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I would like a confirmation to my answer. In this question, faces sharing a common edge cannot be of the same colour.

My way of reasoning started by choosing the colours Red (R), Yellow (Y), Green (G) and Blue (B).

I believe that there is only one way of colouring the cube with two faces red and two faces green (up to rotation naturally). I will call this colouring RG.

So the colourings are RG, RY, RG, YG, YB and GB.

So six different colourings?

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  • $\begingroup$ A cube has six faces, so what do you mean by 2 red and 2 green faces? $\endgroup$
    – Ned
    Dec 11 '14 at 12:49
  • $\begingroup$ You're quite right in asking this. All i meant was two faces R, two faces G, one face Y and one face B. $\endgroup$ Dec 11 '14 at 12:54
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    $\begingroup$ You should specify whether all four colors have to be used. $\endgroup$ Dec 11 '14 at 13:03
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    $\begingroup$ There's more than one way to color the cube with RRGGYB. For example, the red pair could be opposite each other or they could be adjacent. Suggestion: look up Burnside's Lemma, which is a general theorem for enumeration in the presence of symmetries. $\endgroup$
    – Ned
    Dec 11 '14 at 13:28
  • $\begingroup$ @Ned: Yes for Burnside's lemma, but note that two adjacent faces cannot be the same color per first paragraph of Question. $\endgroup$
    – hardmath
    Dec 11 '14 at 13:41
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If all four colors must be used, then your answer is correct, although something needs to be said to prove your claim about the solutions existing and being unique with two pairs of faces of the same color.

Let's say that a coloring of the cube is proper iff no two adjacent faces have the same color.

Essentially we have to show that any coloring of the six cube faces, with no two adjacent faces the same color, must have a "ring" of four faces that alternates between two colors.

Consider a properly colored cube, and ignore for the moment the top and bottom faces. The vertical faces must have at least two colors. If there were four colors on these vertical faces, the top (resp. bottom) faces could not be colored. If there were two colors, then we would have the "ring" of four faces that alternate between those two colors. So consider the possibility of there being three colors on the vertical faces. Then the top and bottom face colors are forced to be the same (by exclusion of the other three colors). Since two of the opposing vertical faces have the same color, this pair together with the top and bottom pair constitute a ring of four faces that alternate between two colors.

Now that we know such an arrangement always exists, we assume that the cube is oriented so that those four faces are vertical. To get all four colors, the top and bottom faces must have the other two colors. Turning the cube upside down shows that both assignments of the last two colors are equivalent (up to orientation of the cube).

Therefore your conclusion is correct: specifying any pair of the four colors to have two pairs of faces each is enough to uniquely identify a proper coloring of the entire cube, so there are $\binom{4}{2} = 6$ proper colorings of the cube that use all four colors.

Note that since three faces are mutually adjacent at each corner of the cube, at least three colors are required for a proper coloring.

If using only three colors is allowed, but still disallowing adjacent faces to have the same color, then the above arguments show that the choice of the three (out of four) colors determines the coloring of the cube.

A simpler argument can be given: Since at least one of the three colors must appear twice, that must be on an opposing pair of faces. Then the remaining four faces must alternate between the other two colors. Hence with three colors opposite faces have the same color, and rotations of the cube show this coloring is unique. Thus we get $\binom{4}{3} = 4$ more colorings.

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