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



## cuber786 (Nov 26, 2008)

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.


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## shafiqdms1 (Nov 26, 2008)

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


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## shelley (Nov 26, 2008)

shafiqdms1 said:


> 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


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## qqwref (Nov 26, 2008)

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.


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## cmhardw (Nov 26, 2008)

cuber786 said:


> 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.


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## Lucas Garron (Nov 26, 2008)

shelley said:


> 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!


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## shelley (Nov 26, 2008)

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.


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## Kenneth (Nov 26, 2008)

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.


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## joey (Nov 26, 2008)

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


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## Kenneth (Nov 26, 2008)

Yep!, but it works


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## joey (Nov 26, 2008)

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


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## Kenneth (Nov 26, 2008)

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.


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## Lucas Garron (Nov 26, 2008)

joey said:


> 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.


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## McWizzle94 (Nov 26, 2008)

Lucas Garron said:


> joey said:
> 
> 
> > Well yes... but considering that's an awful way of doing it.
> ...



Joey, you just got cubically owned


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## Stefan (Nov 26, 2008)

But if you read the original question carefully...


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## Lucas Garron (Nov 26, 2008)

StefanPochmann said:


> But if you read the original question carefully...



But if you read


joey said:


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


carefully...


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## Swordsman Kirby (Nov 26, 2008)

shelley said:


> 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.


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## joey (Nov 26, 2008)

Show me a solution using *commutators*!


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## Stefan (Nov 26, 2008)

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


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## joey (Nov 26, 2008)

StefanPochmann said:


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



Aha! But what about red socks?


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## Stefan (Nov 26, 2008)

Red socks are impossible.


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## joey (Nov 26, 2008)

Okay, I was slightly worried when I saw you had replied. I had feared that you might have been able to.


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## Stefan (Nov 26, 2008)

You're confusing me with these interacting past tense variations whose names I don't know just to teach me a lesson, aren't you?


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## joey (Nov 26, 2008)

It's all subconscious Stefan.. all subconscious.


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## qqwref (Nov 26, 2008)

Hi. I solved it with r plus commutators. I can't quite figure out how to combine these all into one.

r [U' R' U r' U' R U r U' R' U, r2] [u' R2 u R b' R2 b R2 u R2 u' R b R2 b' u' R2 u, r2] [u' R u2 R2 u2 R' u, r']

Anything can be solved "using" commutators, but this case can't be solved with only commutators


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## joey (Nov 26, 2008)

qqwref said:


> Hi. I solved it with r plus commutators. I can't quite figure out how to combine these all into one.
> 
> r [U' R' U r' U' R U r U' R' U, r2] [u' R2 u R b' R2 b R2 u R2 u' R b R2 b' u' R2 u, r2] [u' R u2 R2 u2 R' u, r']
> 
> Anything can be solved "using" commutators, but this case can't be solved with only commutators



Exactly   Obviously, qqwref managed to find/wear and fully utilise his red socks.


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## qqwref (Nov 26, 2008)

Red socks?


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## joey (Nov 26, 2008)

You should read the last few posts P


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## qqwref (Nov 26, 2008)

I did, but I don't get the part where red socks help you solve things :|


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## joey (Nov 26, 2008)

Nor do I, we were just getting silly


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## Kenneth (Nov 27, 2008)

Sock on this: (not the same case but...)

U2 l' U2 r 3D2 r' U2 r 3D2 M U2 l ... 3D2 is a triple layer D-turn = the swaps.

This commutator does a 3-cycle triplets of two centres and one edge each, a J-PLL in the second layer if you like. Obviously it solves edge parity... but also destroys some centres (solvable using commutators).

BTW: change 3D2 to only D2 and skip the last l and you got a nice 3-cycle edges in one layer ("A-PLL").


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## shafiqdms1 (Nov 28, 2008)

joey said:


> StefanPochmann said:
> 
> 
> > Joey, learn English. I can even solve it using red gloves.
> ...





StefanPochmann said:


> Red socks are impossible.



oh crap that is impossible ;p


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