乌龟比赛游戏无法保持乌龟在squ

Question

我和我的朋友目前正在学校编写一个练习，我们遇到了一些困难，如何让海龟们保持正常的轨道。举例来说，在广场的第一面，有两只乌龟在两边赛跑。然而，当他们旋转90度时，其中1或2人会离开侧面，这与练习要求不符。在

以下是我们的代码：

import turtle
from random import randint
def read_int(prompt,first,last):
    x = int(input(prompt))
    while x < first or x > last:
        print("Not in range. Try Again!!!")
        x= int(input(prompt))
    return x

square_count = read_int("Enter your laps between 1 and 10: ",1,10)
print(square_count)
#def t():

window = turtle.Screen()
window.bgcolor('lightblue')
def draw_square(turtle, center, size):
    xPt, yPt = center
    xPt -= size / 2
    yPt += size / 2
    side = 4
    size = 300
    angle = 90
    turtle.speed(0)
    turtle.up()
    turtle.goto(xPt, yPt)
    turtle.down()
    for i in range(side):
        turtle.forward(size)
        turtle.right(angle)

t = turtle.Turtle()
draw_square(t,(0,0),300)
t.shape('turtle')
t.color("red")
t.pensize(5)
t.up()
t.goto(-150, 150)

#def r():
r = turtle.Turtle()
draw_square(r,(0,0),300)
r.shape('turtle')
r.color("yellow")
r.up()
r.pensize(5)
r.goto(-150, 150)

sides = 4
size = 300


count_int = int(square_count)* sides
if count_int > 1:
    for sides in range(count_int):
            i = 0
            e = 0
            while i in range(0, size) or e in range(0, size):
                t_step = randint(1, 5)
                t.forward(t_step)
                i = i + t_step
                r_step = randint(1, 5)
                r.forward(r_step)
                e = e + r_step

            t.right(90)
            r.right(90)





window.exitonclick()

Answer 1

我将采用一种稍微不同的方法——而不是硬编码两个海龟，而是为任意数量的海龟设计，然后只与两只海龟比赛。这里有一个基于任意数量的海龟的重写，其中包括一些style代码的返工，以及@Claudio的一些代码建议：

from random import randint
from turtle import Turtle, Screen
from collections import defaultdict

SIZE = 300
SIDES = 4
ANGLE = 90
MAXIMUM_STRIDE = 5

COLORS = ('red', 'gold', 'green', 'orange', 'blue')

MAX_RACERS = len(COLORS)

def read_int(prompt, first, last):
    x = int(input(prompt))

    while not first <= x <= last:
        print('Not in range! Try Again.')
        x = int(input(prompt))

    return x

def draw_square(turtle, center, size):
    xPt, yPt = center

    xPt -= size / 2
    yPt += size / 2

    turtle.up()
    turtle.goto(xPt, yPt)
    turtle.down()

    for _ in range(SIDES):
        turtle.forward(SIZE)
        turtle.right(ANGLE)

lap_count = read_int('Enter number of laps (between 1 and 10): ', 1, 10)
no_racers = read_int('Enter number of racers (between 2 and {}): '.format(MAX_RACERS), 2, MAX_RACERS)

jockey_colors = COLORS[0:no_racers]

window = Screen()
window.bgcolor('lightblue')

racers = defaultdict(dict)

draw_square(Turtle(visible=False), (0, 0), SIZE)

for color in jockey_colors:
    jockey = Turtle('turtle', visible=False)
    jockey.speed('fastest')
    jockey.color(color)

    jockey.up()
    jockey.goto(-SIZE/2, SIZE/2)
    jockey.showturtle()

    racers[color]['jockey'] = jockey
    racers[color]['sides'] = 0
    racers[color]['position'] = 0

finished = False

while not finished:
    for racer in racers.values():
        jockey = racer['jockey']

        step = randint(1, MAXIMUM_STRIDE + 1)

        if racer['position'] + step > SIZE:

            racer['sides'] += 1

            if racer['sides'] == lap_count * SIDES:
                finished = True
                break

            baby_step = SIZE - racer['position']
            jockey.forward(baby_step)
            jockey.right(ANGLE)
            racer['position'] = 0

            step -= baby_step

        jockey.forward(step)
        racer['position'] += step

window.exitonclick()

Answer 2

你的错误就在你的计划快结束的时候。在

while i in range(0, size) or e in range(0, size):
                t_step = randint(1, 5)
                t.forward(t_step)
                i = i + t_step
                r_step = randint(1, 5)
                r.forward(r_step)
                e = e + r_step

在这样一个场景中，假设i=298，并且t_step选择一个随机整数作为3。乌龟最终会移动超过300的范围，因此它会离开牵引箱。循环不会停止，因为它只意识到当所有代码都已运行时i超出范围。（也就是说，当循环被打破时，乌龟已经离开了盒子。）这就是为什么乌龟会走出盒子。你可以试着用一个“如果”条件来防止这种情况发生。在

Answer 3

你有两种问题。
首先，你应该注意你的海龟走了多少步。因为如果一个正方形的大小是300，那么海龟所走的步数之和可能不等于300。我建议您将随机步骤的生成更改为：

steps = random.randrange(0, 4, 2)

这样一来，一只乌龟只需走0到2步，你就可以确信总数将达到300步。
第二点也许更重要。如果你像你一样构造了竞赛循环，那么只有当另一只乌龟到达了一行的末尾时，一只乌龟才会转向，你应该重新组织代码，使这两个海龟彼此独立。在