P D Tan P D Tan - 3 months ago 80
Java Question

Java: How to check diagonal Connect Four win in 2D array

I have a Connect Four "board" which is a 6*7 2D char array populated with either spaces, X or O. The win condition is met when there are either four Xs or four Os in a row vertically, horizontally or diagonally. I've managed to get the win conditions checked successfully for vertical and horizontal, where the winning char is returned by some methods as below:

private char CheckVerticalWinner(char[][] currentBoard) {
// check vertical (move down one row, same column)
char vWinner = ' ';
for (int col=0; col<7; col++) {
int vCount = 0;
for (int row=0; row<5; row++) {
if (currentBoard[row][col]!=' ' &&
currentBoard[row][col] == currentBoard[row+1][col]) {
vCount++;
System.out.println("VERT "+vCount); //test
} else {
vCount = 1;
}

if (vCount>=4) {
vWinner = currentBoard[row][col];
}
}
}
return vWinner;
}


private char CheckHorizontalWinner(char[][] currentBoard) {
// check horizontal (move across one column, same row)
char hWinner = ' ';
for (int row=0; row<6; row++) {
int hCount = 0;
for (int col=0; col<6; col++) {
if (currentBoard[row][col]!=' ' &&
currentBoard[row][col] == currentBoard[row][col+1]) {
hCount++;
System.out.println("HORIZ "+hCount); //test
} else {
hCount = 1;
}

if (hCount>= 4) {
hWinner = currentBoard[row][col];
}
}
}
return hWinner;
}


I'm just stuck on how to check for diagonal wins, without throwing an ArrayIndexOutOfBoundsException. I know I need to iterate through the 2D array twice, once for forward diagonals and once for backward diagonals that are 4 squares long or more, like in the below diagram:

Diagonals to be checked diagram

Basically, how would I fill in this method to return a winning char?

private char CheckDiagonalWinner(char[][] currentBoard) {

// some iteration here

return dWinner;
}


Any help would be much appreciated!

Answer

The simplest algorithm probably is:

for every direction
    for every coordinate
        check whether the next 3 elements in this direction exist and are the same

in code:

final int maxx = 7;
final int maxy = 6;

char winner(char[][] board) {
    int[][] directions = {{1,0}, {1,-1}, {1,1}, {0,1}};
    for (int[] d : directions) {
        int dx = d[0];
        int dy = d[1];
        for (int x = 0; x < maxx; x++) {
            for (int y = 0; y < maxy; y++) {
                int lastx = x + 3*dx;
                int lasty = y + 3*dy;
                if (0 <= lastx && lastx < maxx && 0 <= lasty && lasty < maxy) {
                    char w = board[x][y];
                    if (w != ' ' && w == board[x+dx][y+dy] 
                                 && w == board[x+2*dx][y+2*dy] 
                                 && w == board[lastx][lasty]) {
                        return w;
                    }
                }
            }
        }
    }
    return ' '; // no winner
}