python实现简单的井字棋
本文实例为大家分享了python实现简单的井字棋的具体代码,供大家参考,具体内容如下
使用python实现井字棋游戏,没有具体算法,只是用随机下棋简单实现:
import random
board = [['+','+','+'],['+','+','+'],['+','+','+']]
def ma(board):
if isempty(board):
a = random.randint(0, 2)
b = random.randint(0, 2)
if board[a][b] != 'X' and board[a][b] != 'O':
print("机器走:")
board[a][b] = 'O'
oput(board)
else:
ma(board)
else:
print("平局")
def oput(board):
print(" 0 1 2")
for i in range(3):
print(i, end=' ')
for j in range(3):
print(board[i][j], end=" ")
print("")
def winput(i,j):
if board[i][j] == 'X':
print("human win")
else:
print("machine win")
return 1
def test(board):
for i in range(3):
for j in range(3):
if board[i][j] != '+':
if j == 0:
if board[i][j] == board[i][j + 1] == board[i][j + 2]:
return winput(i,j)
if i == 0:
if board[i][j] == board[i + 1][j] == board[i + 2][j]:
return winput(i,j)
if i == 0 and j == 0:
if board[i][j] == board[i + 1][j + 1] == board[i + 2][j + 2]:
return winput(i,j)
if i == 2 and j == 0:
if board[i][j] == board[i - 1][j + 1] == board[i - 2][j + 2]:
return winput(i,j)
def isempty(board):
for i in range(3):
for j in range(3):
if board[i][j] == '+':
return True
return False
def main():
print("初始棋盘:")
oput(board)
flag = 0
t = input("human first? Y/N human for X, machine for O\n")
if t == 'Y':
while isempty(board):
print("人走: ")
a, b = map(int, input("输入落子纵横坐标: a,b \n").split(','))
if board[a][b] == '+':
board[a][b] = 'X'
oput(board)
flag = test(board)
if flag == 1:
break
else:
print("落子位置不对")
continue
ma(board)
flag = test(board)
if flag == 1:
break
if isempty(board) == 0 and flag == 0:
print("平局")
break
elif t == 'N':
while isempty(board):
ma(board)
flag = test(board)
if isempty(board) == 0 and flag == 0:
print("平局")
break
if flag == 1:
break
print("人走: ")
a, b = map(int, input("输入落子纵横坐标: a,b \n").split(','))
if board[a][b] == '+':
board[a][b] = 'X'
oput(board)
flag = test(board)
if flag == 1:
break
else:
print("落子位置不对")
continue
if __name__ == "__main__":
main()
结果:
(资源库 www.zyku.net)
原文链接:https://blog.csdn.net/weixin_43871677/article/details/117234869
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