New commutator challenge.

But how is that a direct commutator? I must admit I'm not 100% sure what Per means, but I'm rather certain he doesn't want an outer conjugation.
Surely you realize that [O:[P,Q]] = [[O:P], [O:Q]], no? And any sequence that can be written completely in the form [P,Q] is a direct commutator to me.

So, [R' U' R' U R' U' R, R U2 R'] flips UBL and DFR.


One more challenge: Find a direct 2gen* commutator that flips two edges.

*generated by two moves, not by two move sequences
 
So, [R' U' R' U R' U' R, R U2 R'] flips UBL and DFR.

Wow!!! A mere17 turns with my counting. I guess this one is optimal. But why you call it flip and not twist?? I thought it was "agreed" that corners twist and edges flip ...

Per
 
Wow!!! A mere17 turns with my counting. I guess this one is optimal. But why you call it flip and not twist?? I thought it was "agreed" that corners twist and edges flip ...
Oh yeah, my mistake.

Optimal <U,R> commutator for twisting UBL and DRF corners:
[U R U' R U, R2 U R2]
Neat! (Did you use a computer program to find this?)
 
Neat! (Did you use a computer program to find this?)

Yes, I used GAP to try all possible <U,R> commutators up to 16 moves in length. So I knew this was optimal without allowing for cancellations. JACube gave 16f* as the minimal <U,R> alg for twisting UBL and DRF, so then I also knew it was not possible to get a shorter solution for this case using cancellations.
 
I meant that the commutator must be 2gen, in the sense of being a combination of only two types of moves (although they can be any types), as opposed to the more general sense of being a combination of only two specific sequences/permutations.
 
Ah, my bad. I guess I was confused because it's the same limitation as Per's (or not?), and I thought you were imposing an extra rule.

[M U M U M U M, U] also works. David wasted one move! (Well, not really, thanks to cancellation)
 
[M U M U M U M, U] also works. David wasted one move! (Well, not really, thanks to cancellation)

Not surprised. I just took the pure-4 flip (M' U')4 and threw in a U move knowing I'd get a two-flip. The alg I use for BLD is just [(M' U')4, U'], but since A and A' are identities I perform it as A B A B' to make it faster.

Another way to arrive at your solution Stefan is to take the 4flip (M' U')4 and combine it with the F/B mirror (M U)4, which happens to be its inverse. M' U' M' U' M' U' M' U' M U M U M U M U pretty obviously breaks down into (M' U' M' U' M' U' M') U' (M U M U M U M) U, or [M' U' M' U' M' U' M', U']
 
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