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 Painting high dimension cubes (Posted on 2015-03-05)
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 3n unit cubes, how many will have paint on 0, 1, 2, ..., n of their (n-1) dimensional sides?

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 solution | Comment 2 of 4 |
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 n-dimensional cube's (n-1)-dimensional faces. Two sides will be painted on each of the cubes that lie along one of the (n-2)-dimensional edges (using terminology based on the 3-dimensional cube as an example). Three sides will be painted on each of the cubes that lie on the bound that has (n-3) 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 0-dimensional point is doubled. The number of 1-dimensional sides is doubled plus each of the 0-dimensional points traces out a 1-dimensional line. This continues as each dimension n+1 has its number of k-dimensional units equal to twice that of dimension n plus dimension n's number of (k-1)-dimensional units.

`dimensions    elements' dimensionsofhypercube       0     1     2     3     4     5    6     7      82               4     4     13               8    12     6     14              16    32    24     8     15              32    80    80    40    10     16              64   192   240   160    60    12     17             128   448   672   560   280    84    14     18             256  1024  1792  1792  1120   448   112    16     1`

so for example the 8-dimensional 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 m-dimensional hypercubes on the boundary of an n-dimensional hypercube as:

2^(n-m)*C(n,m)

This checks against the table.

This needs to be translated into a formula for k = 0, 1, ..., n of their (n-1) dimensional sides being painted.

k + m = n
m=n-k

2^(k+m-m)*C(k+m,m) = 2^(k)*C(n,n-k) is then the number of k-painted-sided n-dimensional 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 2015-03-05 15:41:41

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