riffz
Veteran
yeah I know that but when do we do it?
If you mean that A is the 1-move interchange and B is the 3-move insertion, then you want to do B first in the cases where it solves a piece in the cycle.
Take DF -> FU -> RF for example. The commutator used to solve this case would be [M', U' R U], because M' will bring the piece at DF to its solved position. If you did U' R U first, then it would bring the piece at RF to FU, when it should be going to DF.
For the other direction, DF -> RF -> FU, the commutator is [U' R U, M']. You would do U' R U first this time, because you want to bring the piece at RF to FU.
Hopefully that helps... It's one of the things about commutators that's difficult to explain, but once you get it you just GET IT. (At least that's what happened for me.)
I usually do them when I don't want an extra set-up move.
The setup move would cancel with the other part of the commutator anyway. It's just an ugly way to write an 8-mover as a 10-mover:
DF -> RF -> FU
You could retain the form [A, B] (where A is 1 move and B is 3) by doing the setup move U', giving U' [R, U M' U'] U. But this of course is the exact same algorithm as [U' R U, M'] once you cancel the setup moves.
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