Efficient algorithm for counting unique states of tic tac toe

Question

I\'m trying to build a tic tac toe game to demonstrate and experiment with machine learning algorithms, and i\'ve found an interesting problem.

eg: a tic tac toe board c

Answer 1

Math

The trick is to use Polyas enumeration theorem:

http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem

Ignoring the duplicates, there are 9 squares of 3 states (x, o and -), so there are 3^9 = 19683 configurations. You need to define the group which provides actions on the board. For your problem the Dihedral Group D4 seems to work for everything but the juxtapositions. But the juxtapositions are easy since there are 2 whenever it is not a board full of don't cares (all -, the initial configuration).

Computation

While the math lets us count the configurations, it doesn't help identify them. The perhaps simplest solution is to define a board as a tuple: {p1, p2, p3, ..., p9} where each pn is either {0,1,2}. It requires 2 bits per square and there are 9 of them for a total of 18 bits. A board hence can be represented by a 32bit integer or less.

Indexing into boards is easily done by a hash table. There are only the 19000 configurations, so it isn't going to kill us.

Running through all board configurations is best done on radix-3 arithmetic on the 9-tuple above: {0,0,9,...,0}, {0,0,0,..., 1}, {0,0,0,...,1,0} and so on. It is just addition with carry. This generates all boards. Notice how the group action and juxtaposition will transform such a number. There is a limited amount of possibilities (juxta shifts x to o and vice versa, the others moves around positions on the board according to the D4 group. There are 8 such transformations.) You can use Polya to make sure your algorithm got them all.

As suggested, picking the smallest guy from the set is a unique representant of the configuration modulo the actions.