Ok
There are 5 types of cases, ranging from 8-12 moves.
Pure (8)
A9s (9)
Orthogonals (10)
Cyclic Shifts (10)
Columns (11)
Per Specials (12)
We called them this just for the sake of terminology, and having defined vocabulary.
As you can see, the Pures are 8 moves, using
dbeyer.110mb.com/BHcorners.txt as a reference
198 of the 378 cases are optimally 8 moves.
That means there is a 3 move part A, and then an interchanging. Then A'B'
ABA'B' or BAB'A' to solve the cube.
You've cubed enough that you should be able to see most of these pure cases.
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A9s
The concept of all of these cases there isn't an insertion that provides interchangeability. Yet the setup to give interchangeability actually leads to a cancelation with the insertion.
SABA'B'S'
The 1 turn setup and the insertion cancel a move creating a double turn for the first or last move.
URB -> DLB -> BRD
I see that L2 creates interchangeability with the DLB and URB. I also see that doing L' creates interchangeabilitity with the DLB and BRD.
So perhaps a setup of
L', then an insertion of L'U2L, and interchange with D2
S: L'
A: L'U2L
B: D2
SABA'B'S'
L2U2L D2 L'U2L D2 L
The A-perm is a commutator with a cancelation
S: R
A: R F2 R'
B: B'
R2 F2 R' B' R F2 R' B R'
Now the worst and smallest set of cases are the per specials. They are optimally 12 moves. The recognition thereof is Simple. The three stickers interchangeable by double turns with the buffer.
So the URBs 3 double turns are, U2 R2 and B2.
Giving you the ULF, DRF, and DLB The buffer permute to any combination of the other 3 pieces is a per special.
Orthogonals are a varitation of the per specials.
Same pieces, but different sticker permutations.
Buffer and two of the pieces, but all three stickers must be on orthogonal planes.
L/R are parallel,
U/D are parallel,
F/B are parallel,
So you need each sticker to be on a different set of planes.
One for the U/D (buffer)
One for the L/R, and one for the F/B.
The simple solution is do a quarter turn setup to create interchangeability then a commutator and unset. There are no cancelations.
Examples.
UBR -> LFU -> FDR
Do say F as the setup, or perhaps U'
Really there are only 54 cases over 9 moves.
Now the next concept are cyclic shifts.
You are actually getting a 10 mover that looks really wierd.
You're doing something along the lines of taking a 4 move insertion, and double turn interchange. Then you write the alg out but you start on 3 third move of the ABA'B' commutator and loop around to get a weird effect on the cube.
Heres the setup.
All three cubies are on 1 plane.
Say the R face.
That means there is a middle piece.
The target sticker of the middle piece isn't interchangeable with either of the other two target stickers on the other pieces.
But look.
URB -> DBR -> RUF
URB is the middle cubie. Its not interchangeable with anything but look ...
If you do,
F you see that URB and RUF are interchangeable by R2
if you do D' you see that URB and DBR are interchangeable by R2
Now
R2 is the interchanging
FD' R2 DF'D'F R2 F'D
or its inverse cycles the pieces.
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