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Ok... this whole problem started with a 7x7x7 Rubik’s cube. But I’m only telling you that to give some context. Not to ask how to solve the cube.

When solving the cube’s edges you have to math up the colours. Eventually you will get down to the last few edges. I know one algorithm to do this and it works by cycling three pieces around their edges.

(Lower case letter represents the piece colour and upper case represents the edge colour)

So...

Ab Bc Ca can be cycled to Aa Bb Cc and so those three edges are solved. This is a three cycle.

Sometimes though you end up with two edges remaining unsolved, like...

Ab Ba

ie a 2-cycle

From what I have been scribbling down I think it is not possible to use my 3-cycle algorithm to solve this.

You can bring in as many other already solved edges to help. So for instance with one additional edge...

Ab Ba Cc I can get to... Ac Bb Ca and so I still have one solved edge and a 2-cycle.

With two additional edges I can do other moves but still can’t get to a 3-cycle that can be solved.

I don’t know how else to express this but I feel like there is something around the number of cycles and the number of edges I can solve and unsolve but I’m not sure.

Is there a more concrete way of approaching this than just trying different combinations of moves? I feel like it can be solved and tell me that I can’t use my algorithm to solve this.

Hope someone can point me in the right direction.

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A 3-cycle is an even permutation, and combination of even permutations always produces another even permutation.

A single swap is an odd permutation, so it cannot be built out of 3-cycles.

For an odd-sized cube, if the two edge pieces you need to swap are in the middle of their respective edges, then you're out of luck. That cannot be done while still keeping everything else in position.

However for a swap of two non-middle edge pieces there's a way out, namely to do a quarter-turn of a slice of the cube that contains the appropriate family of edge pieces. This is a 4-cycle of the edges, which is odd, so together with the swap you needed to do you now have an even permutation of the edge pieces to do, and that can always be built out of 3-cycles.

Of course, that slice quarter-turn will have upset the centers if you already have those solved. This can be fixed later, because the center pieces (other than the middle centers on an odd-sized cube) come in indistinguishable groups of 4, so one can always get those back in a nice configuration using 3-cycles.

(But slightly easier than to fix up the centers piece by piece is to fix them immediately after the initial quarter-turn, by using commutator 3-cycles that work on entire stripes of center pieces, in the slice you have turned and its mirror partner across the center slice. This will further disrupt some of the edge pieces, but in an even way, so you can still put those together afterwards).

Wikipedia resource regarding parity of permutations

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  • $\begingroup$ Yes! I knew it was going to be representable as something other than Rubik's cubes and stuff and actually had a feeling that something involving parity would show it. I just didn't have the vocabulary or knowledge to search for the correct terms. Thanks very much :D $\endgroup$ – Fogmeister Dec 13 '17 at 9:59

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