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#1
I have this setup.
Now I would need a V-Perm to swap the FLU corner with the RBU corner and the LU edge with the BU edge. I cannot break down the common V-Perm algorithms to a commutator. My question is, is it possible to find a commutator which does basically the same as a V-Perm? If not how would I solve this cube using only commutators?
Er, what exactly do you want to accomplish? Why do you want to use only commutators for this?
(Note that it's impossible to get odd permutations with commutators, so you need to make it an even permutation first. Do a U or a U', and what's left can be solved with a commutator, albeit one that might not be very intuitive.)
edit: One quarter turn + one commutator. Also note that some of the common OLL algs for this case are basically a conjugated U2 move (e.g. [R' U2 Rw U' Rw' : U2]), and the fastest OLLCP alg for this case is also a conjugated U2 move ([Rw' D R2 U R' : U2]).
I am sorry, I think I missed to explain enough what I want to do. I simply looking for a way to solve the cube using only intuition and knowledge about cube theory like commutators without knowing any algorithms - therefore I am not interested in using only a few moves or being extremly fast. I simply want to solve it. I am quite new to this approach. Usually I only need to use easily-to-deduce-commutators like three cycles or orientation swaps. This was the first time I came across this pattern.
I am looking for a way to logically deduce a move which at least swaps the four pieces in a way, that the rest are again simple swaps, three-cycles or a bit of orientation. The two moves are working quite well, but I cannot see how I could have come up with those.
Ok so the "trick" I did not see here is the first U you mentioned. Afterwards I could deduce the rest myself, which looks messy but gets me to my goal.
[R' E' R2 E2 R', U] // flip two edges
[L E L2 E2 L, U] // flip two edges
[R' D R D' R' D R, U] // orient two corners
[L D L' D' L D L', U'] // orient two corners
// We have a V perm now, which can't be solved with only a commutator because the edge pieces are in an odd permutation (as are the corner pieces). So we do a quarter turn first to fix that.
U
// We start with the edges. (This is a plain 3-cycle. If you know how to solve a U perm, you can solve this in a more straightforward manner.)
M D2 M' U M D2 M' U2 M D2 M' U M D2 M'
// Then the corners. First, we swap the two corners in the front while preserving the rest of the U layer, then do a U2 and undo the front corner swap, then another U2 to finish.
[L U' R D2 R' U L', U2]
It seems when you insert a link a moderator need to approve the reply. I already answered you, you just cannot see it now ;-) The thing is missed was the U turn at the start. This is one of the cases where I did not think of the easy solution.
As mentioned by others in this thread, the V perm is an odd permutation of edges and an odd permutation of corners when oriented with the maximum number of pieces solved. However, you can solve either adjacent AUF (they are mirrors) with a commutator:
[R' U L U' R U2' L' U L U2 L', S' U2 S]
Now I would need a V-Perm to swap the FLU corner with the RBU corner and the LU edge with the BU edge. I cannot break down the common V-Perm algorithms to a commutator. My question is, is it possible to find a commutator which does basically the same as a V-Perm? If not how would I solve this cube using only commutators?
Er, what exactly do you want to accomplish? Why do you want to use only commutators for this?
(Note that it's impossible to get odd permutations with commutators, so you need to make it an even permutation first. Do a U or a U', and what's left can be solved with a commutator, albeit one that might not be very intuitive.)
edit: One quarter turn + one commutator. Also note that some of the common OLL algs for this case are basically a conjugated U2 move (e.g. [R' U2 Rw U' Rw' : U2]), and the fastest OLLCP alg for this case is also a conjugated U2 move ([Rw' D R2 U R' : U2]).
I am sorry, I think I missed to explain enough what I want to do. I simply looking for a way to solve the cube using only intuition and knowledge about cube theory like commutators without knowing any algorithms - therefore I am not interested in using only a few moves or being extremly fast. I simply want to solve it. I am quite new to this approach. Usually I only need to use easily-to-deduce-commutators like three cycles or orientation swaps. This was the first time I came across this pattern.
I am looking for a way to logically deduce a move which at least swaps the four pieces in a way, that the rest are again simple swaps, three-cycles or a bit of orientation. The two moves are working quite well, but I cannot see how I could have come up with those.
Ok so the "trick" I did not see here is the first U you mentioned. Afterwards I could deduce the rest myself, which looks messy but gets me to my goal.
[R' E' R2 E2 R', U] // flip two edges
[L E L2 E2 L, U] // flip two edges
[R' D R D' R' D R, U] // orient two corners
[L D L' D' L D L', U'] // orient two corners
// We have a V perm now, which can't be solved with only a commutator because the edge pieces are in an odd permutation (as are the corner pieces). So we do a quarter turn first to fix that.
U
// We start with the edges. (This is a plain 3-cycle. If you know how to solve a U perm, you can solve this in a more straightforward manner.)
M D2 M' U M D2 M' U2 M D2 M' U M D2 M'
// Then the corners. First, we swap the two corners in the front while preserving the rest of the U layer, then do a U2 and undo the front corner swap, then another U2 to finish.
[L U' R D2 R' U L', U2]
It seems when you insert a link a moderator need to approve the reply. I already answered you, you just cannot see it now ;-) The thing is missed was the U turn at the start. This is one of the cases where I did not think of the easy solution.
As mentioned by others in this thread, the V perm is an odd permutation of edges and an odd permutation of corners when oriented with the maximum number of pieces solved. However, you can solve either adjacent AUF (they are mirrors) with a commutator:
[R' U L U' R U2' L' U L U2 L', S' U2 S]
