Stefan's M2 method

彳亍

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 FU can be solved, but how about UF? Or did I miss something?

Right, I forgot those. Here's what I'd do:

For UF and DB one of these:
B' M' (U' R' U) M (U' R U) B M2
U x' M (U R U') M' (U R' U') x U' M2

For UB one of these (they all do the same, I don't understand the
latter two but the first two are cool intuitive, you'll like them):
(U B' R U' B) M2 (B' U R' B U')
(U R' U' r' U' R2 U) M2 (U' R2 U r U R U')
(M U') (M' U) F2 (U' M) (U' M) U2 F2
(M' U') (M' U) (M' U2) (M U) (M U) (M U2) (M' U2) M'

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彳亍

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> But...wait...this scramble had oriented edges, right? well, for me
not, 'cause an edge is oriented for me if it can reach solved position
within R2, L2, U, D, F, B; and for you I think is F2, B2, L, R, U, D..
.or no?

Yes, that's correct. One thing you could do is do a y-rotation after
orientation and before permutation, but that could result in different
headaches :-)

> well, this makes things a bit harder...for example...how would you
solve FD -> LF? I'd do (D U2 F U2 D') M2 (D U2 F' U2 D')...but I think
this is not a good idea...

You mean FL, not LF :-). And nah, you really just use B instead of U.
And an x-rotation helps make it easy to twist. So for LF/FL I'd do:

[solve LF] (U' L' U) M2 (U' L U)
[solve FL] x' (U L2 U') M2 (U L2 U') x

The idea is always:
- one move (out of U/U'/B/B') to get BU out of M
- one move (out of L/L2/L'/R/R2/R') to replace it with the target we
really want
- undo that first of the three moves to get the real target into M

There are of course a few exceptions but this is the general idea.

Cheers!
Stefan

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彳亍

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> hey, that way is good...using B instead of U...maybe that solve my
problems...
>
> let me see...DF -> LB (or FD -> BL) would be:
> x' (L' U L U') M2 (U L' U' L) x
>
> right?
>
> Pedro

Almost right. Notice that the two special treatments (cube rotation
and extra move to get the piece out of the way) combine nicely. So
it's:

[solve BL] (r' U L U') M2 (U L' U' r)

Cheers!
Stefan

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彳亍

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> one more question:
> how do I take care of parity with this method...
> or...
> how would you solve the corners?

In case of odd parity, simply finish edges with this:
(D' L2 D) M2 (D' L2 D)

This will fix the M-slice and leave (UB,UL) swapped, which is what I
need for my regular method for solving corners.

Cheers!
Stefan

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yingmu

又是英文！看来应该好好学学英文了[em06]

yingmu

角块的公式正在寻找中，不过不是M2,是R2或L2了，且步数上一般要12步左右，因为角块方向上分顺逆，所以以一个角块为出发点的话，方向正确的有7个公式，顺方向的有7个，逆公式的有7个，一共21个公式，已经找到几个[em01]

warl0rds

彳亍法  和  M2/R2  相比，改进上是 ？？中间的小循环？
还是？
setup   M2是固定的