I am reasonably competent at using commutators to find my own algorithms on twisty puzzles. Many good tutorials for this exist online.

However, there are also algorithms that are not commutators, because they are odd permutations. These are usually used to solve some kind of parity problem or another. Examples include swapping two corners on a 2x2 Rubik's cube, the "J permutation" on the 3x3, the move that solves the last four edges on a 4x4, and the edge parity move on the 4x4.

I have asked several questions recently about specific examples of these, but each example still feels like a special case and I don't feel I have a deep understanding. For commutators the theory is easy to understand and with a bit of practice I can come up with my own easily, even on a puzzle I haven't seen before, odd permutations it seem much more of a black art and I haven't seen any tutorial or explanation of how to come up with algorithms for them. I am asking for any tips or theory that will allow me to construct my own.

I do know that in general one can often perform a single twist (which on a cube-shaped puzzle is a 4-cycle) and then solve the cube again using commutators, but, for me at least, this tends to result in a long and laborious sequence of moves, rather than a simple algorithm. So I am looking for any theory or techniques by which non-commutator algorithms can be constructed.

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    $\begingroup$ Note that d R F' U R' F d' is not an odd permutation of the 4x4 edges. In fact, it is based on the commutator d (R F' U R' F) d' (F' R U F R'), where (R F' U R' F) flips an edge pair in the u/d slice. It is a 3-cycle of edges (assuming d is a move of just the inner slice) - try it out on a solved 4x4x4 and see. Because face turns don't matter yet when you are just pairing up edges, the final 5 face turns can be skipped, in much the same way as you can skip a U turn before/after doing OLL/PLL. $\endgroup$ – Jaap Scherphuis Jun 6 '18 at 22:46
  • $\begingroup$ @JaapScherphuis thanks, that's very useful information. I had heard it called a parity and assumed it was an odd permutation, but it makes perfect sense that it's a 3-cycle. (Now I also understand why PLL parity on the 4x4 is not really a parity either, since it can be solved with that move.) $\endgroup$ – Nathaniel Jun 8 '18 at 13:48
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    $\begingroup$ (I've removed that part from the question. For future readers, the comment thread is about an algorithm for pairing up the last two edge pairs on the 4x4.) $\endgroup$ – Nathaniel Jun 8 '18 at 13:53

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