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).

Edited: Difference in use of m(0)from problem statement. For this solution m=1 for one point, m=2 for two points, etc.

Solution: n!/((n-m)!*m!) * 2^m

For any hyper-dimension, a point is adjacent to another point if two coordinates toggle by one.  For 4 dimensions, the 8 points are {(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1),(-1,0,0,0),(0,-1,0,0),(0,0,-1,0),(0,0,0,-1)}

Rewrite points as values in 4 dimensions: +t,+x,+y,+z,-t,-x,-y,-z. If the "+", and "-" are variations then:

A vertices is any point: Total 8 points.  m(1) = 4!/((4-1)!*1!) * 2^1 = 8 vertices

An edge is the connection of any two points not in the same dimension: Total 24 edges. m(2)=4!/((4-2)!*2!) * 2^2 = 6 * 4 = 24 (Four dimensions 2 selections = 6 combinations. 2 selections 2 variations = 4)

A surface is the connection of any three points not in the same dimension: Total 32 faces. m(3)=4!/((4-3)!*3!) * 2^3 = 4 * 8 = 32 (Four dimensions 3 selections = 4 combinations. 3 selections 2 variations = 8)

A space is the connection of any four points not in the same dimension: Total 16 spaces.  m(4)=4!/((4-4)!*4!) * 2^4 = 1 * 16 = 16 (Four dimensions 4 selections = 1 combination. 4 selections 2 variations = 16)

A 4D-space is the connection of any five points not in the same dimension: Total 0

Note: A 4D-space would be bounded by five points all connected to each other.  This does not exist, just like in the original two dimensional diamond, there is not one face bounded by three points.

For five dimensions:
m(1) = 5!/(4!*1!)*2^1 = 10 vertices
m(2) = 5!/(3!*2!)*2^2 = 40 edges
m(3) = 5!/(2!*3!)*2^3 = 80 faces
m(4) = 5!/(1!*4!)*2^4 = 80 spaces
m(5) = 5!/(0!*5!)*2^5 = 32 4D-spaces

For six dimensions
m(1) = 6!/(5!*1!)*2^1 = 12 vertices
m(2) = 6!/(4!*2!)*2^2 = 60 edges
m(3) = 6!/(3!*3!)*2^3 = 160 faces
m(4) = 6!/(2!*4!)*2^4 = 240 spaces
m(5) = 6!/(1!*5!)*2^5 = 192 4D-spaces
m(6) = 6!/(0!*6!)*2^6 = 64 5D-spaces

Edited on October 19, 2006, 12:49 pm

Posted by Leming on 2006-10-19 12:43:10