让一只Python只乌龟追逐另一只乌龟

Question

我正在创建一个游戏，用户使用键盘移动一只乌龟以避免另一只乌龟。这是我的代码：

import turtle
playGround = turtle.Screen()
playGround.screensize(500, 500)
playGround.title("Turtle Keys")

run = turtle.Turtle()
follow = turtle.Turtle()
run.shape("turtle")
follow.shape("turtle")
run.color("blue")
follow.color("red")
run.penup()
follow.penup()
run.st()

def k1():
    run.forward(45)
def k2():
    run.left(45)
def k3():
    run.right(45)
def k4():
    run.back(45)
def quitThis():
    playGround.bye()
playGround.onkey(k1, "Up")  # the up arrow key
playGround.onkey(k2, "Left") # the left arrow key
playGround.onkey(k3, "Right") # you get it!
playGround.onkey(k4, "Down")
playGround.onkey(quitThis,'q')
playGround.listen()

我想让红海龟追蓝海龟但是不行

Answer 1

您缺少的是 blue/follow 乌龟的计算机控制运动。我们可以通过添加一个 ontimer() 事件处理程序来做到这一点，该事件处理程序在 towards() 上调用 setheading() 以保持 blue/follow 面向 red/run。加上 blue/follow 上的一点点向前运动。像这样：

from turtle import Turtle, Screen

playGround = Screen()
playGround.screensize(500, 500)
playGround.title("Turtle Keys")

run = Turtle("turtle")
run.speed("fastest")
run.color("blue")
run.penup()
run.setposition(250, 250)

follow = Turtle("turtle")
follow.speed("fastest")
follow.color("red")
follow.penup()
follow.setposition(-250, -250)

def k1():
    run.forward(10)

def k2():
    run.left(45)

def k3():
    run.right(45)

def k4():
    run.backward(10)

def quitThis():
    playGround.bye()

def follow_runner():
    follow.setheading(follow.towards(run))
    follow.forward(1)
    playGround.ontimer(follow_runner, 10)

playGround.onkey(k1, "Up")  # the up arrow key
playGround.onkey(k2, "Left")  # the left arrow key
playGround.onkey(k3, "Right")  # you get it!
playGround.onkey(k4, "Down")
playGround.onkey(quitThis, 'q')

playGround.listen()

follow_runner()

playGround.mainloop()

您可以通过更改 forward() 语句中的移动量来调整 blue/follow 的性能。一旦超过 1，您会惊讶于它赶上 red/run.

的速度有多快

您需要添加代码来检测海龟何时发生碰撞以及随后发生的任何事情。

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