#### cuber786

##### Member

switches DFr->UBr->DFr.

I need it because during 4x4 bld, if I have parity for edges, only DFr and Ubr need to be switched with each other thats why I am asking.

#### shafiqdms1

##### Member
hmm...idk because I cant really make commutators myself.

#### shelley

##### chang
hmm...idk because I cant really make commutators myself.

You could always set it up as the 4x4 "OLL parity" which switches two edges (e.g. l' D2 l and then do the OLL parity to "flip" the UB edge. Just be aware that the alg changes U centers around). Or try this alg Lucas gave me that I never bothered to learn: r' U2 r2 U2 r U2 r U2 l r2' U2 r' U2 r U2 l' U2

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#### qqwref

##### Member
What you're looking for doesn't actually exist: commutators can't solve two-cycles. But if you just want an algorithm, the one shelley gave is fine Way better than what I'd use, anyway.

#### cmhardw

switches DFr->UBr->DFr.
It's impossible to have a commutator do this, because it is an odd cycle. However, I would use this alg if you are allowed to rotate the U center 180 degrees.

D R F' l' U2 l' U2 F2 l' F2 r U2 r' U2 l2 F R' D'

Chris

--edit--
sorry Shelley, I skimmed this thread too quickly, beaten to the punch ;-) Shelley's alg is shorter and more execution friendly than mine, so I recommend to use hers instead.

#### Lucas Garron

##### Member
Or try this alg Lucas gave me that I never bothered to learn: r' U2 r2 U2 r U2 r U2 l r2' U2 r' U2 r U2 l' U2
Hey, you told me you knew it once!

#### shelley

##### chang
Maybe I learned it once, then forgot it. I emailed it to myself so it's in my Gmail archives if I ever need to pull it out.

#### Kenneth

##### Not Alot
It is possible to solve it using commutators. Just do r or r' first then it will work. Problem is that you also have to commutate the edges of the r-slice back into their original places.

#### joey

##### Member
Well yes... but considering that's an awful way of doing it.

#### joey

##### Member
Well, actually. It is *impossible* to solve using commutators!

#### Kenneth

##### Not Alot
OK, then I got a tricky but short one (alg, not commutator) that I just made up:

F2 r (y) M2 (y) U2 l' U2 l U2 r' U2 r (x') U2 r D2 r' U2 (x')

Looks long because of all orientations but it is "only" 16 turns. Remove first and last move, that are setup + restore, and it swaps two diagonal LL edges.

#### Lucas Garron

##### Member
Well yes... but considering that's an awful way of doing it.
Nah, but we've established you can use comms to help solve it. That's solving it using comms.

#### McWizzle94

##### Member
Well yes... but considering that's an awful way of doing it.
Nah, but we've established you can use comms to help solve it. That's solving it using comms.
Joey, you just got cubically owned

#### Stefan

##### Member
But if you read the original question carefully...

#### Swordsman Kirby

##### Member
You could always set it up as the 4x4 "OLL parity" which switches two edges (e.g. l' D2 l and then do the OLL parity to "flip" the UB edge.
Yay something I showed you.

#### joey

##### Member
Show me a solution using *commutators*!

#### Stefan

##### Member
Joey, learn English. I can even solve it using red gloves.

#### joey

##### Member
Joey, learn English. I can even solve it using red gloves.
Aha! But what about red socks?