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).
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
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Posted by Charlie
on 2006-10-19 12:28:33 |
most of a solution
