If you paint the six sides of a 3x3x3 cube and then slice it into 27 unit cubes, how many will have paint on 0, 1, 2, 3 of their square sides?
Generalize to any dimension:
If you paint the sides of a n dimensional cube with sides length 3 and slice it into 3^{n} unit cubes, how many will have paint on 0, 1, 2, ..., n of their (n1) dimensional sides?
Paint on 0 sides will of course be limited to just the one central (hyper)cube.
Those with 1 side painted will be those which are at the center of one of the ndimensional cube's (n1)dimensional faces. Two sides will be painted on each of the cubes that lie along one of the (n2)dimensional edges (using terminology based on the 3dimensional cube as an example). Three sides will be painted on each of the cubes that lie on the bound that has (n3) dimensions.
So the question becomes: how many of each lower dimension bound exist for a hypercube of n dimensions. In going from one dimension to the next, each 0dimensional point is doubled. The number of 1dimensional sides is doubled plus each of the 0dimensional points traces out a 1dimensional line. This continues as each dimension n+1 has its number of kdimensional units equal to twice that of dimension n plus dimension n's number of (k1)dimensional units.
dimensions elements' dimensions
of
hypercube 0 1 2 3 4 5 6 7 8
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
so for example the 8dimensional hypercube will have 1 hypercubelet with no paint, 16 with one hyperface painted, 112 with two, 448 with three, etc.
The Wikipedia article on Hypercube gives the formula for the number of mdimensional hypercubes on the boundary of an ndimensional hypercube as:
2^(nm)*C(n,m)
This checks against the table.
This needs to be translated into a formula for k = 0, 1, ..., n of their (n1) dimensional sides being painted.
k + m = n
m=nk
2^(k+mm)*C(k+m,m) = 2^(k)*C(n,nk) is then the number of kpaintedsided ndimensional cubelets.
Take the case of 3 dimensional cubelets with two painted faces:
2^(2)*C(3,1) = 4*3 = 12.
Edited on March 5, 2015, 3:44 pm

Posted by Charlie
on 20150305 15:41:41 