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I am trying to learn how to solve a Rubik's cube and the 4x4 rubic cube at school is missing 4 squares, that got me thinking: if you randomly switched two squares, would it still be solvable?

I know that the flat picture (example of what I mean) where you slide a square around on the xy plane are solvable omitting if you switch the last two squares when it is in its identity location. For example if cut a picture into 16 squares and then removed the 16th (bottom righthand) and then you switched the 14th and the 15th square then it would be unsolvable.

I really don't know how to say that better mathematicaly

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A Rubik's Cube is $3\times3$. The similar $4\times4$ cube is apparently called Rubik's Revenge or Master Cube. If you're interested in the $4\times4$ version, I think that should be reflected in the title. By the way, please note that frequent edits to the question right after it was posted occur at the time when maximal attention is focussed on the question, and are thus maximally confusing and annoying; please consider preparing your question carefully before posting it to avoid that. – joriki Jan 26 '13 at 15:20
a) Solvability is not affected if you happen to switch squares of the same colour. b) Solvability is definitely lost if you happen to switch (differently coloured) squares of different types (say, one at a corner cube and one inside a face). c) Also if you switch two of the three squares at a vertex. - You may want to be more specific, how random you want to switch. – Hagen von Eitzen Jan 26 '13 at 15:29
my question was just about the more known version, then I was hoping understand the answer that someone gave or go from the hints to solve the answer for a 4x4 cube. – yiyi Jan 26 '13 at 15:29
@HagenvonEitzen at first I was just thinking that any switching is possible, I know that if the colours match, then its still solveable. Didn't realize about the location, just started reading about this. Thanks for the details. – yiyi Jan 26 '13 at 15:32
It should be something like $1/(\alpha\times 27!)$ where $\alpha$ are all the possible initial states of the cube. For a $3\times 3$ Rubik Cube. – Beta Nov 28 '15 at 16:50
up vote 3 down vote accepted

For the standard Rubik cube, the probability is zero if you switch two squares that are actually of different colour:

The Rubik cube has squares at face centers, which don't move at all; edge cubelets, which may rotate and move around, but always stay edge cubelets; vertex cublets, which may rotate and move around, but always stay vertex cubelets.

  • If you switch two face centers, the cube becomes unsolvable because no suitable vertex cubelets will be available for at least one vertex.
  • If you switch the two faces of the same edge cubelet, the cube becomes unsolvable because flipping such a cubelet is not en element of the group.
  • If you switch two faces of the same vertex cubelet, the cube becomes unsolveable because that cubelet no longer fits (wrong orientation of the colours on it)
  • If you switch squares between different edge cubelets, this would require to not alter the set of edge cubelets. E.g., you must not produce a second green-yellow edge. This means that you need to switch between two cubelets sharing a colour (say switch yellow and red between the gren-yellow and the green-red edge cubelet). If I remember correctly$^1$, it is not possible to switch two edge cubelets belonging to the same face while keeping their orientation correct (with respect to the common face)
  • If you switch squares between differnt vertx cubelets, this makes the cube unsolvable. Indeed, knowing two of the squares of a vertex cubelet, its correct placement is already completely determined.

$^1$ This is the only point I'm not 100% sure of and would have to check.

EDIT: After checking at Wikipedia, I got rid of the doubts mentioned in the footnote: Even considering only position, not orientation of cubelets, the group operates only as $A_{12}$, not $S_{12}$ on the edges.

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Note that this answer is very different from the similar quastion: Suppose the cube falls apart and you put it together in random (but of course edges to edges, vertices to vertices), what is the probability that the cube can be solved? Answer: $\frac1{12}$. – Hagen von Eitzen Jan 27 '13 at 12:25
Not only is it impossible to exchange two edge cubies on the same face while keeping their orientation: it is impossible to exchange any two edge cubies (while leaving all other cubies in place), period. That would be an odd permutation of the 20 movable cubies, but everything that can be made from quarter-turns are even permutations. – Henning Makholm Mar 18 '13 at 22:57

I'm posting a new answer, because existing answer deals with 3×3 cube, which is very different in this regard from a 4×4 cube.

To actually answer the question, YES a 4×4 cube will be solvable if you randomly exchange any two centers.

On a 4×4 cube, the parity of centerpieces permutation must always be the same as the parity of the cornerpieces permutation, however, odd and even parity of centers permutation is not distinguishable, as there are 4 identical centers of each color.

This can be demonstrated by a simple commutator-algorithm such as this one. This algorithm exchanges three centerpieces from the mechanical perspective, however, due to the fact that two of those are of the same color, the resulting effect is that only two centerpieces have been exchanged.

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