Pretty good heuristic evaluation rules for big TicTacToe 5x5 board

问题

I have created TicTacToe game. I use minmax algorithm.

When the board is 3x3 I just calculate every possible move for a game till the end and -1 for loss, 0 for tie, 1 for win.

When it comes to 5x5 it can't be done(to many options(like 24^24) so I have created evaluation method which gives: 10^0 for one CIRCLE inline, 10^1 for 2 CIRCLE inline, ..., 10^4 for 5 CIRCLES inline, but it is useless.

Does anybody have better idea for assesment?

Example:

O|X|X| | |
----------
 |O| | | |
----------
X|O| | | |
----------
 | | | | |
----------
 | | | | |

Evaluation -10, 2 circles across once and inline once (+200), 2 crosses inline(-100), and -1 three times and + 1 three times for single cross and circle.

This is my evaluation method now:

public void setEvaluationForBigBoards() {
        int evaluation = 0;
        int howManyInLine = board.length;
        for(; howManyInLine > 0; howManyInLine--) {
            evaluation += countInlines(player.getStamp(), howManyInLine);
            evaluation -= countInlines(player.getOppositeStamp(), howManyInLine);
        }
        this.evaluation = evaluation;
    }

    public int countInlines(int sign, int howManyInLine) {
        int points = (int) Math.pow(10, howManyInLine - 1);
        int postiveCounter = 0;
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[i].length; j++) {
                //czy od tego miejsca jest cos po przekatnej w prawo w dol, w lewo w dol, w dol, w prawo
                if(toRigth(i, j, sign, howManyInLine))
                    postiveCounter++;
                if(howManyInLine > 1) {
                    if(toDown(i, j, sign, howManyInLine))
                        postiveCounter++;
                    if(toRightDiagonal(i, j, sign, howManyInLine))
                        postiveCounter++;
                    if(toLeftDiagonal(i, j, sign, howManyInLine))
                        postiveCounter++;
                }
            }
        }
        return points * postiveCounter;
    }

回答1:

Number of options (possible sequences of moves) after the first move is 24! and not 24^24. It is still a too much high

number so it is correct to implement an heuristic.

Note that answers about good heuristics are necessarily based on the opinion of the writer so I give my opinion but to find

out what is "the best heuristic" you should make the various ideas playing one against the other in the following way:

take the two heuristics A and B that you want to compare
generate at random a starting configuration
let A play with O and B play with X
from the same starting configuration let A play with X and B play with O
take stats of which one wins more

Now my thoughts about good possible heuristics starting points for an nxn playfield with winning sequence length of n:

since the winning condition for a player it to form a straight sequence of its marks my idea is to use as base values the number of possibilities that each player has still available to built such a straight sequence.
in an empty field both O and X have ideally the possibility to realize the winning sequence in several ways:
- horizontal possibilities: n
- vertical possibilities: n
- diagonal possibilities: 2
- total possibilities: 2n+2
in the middle of a round the number of remaining opportunities for a player are calculated as: "the number of rows without opponent's marks + the number of columns without opponent's marks + the number of diagonals without opponent's marks.
instead than calculate all each time it can be considered that:
- after a move of one player the umber of still available possibilities are:
  - unchanged for him
  - equal or lowered for the opponent (if the mark has been placed in a row/col/diagonal where no marks had already been placed by the considered player)
as heuristic i can propose -
- is possible that - k * with k > 1 give better results and in the end this can be related to how a draw is considered with regard to a lose.

One side consideration:

playfield cells are n^2
winning possibilities are 2n+2 if we keep the winning length equal to the field edge size
this give me the idea that the more the size is increased the less interesting is to play because the probability of a draw after a low number of moves (with reference to the playfield area) becomes higher and higher.
for this reason I think that the game with a winning length lower that n (for example 3 independently from the playfield size) is more interesting.
- Named l the wining length we have that the number of possibilities is 2*((n+1-l)*(2n+1-l)) = O(n^2) and so well proportioned with the field area.

来源：https://stackoverflow.com/questions/23982531/pretty-good-heuristic-evaluation-rules-for-big-tictactoe-5x5-board

标签

java

evaluation

tic-tac-toe

heuristics