perplexus.info :: Shapes : Hyper-diamond

Hyper-diamond (Posted on 2006-10-19)

Consider a diamond, with the four corners at (1,0), (-1,0), (0,1), and (0,-1). The three-dimensional equivalent of this diamond would be an octahedron with the additional vertices (0,0,1) and (0,0,-1). The four-dimensional equivalent would have two more vertices (0,0,0,1) and (0,0,0,-1). In general, call the resulting shape an n-hyper-diamond.

If you have an n-hyper-diamond, how many m-dimensional hyper-faces does it have (where n>m≥0)? For example, in the case where n=3, an octahedron, there are 8 faces (m=2), 12 edges (m=1), and 6 vertices (m=0).

Submitted by Tristan

Rating: 4.0000 (1 votes)

Solution: (Hide)

C(n,m+1)*2^(m+1)
The m-dimensional hyper-face must have exactly m+1 different vertices, each of which lies on a different coordinate axis. So we can choose to use any m+1 dimensions out of the total n dimensions, giving C(n,m+1) possibilities. Since each coordinate axis has two possible points on it, we must multiple the number of possibilities by 2^(m+1).

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Solution	Leming	2006-10-19 12:43:10
most of a solution	Charlie	2006-10-19 12:28:33

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