3
$\begingroup$

Considering a single side of a standard 3x3 Rubik's Cube, are all possible color combinations attainable? Or are there certain combinations that can only be made by breaking apart the device or moving stickers?

I know there are unsolvable cubes when considering all 6 faces, but within a solvable cube, are all color combinations attainable on a single face?

To put another way, if I randomly selected the color for each of the 9 squares on one side, could standard solved cube always be shuffled in such a way that one of its faces contained that configuration?

Note: I don't know the answer, and I'm simply curious; it seems a mixture of Math and Rubik's Cube knowledge.

$\endgroup$
  • 2
    $\begingroup$ I would hazard a guess at "if we're only worried about one side, then yes." But I can't say it with any certainty. (The reason that we can't have any combination on ALL sides is because sides are opposite, so you'll never have a white and a yellow edge touching. If we're only worried about one side, it doesn't matter that we can't have a piece on the adjacent side being a certain color) $\endgroup$ – phroureo Nov 9 '17 at 17:06
4
$\begingroup$

Yes.

The single side in question would need to be determined by the color of the middle square of course. Then you can simply slot in any color to the edges and to the corners. This is always possible because:

  1. There are four of each color available for the edges and for the corners
  2. Any edge or corner can be moved to anywhere and any orientation on just one face.

To justify this second claim, consider that once you have chosen the 4 edge pieces and 4 corner pieces you can move them into place using the standard methods for solving a single face.

The restrictions to what you can move where with a cube only apply toward the end where, for example, re-orientating a single piece cannot be done. In this instance, and orientation is possible because the remaining pieces are not restricted and can be orientated either way.

$\endgroup$
  • $\begingroup$ Was about to type almost exactly the same, so +1 from me. If we don't care about any of the other faces/layers, a single face can indeed always have any edge or corner in any orientation. $\endgroup$ – Kevin Cruijssen Nov 9 '17 at 18:53
  • 2
    $\begingroup$ This seems kinda circular to me; your point 2 is more or less the assertion that the answer to the question is "yes". $\endgroup$ – Gareth McCaughan Nov 9 '17 at 19:51
  • $\begingroup$ Not exactly. It's one of the two components of the problem. I haven't justified it further, but the two claims together mean that the overall problem is solvable. Individual moves like that are well-known. $\endgroup$ – Dr Xorile Nov 9 '17 at 22:34
3
$\begingroup$

I take it "all possible color combinations" means: colour each square of a face in any of the six colours; can we make that configuration appear? The answer is

yes, we can

and here's why. First of all,

it's possible to achieve any even permutation of the cubelets, subject to the obvious constraints that (1) faces stay fixed, (2) edges remain edges, and (3) corners remain corners. This is obviously enough to achieve a state in which each cubelet on one face has somewhere on it the colour that we want for its square. (Consider some pair of cubelets not on that face; with them one way around or the other, the permutation taking us there is even.)

Now

we want to get the orientations right. We have a mod-3 constraint on the total "twist" of all the corners, and a mod-2 constraint on the total "flip" of all the edges, and that's all. So pick a corner and an edge that aren't on the target face; by letting these get twisted/flipped in whatever way is necessary, we can achieve whatever twists and flips we want on the target face.

And we're done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.