Question

我正在尝试制作一个tic-tac-toe游戏。不确定约定是什么，但是我使用1表示X，4表示O表示，因为在列，行或对角线上的总和3表示X的胜利，而12的总和表示O的胜利，并且在非获胜场景中无法达到该组合。同样，如果所有行，列和对角线等于1,1,4或1,4,4（即6或9），那么游戏就是平局。

除了胜利条件外，游戏的其余部分工作正常：

#if a row, column, or diagonal sums to 3, you win, if 12, they win 
#a number 1 = X, a number 4 = O
if (    row1 == 3
    or  row2 == 3
    or  row3 == 3
    or  col1 == 3
    or  col2 == 3
    or  col3 == 3
    or  dia1 == 3
    or  dia2 == 3
    ):
    endCondition = True
    whoWon = "You won!"

if (    row1 == 12
    or  row2 == 12
    or  row3 == 12
    or  col1 == 12
    or  col2 == 12
    or  col3 == 12
    or  dia1 == 12
    or  dia2 == 12
    ):
    endCondition = True
    whoWon = "You lost! Computer wins."

#if that was the last spot, then it is a draw
#a losing row is either 1,1,4 or 1,4,4 which equal six and 9 respectively
if (row1 and row2 and row3 and col1 and col2 and col3 and dia1 and dia2) == (6 or 9):
    endCondition = True
    whoWon = "No one. It was a draw."

有任何帮助吗？我知道这是一个简单的错误。

Answer 1

if (row1 and row2 and row3 and col1 and col2 and col3 and dia1 and dia2) == (6 or 9):

这不是针对多个其他值检查多个值的正确语法。 (6 or 9)将始终评估为9。如果任何值为0，(row1 and row2 and ... and dia2)将评估为0;如果没有值为0，则为dia2的值。因此，如果电路板已满且dia2不是9，则条件将错误地评估为False。

相反，请尝试：

if all(x == 6 or x == 9 for x in (row1, row2, row3, col1, col2, col3, dia1, dia2)):

井字问题中的逻辑流程

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