将函数 return 设为布尔值并在 tic/tac/toe 游戏中实现 AI

Question

我制作了一个 tic/tac/toe 游戏，但我想创建一个控制 "O" 而用户控制 "X" 的计算机播放器。

对于我代码中的 clicked 函数，我试图 return 一个布尔值，但我不确定我是否做对了。函数 returns True 如果用户点击操作成功。如果用户点击已经在棋盘区域之外的位置，移动无效，函数应该 return False。我的代码：

import turtle
import time
import random 

pieces = ["_", "_", "_", "_", "_", "_", "_", "_", "_"]
turn = "X"
def drawgame(brd):
    # draw board
    turtle.setup(600, 600)
    turtle.bgcolor("silver")
    turtle.color("white")
    turtle.hideturtle()
    turtle.speed('fastest')
    turtle.width(10)
    turtle.up()

    # Horizontal bars
    turtle.goto(-300, 100)
    turtle.down()
    turtle.forward(600)
    turtle.up()
    turtle.goto(-300, -100)
    turtle.down()
    turtle.forward(600)
    turtle.up()

    # Vertical bars
    turtle.goto(-100, 300)
    turtle.setheading(-90)
    turtle.down()
    turtle.forward(600)
    turtle.up()
    turtle.goto(100, 300)
    turtle.down()
    turtle.forward(600)
    turtle.up()
    turtle.color("blue")
    x, y = -300, 300
    for pos in pieces:
        if pos == "X":
            # Draw X
            turtle.up()
            turtle.goto(x + 20, y - 20)
            turtle.setheading(-45)
            turtle.down()
            turtle.forward(226)
            turtle.up()
            turtle.goto(x + 180, y - 20)
            turtle.setheading(-135)
            turtle.down()
            turtle.forward(226)
            turtle.up()

        elif pos == "O":
            #Draw O
            turtle.up()
            turtle.goto(x + 100, y - 180)
            turtle.setheading(0)
            turtle.down()
            turtle.circle(80)
            turtle.up()
        x += 200
        if x > 100:
            x = -300
            y -= 200


def clicked(board, x, y):
    #sig: list(str), int, int -> NoneType
    global turn, pieces
    turtle.onscreenclick(None)  # disabling handler when inside handler
    column = (x + 300) // 200
    row = (y - 300) // -200
    square = int(row * 3 + column)
    print("User clicked ", x, ",", y, " at square ", square)

    if pieces[square] == "_":
        pieces[square] = turn
        if turn == "X":
            turn = "O"
        else:
            turn = "X"
        drawgame(pieces)
    else:
        print("That square is already taken")
    turtle.onscreenclick(clicked)

def computer_AI(board):
    #sig: list(str) -> NoneType

def gameover(board):
    #sig: list(str) -> bool
    #checks gameover on board if there is a three in a row pattern or not

def handler(x, y):
    #sig: int, int -> NoneType
    if clicked(the_board, x, y):
        drawgame(the_board)
        if not gameover(pieces):
            computer_AI(pieces)
            drawgame(pieces)
            gameover(pieces)


def main():
    #Runs the game 
    turtle.tracer(0,0)
    turtle.hideturtle()
    turtle.onscreenclick(handler)
    drawgame(pieces)
    turtle.mainloop()

main()

我正在尝试实现此输出：

感谢任何帮助。

Answer 1

For my clicked function in my code, I am trying to return a bool but I am not sure if I am doing it right.

clicked() 函数是一个事件处理器，它不会 return 任何东西 到 任何人。我们必须以此为基础进行设计。我在下面重新编写了您的代码，结合了 hander() 和 clicked() 函数，现在 "plays"。即用户先走，为"X"，然后电脑响应为玩家"O":

import turtle
import random

board = ["_", "_", "_", "_", "_", "_", "_", "_", "_"]

def drawgame(board):
    # draw board
    turtle.setup(600, 600)
    turtle.bgcolor("silver")
    turtle.color("white")
    turtle.hideturtle()
    turtle.speed('fastest')
    turtle.width(10)
    turtle.up()

    # Horizontal bars
    turtle.goto(-300, 100)
    turtle.down()
    turtle.forward(600)
    turtle.up()
    turtle.goto(-300, -100)
    turtle.down()
    turtle.forward(600)
    turtle.up()

    # Vertical bars
    turtle.setheading(-90)
    turtle.goto(-100, 300)
    turtle.down()
    turtle.forward(600)
    turtle.up()
    turtle.goto(100, 300)
    turtle.down()
    turtle.forward(600)
    turtle.up()

    turtle.color("blue")
    x, y = -300, 300

    for pos in board:
        if pos == "X":
            # Draw X
            turtle.up()
            turtle.goto(x + 20, y - 20)
            turtle.setheading(-45)
            turtle.down()
            turtle.forward(226)
            turtle.up()
            turtle.goto(x + 180, y - 20)
            turtle.setheading(-135)
            turtle.down()
            turtle.forward(226)
            turtle.up()
        elif pos == "O":
            # Draw O
            turtle.up()
            turtle.goto(x + 100, y - 180)
            turtle.setheading(0)
            turtle.down()
            turtle.circle(80)
            turtle.up()

        x += 200
        if x > 100:
            x = -300
            y -= 200

def handler(x, y):
    # sig: list(str), int, int -> NoneType

    turtle.onscreenclick(None)  # disabling handler when inside handler

    column = (x + 300) // 200
    row = (y - 300) // -200
    square = int(row * 3 + column)

    if board[square] == "_":
        board[square] = "X"
        drawgame(board)
        if not gameover(board):
            computer_AI(board)
            drawgame(board)

            if not gameover(board):
                turtle.onscreenclick(handler)  # allow player to take a turn
    else:
        print("That square is already taken!")
        turtle.onscreenclick(handler)  # allow player to retake turn

def computer_AI(board): # sig: list(str) -> NoneType

    """ stupid robot player, just picks randomly from what's available """

    available = [index for index, character in enumerate(board) if character == "_"]

    if available:
        index = random.choice(available)
        board[index] = "O"

def gameover(board):
    # sig: list(str) -> bool
    # checks game over on board if there is a three in a row pattern or not

    pass  # to be implemented!

    return False

def main():
    # Runs the game
    turtle.hideturtle()
    turtle.onscreenclick(handler)
    drawgame(board)
    turtle.mainloop()

main()

要做的事情：判断游戏是否结束，谁赢了的逻辑还不存在，所以你需要写那个。目前它只是 returns False 表示游戏还没有结束。 computer_AI() 代码根本没有智能，它只是记录所有空心方块并随机选择一个。你需要改进这一点。

将函数 return 设为布尔值并在 tic/tac/toe 游戏中实现 AI

Make function return boolean and implement AI in tic/tac/toe game

python

turtle-graphics

tic-tac-toe

python-3.x