# New challenge!!



## mrCage (Aug 22, 2009)

Hi, I have previoulsy shown that every even position on 3x3x3 is reachable by a series of commutators.

Now i'm looking for a proof or disproof of the following:

Every even position of the 3x3x3 cube is reachable by a *SINGLE *(long) commutator. Good luck!! 

Per


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## Lucas Garron (Aug 22, 2009)

You're mean. 
I spent so long successfully ignoring http://www.speedsolving.com/forum/showthread.php?p=76741#post76741. And I even have Joyner's book now...


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## mrCage (Aug 22, 2009)

Lucas Garron said:


> You're mean.
> I spent so long successfully ignoring http://www.speedsolving.com/forum/showthread.php?p=76741#post76741. And I even have Joyner's book now...


 
Haaa. Well, i dont follow off topic much lately. Sorry - but at least i think this is a better place for such thread

Per


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## stray (Aug 25, 2009)

cuBerBruce'post in that thread: 
Assuming that's what you really meant, Joyner's book Adventures in Group Theory has the answer, but his proof "cheats" by using another theorem.

I'm curious about the proof, but I don't have Joyner's book.

@Lucas Garron: Can you give me the proof or its general idea?

Or someone give me a example how to combine two commutator to one long commutator ?


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## mrCage (Aug 26, 2009)

stray said:


> Or someone give me a example how to combine two commutator to one long commutator ?


 
An example does not constitute a proof (unless you do a full induction proof ... )

Per


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## stray (Aug 28, 2009)

mrCage said:


> stray said:
> 
> 
> > Or someone give me a example how to combine two commutator to one long commutator ?
> ...



I give up, my intuitiveness tell me this will not be true, but I have no idea to disproof it, My math always sucks.

Can you give me the answer now?


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## mrCage (Aug 28, 2009)

PS!

By even position i mean those positions reachable by an even number of quarter turns. There are other possible definitions, but this is the one that i had in mind 

Per


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## Wacky (Aug 28, 2009)

Not totally sure what you mean exactly but:

Suppose such a commutator exists, call it X. Doesn't need to be a commutator.

By pigeonhole principle for some p, X^p = Solved state I, since the cube has a finite number of states you eventually have to cycle back. Let p* be the minimum of these p. So all possible p are multiples of p*, if one isn't then you can get X^(p- np*) = I for p- np* < p*

If solved state+U2 is what you call an even position, then X^q = U2 for some minimum q*. Obviously q* < p* - otherwise X^(q*-p*) = X^q* I^-1 = X^q* = U2, contradiction.

So X^2q = U4 = Solved State I, but 0 < 2q < 2p so the only possibility is 2q = p, q=p/2.

Repeat logic for F2. So X^(p/2) = U2 = F2 contradiction, so X does not exist.

Hope the above is clear / I read the problem correctly.


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## Lucas Garron (Aug 28, 2009)

Wacky said:


> Hope the above is clear / I read the problem correctly.


You didn't.
(Were you just proving that the cube group isn't cyclic, and in particular not generated by a commutator?)

Try this:
∀s(parity of EP & CP are both even ⇒ ∃A,B(A B A' B' = x))
where s, A, B ∈ (legal cube states)


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## Wacky (Aug 29, 2009)

Lucas Garron said:


> Wacky said:
> 
> 
> > Hope the above is clear / I read the problem correctly.
> ...



Figures, and yes, pretty much.

Back to the drawing board...


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