# 5-cycle commutator



## deadalnix (Apr 16, 2010)

Hi,

I'm currently wondering if there is a way to create commutator for 5-cycles. Even if it's not as simple as creating them for 3-cycles.

I have no idea about how to do, but I'm pretty sure this is an interesting way to explore.


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## Kirjava (Apr 16, 2010)

Put two 3cycles together and cancel?

M'F2MF2U'M2UM2


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## deadalnix (Apr 16, 2010)

This is working, for sure, but I'm not very satisfied with a solution like this. I was look for more knowledge here


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## Stefan (Apr 16, 2010)

M U M' U'


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## Kirjava (Apr 16, 2010)

deadalnix said:


> I'm not very satisfied with a solution like this




What do you want, magic?

F2URU'RF2R'URF2R2F2U'


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## Tim Reynolds (Apr 16, 2010)

Kirjava said:


> deadalnix said:
> 
> 
> > I'm not very satisfied with a solution like this
> ...



Well, he did ask for a commutator. Are those commutators? I can't tell.


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## Stefan (Apr 16, 2010)

I guess for commutators, the obvious approach would be to do it like 3-cycle commutators, but instead of [swap1intoArea,turnArea] do [swap*2*intoArea,turnArea]. Here's one:

A = (U M' U')
B = (L' R)
=> (U M' U') (L' R) (U M U') (R' L)

This is actually quite neat, as it allows many variations:
- Switch U and U' in A
- Replace M' with M in A
- Use any L+ R+ as B
That's 2*2*3*3 = 36 algorithms right there, and they're short and easy to understand.

Well, on the other hand, I think M U M' U' plus setups still beats it (except at being a pure commutator).


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## Kirjava (Apr 16, 2010)

Tim Reynolds said:


> Kirjava said:
> 
> 
> > F2URU'RF2R'URF2R2F2U'
> ...




[R2:[R' U' R, F2]][F2,U']


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## Tim Reynolds (Apr 16, 2010)

Wait, what? I'm not sure I get your notation--at least, I can't get it to equal your other alg.

Also, that's not a commutator...a commutator is something of the form [a,b], not the product of two commutators, one of which is conjugated.


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## Simboubou (Apr 16, 2010)

Hum, interresting...

Thus, if you want to cycle UFL->URF->UBL->DFR->DLB, you'll do :

A = U'
B = F B' D2 F' B

==> U' (F B' D2 F' B) U (B' F D2 B F')


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## Kirjava (Apr 16, 2010)

Tim Reynolds said:


> Wait, what? I'm not sure I get your notation--at least, I can't get it to equal your other alg.




[R2:[R' U' R, F2]]

R2 R' U' R F2 R' U R F2 R2

[F2,U']

F2 U' F2 U

Combine & Cancel

R U' R F2 R' U R F2 R2 F2 U' F2 U

Cycle

F2 U R U' R F2 R' U R F2 R2 F2 U'



Tim Reynolds said:


> Also, that's not a commutator...a commutator is something of the form [a,b], not the product of two commutators, one of which is conjugated.




You're being pedantic.


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## deadalnix (Apr 16, 2010)

StefanPochmann said:


> This is actually quite neat
> 
> [...]
> 
> Well, on the other hand, I think M U M' U' plus setups still beats it (except at being a pure commutator).



Actually, this is really neat. I think we have someting here.

For commutator, this is not a requirement actually. I'm more looking for an « understandable » way to create 5-cycles.


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## Stefan (Apr 16, 2010)

cmowla said:


> M' U R *L'* U' M U R' *L* U'.



Can be done shorter, though: F' M F M'


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## mrCage (Apr 19, 2010)

StefanPochmann said:


> I guess for commutators, the obvious approach would be to do it like 3-cycle commutators, but instead of [swap1intoArea,turnArea] do [swap*2*intoArea,turnArea]. Here's one:
> 
> A = (U M' U')
> B = (L' R)
> ...


 
Nice idea Stefan. But there are cases where you would get 2 distinct 3-cycles or 2-cycles, not a 5-cycle.

Example: A=FLEL'F' , B=U (2 2-cycles)

Per


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## Stefan (Apr 19, 2010)

Yes, Per, if you do something entirely different, not fitting my scheme at all, then you can get something my scheme isn't supposed to do. Bravo.


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## mrCage (Apr 20, 2010)

StefanPochmann said:


> Yes, Per, if you do something entirely different, not fitting my scheme at all, then you can get something my scheme isn't supposed to do. Bravo.


 
Hmmm. My example follows your pattern. Or is it something about the swap part i'm missing? My A part simple puts 2 other cubies into the U-layer. Hmm ....

Anyhow i fail to see how one would easily make useful 5-cycles on the fly, ie intuitive. They seem to require much more thinking for the desired effect. Maybe more useful for fewest moves. I have actually constructed 5-cycles on edges for fewest moves ...

Per


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## Stefan (Apr 20, 2010)

mrCage said:


> My A part simple puts 2 other cubies into the U-layer.



First of all, it doesn't look like A = (U M' U') at all. If you didn't mean that, why did you quote it? Secondly... no, it doesn't even put two cubies into the U-layer but only one.


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## mrCage (Apr 20, 2010)

StefanPochmann said:


> mrCage said:
> 
> 
> > My A part simple puts 2 other cubies into the U-layer.
> ...


 
I never said it fits ur EXAMPLE. But it fits the general idea though: [swap*2*intoArea,turnArea]. Right?? Ok fine. If i hadn't used ur notation A for swap*2*intoArea and same with B the confusion wouldn't arise ...


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## Stefan (Apr 20, 2010)

mrCage said:


> I never said it fits ur EXAMPLE.


Well you replied to it! Again: Don't quote it if you're not actually responding to it.



mrCage said:


> But it fits the general idea though: [swap*2*intoArea,turnArea]. Right??


Again: No. One of the two pieces had been in the U layer already and just got moved around inside of it.

There's an easy example using my example, btw, simply don't use M' or M in my A but M2. That would indeed not create a 5-cycle. Which is why I didn't list that option. I *would* have liked to have 54 algs instead of 36.

But I agree it was worth explicitly mentioning that it doesn't always work for the 5-cycle version, because some might not have realized it and because it does always work for for the 3-cycle version.


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