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.
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.
Last edited by macky; 09-14-2011 at 10:22 PM.
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
 Ravi, I haven't ignored your post.
"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].
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.Originally Posted by cmowla
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.
Last edited by macky; 09-15-2011 at 07:24 PM. Reason: needed to repeat "used it"!
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.