我最近参加了cs50人工智能python在线课程,第一个项目是使用minimax算法创建一个tic-tac-toe游戏,我已经尝试过了。但是,当我从他们的网站上运行与zip文件一起提供的runner.py文件时,它会给我一些错误,如以下语句: i=动作[0], 表示“'NoneType'对象不可下标” 你能纠正一下代码吗,或者至少告诉我到底是什么问题 谢谢
import math
import numpy as npy
import sys
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
numpy_board = npy.array(board)
Xno = npy.count_nonzero(numpy_board = X)
Ono = npy.count_nonzero(numpy_board = O)
if Xno > Ono:
return O
elif Ono > Xno:
return X
def actions(board):
"""
Returns set of all possible actions (i, j) available on the board.
"""
Result = set()
for k in range(3):
for l in range(3):
if board[k][l] == EMPTY:
Result.add(board[k][l])
return Result
def result(board, action):
"""
Returns the board that results from making move (i, j) on the board.
"""
i = action[0]
j = action[1]
if board[i][j] != EMPTY:
raise Exception("Invalid Action")
new_player = player(board)
new_board = copy.deepcopy(board)
new_board[i][j] = new_player
return new_board
def winner(board):
"""
Returns the winner of the game, if there is one.
"""
for i in range(3):
if (board[i][0] == board[i][1] == board[i][2] and board[i][0] != EMPTY):
return board[i][0]
if (board[0][0] == board[1][1] == board[2][2] or (board[0][2] == board[1][1] == board[2][0]) and board[1][1] != EMPTY):
return board[1][1]
if (board[0][i] == board[1][i] == board[2][i] and board[0][i] != EMPTY):
return board[1][i]
else:
return None
def terminal(board):
"""
Returns True if game is over, False otherwise.
"""
if winner(board) != None:
return True;
numpy_board = npy.array(board)
empty_no = npy.count_nonzero(numpy_board == EMPTY)
if (empty_no == 0):
return True
else:
return False
def utility(board):
"""
Returns 1 if X has won the game, -1 if O has won, 0 otherwise.
"""
win_player = winner(board)
if (win_player == X):
return 1
elif (win_player == O):
return -1
else:
return 0
def minimax(board):
"""
Returns the optimal action for the current player on the board.
"""
if terminal(board):
return None
currentPlayer = player(board)
if currentPlayer == X:
return max_value(board)[1]
else:
return min_value(board)[1]
def max_value(board):
if terminal(board):
return (utility(board), None)
value = -sys.maxsize-1
optimalAction = None
for action in actions(board):
possibleResult = min_value(result(board, action))
if possibleResult[0] > value:
value = possibleResult[0]
optimalAction = action
if value == 1:
break
return (value, optimalAction)
def min_value(board):
if terminal(board):
return (utility(board), None)
value = sys.maxsize
optimalAction = None
for action in actions(board):
possibleResult = max_value(result(board, action))
if possibleResult[0] < value:
value = possibleResult[0]
optimalAction = action
if value == -1:
break
return (value, optimalAction)
几个问题:
Xno = npy.count_nonzero(numpy_board = X)
中的语法错误。你错过了一个等号。它应该是==
。下一个类似语句中出现相同错误elif Ono > Xno:
中的条件永远不会为真(想想看)。更重要的是,这个条件允许在不输入任何一个块的情况下通过这个if..elif
,从而给出一个None
返回值。要么是轮到X,要么不是。在后一种情况下,总是轮到O了。你不需要第二次测试。因此,请将这一行更正为else:
Result.add(board[k][l])
不添加坐标对,而是添加正方形的内容。这不是你想要的。您想存储坐标。所以这应该是Result.add((k, l))
。注意:对于这样的名称,不要使用Pascal大小写,而是使用camel大小写李>winner
中for
循环将在第一次迭代时退出。它从不执行其他迭代。在第一次迭代中,您不可能知道足够的信息来返回None
。所以删除else: return None
:在这种情况下,循环必须继续。注意:对角线的测试最好移到循环之外,因为重复测试3次是没有意义的。它不依赖于循环变量李>如果你做了这些修正,它应该会起作用
其他一些评论:
player
和terminal
中进行转换都会影响性能李>deepcopy
具有性能成本。考虑使用相同的列表/数组而不复制它。您只需要在递归调用之后添加一个“undo”操作李><> LI>而不是重新创建一组可能的动作,也考虑增量地保持一个集合:在移动时删除该集合中的一个动作,并在回溯时将其放回。这将提高性能李>不要使用此模式:
首先,括号不是必需的,但更重要的是:当您已经有一个布尔表达式(
empty_no == 0
)时,只返回它。不要做这种if..else
的事情:utility
函数,以便对于使用X的win,它返回空闲单元数加1。对于O来说,它将是该值的否定。这样的话,快速获胜是有利的李>相关问题 更多 >
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