Tic Tac Toe perfect AI algorithm: deeper in “create fork” step

Question

I've already read many Tic Tac Toe topics on StackOverflow. And I found the strategy on Wikipedia is suitable for my presentation project:

A player can play perfect tic-tac-toe if they choose the move with the highest priority in the following table[3].

1) Win: If you have two in a row, play the third to get three in a row.

2) Block: If the opponent has two in a row, play the third to block them.

3) Fork: Create an opportunity where you can win in two ways.

4) Block Opponent's Fork:

Option 1: Create two in a row to force the opponent into defending, as long as it doesn't result in them creating a fork or winning. For example, if "X" has a corner, "O" has the center, and "X" has the opposite corner as well, "O" must not play a corner in order to win. (Playing a corner in this scenario creates a fork for "X" to win.)

Option 2: If there is a configuration where the opponent can fork, block that fork.

5) Center: Play the center.

6) Opposite Corner: If the opponent is in the corner, play the opposite corner.

7) Empty Corner: Play an empty corner.

8) Empty Side: Play an empty side.

I've followed these step, and the computer never loses. However, the way it attacks is not perfect. Because I have no clue how to do step 3. Here is what I do in step 3: scan every cell, check if put token on that cell creates a fork, then put it there.

private void step3() // Create Fork.
{
    int[] dummyField = (int[])field.Clone();
    // Try Level 1 Dummy
    for (int i = 0; i < 9; i++)
    {
        if (dummyField[i] != 0) continue;
        dummyField[i] = 2;
        if (countFork(dummyField, 2) >= 2)
        {
            nextCell = i;
            return;
        }
        dummyField[i] = 0;
    }

}

Please give me some advice about this step.

EDIT1: The count fork will count how many forks that computer has (computer's tokens is 2, player tokens is 1, because I used that method for step 4 too, so there is a parameter for token in countFork function).

EDIT2: The reason why I say it is not perfect is this (CPU goes first, and its cells are blue, human cells are red).

As you can see, if I put in the top-side cell, the computer wins. But if I put in the right-side cell, it's a tie, although the computer can still win.

EDIT3: Don't know why, but I commented out the step 3, and the computer plays... perfectly! I'm really surprised! Here is my countFork function (I need to port this code to Alice, which doesn't support 2-dimension array, so I use getNumberFromXY to convert 2-dimension array into 1-dimension):

private int countFork(int[] field, int token)
{
    int result = 0;

    // Vertical
    int cpuTokenCount;
    int spareCell;
    for (int x = 0; x < 3; x++)
    {
        cpuTokenCount = 0;
        spareCell = -1;
        for (int y = 0; y < 3; y++)
        {
            if (field[getNumberFromXY(x, y)] == token)
                cpuTokenCount++;
            else if (field[getNumberFromXY(x, y)] == 0)
                spareCell = getNumberFromXY(x, y);
        }
        if (cpuTokenCount == 2 && spareCell != -1) result++;
    }

    // Horizontal
    for (int y = 0; y < 3; y++)
    {
        cpuTokenCount = 0;
        spareCell = -1;
        for (int x = 0; x < 3; x++)
        {
            if (field[getNumberFromXY(x, y)] == token)
                cpuTokenCount++;
            else if (field[getNumberFromXY(x, y)] == 0)
                spareCell = getNumberFromXY(x, y);
        }
        if (cpuTokenCount == 2 && spareCell != -1) result++;
    }

    // Top-Left To Lower-Right Diagonal
    cpuTokenCount = 0;
    spareCell = -1;
    for (int i = 0; i < 3; i++)
    {
        if (field[getNumberFromXY(i, i)] == token)
            cpuTokenCount++;
        else if (field[getNumberFromXY(i, i)] == 0)
            spareCell = getNumberFromXY(i, i);
    }
    if (cpuTokenCount == 2 && spareCell != -1) result++;

    // Top-Right To Lower-Left Diagonal
    cpuTokenCount = 0;
    spareCell = -1;
    for (int i = 0; i < 3; i++)
    {
        if (field[getNumberFromXY(2 - i, i)] == token)
            cpuTokenCount++;
        else if (field[getNumberFromXY(2 - i, i)] == 0)
            spareCell = getNumberFromXY(2 - i, i);
    }
    if (cpuTokenCount == 2 && spareCell != -1) result++;

    return result;
}

EDIT4: FIXED the bug according to soandos, and updated the code at EDIT 3, now it works perfectly!

Answer 1

I am not sure that it is the most elegant way to do it, but here is a two step way of looking at forks.

If the computer cannot win next turn, and it is not the first or second turn, a fork might be possible (this does not deal with creating the setup for a fork, just finding a fork).

For each of the cells that are empty, fill it, then run your step 1 function (sees if there are two in a row). If it finds two places, congrats, you have a fork. If not, you don't.