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Are all configurations of a Rubik's cube solvable?

  1. Nov 19, 2007 #1
    Obviously I am not speaking of an already solved Rubik's cube which has gone under a series of changes, since undoing every change backwards constitutes itself in a solution. My question is whether or not all configurations are solvable per se, that is if the distribution of the colored squares was really random (without it having been derived from changing an already solved cube), do solutions always exist? Equivalently, can all possible configurations be obtained by changes on an already cube? I'm sure this is more 'computational' rather than mathematical, but I thought this would be the best place to ask.
  2. jcsd
  3. Nov 19, 2007 #2

    Gib Z

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    No, not all preset configurations are solvable. On a normal cube, when already in a solved mode, you are able with little force to twist a corner piece, which is only part of all the configurations you are looking for. Even doing that, there is a 11/12 chance that the resultant cube is unsolvable.

    EDIT: As for your last sentence, this question is actually a active route of research in algebra, specifically group theory. Well not actually ...but the minimum number of moves needed to solve an arbitrary solvable cube. Just 3 months ago they improved the conjecture to 18 moves I think.
    Last edited: Nov 19, 2007
  4. Nov 19, 2007 #3


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    If you played with a Rubik cube before you'll notice that the 6 squares in the centers of the faces will always remain in the centers. So for any permutation these centers should always be of 6 six different colors.
    Same reasoning holds for the 8 corner blocks and the 12 edge blocks. So you have 3 different kinds of blocks: centers, corners and edges.
    So you should for permutations which consist of three seperate ones. One of the centers, one of the edges and one of the corners.
    Since you can't turn the cube inside out (create its mirror image) each of these permutations should be obtainable by rotations of 'the' solved cube. (I assume you take a standard configuration of a solved cube. If you have a solved one, you can interchange two opposite faces and it will still be 'solved', be you can never convert one into the other by legal moves).

    I don't know if that characterized all configurations, probably not.
  5. Nov 19, 2007 #4
    Have a look at this link:
    How do I tell if the cube is unsolvable from a given state?

    Suppose you have solved the first two layers and are about to solve the last layer (orient and permute). Then you can't solve the cube if:

    * One edge piece is flipped in place and all other pieces are correct.
    * Two edge pieces need to be swapped and all other pieces are correct.
    * One corner piece needs rotating and all other pieces are correct.
    * Two corner pieces need to be swapped and all other pieces are correct.
    (Quoted from Mark Jeays website)

    Besides, maybe you know of Macky's video where he loses a piece of the Rubik's cube (I think it's an edge piece) and still manages to solve the cube. Had he put the piece back in with the wrong orientation he would not have been able to solve the cube. But he was lucky and still finished with a remarkable time.
    Last edited: Nov 19, 2007
  6. Nov 19, 2007 #5
    Isn't this a bit like (but obviously much more complicated) that puzzle which is a 4x4 grid of 15 slidable tiles with 1 'gap'. The puzzle is to slide them around so that they are in order (like on a keyboard's number pad). Not all initial configurations can be solved. For instance, if you have 14 and 15 swapped but everything else in the right place, it cannot be solved. Similarly with any other pair of pieces. But, if two pairs are swapped or 3 pieces are cyclicly permuted, it is possible. Something to do with the parity of the moves.

    I don't remember the name of that puzzle but it's a fairly famous one.
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