学进去

井字棋，英文名叫 Tic-Tac-Toe，是一种在 3*3 格子上进行的连珠游戏。游戏需要的工具仅为纸和笔，然后由分别代表 O 和 X 的两个游戏者轮流在格子里留下标记，任意三个标记形成一条直线，则为获胜。现编写程序实现模拟井字棋的人机对弈。
（1）下图函数可实现创建默认的井字棋棋盘（规定双方分别为-1 和 1，空格用 0 表示）。
def create_board(size=3):
       return ____
下列不能填入横线处实现功能的是
A.[[0] * size for _ in range(size)]
B.[[0] * size] * size
C.[[0 for _ in range(size)] for _ in range(size)]
（2）现编写函数实现自动寻找当前状态下的最优下法。请补全代码。
def minimax(board, player, memo):
       key = str(board) + str(player)
       if key in memo: return memo[key]
       if game_over(board): ①____
       if player == 1:
             max_eval = -float('inf')
             for i in range(len(board)):
                    for j in range(len(board)):
                           if board[i][j] == 0:
                                 board[i][j] = player
                                 eval = ②____
                                 board[i][j] = 0
                                 max_eval = max(max_eval, eval)
                                 memo[key] = max_eval
                           return max_eval
       else:
             #处理另一方的状态，代码略
def find_best_move(board, player):
       memo = {}
       best_eval = -float('inf') if player == 1 else float('inf')
       best_move = None
       for i in range(len(board)):
             for j in range(len(board)):
                    if board[i][j] == 0:
                           board[i][j] = player
                                 eval = minimax(board, -1 if player == 1 else 1, memo)
                           board[i][j] = 0
                           if   ③____:
                                 best_eval = eval
                                 best_move = (i, j)
       return best_move
（3）小余同学提出以上代码中 memo 变量应作为返回值传递否则无法实现优化效果。你同意这种说法吗？____（同意/反对）。若同意，请给出解决方案，若反对，请写出你的理由。
（4）以下是主程序代码，请补全。
def print_board(board): #打印棋盘，代码略
def is_winner(board, player): #判断某个玩家是否为赢家，代码略
def game_over(board): #判断游戏是否结束，代码略
def evaluate(board):
       if is_winner(board, 1): return 1
       elif is_winner(board, -1): return -1
       else: return 0
①____
player=1
print_board(board)
while not game_over(board):
       print(f"玩家{'X' if player == 1 else 'O'}的回合：")
       if player==1:
             row, col = ②____
       else:
             row, col=input().split(",")
       board[row][col] = player
       print_board(board)
       player = -1 if player == 1 else 1
       if is_winner(board, 1):
             print("玩家 X 赢了！")
       elif: is_winner(board, -1):
             print("玩家 O 赢了！")
       else:
             print("游戏结束，平局！")