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[英]Koch snowflake rendering time (and how to draw a snowflake using turtle)
[英]How would I go about making it so that each new generation of triangles in this koch snowflake is a different color? (Python + Turtle)
因此,我有一个使用以下两个主要功能生成六角形科赫雪花的程序:
def mvmt(length):
if length <= 10:
voldemort.forward(length)
else:
mvmt(length/3)
voldemort.right(60)
mvmt(length/3)
voldemort.left(120)
mvmt(length/3)
voldemort.right(60)
mvmt(length/3)
def whole(length = 300):
voldemort.hideturtle()
voldemort.penup()
voldemort.goto(-300,-255)
voldemort.pendown()
voldemort.begin_fill()
mvmt(length)
voldemort.left(60)
mvmt(length)
voldemort.left(60)
mvmt(length)
voldemort.left(60)
mvmt(length)
voldemort.left(60)
mvmt(length)
voldemort.left(60)
mvmt(length)
voldemort.end_fill()
我如何使通过迭代过程添加的每组新三角形都是一种新颜色?
我宁愿不使用更改填充颜色的细致过程,然后运行“ voldemort.beginfill()”和“ voldemort.endfill()”。 非常感谢您的帮助。 注意这是使用Turtle模块以python编写的。
一种方法是在mvmt()
克隆乌龟,这样您就可以独立填充一个乌龟,而不会干扰外部的填充。
为了使该递归工作,您必须将当前的turtle作为参数传递给mvmt()
。
那就是我将您的代码更改为:
#!/usr/bin/env python
from __future__ import division
import colorsys
import turtle
def mvmt(original, length):
if length >= 10:
fraction = length / 3
h, s, v = colorsys.rgb_to_hsv(*original.fillcolor())
clone = original.clone()
clone.fillcolor(colorsys.hsv_to_rgb(h + 1/6, s, v))
clone.begin_fill()
mvmt(clone, fraction)
clone.right(60)
mvmt(clone, fraction)
clone.left(120)
mvmt(clone, fraction)
clone.right(60)
mvmt(clone, fraction)
clone.end_fill()
original.forward(length)
def whole(length = 300):
voldemort = turtle.Turtle()
voldemort.speed(0)
voldemort.hideturtle()
voldemort.penup()
voldemort.goto(-200,-255)
voldemort.pendown()
voldemort.fillcolor((1.0, 0.5, 0.5))
voldemort.begin_fill()
mvmt(voldemort, length)
voldemort.left(60)
mvmt(voldemort, length)
voldemort.left(60)
mvmt(voldemort, length)
voldemort.left(60)
mvmt(voldemort, length)
voldemort.left(60)
mvmt(voldemort, length)
voldemort.left(60)
mvmt(voldemort, length)
voldemort.end_fill()
turtle.delay(0)
turtle.speed(0)
whole()
turtle.exitonclick()
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