woutr_be woutr_be - 5 months ago 45
C Question

Four in a row logic

I'm currently working on a basic four in a row game for myself, but I'm rather stuck at the logic behind it.

Currently I have this multi-dimensional array that represents the board

[
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]
]


0
would represent an empty slot, while
1
and
2
represent a player.
So let's say after a while you get this array:

[
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[0, 0, 0, 1, 1, 0, 0],
[0, 0, 1, 2, 2, 2, 0],
[0, 1, 2, 2, 1, 2, 0]
]


How can I write the logic to check if there are four in a row?
Calculating it for horizontal and vertical ones seem rather easy (although still figuring out the best way), but how would I do this for diagonal lines?

Answer Source

Best bet is to probably divide the search space into four:

  • vertical;
  • horizontal;
  • right and down;
  • right and up.

then limit your starting and ending coordinates based on the direction.

For example, let's say your array is board[row=0-5][col=0-6] with board[0][0] at the top left.

First vertical (loops are inclusive at both ends in this pseudo-code):

for row = 0 to 2:
    for col = 0 to 6:
        if board[row][col] != 0 and
           board[row][col] == board[row+1][col] and
           board[row][col] == board[row+2][col] and
           board[row][col] == board[row+3][col]:
               return board[row][col]

This limits the possibilities to only those that don't extend off the edge of the board, a problem most solutions have when they simplistically start by checking each cell and going out in all directions from there. By that, I mean there's no point checking a start row of 3, simply because that would involve rows 3, 4, 5 and 6 (the latter which does not exist).

Similarly, for horizontal:

for row = 0 to 5:
    for col = 0 to 3:
        if board[row][col] != 0 and
           board[row][col] == board[row][col+1] and
           board[row][col] == board[row][col+2] and
           board[row][col] == board[row][col+3]:
               return board[row][col]

For right and down, followed by right and up:

for row = 0 to 2:
    for col = 0 to 3:
        if board[row][col] != 0 and
           board[row][col] == board[row+1][col+1] and
           board[row][col] == board[row+2][col+2] and
           board[row][col] == board[row+3][col+3]:
               return board[row][col]

for row = 3 to 5:
    for col = 0 to 3:
        if board[row][col] != 0 and
           board[row][col] == board[row-1][col+1] and
           board[row][col] == board[row-2][col+2] and
           board[row][col] == board[row-3][col+3]:
               return board[row][col]

Now, you could actually combine those two by making for col = 0 to 3 the outer loop and only doing it once rather than twice but I actually prefer to keep them separate (with suitable comments) so that it's easier to understand. However, if you're addicted to performance, you can try:

for col = 0 to 3:
    for row = 0 to 2:
        if board[row][col] != 0 and
           board[row][col] == board[row+1][col+1] and
           board[row][col] == board[row+2][col+2] and
           board[row][col] == board[row+3][col+3]:
               return board[row][col]
    for row = 3 to 5:
        if board[row][col] != 0 and
           board[row][col] == board[row-1][col+1] and
           board[row][col] == board[row-2][col+2] and
           board[row][col] == board[row-3][col+3]:
               return board[row][col]

Then, if no wins were found in the four possible directions, simply return 0 instead of the winner 1 or 2.

So, for example, your sample board:

row
 0   [0, 0, 0, 0, 0, 0, 0]
 1   [0, 0, 0, 0, 0, 0, 0]
 2   [0, 0, 0, 1, 1, 0, 0]
 3   [0, 0, 0, 1, 1, 0, 0]
 4   [0, 0, 1, 2, 2, 2, 0]
 5 > [0, 1, 2, 2, 1, 2, 0]
         ^
      0  1  2  3  4  5  6 <- col

would detect a winner in the right and up loop where the starting cell was {5,1} because {5,1}, {4,2}, {3,3} and {2,4} are all set to 1.

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