http://games.groups.yahoo.com/group/blindfoldsolving-rubiks-cube/message/933
Ok, so here's my new method for permuting 3x3 edges, I call it "M2
method". It's very simple, fast, and 100% intuitive so you don't need
to learn a single alg for it.
Of course it's a 2-cycle method, one piece getting solved each time.
The piece to be solved is at FD. Since I can't just swap two pieces, I
need to have some side effect. Here it is swapping (FU,BD)+(U,D)+(F,B)
. Note FU and BD are solved relative to the M centers.
Here's an example solve, using scramble #28 from post #754 (http:
//tinyurl.com/lytqj). It's ready to be pasted into this page:
http://thearufam.brinkster.net/cube/wrapplet.asp
-------------------- start --------------------
[Scramble] U2 D2 L' U D2 R2 L2 D R D' B2 U D' F2 R2 U' F2 R2 B2
[solve LU] (L U' L' U) M2 (U' L U L')
[solve LD] (U' L2 U) M2 (U' L2 U)
[solve RF] (U R U') M2 (U R' U')
[solve RU] (R' U R U') M2 (U R' U' R)
[solve BD] M U2 M U2
[solve LF] (U' L' U) M2 (U' L U)
[solve RD] (U R2 U') M2 (U R2 U')
[solve FU] M U2 M U2
[solve RB] (U R' U') M2 (U R U')
[solve LB] (U' L U) M2 (U' L' U)
--------------------- end ---------------------
For M2 I use Joel's finger trick, i.e. L (l' M'), using my left ring
finger to pull the M'. My left middle finger holds the B stickers of
LB+LBD, so my ring finger is free for the M' and my index finger is
free for U'.
Oh, and actually it was a bad choice to have BU already solved. It's
the easiest to solve, it looks like:
[solve BU] M2
So, how do you like it?
Cheers!
Stefan
转来论坛，解决棱块的步数也很少 例子中不过70来步。
研究研究一下
[Scramble] U2 D2 L' U D2 R2 L2 D R D' B2 U D' F2 R2 U' F2 R2 B2
呵，烟头发的java是怎么办到的？步骤放在一起看起来清晰多了。
这M2方法，原理我还是不大明白，再发几个研究一下：
看出一点眉目：
1、每个公式后加一步M2都是标准的棱块三置换公式。
2、左右两边的棱块归位都是从FD那个位置移过去的。
[em08]我也看出些眉目，每个公式都只做了两个块、两个块的对调，然后再找出相应的第二个公式以达到不同的三循环的目的，我想两个公式可以任意搭配，做法相当精妙，这样只要记住少数几个公式搭配后就能成为不同的三循环；不过上面的实例色向上已经全都对了，如果能将色向的问题也考虑进去，这个方法潜力巨大。
简单举例以下：
比如上面提到的，第一个公式是FD和UL对调、UF和DB对调，第二个公式是LD和FD对调、UF和DB对调；这样通过两个公式的搭配就能实现FD-UL-LD-FD的三循环，UF和DB通过两次对调又回原位。
如果角块也能有这样的公式就好了。
[此贴子已经被作者于2006-9-10 14:17:44编辑过]
关于 M2 Method 的讨论
For blindsolving I'd probably use the full-cubie version (that'll be
my name for both permuting and orienting a cubie). It's very similar,
but instead of using U you use B, which should be done as U after cube
rotation, e.g.
[solve UR] x' (U' R U) M2 (U' R' U) x
[solve BR] (l U' R' U) M2 (U' R U l')
> I am not really sure if I understand how the M slice edges are
> handled... But it sure looks cool!
- FD is the buffer and thus ignored
- BU is solved with M2
- FU and BD are solved relative to the M centers
either with (M U2 M U2) or (U2 M' U2 M')
Cheers!
Stefan
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Oh crap... clarification: Of course you don't just use B *instead* of
U but you use it *additionally*.
And actually I'd like to call the full-cubie version "M2" and the
permute-only version "M2p". I just wanted to demonstrate M2p first
cause it's so extremely sweet to execute and might even get the
orient-first guys more interested :-)
Cheers!
Stefan
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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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> 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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> 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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> 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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