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 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).

 See The Solution Submitted by Tristan Rating: 4.0000 (1 votes)

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A hyper-diamond will be the dual of a hypercube in dimension n: that is, it will have a vertex corresponding to each (n-1)-face of the hypercube. When n=2, the hyperdiamond (or diamond in this case) is a square, which is its own dual, as the square is the hypercube (actually hypocube) of dimension n, and has 4 vertices as well as 4 edges (the faces of dimension n-1 = 1).  For octahedra and cubes, the octahedron has 6 vertices corresponding to the 6 faces of a cube, 12 edges corresponding to the 12 edges of the cube and 8 faces, corresponding to the 8 vertices of the cube. So the m-elements on an n-hyper-diamond correspond to the (n-1-m)-elements of a hypercube.

An n-hypercube will have 1 element of dimension n, and the number of elements of lower dimension, d, will equal twice the number of elements of dimension d in an (n-1)-hypercube plus the number of elements of dimension d-1 in the (n-1)-hypercube.  Thus for example a 3-cube's 2-faces number 6, which is twice the number of squares in a square (2*1=2) plus the number of 1-faces (edges) of a square (4). The 3-cube's 12 edges are twice the square's edges (2*4=8) plus the number of the square's vertices (4).

A table of the various dimensioned hypercubes:

` 1     2     1 2     4     4     1 3     8    12     6     1 4    16    32    24     8     1 5    32    80    80    40    10     1 6    64   192   240   160    60    12     1 7   128   448   672   560   280    84    14     1 8   256  1024  1792  1792  1120   448   112    16     1 9   512  2304  4608  5376  4032  2016   672   144    18     110  1024  5120 11520 15360 13440  8064  3360   960   180    20     111  2048 11264 28160 42240 42240 29568 14784  5280  1320   220    22     1`

So a hypocube of dimension 1 has 2 vertices and is one line segment or edge.
A hypocube of dimension 2 has 4 vertices, 4 edges and is one square.
A cube of dimension 3 has 8 vertices, 12 edges, 6 square faces and is one cube.
A hypercube of dimension 4 has 16 vertices, 32 edges, 24 square faces, 8 cubic faces, and is 1 4-cube.
... etc.

For the hyper-diamonds these are taken in reverse: a 3-hyperdiamond has 6 vertices, 12 edges and 8 faces. A 6-hyperdiamond has 12 vertices, 60 2-dimensional faces, 160 3-dimensional faces, etc.

But unlike the hypercubes, where the elements making them up are all hypercubes of differing dimension, such hyperdiamond elements are not hyperdiamonds themselves: the faces of the octahedron are triangles, not diamonds. In fact I don't know what shapes most of them are.

I'm sure there's an actual formula for getting at a value elem(n,m), but the iterative formula above is all I have readily come up with.

The hypercube table above comes from:

DIM elem(1, -1 TO 300)

elem(0, 0) = 1
FOR d = 1 TO 11
PRINT USING "##"; d;
FOR i = 0 TO d
elem(1, i) = elem(0, i) * 2 + elem(0, i - 1)
PRINT USING "######"; elem(1, i);
NEXT
FOR i = 0 TO d
elem(0, i) = elem(1, i)
NEXT
PRINT
NEXT

 Posted by Charlie on 2006-10-19 12:28:33

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