Yes, of course. nxnxn means nxnxn.Does this theory apply to 2x2?
F2 R2 F R2 U' F U2 F R2 U'
not in general:Can [a,b] [c,d] = [e,f] be written in a way such e and f are both constructed only using a,b,c,d?
from: http://en.wikipedia.org/wiki/Commutator_subgroup#CommutatorsHowever, the product of two or more commutators need not be a commutator. A generic example is [a,b][c,d] in the free group on a,b,c,d. It is known that the least order of a finite group for which there exists two commutators whose product is not a commutator is 96; in fact there are two nonisomorphic groups of order 96 with this property.
...How did You find this out!?...
604 btm (Keep in mind that for cube sizes larger than the 3x3x3, my method will at best yield 4 times God's number for the supercube nxnxn of the nxnxn cube being solved. That's assuming we have a supercube optimal solver, which we don't.)
F L2 F D' R2 D F' L2 F D' R2 D F2 U' R2 U2 R2 U' L2 U R2 U' L U' R2 U L u B' F l F' L2 F l' B F' u' B' u L2 u' F D2 F' d2 F D2 B F' D B' d2 B D' U B' U' b2 r2 U B U' B' r2 b' B D l' B' L' R' B l B' R L B D' R U' b' D Bw U b' U' B' U D' b R' b r' b' L b r b2 r' b L' b' r f' L' f r f' L f r' u' r2 f D' b2 D f' r' D' b2 D l r' u R' Uw' r2 U l U' r2 U l2 u R B l2 B r2 u B' l2 B u' r2 l u' b' u l' B2 l u' b u l' U
B2 d' b d B2 d' b' d b l2 f' L f l2 r' f' L' f r f' b2 B' D2 b r b' r' D2 B r b r2 R' f R f' r d2 f R' f' R d2 f B' d2 f2 R' d R f2 d2 L d' L' Bw D2 L f' L' R B' L R' f L' R B R' D2 b' R' U' B F' u' B' F U2 B F' u B' F U' R d' b2 D B' D' B b2 d2 B' D B d' D2 B' d' B U D2 B' d B D' U' r Uw' B' u F u' F' B U' F u F' U2 r' D2 B U' B' D2 B U B' F D U2 L2 D' F D F' L2 U2 F D' F2
If [a,b] and [c,d] are able to be merged together with my method, then for permutations at least, [a,b][c,d] = [c,bd] != [c,db].Can [a,b] [c,d] = [e,f] be written in a way such e and f are both constructed only using a,b,c,d?
By intensive research (brain storming, as I didn't use any existing sources) for 1.5 years. Well, my method itself took about half that time, but it did take that long to both create/fully develop the method and effectively solve every possible case that can arise on the nxnxn cube and to be certain that my proof is correct....How did You find this out!?
I actually thought that myself, especially when I was solving all corner and middle edge orientation cases by hand (and verifying that they are all correct with software that I wrote in Mathematica which is over 200 pages of code and documentation if printed. I needed to write software because there are more than 10 things that can be wrong with a solution)! I never thought I would finish...it's hard to believe it's only been a year and a half since I seriously started figuring all of this out from scratch.TOTAL INSANITY
This general idea is correct for permutations (and is where my proof begins), and I believe at the beginning of this thread, Ravi implied something like this, but it's much more involved than this for permutations of all orbits on the nxnxn cube, and even more so for orientations of corners and middle edges.Every even permutation P can be written as P=(A*B), that B has the same length as A.
So, on cube with applied algorithm P, when i do algorithm A, then on the cube i would have remaining situation B-1 which could be solved with C A-1 C, where alg C is setting up the remaining elements of the cube.
First, I searched for the whole cycle on the scramble, each corners and edges. It's similar with the memorization on 3x3 BLD. (not 3OP, but BH)How do you (all) figure out which moves to do, to form the one commutator that solves a scramble? I can't find it possible.
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