FMC: move replacements - how humans find them?

[bcube]

Hi,

it seems to me that move replacements are found in a way like: try something else, repeat until you get lucky. Is it so or am I missing more systematic/intuitive way for finding them?

F' U R U' D2 L2 F B' R L' F' U L U' F2 R2 B2 L2 U' R2 F2 D2 F2 R2 F' U R

U R D' // EO (3/3)
B' R2 F2 D2 L' R2 B // DR (7/10)
F2 L' D2 L' // HTR (4/14)
U2 L2 U2 // all but 2 edges 2 edges (3/17)


versus the same 13 moves

U R D' // EO (3/3)
B' R2 F2 D2 L' R2 B // DR (7/10)
F2 L' D2 L // HTR (4/14)
R2 D2 R2 U2 // finish (4/18)


Another example: F' U R D' B2 U L' B U F2 U2 L2 B' R2 F D2 B' D2 B2 L2 R F' U R

B U D R // EO (4/4)
U B2 L2 D F // DR (5/9)
D B2 U R2 F2 U R2 U' // HTR (8/17)
R2 U2 L2 U2 // all but last 3 edges (4/21)


versus the same 9 moves

B U D R // EO (4/4)
U B2 L2 D F // DR (5/9)
U R2 U F2 L2 U F2 D' // HTR (8/17)
L2 D2 L2 U2 // finish (4/21)

 

[tseitsei]

1st one:

Insertions. We just know a bunch of algs and insert them in a spot where they cancel a lot of moves.
This is especially effective post-HTR since you only have 6 possible moves and a lot of symmetry.

L' U2 * L2 U2 // all but 2 edges 2 edges (3/17)

Insert at * = U2 L2 R2 D2 R2 L2 this is a very well known 2e2e algorithm.

=> L' U2 U2 L2 R2 D2 R2 L2 L2 U2 //solved

Now just cancel move => L R2 D2 R2 U2

There are some algs to learn for these but less than you would think at first


2nd one:

Slightly different kind of insertion called slice insertion. This can be used to solve the "DR axis layer" usually in 0-1 moves.
What I mean is if you have DR on U/D you can leave E-layer unsolved and then take your solution post DR and insert E-slice moves there to solve it.
Inserting E-slice moves post-DR (assuming DR on U/D) only affects E-slice
This can be done either by brute force or something called slice theory (https://sebastiano.tronto.net/speedcubing/slice-theory/).

I would find this solution by just trying to widen all combinations of U and D layer moves post DR in order and seeing if it solves the cube.

D^ B2 U R2 F2 U R2 U'^ // HTR (8/17)
R2 U2^ L2 U2 // all but last 3 edges (4/21)

insert ^ = w (or if it is easier to think about this way: insert E at first and second ^ and E2 at last ^)

Dw B2 U R2 F2 U R2 Uw' // HTR (8/17)
R2 Uw2 L2 U2 // all but last 3 edges (4/21)

and when written without wide moves you get the solution:

U R2 U F2 L2 U F2 D' // HTR (8/17)
L2 D2 L2 U2 // finish (4/21)