# 7x7 Rubik - Solve sides with commuters and conjugates

I've been practicing to better understand and apply commuters and conjugates in big cubes. I've solved the majority of my 7x7 cube but got to a point where I have two side pieces flipped (so blue-orange is on the blue-yellow spot and viceversa). I've tried making cycles which scramble more of the pieces, but when I resolve it I get to the same point again. Can it be solved just with these tools or do I just need to learn a specific algorithm?

It is not completely clear what your cube looks like, but I am going to assume there are two wing edge pieces that need to be swapped.

A single swap of two edges is an odd permutation while any commutator and conjugation of a commutator is an even permutation. It is therefore impossible to solve with (conjugated) commutators only.

One way to solve it is to perform a single slice move - a quarter turn of a slice containing one of the swapped edges. This is a 4-cycle of such edges (an odd permutation as required) after which you can solve those edges using commutators only. Of course that slice move also disturbs centre pieces, but those can also be fixed using commutators.

One easy way to fix the centres is this:
Hold the cube so that one of the swapped edges is along the UF edge. Let's call the inner slice containing that edge the r slice. Do the move sequence r U2 r U2 r U2 r U2 r. This returns the centres but still does an odd permutation on the edge pieces. You now still have to fix the edges with commutators, but as they are now in an even permutation, that is possible.

Note: I have described this as if there is only a single pair of swapped edges. On a 7x7x7 cube there are two types of wing edges, inner and outer ones. These two types can each have this parity problem independently of each other. Both can be solved the same way.

• I understood a bit more about parity with your answer and now I can solve my odd cubes. Thanks for the explanation. The only thing I don't understand yet is how does this odd cycle donesn't affect the centers. Is it because all the central pieces (both "central corners" and "central sides") are indistinguishable from each other? (meaning the four "central sides" touching the center of the face behave the same way) Commented Feb 11, 2022 at 16:36
• @ArturoVialArqueros On a supercube or picture cube, a cube where all the centres are distinct, there you will see that most undergo odd permutations as well. Only the centres on a face diagonal don't because there are 8 of them undergoing two 4-cycles which is an even permutation. So on a 4x4x4 supercube you can do a 4-cycle of edges without affecting the centres, but not on larger supercubes. This actually makes larger supercubes easier than normal ones because solving the centres correctly first means you will never get a parity problem when pairing up the edge pieces. Commented Feb 11, 2022 at 16:48