Does anybody really understand commutator tricks?
Or are you guys looking at each case and seeing how to set it up to do a commutator, which you all know as an 8 mover?
For example every optimal corner commutator in optimal length is 8-12 moves.
Do you understand how to determine the move count? Do you understand the tricks?
8 movers are rather simple. ABA'B'
A being the insertion (3 moves)
B being the interchanging
A+B+A'+B+ = 8
9 movers have a setup turn. The setup turn actually cancels with either the part A or part B of the commutator.
Such as RB'RF2R'BRF2R2 = R; [B', RF2R'] = R B' RF2F' B RF2R' R'
= R B' RF2R' B RF2R2 <-- Canceling the setup move by merging the two R's making it 9 moves.
There are cases that can only be done in 10 moves. You might look at me, if I showed you some cases, and say oh well I'd do this. It's the same move count, or why not do this. These are called orthogonal cases. These cases are optimally solved by just doing a quarter turn setup, and then solving with an 8 mover. Chris and I call these orthogonal cases because as you look at the cycle you will see three mutually opposite pieces. The pieces as a whole can be interchanged only with double turns.
Such as URB -> ULF -> DRF
Now that specific corner cycle is actually a 12 move count. We are looking just at the pieces, not the sticker permutations for now.
See how the URB and ULF are interchangeable by U2.
The ULF and DRF are interchangeable by F2
The URB and DRF are interchangeable by R2
Now the distinction between the 12 move case, and these 10-mover "orthogonal cases" is the sticker permutation and their positions relative to one another.
There are 6 planes on a cube. The U, D, F, B, L, and R plane.
Each plane has a parallel plane. Yet in free space in order to be perpendicular you need three reference points. To be perpendicular (right angled) in free space you need a sticker from each of the sets of parallel planes. So a target sticker on the U/D plane, a target sticker on the L/R plane, and a target sticker on the F/B plane.
Such as <u>U</u>RB -> <i>L</i>FU -> <b>F</b>DR
There is another case for 10 movers that is pretty sick too. It's called a cyclic shift.
These cases might be recognized as COLL cases if you were to look at the cycles on the U layer.
You have three pieces on one layer. Lets start basic with a U layer cyclic shift.
URB -> RUF -> BLU
You see all three are on the U layer. Notice the middle piece (URB) isn't interchangeable with either of the adjacent pieces. We call this AnI. Adjacent, non-Interchangeable.
Now watch. Both of the middle piece's AnIs on the U layer can actually become interchangeable with setup turns.
L/L' make the BLU interchangeable with U2/B2
F/F' make the RUF interchangeable with R2/U2
Notice the common interchangeability of U2?
U2 is insertion in this case. LF' and F'L are the setups.
The URB is inserted to the RUF
So we need to insert there first
LF' sets the RUF to the ULF, so do U2, undo the setup.
F'L sets the BLU to the ULF, so do U2, undo the setup.
LF' U2 FL' F'L U2 L'F
Now it's called a cyclic shift, because you'll notice if you start at a different point in that alg it becomes
U2 FL'F'L U2 L'FLF' which is just a commutator with a 4 move insertion. Odd how changing the starting point has such a different effect eh?!
There are also columns cases which are 11 moves, there are two ways to handle these. Doing a cyclic shift that has a canceling effect just like in a 9-mover. Or you can do a setup and an 9 mover.
Per Specials are the 12 move cases. There are only 6 of these in a buffer bound system. I showed you an example of one above. All all mutually opposite and on the parallel planes.
Later,
DB
I'll get the link and post it here later.
