# Edges and corners independent?



## mark49152 (May 8, 2013)

I usually think of edges and corners as independent of each other, on 3x3, with constraints on legal configurations applying solely to one or the other. 

If that's true, surely it would be possible to reach a PLL case where the corners are solved, and the edges are solved relative to each other but shifted one place around the cube in the same direction. No such case exists, so I must be wrong about them being independent. 

My question is, how are edge and corner configurations constrained by each other? Can anyone point me at an explanation?


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## ben1996123 (May 8, 2013)

learn about groups and stuff


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## TheOneOnTheLeft (May 8, 2013)

I would assume it's because in PLL, you can only have pair swaps in multiples of 2, 3-cycles of edges or corners. the four edges being cycled one place around the cube would be a 4-cycle, whereas the possible edge-only PLL cases are A perms, which are 3-cycles, and H and Z perms, which are the two possible pair swaps. Watching this (http://www.youtube.com/watch?v=54SGrZbLcoE) gave me a reasonable understanding of why this is, although admittedly it's more of a practical understanding than a genuine mathematical explanation, if that's what you're looking for.


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## Goosly (May 8, 2013)

The number of pairs of swapped pieces must be even.
For example, in a T-permutation, one pair of corners and one pair of edges is swapped. In a H-permutation, 2 pairs of edges are swapped.

_a PLL case where the corners are solved, and the edges are solved relative to each other but shifted one place around the cube in the same direction_
This is not possible because it requires 3 swaps, as you can see here:
ABCD -> BACD -> DACB -> DABC
One letter represents one edge, the red edges are swapped in each step. DABC is one rotation away from ABCD (the case you wanted)
Note that if you do this rotation twice, you get the H-perm case in 6 steps (which is even)


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## MaeLSTRoM (May 8, 2013)

The permutations of corners and edges are actually independent. Its just to do with how the cube group works. When you do a single quarter turn face move, you perform a 4cycle edges and corners. This means that using standard commutators/3cycles the closest you can get to solved after doing this is a 2 cycle of corners and a 2 cycles of edges. But its not really affecting each other directly. Its because they're constrained to the same system.


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## Stefan (May 8, 2013)

MaeLSTRoM said:


> The permutations of corners and edges are actually independent.



No they're not.


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## Christopher Mowla (May 8, 2013)

To the OP, I did an in-depth explanation of this (I don't think it's hard to follow, it's just thorough) in the prerequisite section in my commutator theory document. The document as a whole isn't entirely accurate (but I'm still solving the problem at hand in the My Ultimate Commutator Challenge Thread and will eventually replace the current document), but the prerequisite information is accurate.

As Goosly was hinting at but didn't actually say the term for the relationship between corners and middle edges is parity. In my document, see pages 2-3 and 5-7 (the document has a cover page, so go by the page numbers, not the PDF page numbers). (Obviously you can skip the "Edge Orientations" section on bottom of page 7.)


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## elrog (May 9, 2013)

I had a similar question in mind once, but I figured it out by messing with my 2x2. I was wandering how you can have only 2 corners swapped. The reason is that turning 1 face 90 degrees is a 4-cycle, and any even number of 4-cycles can be solved with 3-cycles. I will use 2-cycles to explain. Remember that a 3-cycle is 2 2-cycles, and any case solvable by 3-cycles is solvable by an even number of 2-cycles.

So, this being said, this is the relationship of corner and edge permutation on the 3x3: Corner and edge permutation can both be solved by an even number of 2-swaps or an odd number of 2-swaps, but you can't have corners needing an even number of 2-swaps to be solved while edges need an odd number of 2-swaps to be solved (and vise-versa). It is also possible to switch both corner and edge permutation from being solvable with an odd number of 2-swaps to needing an even number of 2-swaps with a single 90 degree turn of one face (and vice-cersa).

For a good experimentation of this concept on a 3x3, play around with a commutator form of a J-perm. If you don't know one, heres one: (L U) (B' L' B) (U) (B' L B) (U') (U' L') I highighted my setup moves in blue, while X is green and Y is red. I don't normally think of commutators in X and Y, but I thought it could help. You will realise that the J-perm is actually a 3-cycle of 1x1x2 blocks. This is contradictory to how the j-perm looks as if it is swapping 2-blocks or just swapping 2 corners and 2 edges.

I'm not sure if you can understand half of what I said, but I tried. Hope it helps.

EDIT: Thank you for pointing out the mistake Kirjava. I have corrected it now.


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## mark49152 (May 9, 2013)

Thanks all for the informative answers. I understand the principle now, although there's a lot of detail here that I don't have time to digest right now! THanks.


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## Kirjava (May 9, 2013)

elrog said:


> any case solvable by 3-cycles is solvable by an odd number of 2-cycles



even


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