# Commutator based j-perm



## MiloD (May 4, 2010)

Do a right sided J perm then setup with [U'] so only UR and URB are solved. 

You can see the remaining top layer cubies as 3 corner/edge pairs that need to cycled counter clockwise.(UBL/UB->ULF/UL->UFR/UF)

setup: B
insert: F,d2,F'
interchange: U'

total is:

U',B,[F,d2,F'],U',[F',d2,F],U,B'

Still messes up centers (if you care about that sort of thing) bc the commutator is preceded by U' which essentially changes the parity.

Just thought this was a neat solution to j-perm. probably been done before.

Have a nice day.


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## DavidWoner (May 4, 2010)

I always preferred writing it as [R': (L' d2 L, U)]

Which works out to R' L' d2 L U L' d2 L U' R, which is easily transformed into the popular R' L' U2 R U R' U2 L U' R


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## MiloD (May 4, 2010)

Interesting, I had previously understood that solution as a combo of Petrus's Sune and Niklas. I didn't expect this solution to be equivalent.


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## qqwref (May 4, 2010)

Yeah, and that solution's just a setup from the normal sune+niklas alg. Weird stuff. You probably shouldn't put commas between all the moves in your algs, though, it looks weird.

Another commutator possibility:
F U' (L d' L') U' (L d L') U2 F' + U'
Or, in normal notation and with a rotation:
R U' F U' R' U' R U F' U2 R' U'


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## mrCage (Oct 29, 2010)

Ok. Here is my bump (was reading old threads ....)

How about R2 [D' F2 D,b2] R2 ??

So many options...

Per


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