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Interesting situation, I've never heard of something like this before.
It took me quite a while to realize that, since those sequences of moves both leave the S-slice as it was:

F2 U2 B' U2 B2 = S2 (F2 U2 B' U2 B2) S2 = B2 D2 B' D2 F2

Cool, isn't it?
But I think situations like this are so rare that it's kinda useless to learn how to recognize and exploit them.

My attempt with that scramble:

Spoiler

Scramble: B2 L2 B2 L' B2 L F2 D2 R2 D L2 D' B2 R2 B2 F' L2 R2
Solution: R2 L2 F' U' R2 U L' D2 L D2 L' D L D' F2 U' L2 D L2 D' B2 R2 D' R2 B2 L2 B2 (27 HTM)

Premove: B2
R2 L2 F' //EO (3/3)
U' R2 U //BLocks (3/6)
L' D2 L D2 L' D L //Domino Reduction (7/13)
D' F2 L2 //More Blocks (3/16)
L2 U' L2 D L2 D' * L2 U //Pairs Commutator, 2 moves cancel (6/22)
B2 //Undo Premove (1/23)

At this point I began looking for places to insert those nasty edges cycles like R2 B2 L2 U L2 B2 R2 D; the best I could find was a 2-moves cancellation for a 29 total. With some hint from IF (I've looked at where to insert the cycle):

Interesting situation, I've never heard of something like this before.
It took me quite a while to realize that, since those sequences of moves both leave the S-slice as it was:

F2 U2 B' U2 B2 = S2 (F2 U2 B' U2 B2) S2 = B2 D2 B' D2 F2

Cool, isn't it?
But I think situations like this are so rare that it's kinda useless to learn how to recognize and exploit them.

I've just realized I have badly blundered during my official Mo3, last solve.
(My NISS notation, moves in brackets go on inverse scramble)

Scramble: F2 R2 D F2 L2 D2 F2 D' R' B' D2 B' L B2 R2 U2 B' F L

What I've done:
L R2 U2 //Pseudo 3x2x1 (3/3)
(R2 L2) //3x2x1 (2/5)
(F L F2 R' F B') //Pseudo 2x2x3 + 3 pairs (6/11)
F //Adjust pseudoness (1/12)
R' B' R D2
B2 D' B' D R D R' //Crap (11/23)
R B2 R' B' R B2 L' B R' B' L //More Crap (9/32)

What I should have done
L R2 U2 //Pseudo 3x2x1 (3/3)
(R2 L2) //3x2x1 (2/5)
(F L F2 R' F B') //Pseudo 2x2x3 + 3 pairs (6/11)
F //Adjust pseudoness (1/12)
R' B' R D2
B2 D' B' L' B' L'
L B D' B' D' B D B' L' D' //Not Crap, 4 moves cancel (16/28)

I could have a 27.00 official Mo3 now; I would still be 4th in the world, but...

The hardest scramble: F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2 28 moves by Robert Yau and myself
I found the 2x2x3, he found the rest of the skeleton

Spoiler

Premove D (to help with the end of F2L after 2x2x3)
Scramble: F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2
2x2x3: R' F2 U R2 F L2 D2 L U' D B' (11) This is based on the 2x2x2 R' F2 U R2 F L' U', with some extra moves to get 2x2x3 and to help with the continuation.
F2L-1 + EO: D' F L F' D' L' D'
F2L leaving 3 corners: L B' L2 B
Skeleton: R' F2 U R2 F L2 * D2 L U' D B' D' F L F' D' L' D' L B' L2 B D (23)
Insert L2 D' R' D L2 D' R D at * to cancel 3 moves

The hardest scramble: F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2 28 moves by Robert Yau and myself
I found the 2x2x3, he found the rest of the skeleton

Spoiler

Premove D (to help with the end of F2L after 2x2x3)
Scramble: F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2
2x2x3: R' F2 U R2 F L2 D2 L U' D B' (11) This is based on the 2x2x2 R' F2 U R2 F L' U', with some extra moves to get 2x2x3 and to help with the continuation.
F2L-1 + EO: D' F L F' D' L' D'
F2L leaving 3 corners: L B' L2 B
Skeleton: R' F2 U R2 F L2 * D2 L U' D B' D' F L F' D' L' D' L B' L2 B D (23)
Insert L2 D' R' D L2 D' R D at * to cancel 3 moves

I thought I might as well post my winning set of ("human") solutions for Tom Rokicki's 4x4x4 Fewest Moves Challenge contest here, although I already posted in Tom's thread.

Code:

Scramble 1:
U L F D B2 F2 U2 R' F2 R2 B2 D2 R' D2 L B R D' U' R2 L' F L2 D F2 Uw2 B Uw2 F' R' Fw2 F' Rw2 R' B' R2 Fw2 L' Uw L2 D' B2 U Rw' Fw Uw' F' Uw' U' R B R U
First center: Uw2 F' Rw R' Fw' (5)
2nd center: F' Uw' L' Fw' R2 Fw (11)
3rd center: F Uw' R' Uw' (15)
4th center: Fw2 R2 Fw2 (18)
Finish centers: F2 Uw L Uw' (22)
2 dedges already paired
3rd - 9th dedges: B' (R' U' R) Dw (L U' L') (R' U2 R) (R' D' R) (L D2 L') Dw' (40 - 2 = 38)
10th - 12th dedges: F Dw2 R' D R Dw2 y (44)
3x3x3 phase (explained in terms of 3x3x3)
2x2x1: F' U' (2)
2x2x2: . R D * R (5)
Insert at ".": D (6)
2x2x3: L B L B' (10)
Pseudo F2L minus 1 slot: L' F L F2 (14)
Use premove F' to correct for pseudo F2L minus 1 slot.
Tripod: L2 U L U' L2 (19)
All but 2 corners, 2 edges: U L' U' L (23)
Premove correction: F' (24)
Insert cyclic-shifted J-perm to solve final 4 pieces.
Insert at "*": D L' D' L2 B2 R' U' R B2 L' (34-3 = 31)
OBTM: 44 + 31 = 75
BTM: 42 + 30 = 72
Solution:
Uw2 F' 2R Fw' F' Uw'
L' Fw' R2 Fw F Uw' R' Uw'
Fw2 R2 2F2 Uw L Uw'
B' R' U' R Dw L U' L' R' U2 D' R L D2 L' Dw'
F Dw2 R' D R Dw2 y
F' E' y' R D2 L' D' L2 B2 R' U' R B2
R B L B' L' F L F2
L2 U L U' L2 U L' U' L F'
Scramble 2:
F U R U B' D R U2 R B U2 L D' F' U2 L R B2 D2 R D2 R' B2 Uw2 Rw2 R' D L D2 L2 R Fw2 U L2 B' U Fw' L D2 F2 Rw' Fw2 Rw' Uw' Fw2 Rw D' F U F D
First 2 centers: R L D Fw' D F Uw Rw' D' L' R2 Fw' (12)
3rd-4th centers: F Uw' F R Uw B2 R Uw (20)
Finish centers: R2 Fw2 L2 Fw2 (24)
2 dedges already paired
3rd-6th dedges: Dw' (L U2 L') (L D L') Dw (32-2 = 30)
6th-12 dedges: R2 F R Bw' (D F' D') (R' B2 R) (R' U F' R U') Bw x' y' (46-2 = 44)
3x3x3 phase:
Premoves: D' R' (2)
2x2x2: L' D R U' (6)
2x2x3: D' . B2 D B2 L B * L' D2 (14)
C-E pair extension: D' B D (17-1 = 16)
F2L minus 1 slot: D L2 D' (19-1 = 18)
Tripod: B (19)
Edges: L U' L' U (23)
Premove correction: D' R' (moves already counted)
This leaves 4 corners, solved by two insertions.
Insert at ".": D F' D' B2 D F D' B2 (31-6 = 25)
Insert at "*": F' R' F L' F' R F L (33-2 = 31)
OBTM: 44+31 = 75
BTM: 43+29 = 72
Solution:
R L D Fw' D F Uw Rw' D' L' R2 2F' Uw' F R Uw B2 R Uw R2 Fw2 L2 Fw2
Dw' L U2 D L' Dw R2 F R Bw' D F' D' R' B2 U F' R U' Bw x' y'
L' D R U' F' D' B2 D F B2 L S z' R' F L' F' R F D B D2 L2 D' B
L U' L' E y R'
Scramble 3:
F D B D L' U2 F U2 D' B2 F L B F U2 L' F2 R' U2 F2 L B2 R2 B2 Fw2 Uw2 D' B F' Uw2 B R2 Fw2 Uw2 D' U2 Rw U L B U2 Uw' F' Rw2 R2 Fw F2 Uw' B' L B U L
First 2 centers: R' L' Dw' B2 Lw' F' Dw' L' Dw2 B2 Dw (11)
3rd center: D' Lw F' U B2 Rw (17)
4th center: F Rw' F2 Rw (21)
Finish centers: Uw2 F2 Uw2 (24)
3 dedges already paired
4th-9th dedges: F' (U R') (R' U R) Dw (D' R D) (R' F D' F') (F U' F') Dw' (42-3 = 39)
10th-12th dedges: D Lw' L D L' D' Lw (46)
3x3x3 phase:
1x2x3: F' D R' D' U R' F . (7)
2x2x3: R L2 F R2 (11)
Edges: L2 B2 D' * B D L B2 L' B (20)
A 5-cycle of corners remains.
Insert at ".": F' R' B' R F R' B R (28-4 = 24)
Insert at "*": R2 B L' B' R2 B L B' (32-2 = 30)
OBTM: 46+30 = 76
BTM: 42+27 = 69
Solution:
R' L' Dw' B2 Lw' F' Dw' L' Dw2 B2 2D Lw F' U B2 Rw F Rw' F2 Rw Uw2 F2 Uw2
F' U R2 U R 2D R D R' F D' U' F' 2D' 2L' D L' D' Lw
F' D R' E y R2 B' R F R' B M2 x2 F M2 x2 B2 D' R2 B L' B' R2 B L D L B2 L' B
Scramble 4:
R F D L D2 F D U B2 F2 L' B2 U F2 L D2 U2 L U2 R B2 L B2 Uw2 Rw2 F U2 R' B' Uw2 F2 R Uw2 F' Uw' L2 D2 L B2 R' Uw' D2 Fw' Rw R Uw' L2 B D F D
Full solution:
First center: D' Rw Bw (3)
2nd center (adj.): F' U Rw2 B' Rw' (8)
3rd center: R Fw Dw' F2 Dw Fw' (14)
Finish centers: Fw D' R' Fw D U2 Bw2 (21-2 = 19)
2 dedges already paired
3rd - 4th dedges: Bw2 R' B2 R Bw2 (24-2 = 22)
5th - 10th dedges: D' F (F L F') Rw' (F' R F) (F' L F) (U R2 U') Rw (38-3 = 35)
11th - 12th dedges: D' R' Dw' R' B U' R B' Dw y' (44)
Solving inverse for 3x3x3 phase.
2x2x2: D F' R2 B' (4)
2x2x3: R U2 R2 (7)
F2L - 1: F U F2 (10)
Edges: L' U' B' U2 B2 . L' B' L2 (18)
Insert at ".": F' L' B' L F L' B L (18+8-5 = 21)
Insert at end: L' B L F' L' B' L F (21+8-2 = 27)
Use inverse as solution to 3x3x3 phase.
OBTM: 44+27 = 71
BTM: 41+26 = 67
Solution:
D' Rw 2F z' U Rw2 B' 2R' Fw Dw' F2 2D R' Fw D U2
R' B2 R Bw2 D' F2 L F' Rw' F' R L F U R2 U' Rw D' R' Dw' R' B U' R B' Dw y'
L2 B L B2 U2 B U L F2 U' L F' R2 F L' F' U2 R' B2 L' B' R2 B L S' z D'
Scramble 5:
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
First 2 centers: F' D2 L B' U Uw Lw Fw' L2 Fw' (10)
Paired center pieces: B' F' U Rw (14)
Finish centers: F D2 Rw' D2 Rw2 (19)
4 dedges already paired
5th-7th dedges: U2 Rw2 (F' R F) (U R' U') Rw2 (28)
8th-12th dedges: B' L' Dw (R' U2 R) (R U' R') Dw' x' y2 (38-1 = 37)
3x3x3 phase:
2x2x2: R B' D2 F2 (4)
Siamese 2x2x2's: D' B2 L2 D2 B L B' (11)
F2L minus 1 slot: D' L' F L F' (16)
All but 2 corners, 2 edges: R D' R' . B' D2 B (22)
Insert at ".": R D2 R' D' R D2 L' D R' D' L (33-3 = 30)
(sub-optimal J-perm)
OBTM: 37+30=67
BTM: 37+30=67
Solution:
F' D2 L B' U Uw Lw Fw' L2 Fw' B' F' U Rw F D2 Rw' D2 Rw2
U2 Rw2 F' R F U R' U' Rw2 B' L' Dw R' U2 R2 U' R' Dw' x' y2
R B' D2 F2 D' B2 L2 D2 B L B' D' L' F L F'
R D R' D' R D2 L' D R' D' L B' D2 B

How about trying insertions at the dedge pairing step? My (time consuming) idea for 4x4 FMC that I didn't bother to implement was to reduce into a 3x3, find a short edge solution (e.g. with a EO first approach) and then solve the corners with insertions, as the skeleton would be quite a bit longer than a 3x3 one which might result in nicer insertions (i.e. more cancellations).

How about trying insertions at the dedge pairing step? My (time consuming) idea for 4x4 FMC that I didn't bother to implement was to reduce into a 3x3, find a short edge solution (e.g. with a EO first approach) and then solve the corners with insertions, as the skeleton would be quite a bit longer than a 3x3 one which might result in nicer insertions (i.e. more cancellations).

You're right. There's no need to limit corner insertions to the 3x3x3 phase. And the R U R' type of maneuvering common during dedge pairing could provide some very high cancellations.

And I note time was not really an issue in Tom's competition.

Can someone help me with insertions for this solution:

Originally I had this:

Scramble: U L2 F2 U B2 U R2 U' F2 D' R2 F' L' F2 U B U B' R2 D F'

L' D L U F U' F2 L2 F U' R U // most of 2x2x3
R U R B U2 B2 // F2L-1
U B U B' U B U' B' U B U' B' // To L3C <---- in this step if I had put the 2x2x1 in the UFL position by doing a U2 right after building, I would have had a L4C case with one twisted corner, so I opted to go past that and go for L3C.
U B2 D B' U2 B D' B' U2 B' // L3C (luckily it was just an OCLL, but a comm nonetheless)

L' D L U F U' F2 L2 F U' R U // most of 2x2x3
R U R B U2 B2 // F2L-1
U B U2 B' // to L4C

This is the skeleton to L4C, can someone show me how to insert 2 comms into this? I have tried, but I still just don't understand how to physically see how to cycle the pieces at different points throughout the solve.

Can someone help me with insertions for this solution:

Originally I had this:

Scramble: U L2 F2 U B2 U R2 U' F2 D' R2 F' L' F2 U B U B' R2 D F'

L' D L U F U' F2 L2 F U' R U // most of 2x2x3
R U R B U2 B2 // F2L-1
U B U B' U B U' B' U B U' B' // To L3C <---- in this step if I had put the 2x2x1 in the UFL position by doing a U2 right after building, I would have had a L4C case with one twisted corner, so I opted to go past that and go for L3C.
U B2 D B' U2 B D' B' U2 B' // L3C (luckily it was just an OCLL, but a comm nonetheless)

L' D L U F U' F2 L2 F U' R U // most of 2x2x3
R U R B U2 B2 // F2L-1
U B U2 B' // to L4C

And then I could do a random pure comm to solve one piece:

This is the skeleton to L4C, can someone show me how to insert 2 comms into this? I have tried, but I still just don't understand how to physically see how to cycle the pieces at different points throughout the solve.

First:
L' D L U F U' F2 L2 F U' R U R U R B U2 B2 U B U B' U B U' B' [1] U B U' B' U
[1] = L U' R2 U L' U' R2 U
1 move cancelled. (38 moves total)

Second:
L' D L U F U' F2 L2 F [1] U' R U R [2] U R B U2 B2 U B U2 B'
[1] = L2 U' R U L2 U' R' U
[2] = F' D F U' F' D' F U
7 moves cancelled. (31 moves total)

This is the skeleton to L4C, can someone show me how to insert 2 comms into this? I have tried, but I still just don't understand how to physically see how to cycle the pieces at different points throughout the solve.

If you don't know it yet, this tool finds optimal insertions for you.
If you want to find insertions yourself, I recommend little paper stickers that you glue on the cycling stickers (and write numbers on) to keep track of the pieces. I wouldn't use my main cube for that though, as you might peel your stickers when you remove the paper stickers afterwards. So in your case, for the first insertion put stickers on all four corners (any sticker) and mark the twisted one - there are many choices for a commutator here. Then you might have to remove the marks and use new ones on different stickers to find your second insertion.

D F2 D' B2 D F2 D' B2 U2 F' L' D R D' F2 B2 U L' B' F L2 B2 F2 R2 B' L U D' B' R (30 HTM)

D F2 D' B2 D F2 D' B2 U2 F' L' D //All corners and 3 edges (12)
R D' F2 B2 U L' //3 more edges (6)
B' F L2 B2 F2 R2 //2 more edges (6)
B' L U D' B' R //4 edges and 4 centres (6)

CF, almost sub-30! When I got to this point I thought I had done it, I just needed to delete the first 8 moves (which are a corners- 3-cycle) and insert it later in the solve. 1 cancellation was enough. But I couldn't find any.

Next scramble: L2 F2 D' R2 D B2 L2 U' F2 D F2 L' B' U' F' D F' D2 R' D L2

First, I tried an orient first approach: F' U' R orients corners in only 3 moves, but I couldn't find a good continuation for solving corners.
So I decided to solve corners as I would solve a 2x2x2: U2 F' L' D for the first layer, the I inserted that 8-move commutator at the beginning instead of using a 9 mover. I thought I could insert it later in the solve once I had finished everything else.
So now I have corners and 3 edges.
The rest is pretty straight-forward, with some luck in the last step.
Removing the commutator at the beginning gives a 22 moves skeleton, but sadly insertion finder says that the optimal insertions doesn't cancel any move

I have an interesting scramble to share. It is the 2nd scramble of the current German forum competition. I can't remember having more blocks on a scramble so far.

The scramble: D R' B2 U F U R2 D2 L D2 B U2 F2 U2 R2 F2 U2

My solutions:

Spoiler

1-hour:

F2L-1: F2 U F' B' R D L' D2 R2 U2 R2
L3C: D' B D2 B' D L' F' D' * F L D2

* = D F' U' F D' F' U F

Solution: F2 U F' B' R D L' D2 R2 U2 R2 D' B D2 B' D L' F2 U' F D' F' U F2 L D2

26 moves.

no limit:

Premove: L2
2x2x3: F2 U F' B' R' U2 R2
F2L-1: D2 L' D' L
Domino: F' D' F'
Rest: D' F2 D' F2 D2 F2 D

Solution: F2 U F' B' R' U2 R2 D2 L' D' L F' D' F' D' F2 D' F2 D2 F2 D L2

I thought I might post here as well this corners first solve. I tried this method a few times, but this is the first "good" solve I get.

Scramble (from the Example Solve Game thread): D2 L' B2 R' F2 L2 U2 L2 R' B2 D F U' L B' U' F' U2 B2 F

First, I tried an orient first approach: F' U' R orients corners in only 3 moves, but I couldn't find a good continuation for solving corners.
So I decided to solve corners as I would solve a 2x2x2: U2 F' L' D for the first layer, the I inserted that 8-move commutator at the beginning instead of using a 9 mover. I thought I could insert it later in the solve once I had finished everything else.
So now I have corners and 3 edges.
The rest is pretty straight-forward, with some luck in the last step.
Removing the commutator at the beginning gives a 22 moves skeleton, but sadly insertion finder says that the optimal insertions doesn't cancel any move

I waited many years, for someone else to try CF method, but it is very rare.
I am so happy, see yours excellent solution.
I hope you do many more attempts...

I waited many years, for someone else to try CF method, but it is very rare.
I am so happy, see yours excellent solution.
I hope you do many more attempts...

Thanks
I will try this method again for sure, but I don't think I will use it in one-hour FMC competitions; I'm not so good yet.
But it's very fun to use your method, it's so different from all other popular ones.

D2 L' B2 R' F2 L2 U2 L2 R' B2 D F U' L B' U' F' U2 B2 F

D F2 D' B2 D F2 D' B2 U2 F' L' D //All corners and 3 edges (12)
R D' F2 B2 U L' //3 more edges (6) L2 F2 B2 R2 F B' //2 more edges (6)
B' L U D' B' R //4 edges and 4 centres (6)

Didn't you miss a trick? Didn't check to see if this could be done further. Pretty cool though, seems like a fun method that I would have no chance of getting good solutions with.

U2 R B2 F R' B2 D' R' L B' U2 L' R' D2 B2 F U' F2 R B2 R' B2 U' L R'

D2 F R2 U R' U' // 2x2x3
switch to inverse
R B' D2 B // add square
D R' D' R' F' R' F D R' D' R' F' R' F // sucky tripod
D R' B R B2 D B //L3C

Too tired to physically find an insertion right now, but optimal only cancels one move .
Great start, poor continuation. And also my first solve using NISS!

Didn't you miss a trick? Didn't check to see if this could be done further. Pretty cool though, seems like a fun method that I would have no chance of getting good solutions with.