Python连四游戏获胜函数

Question

除了 win 函数和重力之外，我已经对 Connect Four 的大部分 Python 版本进行了编码。下面是确定游戏是否已经获胜的代码。它不起作用，目前游戏结束的唯一方法是所有名额都被占据。有什么想法吗？

winners = [ [1,2,3,4],[2,3,4,5],[3,4,5,6],[4,5,6,7],[8,9,10,11],[9,10,11,12],[10,11,12,13],[11,12,13,14],
                    [15,16,17,18],[16,17,18,19],[17,18,19,20],[18,19,20,21],[22,23,24,25],[23,24,25,26],[24,25,26,27],
                    [25,26,27,28],[29,30,31,32],[30,31,32,33],[31,32,33,34],[32,33,34,35],[36,37,38,39],[37,38,39,40],
                    [38,39,40,41],[39,40,41,42],[1,8,15,22],[8,15,22,29],[15,22,29,36],[2,9,16,23],[9,16,23,30],[16,23,30,37],
                    [3,10,17,24],[10,17,24,31],[17,23,31,38],[4,11,18,25],[11,18,25,32],[18,24,32,39],[5,12,19,26],[12,19,26,33],
                    [19,25,33,40],[6,13,20,27],[13,20,27,34],[20,28,34,41],[7,14,21,28],[14,21,28,35],[21,28,35,42],[1,9,17,25],
                    [9,17,25,33],[17,25,33,41],[8,16,24,32],[16,24,32,40],[15,23,31,39],[2,10,18,26],[10,18,26,34],[18,26,34,42],
                    [3,11,19,27],[11,19,27,35],[4,12,20,28] ]

    def makeMove(self):
        self.Player = self.game.Player

        if self.game.Player == 'Black':
            self.b.config(image=self.Black, command=self.invalidMove)
            num = self.num
            self.game.moves['Black'].append(num)
            self.game.free.remove(num)
            w = self.won(self.game.Player)
            if self.game.moves['Black'] in winners:
                self.game.turninfo.config(text=self.game.Player+' WON!')
            elif self.game.free == [ ]:
                self.game.turninfo.config(text='Game is a draw!')
            else:
                self.game.Player = 'Red'
                self.game.turninfo.config(text=self.game.Player+"'s Turn")
        elif self.game.Player == 'Red':
            self.b.config(image=self.Red, command=self.invalidMove)
            num = self.num
            self.game.moves['Red'].append(num)
            self.game.free.remove(num)
            w = self.won(self.game.Player)
            if contains(self.game.moves['Red'],winners):
                self.game.turninfo.config(text=Player+' WON!')
                for c in self.game.cells:
                    c.freeze()
            elif self.game.free == [ ]:
                self.game.turninfo.config(text='Game is a draw!')
                for c in self.game.cells:
                    c.freeze()
            else:
                self.game.Player = 'Black'
                self.game.turninfo.config(text=self.game.Player+"'s Turn")

Answer 1

我认为你的问题是你正在检查玩家的移动是否完全符合获胜条件之一。这是可能的，但可能不会经常发生，例如玩家可能会采取 [2, 3, 1, 4] 的动作，这应该算作胜利，但是 [2, 3, 1, 4] 不在获胜者 中，因此您不会获得匹配。此外，如果玩家的步数超过四步，出于同样的原因，您将永远不会获得胜利（因为 [6, 1, 2, 3, 4] 在 中找不到>获胜者 要么）。

您真正想知道的是，对于任何获胜条件，获胜条件的所有动作是否都是由玩家做出的。这可以这样表达：

for winner in winners:
    won = True
    for move in winner:
        if move not in self.game.moves['Black']:
            won = False
    if won:
        # game ending logic
        break

这可以大大缩短为以下内容（但可能更难理解）：

if any(all(m in self.game.moves['Black'] for m in w) for w in winners)):
    # game ending logic