Question

嗨我在编写tic tac toe游戏时遇到了一些麻烦，计算机通过选择0到1之间的数字来随机播放。我尝试了一个代码，但我得到了这样的虚幻情况：

O O O

X X X

O O O

我想在我的代码中加入一些内容，检查每次移动是否都有赢家。如果有人获胜，我希望游戏停止并告诉谁是赢家或者没有人获胜。而且我也不希望输出如下图所示。我只想要一块单板显示器（如果可能的话）。

这是我的代码：

import random

board = []
line= 3
col = 3
for i in range (line) :
    for j in range (col):
        board.append(" ") 


def show():
    print(board[0],'|',board[1],'|',board[2])
    print('--------------')
    print(board[3],'|',board[4],'|',board[5])
    print('--------------')
    print(board[6],'|',board[7],'|',board[8])
    print()

# Verify the possibility of position to take.
B=1
C=2
D=3           
def check(A, B, C, D):
    if board[B] == A and board[C] == A and board[D] == A :
        return True            

# Win conditions
def Winner(G):
    if  check(G, 0, 1, 2):
            True 
    if  check(G, 0, 3, 6):
            True 
    if  check(G, 0, 4, 8):
            True
    if  check(G, 1, 4, 7):
            True             
    if  check(G, 2, 5, 8):
            True         
    if  check(G, 2, 4, 6):
            True     
    if  check(G, 3, 4, 5):
            True 
    if  check(G, 6, 7, 8):
            True  


player1 = "O"
player2 = "X"
play=random.randint(0,1)
if play == 1:
    player1 = "X"
    player2 = "O"

while True:
        random.seed()
        place=random.randint(0,8)
        if board[place] != "O" and board[place] != "X" :
           board[place] = player1
           if Winner(player1)==True:
               print("~~~~~~~ the player ",player1," wins~~~~~~~")           
               break;

           while True:
               random.seed()
               place=random.randint(0,8)

               if board[place] != "O" and board[place] != "X" :
                   board[place] = player2
                   if Winner(player2)==True:
                       print("~~~~~~~ the player ",player2," wins~~~~~~~")           
                       break;

                   break; 

        else: 
            print("This position is already taken")         

        show()

Answer 1

您的Winner()函数不会返回任何内容，因此默认情况下，无论输入是什么，它都会隐式返回False。它应该包含return True而不仅仅是True（根本不做任何事情），例如：

def Winner(G):
    if check(G, 0, 1, 2) or check(G, 0, 3, 6) or check(G, 0, 4, 8) or check(G, 1, 4, 7) or check(G, 2, 5, 8) or check(G, 2, 4, 6) or check(G, 3, 4, 5) or check(G, 6, 7, 8):
        return True
    else
        return False

重点是应return True。剩下的主要是风格偏好 - 我更喜欢单if，有很多or条件。其他人可能更喜欢使用if的许多return True语句。

（从技术上讲，你不需要else部分，因为它无论如何都会有效地返回False。但它会使它更具可读性。你也可以只删除{{ 1}}行，并反向缩进else，这使得它成为整个函数的全部捕获而不是return False。）

Answer 2

代码：

import random
board = []
for x in range(9):
    board.append(" ")
def show():
    print(board[0],'|',board[1],'|',board[2])
    print('--------------')
    print(board[3],'|',board[4],'|',board[5])
    print('--------------')
    print(board[6],'|',board[7],'|',board[8])
def check(A, B, C, D):
    if board[B] == A and board[C] == A and board[D] == A :
        return True 
def Winner(G):
    ret = False
    if  check(G, 0, 1, 2):
            ret = True 
    if  check(G, 0, 3, 6):
            ret = True 
    if  check(G, 0, 4, 8):
            ret = True
    if  check(G, 1, 4, 7):
            ret = True             
    if  check(G, 2, 5, 8):
            ret = True         
    if  check(G, 2, 4, 6):
            ret = True     
    if  check(G, 3, 4, 5):
            ret = True 
    if  check(G, 6, 7, 8):
            ret = True
    return ret
def filled():
    ll = 0
    for x in board:
        if x == " ":
            ll = 11 + 1
    if ll > 0:
        return False
    else:
        return True
done = False
turn = "O"
tie = False
while done == False:
    place = random.randint(0, 8)
    if board[place] == " ":
        board[place] = turn
        if Winner(turn) == True:
            done = True
        else:
            if filled() == True:
                done = True
                tie = True
            else:#change turn
                if turn == "X":
                    turn = "O"
                else:
                    turn = "X"
if tie == True:
    print ("The game was a tie. No winner.")
    show()
else:
    print("~~~~~~~ the player '",turn,"' wins~~~~~~~")
    show()

这应该可以满足您的所有要求。

在每次有效转弯后检查获胜者（）和填充（）。在每次成功循环之后转换循环，以防止重复。

注意：'O'首先，在循环开始之前将更改变量更改为'X'。

Tic Tac Toe WinCheck Def

2 个答案: