Consider a diamond, with the four corners at (1,0), (1,0), (0,1), and (0,1). The threedimensional equivalent of this diamond would be an octahedron with the additional vertices (0,0,1) and (0,0,1). The fourdimensional equivalent would have two more vertices (0,0,0,1) and (0,0,0,1). In general, call the resulting shape an nhyperdiamond.
If you have an nhyperdiamond, how many mdimensional hyperfaces 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 hyperdiamond will be the dual of a hypercube in dimension n: that is, it will have a vertex corresponding to each (n1)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 n1 = 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 melements on an nhyperdiamond correspond to the (n1m)elements of a hypercube.
An nhypercube 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 (n1)hypercube plus the number of elements of dimension d1 in the (n1)hypercube. Thus for example a 3cube's 2faces number 6, which is twice the number of squares in a square (2*1=2) plus the number of 1faces (edges) of a square (4). The 3cube'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 4cube.
... etc.
For the hyperdiamonds these are taken in reverse: a 3hyperdiamond has 6 vertices, 12 edges and 8 faces. A 6hyperdiamond has 12 vertices, 60 2dimensional faces, 160 3dimensional 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 20061019 12:28:33 