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I've always wondered if there was a proof for this, since the beginners method to solving the last layer depends on this (you either have no edges oriented correctly, an L, a line or a cross, and each case has an even amount of oriented/ unoriented edges). When solving it this way, I've never ran into a case where there was an odd number of oriented edges.

Is there a proof showing that this is always the case? If so, could someone prove it/ provide a link?

Julian Jefko
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  • In fact, it's possible to prove more: The total number of edge flips in any configuration is even. – MJD Sep 05 '14 at 03:11
  • There's a corresponding fact about the orientations of the corners: If a corner with a clockwise twist counts $+1$, a corner with a counterclockwise twist counts $-1$, and a corner in its normal position counts 0, the total twists on all 8 corners always adds to 0. – MJD Sep 06 '14 at 05:27

2 Answers2

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As MJD mentioned in the comment, it can be proved that there always has to be an even number (which includes 0) number of edges flipped (and correctly oriented).

If you can prove this, then it will follow that any sub-step state (such as the last layer being unsolved) must have an even number of edges flipped and oriented correctly for all states reachable with legal moves applied to a completely solved cube.

Proof 1

This proof is by far the best proof I have seen so far for edge orientation.

I just uploaded an image to illustrate what Michael Gottlieb meant by a "two 4-cycle" of stickers, if you aren't too savvy with permutation terminology.

Imagine we are doing the move R to a 3x3x3 cube. It will move all pieces in the right slice away from us 90 degrees. If we focus specifically on the edges, we can observe that the move R does the following to the edge stickers.

2 4-cycle of edge stickers is done by the move R

A 4-cycle is an odd permutation because it requires 3 (an odd number) overlapping 2-cycles (transpositions) to be solved/generated. Two disjoint 4-cycles requires 6 (an even number) 2-cycles to solve/generate.

Similarly, a 2-cycle requires 1 (an odd number) 2-cycle to solve. enter image description here

If we can flip a single edge, that implies that we can flip any number of edges (one at a time, for example), and thus for the last layer edges of the 3x3x3, we could have either 1 or 3 edges flipped if we could flip a single edge. Therefore all Michael had to address in his proof was to show that we cannot flip a single edge (which is illustrated in the 2-cycle of stickers image above) to handle all cases.

Proof 2

Of course, the oldest "modern" description of these "cube laws" I know of is Ryan Heise's page, but I personally find Michael Gottlieb's proof for edges to be more satisfying than Ryan's.

Christopher Mowla
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  • Nice proof! Could it be simplified a bit further, though? I was thinking, using similar reason, you could just prove by contradiction (assume that there is a case were an odd amount of edges are flipped. The only way for that to be possible is by either flipping one edge or three edges. In each case, you must flip a single edge (for flipping 3, youd have to flip two and then still flip a single edge), since its not possible to flip a single edge, there are always an even amount of edges flipped)... or is that just saying the same proof in a different way? – Julian Jefko Sep 06 '14 at 18:12
  • That's just stating the consequence of the proof (that we cannot flip a single edge) in an alternate manner than what I did. – Christopher Mowla Sep 06 '14 at 18:37
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This has been analyzed elsewhere in purely group theory terms; this paper on page 7 addresses this issue.

Chas Brown
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  • That analysis suggests that edge flips come in multiples of 4, which is definitely not true. I don't think the frame of reference it's using can possibly be well-defined... – Micah Sep 05 '14 at 03:27
  • Agreed; I'll edit and delink it (the original link was http://www.ryanheise.com/cube/cube_laws.html). The MIT paper remaining is still worth a look. – Chas Brown Sep 05 '14 at 03:38