我试图实现一个最小化算法来创建一个井字形机器人，但我得到RecursionError: maximum recursion depth exceeded in comparison error。我有下面的代码。我已经添加了注释，提到了一个函数应该做什么.我最后你能看看下面的代码吗.谢谢。
X = "X"
O = "O"
EMPTY = None
def initial_state():
"""
Returns starting state of the board.
"""
return [[EMPTY, EMPTY, EMPTY],
[EMPTY, EMPTY, EMPTY],
[EMPTY, EMPTY, EMPTY]]
def player(board):
"""
Returns player who has the next turn on a board.
"""
o_counter = 0
x_counter = 0
for i in board:
for j in i:
if j == 'X':
x_counter += 1
elif j == 'O':
o_counter += 1
if x_counter == 0 and o_counter == 0:
return 'O'
elif x_counter > o_counter:
return 'O'
elif o_counter > x_counter:
return 'X'
def actions(board):
"""
Returns set of all possible actions (i, j) available on the board.
"""
action = []
for i in range(3):
for j in range(3):
if board[i][j] is None:
action.append([i, j])
return action
def result(board, action):
"""
Returns the board that results from making move (i, j) on the board.
"""
p = player(board)
i, j = action
board[i][j] = p
return board
def winner(board):
"""
Returns the winner of the game, if there is one.
"""
i = 1
if board[0][0] == board[1][1] == board[2][2] and (board[0][0] == 'X' or board[0][0] == 'O'):
return board[0][0]
elif board[0][2] == board[1][1] == board[2][0] and (board[0][2] == 'X' or board[0][2] == 'O'):
return board[0][2]
else:
if board[0][0] == board[0][1] == board[0][2] and (board[0][0] == 'X' or board[0][0] == 'O'):
return board[0][0]
elif board[i][0] == board[i][1] == board[i][2] and (board[i][0] == 'X' or board[i][0] == 'O'):
return board[i][0]
elif board[2][0] == board[2][1] == board[2][2] and (board[2][0] == 'X' or board[2][0] == 'O'):
return board[2][0]
elif board[0][0] == board[1][0] == board[2][0] and (board[0][0] == 'X' or board[0][0] == 'O'):
return board[0][0]
elif board[0][i] == board[1][i] == board[2][i] and (board[0][i] == 'X' or board[0][i] == 'O'):
return board[0][i]
elif board[0][2] == board[1][2] == board[2][2] and (board[0][2] == 'X' or board[0][2] == 'O'):
return board[0][2]
def terminal(board):
"""
Returns True if game is over, False otherwise.
"""
check = True
if winner(board) == 'X' or winner(board) == 'O':
return True
elif check:
for i in board:
for j in i:
if j is None:
check = False
return False
if check:
return True
else:
return False
def utility(board):
"""
Returns 1 if X has won the game, -1 if O has won, 0 otherwise.
"""
if winner(board) == 'X':
return 1
elif winner(board) == 'O':
return -1
else:
return 0
def maximum(board):
if terminal(board):
return utility(board)
v = -9999999999999999999999
for action in actions(board):
m = minimum(result(board, action))
if m > v:
v = m
return v
def minimum(board):
if terminal(board):
return utility(board)
v = 9999999999999999999999
for action in actions(board):
m = maximum(result(board, action))
if m < v:
v = m
return v
def minimax(board):
"""
Returns the optimal action for the current player on the board.
"""
return_action = None
curr_player = player(board)
states = actions(board)
temp_board = board.copy()
score = 0
temp_score = 0
for state in states:
i, j = state
if curr_player == 'X':
temp_board[i][j] = curr_player
temp_score = maximum(temp_board)
elif curr_player == 'O':
temp_board[i][j] = curr_player
temp_score = minimum(temp_board)
if curr_player == 'X':
if temp_score > score:
score = temp_score
return_action = state
elif curr_player == 'O':
if temp_score < score:
score = temp_score
return_action = state
return return_action
你的问题是你卡在一个无限条件下，这意味着你一直递归调用函数，直到你达到递归极限。你的问题在于你的玩家函数，以及你如何决定下一个轮到谁。当O在0,0位置下棋，X在0,1位置下棋后，你要决定下一个下棋的是谁。
所以你算一下，O和X都各下了一个令牌。然而你决定谁是下一个的逻辑并不符合这个棋盘状态。
if x_counter == 0 and o_counter == 0:
return 'O'
elif x_counter > o_counter:
return 'O'
elif o_counter > x_counter:
return 'X'
所以当x_counter和y_counter相等但不为0时，你不返回任何东西。这将导致函数返回无，所以你被卡住了，永远不会在0,2的位置放置一个标记。如果O总是在前面，那么任何时候x_counter == o_counter，你应该返回'0'，所以把它改为 "0"。
if x_counter == o_counter:
return 'O'
elif x_counter > o_counter:
return 'O'
elif o_counter > x_counter:
return 'X'