# Coloring of a Cube

• Sep 8th 2008, 02:56 AM
p vs np
Coloring of a Cube
Hey.. needed some help in the analysis of this problem. Here it goes:

"A cube has six distinct colours on its faces. How many such unique cubes are possible? "

PS: The initial looks are lethally deceptive.Please don't get carried away by the simplicity of the statement :P
• Sep 8th 2008, 11:01 AM
ThePerfectHacker
Quote:

Originally Posted by p vs np
Hey.. needed some help in the analysis of this problem. Here it goes:

"A cube has six distinct colours on its faces. How many such unique cubes are possible? "

PS: The initial looks are lethally deceptive.Please don't get carried away by the simplicity of the statement :P

I think it is 6! for such number of colorings divided by number of cube arrangment which is $\displaystyle 6\cdot 5 = 30$. Thus we get $\displaystyle 720/30=24$.
• Sep 8th 2008, 12:03 PM
Opalg
Quote:

Originally Posted by ThePerfectHacker
I think it is 6! for such number of colorings divided by number of cube arrangment which is $\displaystyle 6\cdot 5 = 30$. Thus we get $\displaystyle 720/30=24$.

Haven't you got the 24 and the 30 the wrong way round? There are 24 (orientation-preserving) symmetries of the cube, therefore 6!/24 = 30 possible colourings.

To see this in a totally elementary way, call the six colours A, B, C, D, E and F. We can always place the cube on a table so that the face with colour A is facing downwards. There are 5 possibilities for the colour of the top face. Say this face has colour B. Then we can rotate the cube so that the face with colour C is pointing south (say). There are then 3!=6 distinct ways of colouring the remaining three faces. So the total number of distinct colourings is 5×6=30.
• Sep 8th 2008, 05:02 PM
ThePerfectHacker
Quote:

Originally Posted by Opalg
Haven't you got the 24 and the 30 the wrong way round? There are 24 (orientation-preserving) symmetries of the cube, therefore 6!/24 = 30 possible colourings.

Yes! That is exactly what I wanted to say.

Another way to solve this problem is using Burnside's formula from group theory. Let $\displaystyle G$ be the set of all rotations of the group under the binary operation of composition. There are six faces to choose and for each one there are four more positions (to preserve the order of the cube) thus there are 24 rotations. Thus, $\displaystyle |G| = 24$ and note that $\displaystyle G$ is a group. Let $\displaystyle X$ be the set of all possible arrangements of the cube (even counting similar ones under rotations) thus $\displaystyle |X| = 6! = 720$. Let $\displaystyle G$ act on $\displaystyle X$ in an obvious way i.e. it changes the cube under the rotation. Now two cube representations $\displaystyle x_1,x_2$ are similar if and only if there is a $\displaystyle g$ so that $\displaystyle x_2 = gx_1$ i.e. $\displaystyle x_1,x_2$ are in the same orbit. It remains to count the number of orbits. By the formula this is $\displaystyle \frac{1}{|G|} \sum_{g\in G}|X_g|$. Now $\displaystyle |X_g| = \{x\in X | gx = x \}$. Note if $\displaystyle g$ is not the identity then $\displaystyle |X_g| = 0$ and if $\displaystyle g$ is the identity then $\displaystyle |X_g| = |X| = 720$. Thus, we get $\displaystyle \tfrac{720}{24} = 30$.
• Sep 9th 2008, 01:41 AM
p vs np
Neat solution..
Thank you Mr Opalg.. That was a real cool way to look at the problem.
and of course, thank you Mr PerfectHacker as well :)
• Sep 9th 2008, 03:01 PM
awkward
Quote:

Originally Posted by p vs np
Hey.. needed some help in the analysis of this problem. Here it goes:

"A cube has six distinct colours on its faces. How many such unique cubes are possible? "

PS: The initial looks are lethally deceptive.Please don't get carried away by the simplicity of the statement :P

Place the cube on a tabletop so that color #1 is on top. There are then 5 possibilities for the color of the bottom face and 4! / 4 possible ways to color the side faces (dividing by 4 because of the 4 rotations of the cube, keeping the original face on top).

So there are 5 * 4! / 4 = 30 unique cubes in all.