r2' D2 r2 B r2' D2 R2 B (r2' R2) ... might actually be better than what I was using before since I suck at that case. I too will use this in an official solve for the sheer class.
Still not "understanding" it. I'm guessing that it is actually a weird commutator [r2 d2 r2, F], that somehow does two corner 2 cycles and an edge 5 cycle ... Or maybe there just isn't a simple "explanation" though.
(I have always wondered how it works, myself, so I'm looking forward to an explanation as well.)
Not a commutator, here's how I see the important stuff happening:
Corners: r2 d2 r2 F performs a pair of 2-cycles on F and on B, so repeating it leaves the corners solved.
Edges: r2 d2 r2 swaps UF<->UB and FL<->FR, doing an F move then leaves FD and FL swapped so these will also be solved after r2 d2 r2 F is repeated. Also, since r2 d2 r2 swaps UF and UB and brings FR to FL, doing F leaves these 3-cycles, so repeating the moves results in a 3-cycle.
Yea I learned these specifically for edge insertions in FMC along with these types:
R2 B2 L2 D' L2 B2 R2 U'
usually you can cancel a move or two especially if there are double moves.
I personally just spent some time learning how they influence the edges (how they work). But I don't really have a theoretical knowledge or explanation.
Not a commutator, here's how I see the important stuff happening:
Corners: r2 d2 r2 F performs a pair of 2-cycles on F and on B, so repeating it leaves the corners solved.
Edges: r2 d2 r2 swaps UF<->UB and FL<->FR, doing an F move then leaves FD and FL swapped so these will also be solved after r2 d2 r2 F is repeated. Also, since r2 d2 r2 swaps UF and UB and brings FR to FL, doing F leaves these 3-cycles, so repeating the moves results in a 3-cycle.
Hm... well this is breaking down the permutation into smaller permutations, and I can see how it results in the 3-cycle. But what I was hoping for is an explanation as some clever commutator, or something similar. (For example, I understand the U-perm M2 u M' U2 M' U M2 as [R2 L2 U: [S, U2]]
I know it's not a commutator, what I meant is to look at it as [r2 d2 r2, F] followed by an F2, where the commuator swaps four corners and five edges.
Also the algorithm is in the <R2, L2, U2, D2, F, B> subgroup which is I think called Domino subgroup, so maybe doing the alg on a 2x3x3 makes it easier to understand? Propably not though. ^^
This is the theory behind the algorithm R2 D2 L2 F U2 L2 D2 B and related algorithms.
Speaking from a solved cube viewpoint, R2 D2 L2 isolates the UF edge with respect to all pieces in the back face--with the exception of UB, of course. The move F is analogous to Y in the single piece isolating commutator [X,Y]. The moves U2 L2 D2 restore slices F and S as much as possible AND set up the inverse permutation that R2 D2 L2 sets up in face F in face B. In addition, U2 L2 D2 puts RF in face B.
Now, let me explain this in more detail.
Take note of the formation of pieces in face F after the moves R2 D2 L2 are applied to a solved cube. Write it down on paper. Now write down the "inverse formation" of this formation, should you move them all again with a face quarter turn in the same direction as the move F. Substitute an alternate edge in place of UF in this "inverse formation" to define the 3-cycle you want this algorithm to produce. You will see that the correct "inverse formation" is setup in face B after the move sequence R2 D2 L2 F U2 L2 D2 is applied, where UF is substituted with RF.
EDIT:
In addition, if we want an algorithm of the form:
A
(face quarter turn a)
B
(face quarter turn b)
then the formation of all (well, all pieces except those being put into a cycle) pieces in the face a must be the inverse formation of face b's solved state. This formation must contain most of face b's pieces as well.
I abstracted face a and face b, because we may choose face a and face b. (This algorithm uses face F and face B, but they do not have to be opposite.) As long as the formation of the pieces which are ultimately not going to be affected by the entire algorithm are face b's pieces and are in the inverse permutation of face b's solved state, it's acceptable.
This algorithm uses B as because it's simply the shortest path.
Here's an example 2 corner twist algorithm which uses the U face twice instead of doing the move U, following by move L, R, F, B or D.
The theory behind this 2 corner twist is the same theory behind R2 D2 L2 F U2 L2 D2 B, but this 2 corner twist is simply longer. This is the same theory as single piece isolating commutators as well, only when we "isolate" a piece in a face, instead of maintaining the exact formation of that face, we first create a custom formation of pieces in that face. As long as we're responsible enough to set up the inverse of that custom formation in whichever face we wish to finish off the algorithm with, we will only do a 3-cycle, 2 corner twist, etc. when the algorithm is finished.
As one final illustration, instead of substituting UF with RF, as the moves U2 L2 D2 do in the algorithm R2 D2 L2 F U2 L2 D2 B, let's instead substitute DL in place of RF. R2 D2 L2
F
F2 D U F' L2 U2 R2 B D' U'
B
The move sequence F2 D U F' L2 U2 R2 B D' U' both restores slices F and S as much as possible and sets up the inverse permutation of all pieces from the back face (with the exception of UB) and places DL in the back face.
Similarly, the move sequence (R2 U2 R2 U2 R2 U2 R L D2 L' D2 R' B' D2 B)' in the 2 corner twist algorithm (1) restores slices E and D as much as possible AND (2) sets up the inverse permutation which (R2 U2 R2 U2 R2 U2 R L D2 L' D2 R' B' D2 B) set up in slice U.
Okay, so removing rotations and then translating into a legible format gives us this: U' L' B L U' L' B' L U' B U' B' U'
Basically what I first noticed: U' (L' B L) U' (L' B' L) U' (B U' B') U' which is full of tiny conjugates separated by U' moves.
U' [L':B] U' [L':B'] U' [B:U'] U'
I think that the 4 U' moves between these 3 move conjugates somehow cancel to provide a full face rotation while they're swapping the affected pieces around but in different phases of the algorithm? Will take a closer look when I have a real cube in my hands though Ollie
Yes they are the same. I used the term "formation" sometimes because it's easier to look at the colors of the stickers of the cube itself to recognize the permutation (in cases when it does not appear to be that "scrambled" and is probably easy to recognize in comparison to finding a permutation representation).
