# New commutator challenge.



## mrCage (Jun 1, 2011)

Find a way to twist (rotate) 2 corners with no side effect using only 2 faces of the cube. This is not too hard by easy modification of a well known alg. 

Now do it as a direct commutator using only 2 faces!!

Per


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## d4m4s74 (Jun 1, 2011)

right now without my cube on me I can only think of three faces, I'll be back later


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## Cubenovice (Jun 1, 2011)

Spoiler



I suppose the well know alg would be Sune / Antisune combo's. RU all the way to twist two corners:

Sune U2 ASune U2 to give headlights on L:
R U R' U R U2 R' U2 R' U' R U' R' U2 R U2

But since the Sunes in itself are commutators with cancelling moves I suppose you are looking for something different with your "direct" commutator?


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## Stefan (Jun 1, 2011)

Spoiler



[U2 R U R' U R U2 R', R']


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## PM 1729 (Jun 1, 2011)

Spoiler



Easy way using sune/antisune:
RU2R'U'RUR'U'RU'R' U RUR'URU2R'U


What exactly do you mean by direct commutator?

Edit:
Oh I see. Nice one Stefan . Looks neater written like that


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## mrCage (Jun 1, 2011)

Stefan said:


> [U2 R U R' U R U2 R', R']


 
I knew you could do it Stefan. How about twisting UFR and UBL with a direct commutator using only U and R ??
(Or even UFD and UBL.)

Per


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## Stefan (Jun 1, 2011)

Please use commutator notation so we can easily see it's indeed a commutator, and how. Or in your cases, *you* can see it's *not* a commutator, and refrain from posting an invalid answer in the first place.



mrCage said:


> I knew you could do it Stefan. How about twisting UFR and UBL with a direct commutator using only U and R ??





Spoiler



[R2 U' R' U R' U' R2 U, U2]





mrCage said:


> (Or even UFD and UBL.)


 
UFD?


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## mrCage (Jun 1, 2011)

I meant URD and UBL Stefan. My bad !!

Per


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## Kirjava (Jun 1, 2011)

lol URD


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## Stefan (Jun 1, 2011)

lol I thought about highlighting the problem but thought "nah..."


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## vcuber13 (Jun 1, 2011)

U2 R [U2 R U R' U R U2 R', R'] R' U2?

i think that would be
[U2 R, [U2 R U R' U R U2 R', R']]


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## Stefan (Jun 1, 2011)

Stefan said:


> [R2 U' R' U R' U' R2 *U*, U2]



That U isn't really needed, it can be left out or changed to U' or U2 and all four variations have *exactly the same effect*:
[R2 U' R' U R' U' R2, U2]
[R2 U' R' U R' U' R2 U', U2]
[R2 U' R' U R' U' R2 U2, U2]


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## Stefan (Jun 1, 2011)

vcuber13 said:


> U2 R [U2 R U R' U R U2 R', R'] R' U2?
> 
> i think that would be
> [U2 R, [U2 R U R' U R U2 R', R']]


 
Sigh.
[U2 R, [U2 R U R' U R U2 R', R']]


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## vcuber13 (Jun 1, 2011)

[U2 R: [U2 R U R' U R U2 R', R']]


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## Stefan (Jun 1, 2011)

But how is that a direct commutator? I must admit I'm not 100% sure what Per means, but I'm rather certain he doesn't want an outer conjugation.

Here's one for UBL and DFR:
[[R2 U' R' U R' U' R2, U2], R2 U' R2]


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## Lucas Garron (Jun 1, 2011)

Stefan said:


> [U2 R U R' U R U2 R', R']


Aw, I get here an hour late, and you've already spoiled all the fun of being first. I literally thought of posting [U2 R U R' U R U2 R', R] immediately, but apparently I'm not original enough.


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## Cubenovice (Jun 1, 2011)

Some nice stuff here.

Now if I had only put the U2 at the beginning I could have claimed the solution ;-)

BTW: shoudn't we all use spoilers to keep the challenge alive for people who come fresh into the thread?


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## Stefan (Jun 1, 2011)

Sorry, Lucas 

Another approach for UBL and DFR:
[(R U R' U R U2 R') U (R2 U' R' U' R U R U R U' R) U, U R U']



Cubenovice said:


> BTW: shoudn't we all use spoilers to keep the challenge alive for people who come fresh into the thread?


 
Alright... I've spoilered my solutions on the first page. If people read further before doing it themselves, it's their own fault.


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## mrCage (Jun 1, 2011)

Stefan said:


> But how is that a direct commutator? I must admit I'm not 100% sure what Per means, but I'm rather certain he doesn't want an outer conjugation.
> 
> Here's one for UBL and DFR:
> [[R2 U' R' U R' U' R2, U2], R2 U' R2]




I wonder if this is optimal though??

Per


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## Stefan (Jun 1, 2011)

I doubt it. But beat me


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## qqwref (Jun 1, 2011)

Stefan said:


> But how is that a direct commutator? I must admit I'm not 100% sure what Per means, but I'm rather certain he doesn't want an outer conjugation.


Surely you realize that [O:[P,Q]] = [[O], [O:Q]], no? And any sequence that can be written completely in the form [P,Q] is a direct commutator to me.

So, [R' U' R' U R' U' R, R U2 R'] flips UBL and DFR.


One more challenge: Find a direct 2gen* commutator that flips two edges.

*generated by two moves, not by two move sequences


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## Kirjava (Jun 1, 2011)

qqwref said:


> [O:[P,Q]] = [[O], [O:Q]]



Mind = blown.



qqwref said:


> *generated by two moves, not by two move sequences


 
[F',U]


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## mrCage (Jun 1, 2011)

qqwref said:


> So, [R' U' R' U R' U' R, R U2 R'] flips UBL and DFR.


 
Wow!!! A mere17 turns with my counting. I guess this one is optimal. But why you call it flip and not twist?? I thought it was "agreed" that corners twist and edges flip ...

Per


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## cuBerBruce (Jun 1, 2011)

Optimal <U,R> commutator for twisting UBL and DRF corners:


Spoiler



[U R U' R U, R2 U R2]


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## qqwref (Jun 1, 2011)

mrCage said:


> Wow!!! A mere17 turns with my counting. I guess this one is optimal. But why you call it flip and not twist?? I thought it was "agreed" that corners twist and edges flip ...


Oh yeah, my mistake.



cuBerBruce said:


> Optimal <U,R> commutator for twisting UBL and DRF corners:
> 
> 
> Spoiler
> ...


Neat! (Did you use a computer program to find this?)


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## cuBerBruce (Jun 1, 2011)

qqwref said:


> Neat! (Did you use a computer program to find this?)


 
Yes, I used GAP to try all possible <U,R> commutators up to 16 moves in length. So I knew this was optimal without allowing for cancellations. JACube gave 16f* as the minimal <U,R> alg for twisting UBL and DRF, so then I also knew it was not possible to get a shorter solution for this case using cancellations.


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## Stefan (Jun 1, 2011)

qqwref said:


> Surely you realize that [O:[P,Q]] = [[O], [O:Q]], no?



I do now 



qqwref said:


> So, [R' U' R' U R' U' R, R U2 R'] flips UBL and DFR.



Now that's just witchcraft. How did you find that?



qqwref said:


> Find a direct 2gen* commutator that flips two edges.
> *generated by two moves, not by two move sequences



Mmh, I understand neither the question nor Thom's answer :-(


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## DavidWoner (Jun 1, 2011)

qqwref said:


> [O:[P,Q]] = [[O], [O:Q]]



O_O



qqwref said:


> One more challenge: Find a direct 2gen* commutator that flips two edges.
> 
> *generated by two moves, not by two move sequences


 


Spoiler



(M' U' M' U' M' U' M' U', U)


ez


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## qqwref (Jun 1, 2011)

Stefan said:


> Now that's just witchcraft. How did you find that?


[R' U' R' U R' U' R, R U2 R'] is an R setup into your [R2 U' R' U R' U' R2, U2].

Nice job Woner


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## Stefan (Jun 1, 2011)

qqwref said:


> [R' U' R' U R' U' R, R U2 R'] is an R setup into your [R2 U' R' U R' U' R2, U2].



Ok, thanks. I think I get it now, but it's still witchcraft!!!



qqwref said:


> Nice job Woner



What the ... ? He used a sequence. I don't get it.


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## qqwref (Jun 1, 2011)

I meant that the commutator must be 2gen, in the sense of being a combination of only two types of moves (although they can be any types), as opposed to the more general sense of being a combination of only two specific sequences/permutations.


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## Stefan (Jun 1, 2011)

Ah, my bad. I guess I was confused because it's the same limitation as Per's (or not?), and I thought you were imposing an extra rule.

[M U M U M U M, U] also works. David wasted one move! (Well, not really, thanks to cancellation)


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## Kirjava (Jun 1, 2011)

Stefan said:


> Mmh, I understand neither the question nor Thom's answer :-(


 
I was being silly, he didn't say it had to be a pure flip


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## DavidWoner (Jun 2, 2011)

Stefan said:


> [M U M U M U M, U] also works. David wasted one move! (Well, not really, thanks to cancellation)


 
Not surprised. I just took the pure-4 flip (M' U')4 and threw in a U move knowing I'd get a two-flip. The alg I use for BLD is just [(M' U')4, U'], but since A and A' are identities I perform it as A B A B' to make it faster.

Another way to arrive at your solution Stefan is to take the 4flip (M' U')4 and combine it with the F/B mirror (M U)4, which happens to be its inverse. M' U' M' U' M' U' M' U' M U M U M U M U pretty obviously breaks down into (M' U' M' U' M' U' M') U' (M U M U M U M) U, or [M' U' M' U' M' U' M', U']


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## riffz (Jun 8, 2011)

qqwref said:


> Surely you realize that [O:[P,Q]] = [[O], [O:Q]], no?


 
Holy ****.

And I got the same answer as Woner for edge 2-flip.


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