TicTacToe strategic reduction

Question

I decided to write a small program that solves TicTacToe in order to try out the effect of some pruning techniques on a trivial game. The full game tree using minimax to solve it only ends up with 549,946 possible games. With alpha-beta pruning, the number of states required to evaluate was reduced to 18,297. Then I applied a transposition table that brings the number down to 2,592. Now I want to see how low that number can go.

The next enhancement I want to apply is a strategic reduction. The basic idea is to combine states that have equivalent strategic value. For instance, on the first move, if X plays first, there is nothing strategically different (assuming your opponent plays optimally) about choosing one corner instead of another. In the same situation, the same is true of the center of the walls of the board, and the center is also significant. By reducing to significant states only, you end up with only 3 states for evaluation on the first move instead of 9. This technique should be very useful since it prunes states near the top of the game tree. This idea came from the GameShrink method created by a group at CMU, only I am trying to avoid writing the general form, and just doing what is needed to apply the technique to TicTacToe.

In order to achieve this, I modified my hash function (for the transposition table) to enumerate all strategically equivalent positions (using rotation and flipping functions), and to only return the lowest of the values for each board. Unfortunately now my program thinks X can force a win in 5 moves from an empty board when going first. After a long debugging session, it became apparent to me the program was always returning the move for the lowest strategically significant move (I store the last move in the transposition table as part of my state). Is there a better way I can go about adding this feature, or a simple method for determining the correct move applicable to the current situation with what I have already done?

Answer 1

My gut feeling is that you are using too big of a hammer to attack this problem. Each of the 9 spots can only have one of two labels: X or O or empty. You have then at most 3^9 = 19,683 unique boards. Since there are 3 equivalent reflections for every board, you really only have 3^9 / 4 ~ 5k boards. You can reduce this by throwing out invalid boards (if they have a row of X's AND a row of O's simultaneously).

So with a compact representation, you would need less than 10kb of memory to enumerate everything. I would evaluate and store the entire game graph in memory.

We can label every single board with its true minimax value, by computing the minimax values bottom up instead of top down (as in your tree search method). Here's a general outline: We compute the minimax values for all unique boards and label them all first, before the game starts. To make the minimax move, you simply look at the boards succeeding your current state, and pick the move with the best minimax value.

Here's how to perform the initial labeling. Generate all valid unique boards, throwing out reflections. Now we start labeling the boards with the most moves (9), and iterating down to the boards with least moves (0). Label any endgame boards with wins, losses, and draws. For any non-endgame boards where it's X's turn to move: 1) if there exists a successor board that's a win for X, label this board a win; 2) if in successor boards there are no wins but there exists a draw, then label this board a draw; 3) if in successor boards there are no wins and no draws then label this board a loss. The logic is similar when labeling for O's turn.

As far as implementation goes, because of the small size of the state space I would code the "if there exists" logic just as a simple loop over all 5k states. But if you really wanted to tweak this for asymptotic running time, you would construct a directed graph of which board states lead to which other board states, and perform the minimax labeling by traversing in the reverse direction of the edges.

Answer 2

Out of curiosity, I wrote a program to build a full transposition table to play the game without any additional logic. Taking the 8 symmetries into account, and assuming computer (X) starts and plays deterministic, then only 49 table entries are needed!

1 entry for empty board

5 entries for 2 pieces

21 entries for 4 pieces

18 entries for 6 pieces

4 entries for 8 pieces