博弈树搜索极大极小算法实现一字棋 python代码有注释
时间: 2023-11-12 19:37:17 浏览: 218
下面是一个简单的博弈树搜索极大极小算法的实现,以一字棋(Tic-Tac-Toe)为例,使用Python编写,带有详细注释:
```python
# 初始化游戏棋盘
board = [
['-', '-', '-'],
['-', '-', '-'],
['-', '-', '-']
]
# 定义博弈树节点类
class Node:
def __init__(self, state, player):
self.state = state # 当前状态
self.player = player # 当前玩家
self.children = [] # 子节点列表
self.value = None # 节点价值
# 添加子节点
def add_child(self, child):
self.children.append(child)
# 判断是否为叶子节点
def is_leaf(self):
return len(self.children) == 0
# 判断游戏是否结束
def is_game_over(board):
# 判断每一行是否有相同的符号
for i in range(3):
if board[i][0] != '-' and board[i][0] == board[i][1] == board[i][2]:
return True
# 判断每一列是否有相同的符号
for j in range(3):
if board[0][j] != '-' and board[0][j] == board[1][j] == board[2][j]:
return True
# 判断两个对角线是否有相同的符号
if board[0][0] != '-' and board[0][0] == board[1][1] == board[2][2]:
return True
if board[0][2] != '-' and board[0][2] == board[1][1] == board[2][0]:
return True
# 如果棋盘满了,且没有任何一方获胜,则判定为平局
for i in range(3):
for j in range(3):
if board[i][j] == '-':
return False
return True
# 计算节点价值
def calculate_value(board):
if is_game_over(board):
# 如果游戏结束,则返回相应的价值
if has_won(board, 'X'):
return 1 # X获胜,价值为1
elif has_won(board, 'O'):
return -1 # O获胜,价值为-1
else:
return 0 # 平局,价值为0
else:
# 如果游戏还没结束,则返回None
return None
# 判断某个玩家是否获胜
def has_won(board, player):
# 判断每一行是否有相同的符号
for i in range(3):
if board[i][0] == player and board[i][1] == player and board[i][2] == player:
return True
# 判断每一列是否有相同的符号
for j in range(3):
if board[0][j] == player and board[1][j] == player and board[2][j] == player:
return True
# 判断两个对角线是否有相同的符号
if board[0][0] == player and board[1][1] == player and board[2][2] == player:
return True
if board[0][2] == player and board[1][1] == player and board[2][0] == player:
return True
return False
# 极大极小算法
def minimax(node, depth, maximizing_player):
if depth == 0 or node.is_leaf():
return node.value # 如果到达最大深度或者叶子节点,则返回节点价值
if maximizing_player:
value = float('-inf') # 初始化价值为负无穷大
for child in node.children:
value = max(value, minimax(child, depth - 1, False)) # 递归搜索子节点,取最大值
return value
else:
value = float('inf') # 初始化价值为正无穷大
for child in node.children:
value = min(value, minimax(child, depth - 1, True)) # 递归搜索子节点,取最小值
return value
# 创建博弈树
root = Node(board, 'O')
def create_tree(node, depth, maximizing_player):
value = calculate_value(node.state)
if value is not None:
node.value = value
return
for i in range(3):
for j in range(3):
if node.state[i][j] == '-':
new_board = [row[:] for row in node.state] # 复制当前状态
new_board[i][j] = node.player # 在新状态中放置当前玩家的符号
new_player = 'X' if node.player == 'O' else 'O' # 切换玩家
new_node = Node(new_board, new_player) # 创建新节点
node.add_child(new_node) # 将新节点加入子节点列表
create_tree(new_node, depth + 1, not maximizing_player) # 递归创建子树
create_tree(root, 0, True)
# 计算根节点的价值
root.value = minimax(root, 9, True)
# 输出结果
if root.value == 1:
print('X wins!')
elif root.value == -1:
print('O wins!')
else:
print('Tie!')
```
在这个例子中,我们首先定义了一个`board`变量来表示游戏棋盘,然后定义了一个`Node`类来表示博弈树中的节点。然后我们实现了`is_game_over`函数用于判断游戏是否结束,以及`calculate_value`函数用于计算节点的价值。接下来我们实现了`has_won`函数用于判断某个玩家是否获胜,以及`minimax`函数用于进行极大极小搜索。最后,我们创建了博弈树,并计算了根节点的价值,输出了结果。
阅读全文