insable

2011-04-12

2011-01-16

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

Custom Search API

标签: Science and Technology

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2010-12-30

解决一种MSN运行缓慢的情况

标签: Science and Technology

最近加装虚拟机软件，带来了副作用，MSN慢到无法接受，
比如随便打开一个聊天窗口，大概要两、三分钟，
按下表情符号接近于就直接当掉，要过相当长的时间才会恢复响应，
或者是打字会奇慢无比，过几秒才会出现
点下去居然也要一段时间窗口才会出现…

解决办法就是
把局域网络IP设为自动取得
原因就是MSN在启动对话Session的时候
会先判别是不是内网的使用者
(不需要经过因特网,直接透过局域网传播信息)
但若设了一个假的局域网络IP（虚拟机软件的原因,ip需要做相应的调整）
封包送出去便卡在那边等待回传信息，
但因为是假的IP,所以不会有回传信息出现,直到Timeout以后他才会改用因特网的方式通讯..
所以使用者其实是在一直等待Timeout，速度必然会很慢。

2010-10-25

I will persist. I will win

标签: short essay

In the Orient young bulls are tested for the fight arena in a certain manner. Each is brought to the ring and allowed to attack a picador who pricks them with a lance. The bravery of each bull is then rated with care according to the number of times he demonstrates his willingness to charge in spite of the sting of the blade. Henceforth will I recognize that each day I am tested by life in like manner. If I persist, if I continue to try, if I continue to charge forward, I will succeed.

在古老的东方，挑选小公牛列竞技场格斗有一定的程序、它们被带进场地，向手持长矛的斗牛士攻击，裁判以它受激后再向斗牛士进攻的次数多寡来评定这只公牛的勇敢程度。从今往后。我须承认，我的生命每天都在接受类似的考验。如果我坚韧不拔，勇往直前，迎接挑战。那么我一定会成功。