# How to generate a combination of vectors that are as different in direction as possible? [closed]

I want to generate a combination of vectors of $N$ dimensions so that their directions are as different as possible if you go down the list.

So, if $N=2$, the first would be $[1, 0]$, the second would be $[-1, 0]$, the third would be $[0, 1]$, the fourth would be $[0, -1]$, and so forth, but extend this to allow for any number of dimensions.

No table lookup, please. I need an algorithm that can do this.

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Don't you just want a list of vectors that have 1 or -1 is some position and 0 everywhere else? Or can there be more vectors that (twice the) dimensions? – Karolis Juodelė Oct 3 '12 at 16:57
There can be more vectors than dimensions. It basically needs to be a complete combination, but sorted in a way where the difference in direction between the vectors as an aggregation will always be maximized while going through the list top to bottom. – user644337 Oct 3 '12 at 17:01
Well, the first and the third have the same direction ;-) And the "so forth" in the example leaves me wondering... does the sequence (for N=2) continues? – leonbloy Oct 3 '12 at 17:07
The first and third are not the same. One points down and the other points right. The 5th one in the sequence could be [1, 1], [-1, -1], [1, -1], or [-1, 1] – user644337 Oct 3 '12 at 21:56

## closed as not a real question by Jasper Loy, Thomas, tomasz, Nate Eldredge, NorbertOct 6 '12 at 22:05

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

What about these $2n$ vectors: $(\pm 1,0,0,\dots,0)$, $(0,\pm 1,0,\dots,0)$, $(0,0,\pm 1,\dots,0)$, ..., $(0,0,0,\dots,\pm 1)$?
 What is the number of vectors as a function of $N$? – Stefan Smith Oct 3 '12 at 22:30