 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Colored Blocks (Posted on 2005-12-12) You want to make up a set of 27 1-cm 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 3-cm 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 again--the same set of colorations for the 1-cm 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 1-cm cubes?

Then consider the same problem with 64 1-cm cubes using four colors this time, fitting together to make any one of four 4-cm solid-color-on-the-outside cubes.

Is there a method that will work for n^3 unit cubes with n colors?

 See The Solution Submitted by Brian Smith No Rating Comments: ( Back to comment list | You must be logged in to post comments.) re: ideas solution of 3x3x3 | Comment 4 of 5 | (In reply to ideas by Charlie)

Using the letters A, B and C as in my previous post (as types of 1-cm cubes with different combinations of colors), but with subscripts (in parentheses) to indicate the colors present,

A(RB) is the type of 1-cm 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 2005-12-12 14:31:17 Please log in:

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