井字棋
from copy import deepcopyclass Board:
def __init__(self):
self.state = [, , ]
self.flag = 0
def __str__(self):
new_lst = deepcopy(self.state)
for i in range(len(new_lst)):
for j in range(len(new_lst)):
if new_lst == 0:
new_lst = "-"
elif new_lst == 1:
new_lst = "X"
else:
new_lst = "O"
string = "{:^5}|{:^5}|{:^5}\n{:_^5}|{:_^5}|{:_^5}\n{:^5}|{:^5}|{:^5}\n{:^5}|{:^5}|{:^5}\n{:_^5}|{:_^5}|{:_^5}\n{:^5}|{:^5}|{:^5}\n{:^5}|{:^5}|{:^5}\n{:^5}|{:^5}|{:^5}\n"
return string.format(*new_lst, "_", "_", "_", " ", " ", " ", *new_lst, "_", "_", "_", " ", " ", " ", *new_lst, " ", " ", " ")
def reset(self):
self.state = [, , ]
self.flag = 0# 采用法一的时候必须将flag置为0
@property
def is_valid(self):
count_X = 0
count_O = 0
if len(self.state) != 3:
return False
for line in self.state:
if len(line) != 3:
return False
if not(isinstance(line, int) and isinstance(line, int) and isinstance(line, int) and line in and line in and line in ):
return False
for each in line:
if each == 1:
count_X += 1
elif each == 2:
count_O += 1
if abs(count_X - count_O) > 1:
return False
return True
@property
def round(self):
count = 0
for line in self.state:
for each in line:
if each != 0:
count += 1
return count
def __getitem__(self, key):
return self.state
def __setitem__(self, key, value):
if self.state != 0:
raise ValueError
# 交替添加就叫一致或者连续
# 法一
if value == 1 and self.flag != -1:
self.state = value
self.flag = -1
elif value == 2 and self.flag != 1:
self.state = value
self.flag = 1
else:
raise ValueError
# 法二
# 如果在添加之后，棋盘不合法，则valueerror
@property
def winner(self):
# 横竖
for i in range(3):
if self.state == self.state == self.state and self.state != 0:
return self.state
elif self.state == self.state == self.state and self.state != 0:
return self.state
# 对角线
if self.state == self.state == self.state and self.state != 0:
return self.state
# 副对角线
elif self.state == self.state == self.state and self.state != 0:
return self.state
return 0
@property
def is_finished(self):
if self.winner == True or self.round == 9:
return True
class Position:
def __init__(self, row, col):
if row in and col in :
self.row = row
self.col = col
else:
raise ValueError
import random
# 井字棋核心算法
class Agent:
def make_move(self, board):
iswin = self.get_win_or_lost(board, 1)
# 如果自己能赢直接赢
if iswin != None:
return iswin
islost = self.get_win_or_lost(board, 2)
# 如果对方能赢则堵住
if islost != None:
return islost
# 预判一下自己能否必胜
will_win = self.I_will_win_or_lost(board, 1)
if will_win != None:
return will_win
# 预判一下自己是否必输
will_lost = self.I_will_win_or_lost(board, 2)
if will_lost != None:
return will_lost
# 如果中间位置没有棋子则下在中间，因为放在中间赢得概率最大
if board == 0:
return Position(1, 1)
if board.round == 1:
# 当别人先手时，如果中间位置被占，那么就在(0, 0)下子
if board != 0:
return Position(0, 0)
# 其他情况随机放子，基本不影响赢的概率
row = random.randint(0, 2)
col = random.randint(0, 2)
while board != 0:
row = random.randint(0, 2)
col = random.randint(0, 2)
return Position(row, col)
# 去试一试自己要赢还是要输
def get_win_or_lost(self, board, player):
empty_position = []
for i in range(3):
for j in range(3):
if board == 0:
empty_position.append(Position(i, j))
for each in empty_position:
board.state = player
if board.winner == player:
board.state = 0
return each
board.state = 0
def I_will_win_or_lost(self, board, player):
empty_position = []
for i in range(3):
for j in range(3):
if board == 0:
empty_position.append(Position(i, j))
for each in empty_position:
board.state = player
if self.get_win_lost_count(board, player) > 1:
board.state = 0
return each
board.state = 0
def get_win_lost_count(self, board, player):
empty_position = []
for i in range(3):
for j in range(3):
if board == 0:
empty_position.append(Position(i, j))
count = 0
for each in empty_position:
board.state = player
if board.winner == player:
board.state = 0
count += 1
board.state = 0
return count
count_a = 0
count_b = 0
agent1 = Agent()
# agent2 = Agent()
# 完全随机下棋的玩家
class agent_random:
def make_move(self, board):
row = random.randint(0, 2)
col = random.randint(0, 2)
while board != 0:
row = random.randint(0, 2)
col = random.randint(0, 2)
return Position(row, col)
agent2 = agent_random()
board = Board()
# 先手一定不会输
# for i in range(5000):
# while board.winner == 0:
# board = 1
# if board.winner != 0:
# # print("agent1 win")
# # print(board)
# count_a += 1
# break
# if board.is_finished:
# break
# board = 2
# if board.winner != 0:
# print("agent2 win")
# print(board)
# count_b += 1
# break
# if board.is_finished:
# break
# board.reset()
# board.flag = 0
# # print("Finished %s times" % (i+1))
# 后手获胜概率很大
for j in range(5000):
while board.winner == 0:
board = 1
if board.winner != 0:
print("agent2 win")
print(board)
count_b += 1
break
if board.is_finished:
break
board = 2
if board.winner != 0:
# print("agent1 win")
# print(board)
count_a += 1
break
if board.is_finished:
break
board.reset()
board.flag = 0
print(count_a, count_b)
代码太臃肿了
时间复杂度还是比较高的
不过好在还是可以跑起来
测试了5000次
先手的话应该是必胜
后手获胜概率也很大
记录一下 好厉害的样子，good
人机对战，人类输入想要下子的位置，比如输入0 0，代表在左上角下子，其他情况类似
board = Board()
agent = Agent()
# 机器先手
# while True:
# board = 1
# print(board)
# if board.winner != 0:
# break
# if board.is_finished:
# break
# while True:
# try:
# board)] = 2
# break
# except:
# pass
# print(board)
# if board.winner != 0:
# break
# if board.is_finished:
# break
# board.reset()
# board.flag = 0
# 玩家先手，有必胜的策略
# 玩家先放(0, 0)，机器会放在中间,玩家放在(2, 2),机器会放在(0, 2),玩家放在(2, 0)，人类获胜
while True:
while True:
try:
board)] = 2
break
except:
pass
print(board)
if board.winner != 0:
break
if board.is_finished:
break
board = 1
print(board)
if board.winner != 0:
break
if board.is_finished:
break
board.reset()
board.flag = 0
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