brought back a variation of the “Rock, Paper, Scissors” game from the Mathcamp. They call it “Rock, Paper, Scissor.” In this variation one of the players is not allowed to play Scissors. The game ends as soon as someone wins a turn.

Can you suggest the best strategy for each player?

They also invented their own variation of the standard “Rock, Paper, Scissors.” In their version, players are not allowed to play the same thing twice in a row.

If there is a draw, then it will remain a draw forever. So the game ends when there is a draw. The winner is the person who has more points.

They didn’t invent a nice name for their game yet, so I am open to suggestions.



Let me describe a variation of Nim that is at the same time a variation of Chomp. Here’s a reminder of what Nim and Chomp are.

In the game of Nim, there are several piles of matches and two players. Each of the players, in turn, can take any number of matches, but those matches must come from the same pile. The person who takes the last match wins. Some people play with a different variation in which the person who takes the last match loses.

Nim-Chomp Chocolat

Mathematicians do not differentiate between these two versions since the strategy is almost the same in both cases. The classic game of Nim starts with four piles that have 1, 3, 5 and 7 matches. I call this configuration “classic” because it is how Nim was played in one of my favorite movies, “Last Year at Marienbad”. Recently that movie was rated Number One by Time Magazine in their list of the Top 10 Movies That Mess with Your Mind.

In the game of Chomp, also played by two people, there is a rectangular chocolate bar consisting of n by m squares, where the lowest left square is poisoned. Each player in turn chooses a particular square of the chocolate bar, and then eats this square as well as all the squares to the right and above. The player who eats the poisonous square loses.

Here is my Nim-Chomp game. It is the game of Nim with an extra condition: the piles are numbered. With every move a player is allowed to take any number of matches from any pile, with one constraint: after each turn the i-th pile can’t have fewer matches than the j-th pile if i is bigger than j.

That was a definition of the Nim-Chomp game based on the game of Nim, so to be fair, here is a definition based on the game of Chomp. The game follows the rules of Chomp with one additional constraint: the squares a player eats in a single turn must all be from the same row. In other words, the chosen square shouldn’t have a square above it.

The game of Nim is easy and its strategy has been known for many years. On the other hand, the game of Chomp is very difficult. The strategy is only known for 2 by m bars. So I invented the game of Nim-Chomp as a bridge between Nim and Chomp.


Latin Squares Game

I just invented a two-player game. To start, you have an empty n by n board. When it’s your turn you must write an integer between 1 and n into an empty cell on the board. Your integer has to differ from the integers that are already present in the same row or column. If you finish filling up the board, you will get a Latin square and the game will be a tie. The person who doesn’t have a move loses. What is the best strategy?

Let’s see what happens if n is 2. The first player puts any number in one of the four corners of the 2 by 2 board. The second player wins by placing a different number in the opposite corner.

I played this game with my son Sergei Bernstein on a 3 by 3 board. We discovered that the first player can always win. Since this game is so much fun, I’ll leave it to the reader to play it and to find the winning strategy for the first player.

Can you analyze bigger boards? Remember that this game has many symmetries. You can permute rows and columns. Also, you can permute numbers.

While we were playing Sergei invented two theorems.

Sergei’s theorem 1. If n is odd the first player can guarantee a tie.

Proof. In the first move the first player writes (n+1)/2 in the center cell. If the second player puts number x in any cell the first player puts number n+1−x into the cell that is rotationally symmetric to the second player’s cell with respect to the center. With this strategy the first player will always have a legal move.

Sergei’s theorem 2. If n is even the second player can guarantee a tie.

Proof. If the first player puts number x in any cell the second player puts number n+1−x into the cell that is vertically symmetric to the first player’s cell with respect to the vertical line of symmetry of the board. With this strategy the second player will always have a legal move.

As you play the game, let me know if you develop any theorems of your own.


Langton’s Ant’s Life

Langton’s ant travels on the infinite square grid, colored black and white. At each time step the ant moves one cell forward. The ant’s direction changes according to the color of the cell he moves onto. The ant turns 90 degrees left if the cell is white, and 90 degrees right if the cell is black. After that, the cell he is on changes its color to the opposite color.

There is a symmetry of time and space for this ant. If at any point of the ant’s travel, someone interferes and reverses the ant’s direction in between the cells, the ant and the grid will traverse the steps and stages back to the starting point.

Let’s give this ant a life. I mean, let’s place him inside the Game of Life invented by John H. Conway. In addition to the Langton’s ant’s rules, I want the cells to change colors according to the rules of the Game of Life.

Let me remind you of the rules of Conway’s Game of Life. We call black cells live cells and white cells dead cells. Black is life and white is death. The cell has eight neighbors — horizontal, vertical, diagonal. At each time step:

  • A cell dies of agoraphobia, if it has more than three neighbors.
  • A cell dies of boredom, if it has less than two neighbors.
  • A dead cell can be born again, if it has exactly three neighbors.
  • Otherwise, the cell’s status doesn’t change.

So, our ant will be traveling in this dying and reproducing population and correcting nature’s mistakes. He revives dead cells and kills live cells.

There is an ambiguity in this ant’s life description. The life can happen at two different moments. In the first ant’s world, the ant jumps from one cell to the next, and while he is in the air, the cells have time to copulate, give birth and die. Upon landing, the ant changes direction and uses his magic wand to change the life status of its landing cell. In the second ant’s world, the ant moves to the destination cell, changes its own direction and the status of the cell and then takes a smoke. All the fun, sex and death happen while he is enjoying his cigarette.

The ant’s life has symmetry in a way that is similar to the symmetry of the ant without life. If we reverse the ant’s direction back and also switch his life-style from the first to the second or vice versa, then the ant and the grid will go backwards in their states.

The parameters for the Langton’s ant were chosen to make the ant’s behavior interesting. The parameters of the Game of Life were chosen to make the Game of Life’s behavior interesting. To make the ant’s life fascinating, we might want to modify the ant’s behavior or the Life’s rules. The synergy of the ant and the Life might be intriguing only if the ant changes its behavior and the Life changes its rules.

Let’s experiment and discover how we need to change the rules in order to make the ant’s life interesting.


Evolutionarily Stable Strategy

Robert Calderbank and Ingrid Daubechies jointly taught a course called “The Theory of Games” at Princeton University in the spring. When I heard about it I envied the students of Princeton — what a team to learn from!

Here is a glimpse of this course — a problem on Evolutionarily Stable Strategy from their midterm exam with a poem written by Ingrid:

On an island far far away, with wonderful beaches
Lived a star-bellied people of Seuss-imagin’d Sneetches.

Others liked it there too — they loved the beachy smell,
From their boats they would yell “Can we live here as well?”
But it wasn’t to be — steadfast was the “No” to the Snootches:
For their name could and would rhyme only with booches …

Until with some Lorxes they came!
These now also enter’d the game;
A momentous change this wrought
As they found, after deep thought.

Can YOU tell me now
How oh yes, how?
In what groupings or factions
Or gaggles and fractions
They all settled down?

Sneetches and Snootches only:

  Sneetches Snootches
Sneetches 4 3
Snootches 3 2

Sneetches and Snootches and Lorxes:

  Sneetches Snootches Lorxes
Sneetches 4 3 8
Snootches 3 2 16
Lorxes 8 16 -60

Find all the ESSes, in both cases.


The Polynomial Game

This puzzle is a generalization of a problem from the 1977 USSR math Olympiad:

At the beginning of the game you are given a polynomial, which has 1 as its leading coefficient and 1 as its constant term. Two people play. On your turn you assign a real value to one of the unknown coefficients. The person that goes last wins if the polynomial has no real roots at the end. Who wins?

It is clear that if the last person’s goal is for the polynomial to have a root, then the game is trivial: in this case, he can always make 1 a root with the last move. Also, an odd degree polynomial always has a real root. Therefore, to make the game interesting we should assume that the degree of the polynomial is even.

Though I can’t imaging myself ever being interested in playing this game, figuring out the strategy is a lot of fun.