Today, I got my hands on a Rubik's cube with text on it. It looks like this:

Rubik's cubes

Now, I would love to know whether solving the right cube will also always correctly align the text on the cube or whether it's possible to solve the cube in a way that the colors match correctly but the text is misaligned.

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    $\begingroup$ No. What you have there is known in the community as a ‘supercube’. After solving for the usual position and orientation for corner and edge pieces, an additional step is required to to orient the centres of the faces. $\endgroup$ – Zhen Lin Sep 9 '11 at 14:39
  • $\begingroup$ A maths professor once told us that 'there are no silly questions'. I asked him your exact question. He answered "that's not a silly question", paused, then said "no, that is a silly question" $\endgroup$ – Colonel Panic Aug 20 '12 at 19:08

whether it's possible to solve the cube in a way that the colors match correctly but the text is misaligned

Yes. But the total amount of misalignment (if I remember correctly; it's been a while since I played with a Rubik's cube) must be a multiple of $\pi$ radians. So if only one face has the center piece misaligned, it must be upside down. On the other hand it is possible to have two center pieces simultaneously off by quarter-turns. (This is similar to the fact that without taking the cube apart, you cannot change the orientation of an edge piece (as opposed to center piece or corner piece) while fixing everything else.)

(I don't actually have a group-theoretic proof for the fact though; this is just from experience.)

Edit: Henning Makholm provides a proof in the comments

Here's the missing group-theoretic argument: Place four marker dots symmetrically on each center, and one marker at the tip of each corner cubie, for a total of 32 markers. A quarter turn permutes the 32 markers in two 4-cycles, which is even. Therefore every possible configuration of the dots is an even permutation away from the solved state. But misaligning one center by 90° while keeping everything else solved would be an odd permutation and is therefore impossible.

  • $\begingroup$ I vaguely remember this as well. This page gives two algorithms for solving centres which corroborate your experience. $\endgroup$ – Zhen Lin Sep 9 '11 at 15:25
  • $\begingroup$ I know the second algorithm there. I also have another algorithm which is even less efficient (requiring around 30 steps). But I didn't know the first algorithm posted. Thanks! $\endgroup$ – Willie Wong Sep 9 '11 at 17:37
  • $\begingroup$ @Henning: Thanks! Would you mind if I copy that comment to the answer body to be more visible? $\endgroup$ – Willie Wong Apr 9 '12 at 15:23
  • $\begingroup$ @Willie: Go right ahead. Didn't want to bump this old question unilaterally. $\endgroup$ – hmakholm left over Monica Apr 9 '12 at 15:28
  • $\begingroup$ @WillieWong: an easier way to prove this is to do the y-perm or the t-perm twice, and in both cases, the top center piece will turn a half-rotation. $\endgroup$ – Jason Chen May 8 '14 at 0:08

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