使用海龟图形和递归的希尔伯特曲线 [英] Hilbert curve using turtle graphics and recursion

问题描述

我正在尝试使用 python 海龟图形和递归来实现 L 系统生成的希尔伯特曲线.我的代码似乎适用于递归 n=1 和 n=2 的前两个级别，但除此之外，图形只是纠缠不清(尽管我能够观察其中的更多模块)，我似乎无法理解这里可能有什么问题，我是否需要一些中间步骤来重新生成 Hilbert 模块以进行更深层次的递归?请看我下面的代码，它比较简单:

导入海龟def Hilbert_curve(A,rule,t,n):如果n>=1:如果规则:t.left(90)Hilbert_curve(A,not rule,t, n-1)t.forward(A)右(90)Hilbert_curve(A, rule,t, n-1)t.forward(A)Hilbert_curve(A,rule,t, n-1)右(90)t.forward(A)Hilbert_curve(A,not rule,t, n-1)t.left(90)别的:右(90)Hilbert_curve(A,rule,t, n-1)t.forward(A)t.left(90)Hilbert_curve(A,not rule,t, n-1)t.forward(A)Hilbert_curve(A,not rule,t, n-1)t.left(90)t.forward(A)Hilbert_curve(A, rule,t, n-1)右(90)定义主():A=10t=turtle.Turtle()my_win=turtle.Screen()n=2规则=真Hilbert_curve(A,rule,t,n)my_win.exitonclick()主要的()

I'm trying to implement an L-System generated Hilbert curve ,making use of python turtle graphics and recursion. My code seems to be working for the first two levels of recursion n=1 and n=2 but beyond that , the graphics just entangled (although I´m able to observe further modules within them), and I can´t seem to grasp what might be wrong here, do I need some intermediate steps to regenerate the Hilbert modules for deeper levels of recursion? Please see my code below , its relatively simple:

import turtle

def Hilbert_curve(A,rule,t,n):

    if n>=1:
        if rule:
            t.left(90)
            Hilbert_curve(A,not rule,t, n-1)
            t.forward(A)
            t.right(90)
            Hilbert_curve(A, rule,t, n-1)
            t.forward(A)
            Hilbert_curve(A,rule,t, n-1)
            t.right(90)
            t.forward(A)
            Hilbert_curve(A,not rule,t, n-1)
            t.left(90)
        else:
            t.right(90)
            Hilbert_curve(A,rule,t, n-1)
            t.forward(A)
            t.left(90)
            Hilbert_curve(A,not rule,t, n-1)
            t.forward(A)
            Hilbert_curve(A,not rule,t, n-1)
            t.left(90)
            t.forward(A)
            Hilbert_curve(A, rule,t, n-1)
            t.right(90)

def main():
    A=10
    t=turtle.Turtle()
    my_win=turtle.Screen()
    n=2
    rule=True
    Hilbert_curve(A,rule,t,n)
    my_win.exitonclick()

main()

Hilbert n=2

Hilbert n=3

解决方案

The problem is with your else clause. The rule is already inverted coming in to the function, so you need to treat the rule the same as the then clause:

    else:
        t.right(90)
        Hilbert_curve(A, not rule, t, n - 1)
        t.forward(A)
        t.left(90)
        Hilbert_curve(A, rule, t, n - 1)
        t.forward(A)
        Hilbert_curve(A, rule, t, n - 1)
        t.left(90)
        t.forward(A)
        Hilbert_curve(A, not rule, t, n - 1)
        t.right(90)

However, if we change rule from a boolean to a number, parity, that's either 1 or -1, and then multiply parity by the angle, we can eliminate one of the two clauses of the orignal if statement:

from turtle import Screen, Turtle

def hilbert_curve(turtle, A, parity, n):

    if n < 1:
        return

    turtle.left(parity * 90)
    hilbert_curve(turtle, A, - parity, n - 1)
    turtle.forward(A)
    turtle.right(parity * 90)
    hilbert_curve(turtle, A, parity, n - 1)
    turtle.forward(A)
    hilbert_curve(turtle, A, parity, n - 1)
    turtle.right(parity * 90)
    turtle.forward(A)
    hilbert_curve(turtle, A, - parity, n - 1)
    turtle.left(parity * 90)

screen = Screen()

yertle = Turtle()
yertle.speed('fastest')  # because I have no patience

hilbert_curve(yertle, 10, 1, 4)

screen.exitonclick()

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