python - n 维井字棋对角线检查器(预测谁将一举获胜)
问题描述
我得到了井字游戏的状态。这是一个*n 游戏,获胜者是可以拥有“a”个连续 X 或 Os(也给出了 a)的人。我应该预测天气 X 或 O 可以在下一场比赛中获胜。例如,如果状态是这样的:
X X O X O
X O X O X
O X O - -
- - - - -
X O X O X
输出是:
O
我对对角线检查有问题。我写了一个代码。
X = 'X'
O = 'O'
DASH = '-'
def o_left_diagonal_checker_winner(game_board, n, a):
x_counter = 0
for j in range(n):
for i in range(1, n):
if i - 1 >= 0 and n - i >= 0:
if game_board[i - 1][n - i] == X:
x_counter += 1
if n - i - 2 >= 0 and i - 2 >= 0 and x_counter == a - 1 and game_board[n - i - 2][i - 2] == DASH:
return O
else:
x_counter = 0
这部分代码将从左上到右检查游戏。但它不起作用。如果您有更好的算法,请告诉我,我不能使用库。
解决方案
考虑这个替代解决方案。
假设您有一个代表您的游戏板的数组。对于每个位置的不同状态,值为 +1,-1,0。然后取 ad*d 大小的子数组并沿对角线求和。行和列。如果任何值 = -2 或 2,则玩家 +1 或 -1 有机会在下一个状态下获胜。d 将是您的数组的大小。这是一个获胜条件为 3 的 5*5 棋盘示例。因此,简单地用获胜可能性掩盖棋盘将预测下一个获胜者。
import numpy as np
board = np.array([[-1., -1., 1., 1., -1.],
[-1., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 1.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.]])
win_length = 3
#lets assume +1 = X, -1 = 0 and 0 = nothing
# Also assume that X always starts
# You could also simply replace priority with who is moving next
def who_wins_next(board,a):
priority = sum(sum(board))
if priority >0: # If else case to automatically determine who is moving next. Alternatively you can add another input to who is moving next to replace this
target = 2
text = "X (1)"
else:
target = -2
text = "O (-1)"
width,height = board.shape
for i in range(width-a+1):
for j in range(height-a+1):
sub = board[i:i+a,j:j+a]
diagonal_forward = sum(sub[i][i] for i in range(a))
diagonal_backward = sum(sub[i][a-i-1] for i in range(a))
row_sums = np.sum(sub,axis=1) #Using numpy sum with axis to get an array of row sums
col_sums = np.sum(sub,axis=0) # Likewise for columns
if diagonal_forward == target or diagonal_backward == target or target in list(row_sums) or target in list(col_sums):
#Only need to know if win is possible. Not how.
return text + " can win in next step"
return text + " cannot win in next step"
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