[英]I'm trying to implement a minimax algorithm to create a tic-tac-toe bot, but i'm getting a recursion error
我正在嘗試實現一個極小極大算法來創建一個井字游戲機器人,但我得到了 RecursionError: maximum recursion depth exceeded in comparison 錯誤。 我有下面的代碼。 我添加了提到 function 應該做什么的評論。 我最后你能不能看一下下面的代碼。 謝謝
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
您的問題是您陷入無限狀態,這意味着您不斷遞歸調用 function 直到達到遞歸限制。 您的問題在於您的玩家 function 以及您如何決定下一個輪到誰。 在 O 播放 position 0,0 和 X 播放 position 0,1 之后,您然后嘗試決定誰是下一個播放
所以你數了數,O 和 X 都各放置了 1 個令牌。 但是,您決定下一個是誰的邏輯不適合此板 state。
if x_counter == 0 and o_counter == 0:
return 'O'
elif x_counter > o_counter:
return 'O'
elif o_counter > x_counter:
return 'X'
因此,當 x_counter 和 y_counter 相等但不為 0 時,您不會返回任何內容。 這導致 function 返回 None 所以你卡住了,然后永遠不要在 position 0,2 中放置令牌。 如果 O 總是先走,那么任何時候 x_counter == o_counter 你應該返回 '0' 所以把它改成
if x_counter == o_counter:
return 'O'
elif x_counter > o_counter:
return 'O'
elif o_counter > x_counter:
return 'X'
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.