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python - 无限递归调用极小极大算法

转载 作者:行者123 更新时间:2023-12-02 18:19:44 27 4
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我最近实现了使用极小极大算法的4X4 tic tac toe 游戏的代码。然而,我的 minimax 函数正在递归地调用自身无限否。次。

初始棋盘 (4X4) 井字游戏 ->

board = np.array([['_','_','_', '_'],['_','_','_', '_'],['_','_','_', '_'],['_','_','_', '_']])

计算机轮流代码 ->

import math
from utility import *

def minimax(board, spotsLeft, maximizing):
bestIdx = (0,0)
p1 = 'X'
p2 = 'O'
res, stat = checkGameOver(board, spotsLeft)
if res==True:
if stat == 'X': return 17-spotsLeft, bestIdx
elif stat == 'O': return -17+spotsLeft, bestIdx
else: return 0, bestIdx

if maximizing:
# return max score
bestMove = -math.inf
for i in range(4):
for j in range(4):
if board[i][j] == '_':
board[i][j] = p1
score, idx = minimax(board, spotsLeft-1, False)
print(score, idx)
board[i][j] = '_'
if bestMove<score:
bestMove = score
bestIdx = (i,j)
return bestMove, bestIdx
else:
# return min score
bestMove = math.inf
for i in range(4):
for j in range(4):
if board[i][j] == '_':
board[i][j] = p2
score, idx = minimax(board, spotsLeft-1, True)
print(score, idx)
board[i][j] = '_'
if bestMove>score:
bestMove = score
bestIdx = (i,j)
return bestMove, bestIdx


def ai(board):
spotsLeft = 16
p1 = 'X' # Computer
p2 = 'O' # Player
turn = p1
while True:
res, stat = checkGameOver(board, spotsLeft)
if res==False:
if turn == p1:
# AI
boardCopy = board[:]
score, idx = minimax(boardCopy, spotsLeft, True)
board[idx[0]][idx[1]] = turn
turn = p2
spotsLeft-=1
else:
# Human
inp = int(input(f"Your turn: "))
i, j = spotToIdx(inp)
if board[i][j]=='_':
board[i][j] = turn
turn = p1
spotsLeft-=1
else: return stat
printBoard(board)

在上面的代码中spotsLeft 是船上的空位,checkGameOver 返回“X”(如果玩家 X 获胜),返回“O”(如果玩家 O 获胜)并返回“T”(如果比赛平局)

checkGameOver 函数 ->

def checkWin(board):
# Check Row
for i in board:
if len(set(i)) == 1 and i[0]!='_':
return i[0]

# Check Column
for i in range(4):
if (board[0][i] == board[1][i] == board[2][i] == board[3][i]) and board[0][i]!='_':
return board[0][i]

# Check Diagonals
if (board[0][0]==board[1][1]==board[2][2]==board[3][3]) and board[0][0]!='_':
return board[0][0]
if (board[0][3]==board[1][2]==board[2][1]==board[3][0]) and board[0][3]!='_':
return board[0][3]

# No One Won Yet
return -1


def checkGameOver(board, spotsLeft):
# t -> Game Tied
# x -> Player X won
# o -> Player O won

# if tied - all spots filled
if spotsLeft == 0:
return True, 'T'

# if any player won the game
result = checkWin(board)
if result!=-1:
return True, result

return False, -1

最佳答案

我认为你的代码很好,并且可以做你想做的事情。该问题很可能是由于问题的复杂性造成的,对于较低维度的井字游戏,它会工作得很好。

让我们首先简化并查看 2x2 的情况。对于第一轮,您从 ai 函数调用 minimax,该函数将在第一级调用自身 4 次。在下一级别,这些调用中的每一个也将调用 minimax,但比上一级别少一次。将其作为列表:

  • 级别 0(ai 函数):1 次调用 minimax
  • 级别 1:4 次调用
  • 第 2 级:4x3 调用(上述 4 个调用中的每一个调用都会进行 3 个新调用)
  • 第 3 级:4x3x2 调用
  • 第 4 级:4x3x2x1 调用

现在使用 n 阶乘表示法作为 n!,我们可以将调用总数计算为:

4!/4! + 4!/(4-1)! + 4!/(4-2)! + 4!/(4-3)! + 4!/(4-4!)

这是 0n 之间 kn!/(n-k)! 之和(包括),n以板上的单元数开始。这里的结果是 2x2 板对 minimax 的调用 65 次。

放入Python代码:

def factorial(n):
if n > 1: return n*factorial(n-1)
else: return 1

def tictactoeComplexity(gridsize):
return sum([factorial(gridsize)/factorial(gridsize-k) for k in range(gridsize+1)])

print(tictactoeComplexity(2*2)) # result is 65

现在让我们检查一下 3*3 板:

print(tictactoeComplexity(3*3)) # result is 986,410

仅仅从 2x2 到 3x3,我们就从大约 100 增加到大约 1,000,000。您可以猜测 4x4 棋盘的结果:

print(tictactoeComplexity(4*4)) # result is 56,874,039,553,217

所以你的程序做了你要求它做的事情,但你要求的太多了。

关于python - 无限递归调用极小极大算法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/71031045/

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