[video] Permutation of last layer without algorithms

Euh, how would you explain them?

Well cmowla did an excellent job explaining the A perm so I'll just do H perm.

M2 U M2 U2 M2 U M2 or [M2 U: [M2, U2]] which follows the form [c: [a, b]] which when expanded reads c a b a' b' c'.

When you break down that commutator notation it would read - M2 U (M2 U2 M2 U2) U' M2

The M2 U2 M2 U2 is the simple 2-2 swap you used in your H perm as well, it swaps DF-DB and UF-UB. In H perm we want to swap UF-UB and UL-UR. The first setup move, M2, separates the 2-2 swaps into different layers, moving the UF-UB swap to DB-DF, while leaving UL-UR unchanged. Then the second setup move, U, moves UL-UR to UB-UF. After just two moves, compared to your alg's three, we've setup to the simple [M2, U2] commutator. After doing [M2, U2] you'll still need to undo the conjugation by doing U' M2, however the last U2 of [M2, U2] cancels into the U' of the setdown to just leave U.

M2 U M2 U2 M2 U M2
 
Is it really shorter than combining with 3-cycle of edges ?
Mine is 20 STM, and yours is 24 STM (STM = slice turn moves, where you count E2, R2, R, E, etc. as one move each).

Because I believe you would handle that case (I didn't go back to the video to see) like so:

(L' R' D2 L R U' R' U L' D2 L U' R U) (L' R' E2 L U L' E2 L U' R) (24 STM)

But if you cancel moves in mine, then it is 19 STM

cmowla said:

Since M2 L R = (L R x2)',

M2 U M2 U' L' R' x2
d2 L' U' L d2 L' U R'
M2 U M2 U' M2


But, I don't really recommend doing the alg I did because it's much easier to do two shorter algorithms rather than one, especially that they are this close in moves. I just wanted to see your idea apply to a more complex case without just trivially breaking that up into 2 steps. When I heard "PLL", I was actually expecting you to be able to intuitively handle every one of the PLL cases in one step.

Also, if you adjust my alg (which IS based off of your preferred type of commutator), you can get an R Perm if you add one move at the end. The following red moves are inverted, and the blue move is the one which was added:

M2 U M2 U M2
L R d2 L' U' L d2 L' U R'
M2 U' M2 U' M2
U'
Link

Perhaps you could experiment with doing both of them at once to generate the rest of the odd permutation PLLs directly.
 
OK, this is the first time I've picked up a 3x3 in a few days, I tried to do an A perm as you demonstrated it in your video, having only seen it that one time (when you first posted it), and I did it correct first try, without having to even think about it.
So I'm quite convinced that you're video is very good for making alg memorisation easy.
 
Thanks Godmil :)

@cmowla: Yeah I'm getting a little bit lost with conjugating big commutators.

But I found one that is interesting :

(L' U' L U) + (U R U' L d2 L' U L d2 L' R') + (U' L' U L)
(conjugate) + (J perm) + (deconjugate)

Which gives Y perm.

It can be simplified to
L' U' L + L U' R d2 R' U R d2 R' L' U + L' U L
(conjugate without top align) + (rotated J perm with top rotation after) + (deconjugate)

which of course simplifies into
L' U' L2 U' R d2 R' U R d2 R' L' U L' U L

or, avoiding d2 :
L' U' L2 U' R U2 L' U L U2 R' L' U L' U L (16 moves)

which can be done for example as :

(L' U' L) (L U') (R U2) (L' U L) U2 R' (L' U) (L' U L)

When I heard "PLL", I was actually expecting you to be able to intuitively handle every one of the PLL cases in one step.
I woud like too :)
 
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Mine is 20 STM, and yours is 24 STM (STM = slice turn moves, where you count E2, R2, R, E, etc. as one move each).

Because I believe you would handle that case (I didn't go back to the video to see) like so:

(L' R' D2 L R U' R' U L' D2 L U' R U) (L' R' E2 L U L' E2 L U' R) (24 STM)
No, it would be
(R' (U L' D2 L U' L' D2 L) R)(F2 U') (M' U2 M) (U' F2)

Also, if you adjust my alg (which IS based off of your preferred type of commutator), you can get an R Perm if you add one move at the end. The following red moves are inverted, and the blue move is the one which was added:

M2 U M2 U M2
L R d2 L' U' L d2 L' U R'
M2 U' M2 U' M2
U'
Link
That's a bit complicated for me. But thinking about it, I found this one to do the R perm :
(M2 U2 M2)(U L' R' d2 R U R' d2 R U' L)(M2 U2 M2)

Perhaps you could experiment with doing both of them at once to generate the rest of the odd permutation PLLs directly.
Sure. Why not.

I can't find the T perm and the F perm. Well for example, the F perm may be with a 7 move long conjugate, which is 14 moves, which is much longer than doing a 3-cycle of edges. But anyway, finding conjugates for two adjacent corners only is interesting only if the conjugate is 3 moves long or less, because the 3-cycle of edges is 7 moves long.

So it may be more interesting for opposite corners for V or N to find something with a J perm.

I noticed that for V, it's faster to exchange adjacent pairs (J) then rotate corners, instead of exchanging opposite pairs (N) then rotate edges, because J is shorter.
 
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Here is a video to explain how to create algorithms using commutators :
 
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