I just thought I would mention a kind of interesting situation on one of my "computer" solves.
It was for scramble #5.
Scramble:
F U F U F2 R2 D' R2 B2 D B2 D2 B2 R2 U' F R F' B' D' R2 D' L'
Uw2 F' Uw2 U2 Rw2 Fw2 B D F' B D' L2 B2 Rw' Fw2 R' U B U' Rw' Uw R' Fw B U L D R
Reduction:
U' L2 2F' 2U' L2 F 2L' 2F
D B Uw2 L U' Lw2 U Lw2 F2 U' Uw2 Fw2 Lw2
Fw R' F R Fw2 B' U B U' Fw y2
The reduction created a logical 3x3x3 state with a prebuilt 2x2x1 block. It also had a prebuilt C-E pair that would extend
that 2x2x1 block to a 3x2x1 block. In fact, applying:
D' L' F' L2 D2
builds a
2x2x3 block (with puzzle being viewed as a 3x3x3) with only
5 moves!
U' L2 2F' 2U' L2 F 2L' 2F
D B Uw2 L U' Lw2 U Lw2 F2 U' Uw2 Fw2 Lw2
Fw R' F R Fw2 B' U B U' Fw y2 D' L' F' L2 D2
Now, the big question is: why did that happen on a "computer" solve instead of on a "human" solve where I could actually take advantage of it?
EDIT: So far, I've found (without computer searching) a 29-move solution for the above reduction. While I know that F2L minus 1 slot can be achieved in 10 moves, I used 12 in this solution.
Skeleton:
2x2x3: D' L' F' L2 D2 (5)
F2L minus 1 slot: F . R2 * F' U2 R U2 R (12)
Edges: R U R' U' F' U' F U (20-1 = 19)
This left a 5-cycle of corners.
I found pairs of 3-cycle insertions that would cancel 2+2 moves or 3+1 moves. Thus, it looked to me like I would get a 19+8+8-4 = 31-move solution. However, the insertions cancelling 2 moves each were so close together, that there was a move from the 2nd insertion that cancelled out a move from the first insertion. This extra double-cancellation brought down the final move count to 29f (also 29s).
Insert at ".": D L' D' R2 D L D' R2 (27-2 = 25)
Insert at "*": D B2 D' F' D B2 D' F (33-2-2 = 29)
Final solution: D' L' F' L2 D2 F D L' D' R2 D L B2 D' F' D B2 D' U2 R U2 R2 U R' U' F' U' F U (29f/29s)
