生成给定幅度的随机5d向量 [英] Generate random 5d vector of given magnitude
问题描述
给定五维空间，我想生成100个向量，所有向量都具有固定的幅度= M，其中组件值是随机分布的。
Given a five dimensional space, I would like to generate 100 vectors, all with a fixed magnitude=M, where the components values are randomly distributed.
我是最初考虑从单位矢量开始，然后应用旋转矩阵，随机参数为10自由度......这会起作用吗？以及如何？
I was originally thinking of starting off with a unit vector and then applying a rotation matrix, with random parameters for the 10 degrees of freedom ... Would this work? and how?
在Javascript中执行此操作的任何好方法......？
Any nice way of doing this in Javascript...?
欢呼任何指针！
推荐答案
这是我将使用的蒙特卡罗算法（我不知道Javascript是否足以在顶部编码我的头脑）：
Here is the Monte Carlo algorithm that I would use (I do not know Javascript well enough to code in it off the top of my head):
-
为五个维度中的每一个生成-1到1范围内的随机值。
Generate random values in the range from -1 to 1 for each of the five dimensions.
计算幅度M，如果M = 0或M> 1则拒绝这些值并返回步骤＃1。
Calculate the Magnitude M, if M=0 or M>1 then reject these values and return to step #1.
将矢量标准化为幅度为1（将每个维度除以M）。
Normalize the vector to have a Magnitude of 1 (divide each dimension by M).
这应该为你提供均匀分布在5维超球面上的随机单位向量。
That should give you random unit vectors evenly distributed over the 5-dimensional super-sphere surface.
有人提出这样的问题：如果M> 1，为什么拒绝向量？
答案：这样最终的向量将会均匀分布在单元5-sphe的表面上重新
Answer: So that the final vectors will be uniformly distributed across the surface of the unit 5-sphere.
推理：我们在第一步生成的是一组随机向量均匀分布在单元5立方体的体积内。这些矢量中的一些也在单位5球的体积内，并且其中一些在该体积之外。如果归一化，5球内的矢量均匀分布在其表面上，然而，其外部的矢量根本不均匀分布。
Reasoning: What we are generating in the first step is a set of random vectors that are uniformly distributed within the volume of the unit 5-cube. Some of those vectors are also within the volume of the unit 5-sphere and some of them are outside of that volume. If normalized, the vectors within the 5-sphere are evenly distributed across its surface, however, the ones outside it are not at all evenly distributed.
想象一下：就像普通的三维单位立方体和单位球，甚至单位正方形和单位圆一样，单位5球体完全包含在单位5立方体内，仅接触在五个正单位维度轴点：
Think about it like this: Just as with a normal 3-dimensional Unit Cube and Unit Sphere, or even the Unit Square and the Unit Circle, the Unit 5-Sphere is wholly contained within the Unit 5-Cube, which touches only at the five positive unit dimension axis points:
(1,0,0,0,0)
(0,1,0,0,0)
(0,0,1,0,0)
(0,0,0,1,0)
(0,0,0,0,1)
及其对应的负单位轴点。这是因为这些是立方体表面上的唯一点，其大小（距离原点的距离）为1，在所有其他点，5立方体的表面与原点的距离更大比1。
and their corresponding negative unit axis points. This is because these are the only points on the surface of the cube that have a magnitude (distance from the origin) of 1, at all other points, the 5-cube's surface has a distance from the origin that is greater than 1.
这意味着（0,0,0,0,0）和（1,1,1之间还有更多的点数， 1,1）比（0,0,0,0,0）和（1,0,0,0,0）之间的差距大。事实上关于SQRT（5）或aprx。 2.25倍以上。
And this means that there are many more points between (0,0,0,0,0) and (1,1,1,1,1) than there are between (0,0,0,0,0) and (1,0,0,0,0). In fact about SQRT(5) or aprx. 2.25 times more.
这意味着如果你在单位5-cube中包含了所有的向量，你最终会得到两倍以上的数量结果随机映射到约（0.44,0.44,0.44,0.44,0.44）而不是（1,0,0,0,0）。
And that means that if you included all of the vectors in the unit 5-cube, you would end up with more than twice as many results "randomly" mapping to about (0.44,0.44,0.44,0.44,0.44) than to (1,0,0,0,0).
对于那些挑战（没有基础，恕我直言）这导致5-D球体表面均匀分布的人，请参阅维基百科文章： http://en.wikipedia.org/wiki/N-sphere#Alternatives
For those who are challenging (without foundation, IMHO) that this results in a uniform distribution across the surface of the 5-D Sphere, please see the Wikipedia article: http://en.wikipedia.org/wiki/N-sphere#Alternatives
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