问题
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.
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?
Any nice way of doing this in Javascript...?
cheers for any pointers!
回答1:
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):
Generate random values in the range from -1 to 1 for each of the five dimensions.
Calculate the Magnitude M, if M=0 or M>1 then reject these values and return to step #1.
Normalize the vector to have a Magnitude of 1 (divide each dimension by M).
That should give you random unit vectors evenly distributed over the 5-dimensional super-sphere surface.
The question has been asked: "Why reject the vector if M>1?"
Answer: So that the final vectors will be uniformly distributed across the surface of the unit 5-sphere.
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.
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)
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.
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.
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).
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
回答2:
The problem with sampling from a unit hypercube in 5-dimeensions and then re-scaling, is points in some directions (towards the corners of the hypercube) will be over sampled.
But if you use a rejection scheme, then you lose too many samples. That is, the volume of a unit hypercube in 5-d is pi^2*(8/15) = 5.26378901391432. Compare that to the volume of a unit hyper-cube in 5 dimensions that will just contain the sphere. That hypercube has volume 32. So if you reject points falling outside the sphere, you will reject
1 - 5.26378901391432/32 = 0.835506593315178
or roughly 83.5% of the points get rejected. That means you will need to sample roughly 6 points on average before you do find a sample that is inside the 5-sphere.
A far better idea is to sample using a unit normal sample, then rescale that vector of points to have unit norm. Since the multi-variate normal distribution is spherically symmetric, there is no need for rejection at all.
回答3:
Here are some approaches, it is for unit vectors but you can just multiply by M:
http://burtleburtle.net/bob/rand/unitvec.html
回答4:
I'd recommend assigning random numbers between -1 and +1 for each element. Once all elements for a vector have been assigned, then you should normalize the vector. To normalize, simply divide each element by the magnitude of the overall vector. Once you've done that, you've got a random vector with a magnitude of 1. If you want to scale the vectors to magnitude M, you just need to multiply each element by M.
回答5:
Picking without rejections (6 times faster)
From Mathworld and a link posted by Bitwise:
If you choose n numbers with a normal distribution, treat those as coordinates, and normalize the resulting vector into a unit vector, this chooses a uniformly distributed unit vector in n dimensions.- Burtleburtle (ht Bitwise)
As Bitwise points out, that should be multiplied by the desired magnitude.
Note that there is no rejection step here; the distributions themselves have dealt with the necessary bias towards zero. Since the normal distribution and a semi-circle are not the same shape, I wonder if RBarryYoung's answer really gives a uniform distribution on the surface of the sphere or not.
Regarding uniform hypercube picking with hypersphere rejection (RBarryYoung's answer)
The Mathworld article includes a description of using 4 random numbers picked from uniform distributions to calculate random vectors on the surface of a 3d sphere. It performs rejection on the 4d vector of random numbers of anything outside the unit hypersphere, as RBarryYoung does, but it differs in that it (a)uses an extra random number and (b)performs a non-linear transform on the numbers to extract the 3d unit vector.
To me, that implies that uniform distributions on each axis with hypersphere rejection will not achieve a uniform distribution over the surface of the sphere.
来源：https://stackoverflow.com/questions/12591241/generate-random-5d-vector-of-given-magnitude