7

I was using the beginner method to deliberately solve a 3X3 cube with one edge piece flipped. But, by the time I got to the top I discovered that I couldn't solve it with only one edge flipped. I had to also flip an edge on the top layer. Why is that?

enter image description here

crownjewel82
  • 173
  • 1
  • 4
  • That question specifically asks for a mathematical explanation. I honestly don't remember that much about group theory and I was wanting a more plain language answer. Is that a good enough reason to keep this question open? – crownjewel82 Jun 01 '14 at 13:06
  • 1
    Not a dupe. Edge =/= corner. – clickbait Jun 09 '18 at 18:26

1 Answers1

3

12 edges on a cube. 24 positions an edge piece can be in once you consider flipping. If you consider one move to be a single quarter turn, then you can move an edge to exactly 12 of those 24 positions in an even number of moves, and to the other 12 only in an odd number of moves. You should be able to convince yourself of this by trial and error, but if you want proof you could come up with names for the 24 positions and check that each possible move only moves edges from a position in the even set to one in the odd set or vice versa.

Now consider what a single move does. It moves each of four edge pieces from one set of positions to the other. This means if you started with (say) 6 in one position-set and 6 in the other, you now have 2-10 or 4-8 or 6-6 or 8-4 or 10-2, but you can't have an odd number of pieces that changed position-sets.

In order to flip a single edge piece, you need to move a single piece to the other position set without moving any of the others. But you can only switch the position-sets of an even number of pieces at a time.

greg m
  • 2,164
  • 1
  • 10
  • 6
  • While this answer is a good intuitive explanation, I'm not completely certain it's a sufficient one. For instance, on the tetraminx, three pieces change positions, but one edge still cannot be flipped. –  May 31 '14 at 05:26
  • on the tetraminx you need to make an even number of moves to get the non-edge pieces back to where you want them – greg m May 31 '14 at 06:47
  • 3
    no that doesn't explain it, sorry. i've never seen a tetraminx before so bear with me. but it seems like you can make three moves and end up where you started.

    anyway, just because you need a different explanation for the tetraminx's property, doesn't mean my explanation doesn't hold for the cube. (though for sure i could be mistaken)

     – greg m May 31 '14 at 06:53
  • That's fair, but it isn't always the case. Your writeup of the intuitive proof is good, but it doesn't necessarily hold true as a line of reasoning. It is still possibly/probably true, though! –  May 31 '14 at 07:05
  • sure, any proof can contain a mistake. but i presume you can't see anything wrong with this one as it applies to the cube? – greg m May 31 '14 at 07:20
  • Well... the point I was making is that it isn't always the case that even flips are generated by even piece movements. I suppose a better example would be the rex cube, which meets your initial requirement, but can actually have a single pair swapped. –  May 31 '14 at 07:25
  • not sure what you mean by my initial requirement, but obviously the same logic can't be applied to puzzles where three moves take you back to the start - you won't be able to find the two subsets of edge positions my proof relies on – greg m May 31 '14 at 07:36
  • @Emrakul Can you explain how those other puzzles are relevant given that I specifically asked about the 3x3 cube? – crownjewel82 Jun 01 '14 at 13:17
  • @crown I was using them as a counterexample for one of the assumptions in the answer. –  Jun 01 '14 at 16:38
  • @Emrakul Is the assumption incorrect for the 3x3 cube? – crownjewel82 Jun 02 '14 at 14:56
  • @crown It means that, despite being logical, there's nothing to suggest that it holds. –  Jun 02 '14 at 15:09
  • @Emrakul Holds for what? The 3x3 cube or for any possible puzzle of this type? – crownjewel82 Jun 02 '14 at 15:15
  • @crown For any puzzle; the logic sounds like it works, but it's not a supported argument. –  Jun 02 '14 at 16:00