You want to make up a set of 27 1cm cubes with their faces variously colored red, yellow and blue. You want to do this in such a way that you can form any of three 3cm cubes: one that is all red on the outside, one that is all yellow on the outside or one that is all blue on the outside. You can't repaint the original 27 cubes againthe same set of colorations for the 1cm cubes must work regardless of whether you want the outside to be red, yellow or blue. How must you color the faces of the 27 1cm cubes?
Then consider the same problem with 64 1cm cubes using four colors this time, fitting together to make any one of four 4cm solidcolorontheoutside cubes.
Is there a method that will work for n^3 unit cubes with n colors?
(In reply to
ideas by Charlie)
Using the letters A, B and C as in my previous post (as types of 1cm cubes with different combinations of colors), but with subscripts (in parentheses) to indicate the colors present,
A(RB) is the type of 1cm cube that has 3 reds meeting at a corner and 3 blues meeting at the opposite corner, and similarly for two other color combinations of type A.  A(BR) is the same as A(RB).
B(RBY) is the type with 3 reds meeting at a vertex plus two blues and a yellow, and similarly for 5 other subtypes of type A. (Permutations count here).
C comes in only one variety: 2 adjacent faces of each color.
To form a yellow cube, for example:
A(YB) + A(YR) + B(YRB) + B(YBR) = 8 corners
B(RYB) + B(BYR) + C = 12 edges
B(RBY) + B(BRY) = 6 faces
A similar set of 3 equations would apply for red or blue cube construction. So far there are 9 equations.
There are ten subtypes of cubes A(BR), A(RY), A(BY), B(RBY), B(RYB), B(BRY), B(BYR), B(YRB), B(YBR) and C. The tenth equation is that all these must add to 27.
Due to the symmetry of the equations, the three A types must all have the same number, as must the six B types. So we can treat A(YB) for example as A/3. That leads to the above equations translating to the following:
2 A/3 + B/3 = 8
B/3 + C = 12
B/3 = 6
so that
B=18, A=3 and C = 6, and there are 3 B's of each subtype, 1 of each subtype of A and 6 C's.
Edited on December 12, 2005, 2:35 pm

Posted by Charlie
on 20051212 14:31:17 