[英]Minimax algorithm for Tictactoe in Python
我最近参加了 CS50 AI python 课程,要做的项目之一是为 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,您需要做的就是对模板的给定功能进行编程,我相信我做对了,除了算法部分,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 = min_value(result(board, action)) #FIXED
if k > v:
v = k
best_move = action
else:
v = math.inf
for action in actions(board):
k = max_value(result(board, action)) #FIXED
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 #FIXED
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 #FIXED
最后一部分是 minimax(board) function 所在的位置,它应该采用当前的 state 并根据 AI 是玩家“X”还是“O”来计算可能的最佳移动(它可以是任何两个),“X”玩家试图最大化分数,“O”应该使用实用程序(板)function 最小化分数,该实用程序返回 1 表示 X 获胜,-1 表示“O”获胜或 0 表示平局。 到目前为止,人工智能的动作并不是最优的,当我不应该的时候,我可以轻松地战胜它,因为在最好的情况下,我应该得到的只是平局,因为人工智能应该计算出那个时候所有可能的动作。 但是不知道怎么回事...
首先是关于调试的一句话:如果您要打印在递归调用中完成的计算,您可以跟踪问题的执行并快速找到答案。
但是,您的问题似乎在树的顶部。 在你的 minimax 调用中,如果当前玩家是 X,你调用 state 的每个孩子的 max_value,然后取该结果的最大值。 但是,这在树的顶部应用了两次最大 function。 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 function。
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。
@harsh-kothari, cs50 项目页面说
重要的是,原始棋盘应保持不变:因为 Minimax 最终将需要在计算过程中考虑许多不同的棋盘状态。 这意味着简单地更新电路板本身的单元并不是结果 function 的正确实现。 在进行任何更改之前,您可能需要先对板进行深层复制。
为了避免子列表被修改,我们使用 deepcopy 而不是 copy
将操作(板)代码更改为此
possibleActions = set()
for i in range(0, len(board)):
for j in range(0, len(board[0])):
if board[i][j] == EMPTY:
possibleActions.add((i, j))
return possibleActions
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