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2011-01-16

科赫雪花/コッホ雪片/Koch snowflake

外面开始下雪了，入冬的第一场有规模的雪。

科赫曲线是一种分形。其形态似雪花，又称科赫雪花、雪花曲线。其豪斯多夫维是

log4 / log3

。

它最早出现在Helge von Koch的论文《关于一条连续而无切线，可由初等几何构作的曲线》（1904年，法语原题：Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire）。

科赫曲线是 de Rham曲线的特例。

给定线段AB，科赫曲线可以由以下步骤生成：

将线段分成三等份（AC,CD,DB）
以CD为底，向外（内外随意）画一个等边三角形DMC
将线段CD移去
分别对AC,CM,MD,DB重复1~3。

生成过程演示

科赫雪花是以等边三角形三边生成的科赫曲线组成的。科赫雪花的面积是 $\frac{2\sqrt{3}(s^2)}{5}$ ，其中

s

是原来三角形的边长。每条科赫曲线的长度是无限大，它是连续而无处可微的曲线。

コッホ曲線（コッホきょくせん、英語: Koch curve）はフラクタル図形の一つ。スウェーデンの数学者ヘルゲ・フォン・コッホ (Helge von Koch) が考案した。線分を3等分し、分割した2点を頂点とする正三角形の作図を無限に繰り返すことによって得られる図形である。1回の操作で線分の長さが 4/3 倍になるので、操作を無限に繰り返して得られるコッホ曲線の長さは無限大である。完全なものは作図することができない。コッホ曲線は相似比が1/3の4個のセグメントから成っているので、フラクタル次元はlog4 / log3 = 1.26186...次元である。

コッホ雪片(コッホせっぺん)は、上記のコッホ曲線をつなぎ合わせ、始点と終点を一致させたものである。

コッホ雪片は、有限の面積であるにもかかわらず、無限の周囲を持つ。

Koch snowflake

The Koch curve can be constructed by starting with an equilateral triangle, then recursively altering each line segment as follows:

divide the line segment into three segments of equal length.
draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
remove the line segment that is the base of the triangle from step 2.

After one iteration of this process, the result is a shape similar to the Star of David.

The Koch curve is the limit approached as the above steps are followed over and over again.

Properties

The Koch curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage. Hence the total length increases by one third and thus the length at step n will be (4/3)ⁿ of the original triangle perimeter: the fractal dimension is log 4/log 3 ≈ 1.26, greater than the dimension of a line (1) but less than Peano's space-filling curve (2).

The Koch curve is continuous everywhere but differentiable nowhere.

Taking s as the side length, the original triangle area is $\frac{s^2\sqrt{3}}{4}$ . The side length of each successive small triangle is 1/3 of those in the previous iteration; because the area of the added triangles is proportional to the square of its side length, the area of each triangle added in the nth step is 1/9th of that in the (n-1)th step. In each iteration after the first, 4 times as many triangles are added as in the previous iteration; because the first iteration adds 3 triangles, the nth iteration will add $3 \cdot 4^{n-1}$ triangles. Combining these two formulae gives the iteration formula:

$A_{n+1} = A_n + \frac{3 \cdot 4^{n-1}}{9^n}A_0\quad n \ge 1 \, ,$

where

A 0

is area of the original triangle. Substituting in

$A_1 = \frac{4}{3} A_0 \, ,$

and expanding yields:

$A_{n+1} = \frac{4}{3} A_0 + \sum_{k=2}^n \frac{3 \cdot 4^{k-1}}{9^k} A_0 = \left (\frac{4}{3} + \frac{1}{3} \sum_{k=1}^n \frac{4^{k}}{9^k} \right ) A_0 \, .$

In the limit, as n goes to infinity, the limit of the sum of the powers of 4/9 is 4/5, so

$\lim_{n \rightarrow \infty} A_n = \left( \frac{4}{3} + \frac{1}{3} \cdot \frac{4}{5} \right ) A_0 = \frac{8}{5} A_0 \, .$

So the area of a Koch snowflake is 8/5 of the area of the original triangle, or $\frac{2s^2\sqrt{3}}{5}$ . Therefore the infinite perimeter of the Koch triangle encloses a finite area.

Excel VBA Project:http://goo.gl/2vinu

2011-01-15

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