python实现人机对战的井字棋游戏
时间:2022-06-18 12:38:00|栏目:Python代码|点击: 次
本文实例为大家分享了python实现人机对战井字棋的具体代码,供大家参考,具体内容如下
游戏简介:在九宫格内进行,如果一方抢先于另一方向(横、竖、斜)连成3子,则获得胜利。游戏中输入方格位置代号的形式如下:
设计前的思路:
游戏中,board棋盘存储玩家、计算机的落子信息,未落子处未EMPTY。由于人机对战,需要实现计算机智能性,下面是为这个计算机机器人设计的简单策略:
如果有一步棋可以让计算机机器人在本轮获胜,那就选那一步走。
否则,如果有一步棋可以让玩家在本轮获胜,那就选那一步走。
否则,计算机机器人应该选择最佳位置来走。
最佳位置就是中间,其次是四个角
定义第一个元组best_weizhi存储最佳方格位置:
按优劣顺序排序的下棋位置
best_weizhi= (4, 0, 2, 6, 8, 1, 3, 5, 7)
井字棋盘输赢判断规则只有8种方式。每种获胜方式都被写成一个元组,利用嵌套元组表达:
win = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6),(1, 4, 7), (2, 5, 8), (0, 4, 8), (2, 4, 6))
代码:
#全局常量 best_weizhi= (4, 0, 2, 6, 8, 1, 3, 5, 7) win = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6),(1, 4, 7), (2, 5, 8), (0, 4, 8), (2, 4, 6)) X = "X" O = "O" EMPTY = " " #定义函数产生一个新的棋盘 def new_board(): board = [] for square in range(9): board.append(EMPTY) return board #询问该谁下棋 def ask_yes_no(question): response = None #如果输入不是"y", "n",继续重新输入 while response not in ("y", "n"): response = input(question).lower() return response #询问谁先走,先走方为X,后走方为O #函数返回电脑方、玩家的角色代号 def pieces(): go_first = ask_yes_no("玩家你是否先走 (y/n): ") if go_first == "y": print("\n玩家你先走.") human = X computer = O else: print("\n电脑先走.") computer = X human = O return computer, human #显示棋盘 def display_board(board): board2=board[:] #创建副本,修改不影响原来列表board for i in range(len(board)): if board[i]==EMPTY: board2[i]=i print("\t", board2[0], "|", board2[1], "|", board2[2]) print("\t", "---------") print("\t", board2[3], "|", board2[4], "|", board2[5]) print("\t", "---------") print("\t", board2[6], "|", board2[7], "|", board2[8], "\n") #输入你想下的位置数字 def ask_number(question, low, high): response = None while response not in range(low, high): response = int(input(question)) return response #产生可以合法走棋位置序列(也就是还未下过子位置) def legal_moves(board): moves = [] for i in range(9): if board[i] == EMPTY: moves.append(i) return moves #判断输赢 def winner(board): for row in win: if board[row[0]] == board[row[1]] == board[row[2]] != EMPTY: winner = board[row[0]] return winner #返回赢方 #棋盘没有空位置 if EMPTY not in board: return "True" #"平局和棋,游戏结束" return False #人走棋 def human_move(board, human): legal = legal_moves(board) move = None while move not in legal: move = ask_number("你走那个位置? (0 - 9):", 0, 9) if move not in legal: print("\n此位置已经落过子了") return move #电脑走棋 def computer_move(board, computer, human): # make a copy to work with since function will be changing list board = board[:] #创建副本,修改不影响原来列表board #按优劣顺序排序的下棋位置best_weizhi # 如果电脑能赢,就走那个位置 for move in legal_moves(board): board[move] = computer if winner(board) == computer: print("电脑下棋位置是" ,move) return move # 取消走棋方案 board[move] = EMPTY # 如果玩家能赢,就堵住那个位置 for move in legal_moves(board): board[move] = human if winner(board) == human: print("电脑下棋位置是" ,move) return move #取消走棋方案 board[move] = EMPTY #不是上面情况则,也就是这一轮时都赢不了则 #从最佳下棋位置表中挑出第一个合法位置 for move in best_weizhi: if move in legal_moves(board): print("电脑下棋位置是" ,move) return move #转换角色 def next_turn(turn): if turn == X: return O else: return X #主方法: def main(): computer, human = pieces() turn = X board = new_board() display_board(board) while not winner(board): if turn == human: move = human_move(board, human) board[move] = human else: move = computer_move(board, computer, human) board[move] = computer display_board(board) turn = next_turn(turn) the_winner = winner(board) #游戏结束输出输赢信息 if the_winner == computer: print("电脑赢!\n") elif the_winner == human: print("玩家赢!\n") elif the_winner == "True": #"平局" print("平局,和棋,游戏结束\n") # start the program main() input("按任意键退出游戏.")
在最后附上结果图:
至此一个简单的井字棋就完成了。