我尝试使用minimax算法制作一个不会在井字游戏中丢失的程序。但是,在某些情况下,它会失败。例如,在井字游戏板上还剩下两个点时(在某些情况下),该程序停止播放并询问用户两个连续的输入。而且,在某些情况下,如果计算机明显赢得了胜利,那就不是对移动的正确选择。
这是一项任务,非常感谢您今天提供的任何帮助。
非常感谢!
编辑: 请注意,该代码允许用户覆盖先前的动作。我将尽快解决此问题。 但是,即使我不覆盖以前的机会,也不会得到结果。我已经测试了代码,问题似乎出在minimax函数中,但是我保留了整个代码,以防万一我错了,真正的问题出在其他地方。
编辑2:对不完整的帖子表示抱歉!重现此问题的测试用例如下。在我进入移动位置(位置5)后,程序停止播放并要求我玩所有机会。
Would you like to go first (Y/N)?: n
. . .
. . .
. . .
x . .
. . .
. . .
Enter your choice (1-9): 5
x . .
. o .
. . .
x x .
. o .
. . .
Enter your choice (1-9): 7
x x .
. o .
o . .
x x .
. o .
o . .
Enter your choice (1-9):
此外,我知道我的代码很混乱而且很业余-但是尽管使用了全局变量,但我应该能够使其工作。如果您可以帮我解决这个问题,我会全部清理干净。再次感谢!
编辑3:另一个测试案例: 您要先(是/否)吗?:y
. . .
. . .
. . .
Enter your choice (1-9): 5
. . .
. o .
. . .
x . .
. o .
. . .
Enter your choice (1-9): 3
x . o
. o .
. . .
x . o
. o .
x . .
Enter your choice (1-9): 2
x o o
. o .
x . .
x o o
. o .
x . .
Enter your choice (1-9): 6
x o o
. o o
x . .
x o o
. o o
x . .
Enter your choice (1-9): 9
You win!
我的代码在Python 3.6中,如下所示:
move = -1
n = 0
def evaluateBoard(board):
global n
#Checking for rows
cnt = 0
for i in range(n):
res = 0
for j in range(n):
res += board[cnt * n + j]
cnt += 1
if res == n:
return 1
elif res == -n:
return -1
#Checking for columns
for i in range(n):
res = 0
for j in range(n):
res += board[i + n * j]
if res == n:
return 1
elif res == -n:
return -1
#Checking for diagonals
res = res2 = 0
for i in range(n):
res += board[i * (n + 1)]
res2 += board[(i + 1) * (n - 1)]
if n in [res, res2]:
return 1
elif -n in [res, res2]:
return -1
return 0
def checkNonTerminal(board):
for pos in board:
if pos == 0:
return 1
return 0
def getScore(board, depth):
if evaluateBoard(board) == 1:
return 10 - depth
elif evaluateBoard(board) == -1:
return depth - 10
else:
return 0
def minimax(board, turn, depth):
if evaluateBoard(board) == 0 and checkNonTerminal(board) == 0:
return getScore(board, depth)
global move
moves = list()
scores = list()
for square, pos in enumerate(board):
if pos == 0:
#print(board)
new_board = board.copy()
new_board[square] = turn
moves.append(square)
#print("Moves:", moves, "depth:", depth, "turn:", turn, checkNonTerminal(new_board) == 0)
if evaluateBoard(new_board) in [1, -1] or checkNonTerminal(new_board) == 0:
return getScore(new_board, depth)
scores.append(minimax(new_board, turn * -1, depth + 1))
#print("scores", scores)
if turn == 1:
move = moves[scores.index(max(scores))]
return max(scores)
elif turn == -1:
move = moves[scores.index(min(scores))]
return min(scores)
def displayBoard(board):
global n
for i in range(n):
for j in range(n):
if board[n*i+j] == 1:
print("x", end = " ")
elif board[n*i+j] == -1:
print("o", end = " ")
else:
print(".", end = " ")
print()
def main():
global n
global move
n = 3
first_turn = input("Would you like to go first (Y/N)?: ")
if first_turn in ['Y', 'y']:
first_turn = -1
cnt = 1
else:
first_turn = 1
cnt = 2
board = [0] * 9
while evaluateBoard(board) == 0 and checkNonTerminal(board) == 1:
displayBoard(board)
if cnt % 2 == 0:
score = minimax(board, 1, 0)
print(score)
board[move] = 1
else:
choice = eval(input("Enter your choice (1-9): "))
board[choice - 1] = -1
cnt += 1
if evaluateBoard(board) == 1:
print("You lose!")
elif evaluateBoard(board) == -1:
print("You win!")
else:
print("It's a draw!")
main()
答案 0 :(得分:2)
如果您选中的第一招是游戏玩家,那么您将返回而未设置招数。可能是您的获胜逻辑失败以及转弯失败的原因。
多说一些程序员的话:
您的递归终止条件被过早触发,您也需要处理这种情况!
for square, pos in enumerate(board):
if pos == 0:
#print(board)
new_board = board.copy()
new_board[square] = turn
moves.append(square)
#print("Moves:", moves, "depth:", depth, "turn:", turn, checkNonTerminal(new_board) == 0)
if evaluateBoard(new_board) in [1, -1] or checkNonTerminal(new_board) == 0:
return getScore(new_board, depth) <----here
scores.append(minimax(new_board, turn * -1, depth + 1))
#print("scores", scores)
太忙而无法检查,但是我相信您也可以在该位置设置move变量-如果您只是弹出递归堆栈,以后它将被覆盖。
PS,这是使用适当的返回变量而不是仅设置全局变量的另一个原因;)