我最近注册了CS50 AI python课程，该项目要做的一个工作是为tictactoe游戏实现minimax算法。我寻求帮助并搜索了stackoverflow，但没有找到可以帮助我的答案。它的图形部分已经实现，您所需要做的就是对模板的给定功能进行编程，我相信除了算法部分的唯一例外，我的理解是正确的，这些功能如下：
import math
import copy
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.
"""
if board == initial_state():
return X
xcounter = 0
ocounter = 0
for row in board:
xcounter += row.count(X)
ocounter += row.count(O)
if xcounter == ocounter:
return X
else:
return O
def actions(board):
"""
Returns set of all possible actions (i, j) available on the board.
"""
possible_moves = []
for i in range(3):
for j in range(3):
if board[i][j] == EMPTY:
possible_moves.append([i, j])
return possible_moves
def result(board, action):
"""
Returns the board that results from making move (i, j) on the board.
"""
boardcopy = copy.deepcopy(board)
try:
if boardcopy[action[0]][action[1]] != EMPTY:
raise IndexError
else:
boardcopy[action[0]][action[1]] = player(boardcopy)
return boardcopy
except IndexError:
print('Spot already occupied')
def winner(board):
"""
Returns the winner of the game, if there is one.
"""
columns = []
# Checks rows
for row in board:
xcounter = row.count(X)
ocounter = row.count(O)
if xcounter == 3:
return X
if ocounter == 3:
return O
# Checks columns
for j in range(len(board)):
column = [row[j] for row in board]
columns.append(column)
for j in columns:
xcounter = j.count(X)
ocounter = j.count(O)
if xcounter == 3:
return X
if ocounter == 3:
return O
# Checks diagonals
if board[0][0] == O and board[1][1] == O and board[2][2] == O:
return O
if board[0][0] == X and board[1][1] == X and board[2][2] == X:
return X
if board[0][2] == O and board[1][1] == O and board[2][0] == O:
return O
if board[0][2] == X and board[1][1] == X and board[2][0] == X:
return X
# No winner/tie
return None
def terminal(board):
"""
Returns True if game is over, False otherwise.
"""
# Checks if board is full or if there is a winner
empty_counter = 0
for row in board:
empty_counter += row.count(EMPTY)
if empty_counter == 0:
return True
elif winner(board) is not None:
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 minimax(board):
current_player = player(board)
if current_player == X:
v = -math.inf
for action in actions(board):
k = max_value(result(board, action))
if k > v:
v = k
best_move = action
else:
v = math.inf
for action in actions(board):
k = min_value(result(board, action))
if k < v:
v = k
best_move = action
return best_move
def max_value(board):
if terminal(board):
return utility(board)
v = -math.inf
for action in actions(board):
v = max(v, min_value(result(board, action)))
return v
def min_value(board):
if terminal(board):
return utility(board)
v = math.inf
for action in actions(board):
v = min(v, max_value(result(board, action)))
return v
最后一部分是minimax（board）函数所在的位置，它应该采用当前的棋盘状态并根据AI是玩家'X'或'O'来计算最佳移动方式（可以是两者中的任何一个），“ X”玩家尝试使得分最大化，而“ O”则利用功能（板）功能将得分最小化，该函数将为X获胜返回1，为“ O”获胜返回-1或为平局获得0 。到目前为止，人工智能的移动并不是最佳的，我可以在不应该的情况下轻松地与之抗衡，因为在最佳情况下，我应该得到的只是平局，因为人工智能应该计算出此时的所有可能移动。但是我不知道怎么了...
关于调试的第一句话：如果您要打印递归调用中完成的计算，您可以跟踪问题的执行情况并快速找到答案。
但是，您的问题似乎在树的顶端。在您的minimax通话中，如果当前玩家为X，则在状态的每个子级上调用max_value，然后取该结果的最大值。但是，这会将max函数两次应用到树的顶部。游戏中的下一个玩家是O，因此您应该为下一个玩家调用min_value函数。
因此，在minimax调用中，如果current_player为X，则应调用min_value；如果current_player为O，则应调用max_value。