Koch Curve

本文介绍如何使用Python编写一个程序,通过递归调用生成n级Koch曲线。从0到100的区间作为起始线段,根据给定的整数n决定复杂度,输出科赫曲线上的每个点,精确到小数点后8位。程序涉及几何变换和递归算法,适合学习算法和图形绘制。

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Koch Curve(科赫曲线)

Write a program which reads an integer n and draws a Koch curve based on recursive calles of depth n.

The Koch curve is well known as a kind of fractals.

You can draw a Koch curve in the following algorithm:

Divide a given segment (p1, p2) into three equal segments.
Replace the middle segment by the two sides of an equilateral triangle (s, u, t) of the same length as the segment.
Repeat this procedure recursively for new segments (p1, s), (s, u), (u, t), (t, p2).
在这里插入图片描述

You should start (0, 0), (100, 0) as the first segment.

Input
An integer n is given.

Output
Print each point (x, y) of the Koch curve. Print a point in a line. You should start the point(0, 0), which is the endpoint of the first segment and end with the point (100, 0), the other endpoint so that you can draw the Koch curve as an unbroken line. Each solution should be given as a decimal with an arbitrary number of fractional digits, and with an absolute error of at most 10-4.

Constraints
0 ≤ n ≤ 6
Sample Input 1
1
Sample Output 1
0.00000000 0.00000000
33.33333333 0.00000000
50.00000000 28.86751346
66.66666667 0.00000000
100.00000000 0.00000000
Sample Input 2
2
Sample Output 2
0.00000000 0.00000000
11.11111111 0.00000000
16.66666667 9.62250449
22.22222222 0.00000000
33.33333333 0.00000000
38.88888889 9.62250449
33.33333333 19.24500897
44.44444444 19.24500897
50.00000000 28.86751346
55.55555556 19.24500897
66.66666667 19.24500897
61.11111111 9.62250449
66.66666667 0.00000000
77.77777778 0.00000000
83.33333333 9.62250449
88.88888889 0.00000000
100.00000000 0.00000000
Notes

	double th=M_PI*60.0/180;  //把单位从度变为弧度,M_PI就是Π
	s.x=(2.0*a.x+1.0*b.x)/3.0; //求s点的坐标 
	s.y=(2.0*a.y+1.0*b.y)/3.0;
	t.x=(1.0*a.x+2.0*b
### 绘制Koch雪花分形图 要使用Python绘制Koch雪花分形图,可以通过`turtle`库实现。以下是详细的说明以及代码示例。 #### 使用Turtle库绘制Koch曲线 Koch曲线是由线段逐步替换形成的复杂几何图案。通过递归函数可以轻松实现这一目标。每一段直线被分为三部分,并在中间部分构建一个新的三角形[^2]。 ```python import turtle def koch_curve(t, iterations, length, shortening_factor, angle): if iterations == 0: t.forward(length) else: iterations -= 1 length /= shortening_factor koch_curve(t, iterations, length, shortening_factor, angle) t.left(angle) koch_curve(t, iterations, length, shortening_factor, angle) t.right(2 * angle) koch_curve(t, iterations, length, shortening_factor, angle) t.left(angle) koch_curve(t, iterations, length, shortening_factor, angle) # 初始化画布和海龟对象 screen = turtle.Screen() t = turtle.Turtle() # 设置参数 iterations = 4 # 迭代次数 length = 300 # 初始长度 shortening_factor = 3 # 长度缩短因子 angle = 60 # 转角角度 # 移动到起始位置 t.penup() t.backward(length / 2) t.pendown() # 开始绘制Koch曲线 koch_curve(t, iterations, length, shortening_factor, angle) # 关闭窗口 screen.exitonclick() ``` 上述代码定义了一个名为`koch_curve`的递归函数,用于绘制单条Koch曲线的一部分。通过调整迭代次数和其他参数,可以获得不同层次的细节效果。 #### 构建完整的Koch雪花 为了形成完整的Koch雪花,需要将三条Koch曲线组合成一个封闭的形状。这可以通过三次调用`koch_curve`函数完成,每次旋转一定的角度以连接下一条边。 ```python def draw_koch_snowflake(iterations, length, shortening_factor=3, angle=60): screen = turtle.Screen() t = turtle.Turtle() t.speed(0) # 加快速度 for _ in range(3): # 循环三次以构成雪花 koch_curve(t, iterations, length, shortening_factor, angle) t.right(120) # 每次转过120° draw_koch_snowflake(4, 300) screen.exitonclick() ``` 此代码片段扩展了之前的逻辑,在循环中依次绘制三个方向上的Koch曲线,从而组成最终的Koch雪花图形。 --- ###
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