# Thread: My ultimate commutator challenge

1. Originally Posted by Kirjava
reThinking the Cube; are you implying that the Sune produces an odd number of swaps in each orbit?
NO, and that also implies that Sune+U cannot be written as a direct commutator, but Sune+U2 can.

2. There seems to be a confusion as to which sequence exactly is a sune: R' D' R D' R' D2 R or R' D' R D' R' D2 R D2.
Both permutations can be achieved be a singular commutator.

Per

3. Originally Posted by reThinking the Cube
NO
Might be worth elaborating on what you mean in the future then. If your post just consists of two quotes by other people it's quite difficult to understand what you're going on about.

4. Originally Posted by mrCage
There seems to be a confusion as to which sequence exactly is a sune: R' D' R D' R' D2 R or R' D' R D' R' D2 R D2.
Both permutations can be achieved be a singular commutator.

Per
When a speedcuber is talking about a Sune, it's generally assumed that R U R' U R U2 R'. The inverse is an anti-Sune, and both algs have 48 symmetries, of which your former is one.

5. Originally Posted by Ravi
Then, as glazik pointed out, every element of A_n is a commutator, so everything in H_perm (and therefore H) should be a single commutator.
Wait, now I can't remember why I thought it follows that everything in H is a single commutator. Orientations are easy to write as products of commutators, and maybe even single commutators... but can they just be inserted into an existing permutation commutator? I'm too tired to make sense of this right now. It should at least still be true that H = G' and that everything in H_perm is a single commutator.

6. cmowla, I didn't work out your example, but the question is whether or not it is possible to write the product of any two commutators as a single commutator. So we want to show either
(T) that the product of any two commutators can be written as a single commutator, or that
(F) some product of two commutators cannot be written out as a single commutator.
Your general approach merely shows one special case where merging is possible (by simple considerations, it seems); I'm sure you and others could come up with many more. But unless it suggests a general approach towards a constructive proof of (T), it does nothing to resolve our question.

And Per, I hope you realize that being good at FMC doesn't say much about being "brainy" on cube theory.

7. Originally Posted by macky
And Per, I hope you realize that being good at FMC doesn't say much about being "brainy" on cube theory.
Yes i know! I'm, still a bit uncertain how i would approach my own challenge.

Here are s few clues i would follow:
if 2 comms can always be combined into one, then so can any any number of comms. Proof by induction.

If we limit to one orbital it's not all that hard so see that an even permutation can be achieved by comms. Can we somehow "parallellize" the comm's first parts to schieve a general proof?? Hence combinig all orbitals involved.

I wish Jaap would come with some input

Per

8.  Ravi, I haven't ignored your post.

Originally Posted by cmowla
I don't see why this would be necessary to prove that any even permutation can be written as a single commutator because my approach doesn't limit the length of the commutators combined (and hence the two commutators could theoretically cover every permutation when combined). There many commutators which cause the same permutation of pieces as other commutators but are obviously composed of different moves.
Originally Posted by cmowla
Originally Posted by TMOY
The really nontrivial part would be to prove that 2 comms can actually be merged into one.
Do you mean that a composite commutator composed of two commutators has the same result as if the two commutators are performed one after the other in either order?
because it wasn't clear to me from this that you understood what TMOY said.

"Merging," as TMOY or whoever first used it used it, means writing some [A, B][C, D] in the form [E, F] for some E and F. You didn't define anything; you gave one set of conditions (B = D etc) under which merging is possible, as [A, B][C, B] = [AC, B].

Originally Posted by cmowla
If you look at my definition, isn't that what is meant by merging commutators? In what other combination can two commutators be merged into one, besides the trivial case when the same exact commutator is applied to more than one orbit of wings or big cube center pieces?
That's part of TMOY's question. In fact, if you can find even one product of two commutators that cannot be merged (provably, not just because the obvious methods don't work or "according my experience and theory"), then that's an even permutation that cannot be written as a single commutator and a counterexample to Per's question.

But ok, you seem to be trying to answer Per's question directly without inducting, by writing any even permutation first in the specific form you brought up, as [A, B][C, B] with A, B, C satisfying certain conditions. Your example does not seem to indicate a general approach for doing this. Maybe you can; you've certainly worked more closely with commutators than most cubers (including me). Which is why I hope that you in particular understand the obvious reductions suggested in the first posts, so that you can put your intuition to best use.

9. Originally Posted by cmowla
those particular two commutators cannot be merged into one (well, not to one that yields the same permutation as the product of the two commutators).
*Any* two can be merged into one (if I understood Bruce/Joyner correctly). You seem to misunderstand what's meant with "merging" - you don't have to reuse "the moves from" the given commutators to build your output commutator. Think about the cube group elements, not about some move sequences representing them.

10. If I understand Bruce/Joyner correctly, their result is not sufficient to prove Per's challenge. It only implies that any non-disorienting even permutation of the cube (that is any even permutation A such that A^n has no pieces in place but disoriented for any n) can be written as a commutator, you still have to deal with the disorienting ones.

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