Tic Tac Toe Python的Minimax算法
作者：互联网
我有点理解minimax算法是如何为Tic Tac Toe python工作的,但我不知道如何在Python中实际编码…这是我到目前为止所做的：
from copy import deepcopy
class TicTacToeBrain :
def __init__(self, player = "x") :
self._squares = {}
self._copySquares = {}
self._winningCombos = (
[0, 1, 2], [3, 4, 5], [6, 7, 8],
[0, 3, 6], [1, 4, 7], [2, 5, 8],
[0, 4, 8], [2, 4, 6])
def createBoard(self) :
for i in range(9) :
self._squares[i] = None
print(self._squares)
def showBoard(self) :
print(self._squares[0], self._squares[1], self._squares[2])
print(self._squares[3], self._squares[4], self._squares[5])
print(self._squares[6], self._squares[7], self._squares[8])
def getAvailableMoves(self) :
self._availableMoves = []
for i in range(9) :
if self._squares[i] == None :
self._availableMoves.append(i)
return self._availableMoves
def makeMove(self, position, player) :
self._squares[position] = player
self.showBoard()
def complete(self) :
if None not in self._squares.values() :
return True
if self.getWinner() != None :
return True
return False
def getWinner(self) :
for player in ("x", "o") :
for combos in self._winningCombos :
if self._squares[combos[0]] == player and self._squares[combos[1]] == player and self._squares[combos[2]] == player :
return player
if None not in self._squares.values() :
return "tie"
return None
def getEnemyPlayer(self, player) :
if player == "x" :
return "o"
return "x"
def minimax(self, node, player, depth = 0, first = True) :
if first :
best = 0
self._copySquares = deepcopy(self._squares)
if node.complete() :
if node.getWinner() == "x" :
self._squares = self._copySquares
return -1 - depth
elif node.getWinner() == "tie" :
self._squares = self._copySquares
return 0
elif node.getWinner() == "o" :
self._squares = self._copySquares
return 1 + depth
best = None
for move in node.getAvailableMoves() :
depth += 1
node.makeMove(move, player)
print()
val = self.minimax(node, node.getEnemyPlayer(player), depth, first = False)
print(val)
if player == "o" :
if val > best :
best = val
else :
if val < best :
best = val
return best
print()
print()
def printCopy(self) :
print(self._copySquares)
但是,它从来没有打印出所有的场景….有人请帮助！这是周一的一个项目.
解决方法:
一些问题：
>执行在第一次迭代时返回for循环并返回：这是不成熟的,因为你永远不会测试任何其他可用的移动.该返回应该在循环之后发生.
>在for循环的每次迭代中增加深度值是错误的.相反,将深度1传递给递归调用,这样当您从那里返回时,您将继续处于相同的深度.
>在递归调用之前完成的移动必须在它之后立即收回,否则for循环的下一次迭代将不会从相同的位置开始.
>最好的值需要在每次调用minimax方法时初始化,而不仅仅是在递归树的顶部.此初始值不应为0,因为当前用户的最佳值可能低于0.因此,您需要将其初始化为极差值.
> minimax方法不返回最佳移动,仅返回评估值.由于该方法的整个目的是告诉你应该播放哪个动作,你需要两个.因此,让方法返回一个包含两个值的元组：评估值和生成该值的移动.
一些非关键问题：
>当你想要延迟不可避免的损失,或者加速强制胜利时,当玩家获胜时计算价值的公式应该越接近于0,而不是距离越近.因此,该公式需要改变.
>由于您应该通过取回移动来恢复电路板,因此无需使用复制板和复制方块.如果所有编码都很好,则在minimax方法的最高调用完成后,该板应该处于与该调用之前完全相同的状态.
>如果不对空方块使用None,而对单个字符(如“.”)使用None,则会打印得更好.因此,无论您在哪里引用空方格值,都要放置该字符.
>你在这里和那里有print()以分离输出.将一个放在showBoard方法中,其余的代码可以不用它们.
>鉴于以上几点,您不需要节点,也不需要minimax方法的第一个参数.
这是一个评论,更正的版本.我把原来的线留在原处,但在需要的地方将它们评论出来.
# *** not needed:
# from copy import deepcopy
class TicTacToeBrain :
def __init__(self, player = "x") :
self._squares = {}
self._copySquares = {}
self._winningCombos = (
[0, 1, 2], [3, 4, 5], [6, 7, 8],
[0, 3, 6], [1, 4, 7], [2, 5, 8],
[0, 4, 8], [2, 4, 6])
def createBoard(self) :
for i in range(9) :
# *** use a single character, ... easier to print
self._squares[i] = "."
print(self._squares)
def showBoard(self) :
# *** add empty line here, instead of in minimax
print ()
print(self._squares[0], self._squares[1], self._squares[2])
print(self._squares[3], self._squares[4], self._squares[5])
print(self._squares[6], self._squares[7], self._squares[8])
def getAvailableMoves(self) :
self._availableMoves = []
for i in range(9) :
# *** see above
if self._squares[i] == "." :
self._availableMoves.append(i)
return self._availableMoves
def makeMove(self, position, player) :
self._squares[position] = player
self.showBoard()
def complete(self) :
# *** see above
if "." not in self._squares.values() :
return True
if self.getWinner() != None :
return True
return False
def getWinner(self) :
for player in ("x", "o") :
for combos in self._winningCombos :
if self._squares[combos[0]] == player and self._squares[combos[1]] == player and self._squares[combos[2]] == player :
return player
# *** see above
if "." not in self._squares.values() :
return "tie"
return None
def getEnemyPlayer(self, player) :
if player == "x" :
return "o"
return "x"
# *** no need for `node` argument, nor `first`
# *** use `self` instead of `node` in all this method
def minimax(self, player, depth = 0) :
# *** not needed
# if first :
# best = 0
# *** not needed
# self._copySquares = deepcopy(self._squares)
# *** always start with initilisation of `best`, but with worst possible value
# for this player
if player == "o":
best = -10
else:
best = 10
if self.complete() :
if self.getWinner() == "x" :
# *** don't do this, you may still need the position to try other moves
# self._squares = self._copySquares
# *** value should be closer to zero for greater depth!
# *** expect tuple return value
return -10 + depth, None
elif self.getWinner() == "tie" :
# self._squares = self._copySquares
# *** expect tuple return value
return 0, None
elif self.getWinner() == "o" :
# self._squares = self._copySquares
# *** value should be closer to zero for greater depth!
# *** expect tuple return value
return 10 - depth, None
# *** Execution can never get here
# best = None
for move in self.getAvailableMoves() :
# *** don't increase depth in each iteration, instead pass depth+1 to
# the recursive call
# depth += 1
self.makeMove(move, player)
# *** pass depth+1, no need for passing `node` nor `first`.
# *** expect tuple return value
val, _ = self.minimax(self.getEnemyPlayer(player), depth+1)
print(val)
# *** undo last move
self.makeMove(move, ".")
if player == "o" :
if val > best :
# *** Also keep track of the actual move
best, bestMove = val, move
else :
if val < best :
# *** Also keep track of the actual move
best, bestMove = val, move
# *** don't interrupt the loop here!
# return best
# *** this is dead code:
# print()
# print()
# *** Also keep track of the actual move
return best, bestMove
def printCopy(self) :
print(self._copySquares)
以下是如何使用该类的示例：
game = TicTacToeBrain()
game.createBoard()
game.makeMove(4, "o")
game.makeMove(3, "x")
val, bestMove = game.minimax("o")
print('best move', bestMove) # --> 0 is a winning move.
看它在eval.in上运行……等待它.
有些事你还可以改进
我不会为此提供代码,但您可以：
>在self.player中跟踪它的转弯.这样你就不必将玩家作为参数传递给minimax,这样可以避免错误.它也使构造函数参数有用 – 目前你不用它做任何事情.
>添加一个方法bestMove,它只会调用minimax,但只返回最佳移动,而不是值.这将更容易管理.
>使用alpha-beta修剪,以便在显然无法提高递归树中已实现的值时停止评估其他移动.
标签：minimax,python,algorithm 来源： https://codeday.me/bug/20190823/1700276.html