python 2.7中的简单TicTacToe

Question

我试图为一个类项目创建一个简单的tic-tac-toe游戏，但是我不确定如何完成剩下的代码。我真的很感激任何关于如何完成剩下的工作的见解或意见。这段代码目前在技术上是可行的，但它所做的只是显示一个充满none和“游戏是一个平局”。

PLAYERS = ["X", "O"]

def display_board(board):
        print board[0][0:3]
        print board[1][0:3]
        print board[2][0:3]

def create_empty_board():
        return [[None, None, None], [None, None, None], [None, None, None]]

def board_is_full(board):
        for row in board:
                if None not in board:
                        return True


def winner(board):
        if board[0][0] == board[0][1] == board[0][2] != None:
                return board[0][0]
        elif board[1][0] == board[1][1] == board[1][2] != None:
                return board[1][0]
        elif board[2][0] == board[2][1] == board[2][2] != None:
                return board[2][0]
        elif board[0][0] == board[1][0] == board[2][0] != None:
                return board[0][0]
        elif board[0][1] == board[1][1] == board[2][1] != None:
                return board[0][1]
        elif board[0][2] == board[1][2] == board[2][2] != None:
                return board[0][2]
        elif board[0][0] == board[1][1] == board[2][2] != None:
                return board[0][0]
        else:
                return None

def game_over(board):
        if board_is_full(board) == True:
                return True
def player_turn(board, playerid):
    """ Ask the player to select a coordinates for their next move. The player needs to select a row and a column. If the coordinates the player selects are outside of the board, or are already occupied, they need to be prompted to select coordinates again, until their input is valid."""
    return 1, 1 # by default, return row 1, column 1 as the player's desired location on the board; you need to implement this

def play():
    """ This is the main function that implements a hot seat version of Tic Tac Toe."""
    # the code below is just an example of how you could structure your play() function
    # if you implement all the functions above correctly, this function will work
    # however, feel free to change it if you want to organize your code differently
    board = create_empty_board()
    display_board(board)
    current_player = 0
    while not game_over(board):
        board = player_turn(board, PLAYERS[current_player])
        current_player = (current_player + 1) % len(PLAYERS)
        display_board(board)
    who_won = winner(board)
    if who_won is None:
        print "The game was a tie."
    else:
        print "The winning player is", who_won

if __name__ == "__main__":
    play()

Answer 1

你的第一个问题是你的board_is_full是错误的。对于每一行，不是检查而是检查None not in board。这总是正确的；board是三个列表的列表，因此None永远不会在其中。你要检查每一行，而不是整个董事会。但是只要if None not in row是不对的，只要任何行都满了，而不是所有行都满了，就不会是这样。因此，当您找到一个非整行时，您希望返回False；如果您到达结尾时没有找到任何行，则返回True。你应该自己能做到。

下一个问题是您还没有实现player_turn。对于真正的黑客实现，您只需使用return input('row, col: ')。除非你能理解它为什么起作用，否则你不应该把它作为你的家庭作业（如果你真的理解它为什么起作用，你就不会想用它了）。而且，它不是很友好的用户-它不告诉你轮到谁了，或给你看板，或任何东西。你把这些东西作为参数，这样你就可以使用它们。不管怎么说，这一行已经足够好了，现在可以让你进入下一个问题，所以你可以稍后再来。

下一个问题是player_turn返回一对数字，而您只是将这对数字赋给board。你不能那样做。你想要的是这样的东西：

row, col = player_turn(board, PLAYERS[current_player])
board[row][col] = PLAYERS[current_player]

即使有人已经赢了，游戏也会让你一直玩到棋盘满为止。但事实上，如果棋盘满了，你会想结束比赛，或有赢家。（事实上，理想情况下，你想在不可能获胜的情况下尽快结束比赛，但这有点复杂。只需将前面的句子转换为while语句的替换语句，就可以完成简单的操作

这将使您能够找到任何其他小错误，编写适当的player_turn函数，寻找简化现有代码的方法，等等