# Need commutator that rotates centers 180 degrees on D face.



## cuber786 (Nov 20, 2008)

Hello all, I am very bad with making commutators. So can someone please give me a commutator for this case?:

1.)rotate centers on D face 180 degrees (So top left center on D switches with bottom right,etc).

I also need a commutator for this case:
2) Rotate centers on U face counterclockwise(So bottom left center on U would then be bottom right,etc).

Thanks in advance.

Edit: Oh crap My bad I forgot to mention that this was for 4x4. Sorry about that


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## ShadenSmith (Nov 20, 2008)

Rotate the D center 180 degrees? 

x2 (2 T-Perms) x2



I think you're asking for the wrong thing. I'm not sure what exactly you're asking, actually. It sounds like something for a >3 cube because you mentioned centers, but without specifying I really have no idea. More specific please?


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## cuber786 (Nov 20, 2008)

oops I meant for the 4x4 sorry.


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## ThePizzaGuy92 (Nov 20, 2008)

in that case...

x2 (2 T-Perms) x2

lol


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## masterofthebass (Nov 20, 2008)

or... 

x2 RL U2 R'L' U RL U2 R'L' U x2

There's also a nicer one:

(R' U' R U')*5


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## ConnorCuber (Nov 20, 2008)

masterofthebass said:


> or...
> 
> x2 RL U2 R'L' U RL U2 R'L' U x2
> 
> ...



In the second you for got x2 surrounding the alg


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## deadalnix (Nov 20, 2008)

(RLD2R'L'D)2 is the simpliest.


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## Stefan (Nov 20, 2008)

You guys are all missing the point. He explicitly said he needs *commutators* and you're not giving him any.

Here's one (lower case meaning inner slice only): [ r l u2 r' l', U ]

That was trivial, now someone please find a pure single-center rotation commutator for the 3x3 or prove that it's impossible. I just thought about it for a bit and am not sure.


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## TMOY (Nov 20, 2008)

For the 3^3 it's impossible. That's because the trivial permutation is even; if you start and end on the solved state you will always have performed an even number of quarter-turns inbetween.
For the 4^3 it's impossible too, for similar parity reasons: the permutations of corners and center pieces must always have the same parity.


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## Stefan (Nov 20, 2008)

TMOY, I'm obviously allowing 180 degrees turns. Try again.


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## AvGalen (Nov 20, 2008)

StefanPochmann said:


> You guys are all missing the point. He explicitly said he needs *commutators* and you're not giving him any....


That's because he didn't specify why he needed commutators. Everyone seemed to assume (and I think rightfully) that he didn't really need commutators.

Maybe the topic starter can explain why he needs commutators.

(and I personally do x2 (2*J-Perm) x2 instead of x2 (2*T-Perm) x2


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## Stefan (Nov 20, 2008)

AvGalen said:


> Everyone seemed to assume [...] that he didn't really need commutators.



Yeah I assumed that as well. I think what he needed was someone to make him aware he's misusing terminology.

Now c'mon, solve my riddle. That's much more interesting than providing yet another trivial "well the way I do it" answer.


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## TMOY (Nov 20, 2008)

OK, I've got a solution for the 4^3. Lowercase letters are still inner-slice moves:
[l u r u r' u l', U]

Edit: For the 3^3 it is actually impossible. Say you have a commutator [A, B]. If A rotates the centers on themselves a given number of times, A' will execute the same rotations backwards, not necessarily on the same centers if you allow double layer and slice moves, but the overall sum will always be zero. Same for B and B'. So you cannot perform a 180 degree rorarion on one single center using only a commutator.


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## cubeRemi (Nov 20, 2008)

StefanPochmann said:


> TMOY, I'm obviously allowing 180 degrees turns. Try again.



2 faces, 180:

(M2U2M2U2) (S) (U2M2U2M2) (S')

I did not found a comm. for just one face (yet).


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## AvGalen (Nov 20, 2008)

StefanPochmann said:


> AvGalen said:
> 
> 
> > Everyone seemed to assume [...] that he didn't really need commutators.
> ...



First time you said "now someone *please *find..."
Second time you said "Try again."
Now you say: "Now c'mon, solve my riddle"

First you go out and do the Smith experiment


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## Stefan (Nov 20, 2008)

Ok thanks TMOY, that's a good explanation and I agree with it (just adding that counting the rotations ought to be done modulo 4 (or modulo 360 degrees)).


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## cubeRemi (Nov 20, 2008)

and this sort of commutator: A B C A B C, A B C B C A ore A B C A C B 

ABCDEFGABCDEFG

what is the fewest number of letters needed? 

Remi


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## deadalnix (Nov 20, 2008)

I don't know any commutator technique that can perform changes on only one piece :S

I'm not sure it is possible.


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## Stefan (Nov 20, 2008)

deadalnix said:


> I don't know any commutator technique that can perform changes on only one piece



Just to prove there are puzzles where it's possible: On a 1x1x3 do [ E, x2 ]

On a 3x3x3 we now know it's impossible because its centers are out of the question and all other pieces can't be changed alone anyway. It also easily extends to all NxNxN cubes (except 1x1x1). Don't know about other puzzles.


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## TMOY (Nov 20, 2008)

deadalnix said:


> I don't know any commutator technique that can perform changes on only one piece :S


Pop a corner and hold it with the stickers backwards, twist it clockwise, fix the pop, twist the corner counterclockwise.


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## rjohnson_8ball (Nov 20, 2008)

deadalnix said:


> (RLD2R'L'D)2 is the simpliest.



That is the one I use. If you perform the commutator [ [R L, D2], D ]
and apply an extra D2 afterward, you get
((R L D2 L' R' D2) D (D2 R L D2 L' R') D' )(D2)
which is (R L D2 L' R' D)*2. Did I do it okay?


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## deadalnix (Nov 21, 2008)

Like this, it's a double permutation of group composed by one corner and one edge. It's an elegant way to see it as a commutaor !


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## Stefan (Nov 21, 2008)

For understanding, I like this best: [ L R d2 R' L', U ]
It's also a double swap of corner-edge pairs (in the U layer) and more direct.


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## rjohnson_8ball (Nov 21, 2008)

StefanPochmann said:


> For understanding, I like this best: [ L R d2 R' L', U ]
> It's also a double swap of corner-edge pairs (in the U layer) and more direct.



Would that need a U2 applied afterward?
((L R d2 R' L') U (L' R' d2 R L) U') U2


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## Stefan (Nov 22, 2008)

Yes of course. Just wanted to show a more understandable way for the pair swaps.

Oh and you made a mistake unwinding my commutator notation alg. Your alg does something different, resulting in a nice pattern. Can be done shorter, though.


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