Please help! Need a 4x4 commutator that does the following...

[cuber786]

Please help me, I really need a commutator that does the following on the 4x4:

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.

Thanks in advance.

 

[shafiqdms1]

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

 

[shelley]

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

Thanks. That was really helpful.

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

 

[qqwref]

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]

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]

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]

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]

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

 

[Kenneth]

Yep!, but it works

 

[joey]

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

 

[Kenneth]

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]

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]

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]

But if you read the original question carefully...

 

[Lucas Garron]

But if you read the original question carefully...

But if you read

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

carefully...

 

[Swordsman Kirby]

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]

Show me a solution using *commutators*!

 

[Stefan]

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

 

[joey]

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

Aha! But what about red socks?