Tic-Tac-Toe AI

This is a discussion on Tic-Tac-Toe AI within the Game Programming forums, part of the General Programming Boards category; I have written a progam that allows two people to play tictactoe, how can I modify it so that the ...

  1. #1
    Registered User
    Join Date
    Feb 2003
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    6

    Tic-Tac-Toe AI

    I have written a progam that allows two people to play tictactoe, how can I modify it so that the second player becomes the computer?
    ************source code***********8

    #include <stdio.h>
    #define SIZE 3

    void display( int[][] );
    int winner( int[][] );

    int main()
    {
    int win = -1, i, j, col, row, player = 0;
    int array[SIZE][SIZE];

    printf( "TIC TAC TOE\n\n" );

    // initalize the array
    for ( i = 0; i < SIZE; i++ )
    for( j = 0; j < SIZE; j++ )
    array[i][j] = -3;

    while (win == -1) {

    display( array );
    printf( "Enter row(0 to quit): " );
    scanf( "%d", &row ); // read in row
    if (row == 0 )
    break;
    printf( "Enter column: ");
    scanf( "%d", &col ); // read in col
    if ( array[row-1][col-1] == -3 ) { // check if square is empty
    array[row-1][col-1] = (player); // assign player# to square
    player = (player+1)%2;
    }
    else
    printf( "Invalid entry\n" );

    win = winner( array );
    if ( win == 0 ) {
    display( array );
    printf( "X is the winner\n" );
    }
    if ( win == 1 ) {
    display( array );
    printf( "O is the winner\n" );
    }
    }

    return 0;
    }

    void display( int board[SIZE][SIZE] )
    {
    int i, j;

    printf("1 2 3\n\n" ); // print first line

    for( i = 0; i < SIZE; i++ ) {
    for( j = 0; j < SIZE; j++ ) {
    if ( board [i][j] == 0 ) // print X
    printf( "X " );
    if ( board [i][j] == 1 ) // print O
    printf( "O " );
    if ( board [i][j] == -3 ) // print blank
    printf( " " );
    }
    printf( "%d\n\n", i+1); // print row number
    }
    }

    int winner( int board[SIZE][SIZE] )
    {
    int i, j, total = 0, win = -1;

    // check horizontal
    for( i = 0; i < SIZE; i++ ) {
    for( j = 0; j < SIZE; j++ )
    total += board[i][j];
    if ( total == 0 )
    win = 0;
    if ( total == 3 )
    win = 1;
    total = 0;
    }

    // check vertical
    for( j = 0; j < SIZE; j++ ) {
    for( i = 0; i < SIZE; i++ )
    total += board[i][j];
    if ( total == 0 )
    win = 0;
    if ( total == 3 )
    win = 1;
    total = 0;
    }

    // check diagonals
    for( j = 0; j < SIZE; j++ )
    total += board[j][j];
    if ( total == 0 )
    win = 0;
    if ( total == 3 )
    win = 1;

    total = board[0][2] + board[1][1] + board[2][0];
    if ( total == 0 )
    win = 0;
    if ( total == 3 )
    win = 1;

    return win;
    }


    **************************

  2. #2
    Registered User Vber's Avatar
    Join Date
    Nov 2002
    Posts
    807
    My way to do it is this, I'll check all the possibilities of the computer (places where I still can put a piece) and give them points, like, if in a certain place I'll put a X and in the same line I have another X and don't have any O on the same line I'll give 100 points, and like this works..., at the end, the most valuable play, I'll do it.

    Also use code tags

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