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Hyper-diamond (Posted on 2006-10-19) Difficulty: 3 of 5
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    
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Solution most of a solution | Comment 1 of 2

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     1
10  1024  5120 11520 15360 13440  8064  3360   960   180    20     1
11  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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