[转帖]Fast Edge Cycles

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http://games.groups.yahoo.com/group/blindfoldsolving-rubiks-cube/message/1568

跟我的 三循环同时解决位置和方向 殊途同归  ：） 

先转过来，有进展随时报告。

Hey guys, I've been working on getting my BLD times down a lot.
I'm learning new things daily, and working on concepts.

The fastest Cycle I could come up with was this ... seriously too.

M'U2MU2, and any isomorphic M Slice Case thereof.

This would be a two step for solving the edges EO and EP.
It is possible to do direct EP, but setups take a little more
thought.

Dir = Direct = No EO
Indir = Indirect = An EO step
2x = Permuting 2 Edges per Cycle
Slice Cycles can be used for, Dir, 2xDir, Indir, and 2xIndir on the
edges. I've achieved sub-10 BLD w/ 2xIndir, I'll try the other
three options.

Ahhh I think I just realized how this can work. Well, we're doing
all these cycles, if we do Direct or Indirect Solving piece by piece,
It is possible to solve the edges with this method:
Set, M'U2MU2, Reset
Set, U2M'U2M, Reset
Set, M'U2MU2, Reset
Set, U2M'U2M, Reset
etc.

Anybody interested, please look at it. Thanks

Later,
Daniel Beyer

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另一贴

http://games.groups.yahoo.com/group/blindfoldsolving-rubiks-cube/message/1204

Hey guys,

Most of you know I have been working on solving in three cycles
Pochmann style (cycling the stickers). This can solve 2 edges at a
time. I have been making 'procedures'/strategy's in order to make
this easy (strategy's that describe how to handle every type of
situation that can occor). While making these strategy's at first,
I ended up describing almost all 440 cases I can get with this
method... (That is, every three cycle of stickers that start at UR).
So now I am almost done making a simpler procedure, that will have a
bit more, but easier setup moves (usually 2 or so).

Now, since I started making these tables, I didn't try a lot BLD
solves... But today, Craig asked me to race him with the webcam...
At first, I didn't want to, because I didn't feel like I was up for
it... But he convinced me to do it anyway... And I got a PB of
2:20.91!!! Which is extremely good for me..

The cool thing about this solve was, that during memorization, I
didn't only memorise the letters, but I was also thinking to
myself: 'Here I can use that alg, there I will use that commutator'.
This was in fact also a way of memorization! And it made
memorisation go a lot easier than usual.

To give you guys an idea of how a nice solve with three cycles
lookes like, here is the sramble, with my solution:

Scramble:
U F L2 D2 F U R B F2 U' D2 F D L F2 U' L U R2 U2 B R' D2 R L2

Defenitions:
C := (R U' R' U' R U R' F') (R U R' U' R' F R) (Corner alg)
J := (L' U) (R U') (L U2') (R' U) (R U2 R')
T := RUR'U' R'F R2U'RU' RUR'F'

Corners:
1. DF' C FD'
2. F'R' C RF
3. R2D' C DR2
4. F'D C D'F
5. D C D'
6. D' C D

Edges:
1. D' MU'M' D MUM'
2. J
3. R2 U' R' U' R U R U R U' R
4. URU'R' (E2) RUR'U' (E2)
5. B2 L'U2L (E) L'U2L (E') B2
6. D2 L2 T L2 D2

The edges were solve in about 72 moves (counting slice moves as 1).
4 three cycles solved 8 edges .

Well.. That's it for now!

Have fun!

yingmu

学学大烟头也来研究研究

[Scramble] U F L2 D2 F U R B F2 U' D2 F D L F2 U' L U R2 U2 B R' D2 R L2

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QUOTE:

以下是引用ggwfnh在2006-9-12 10:59:58的发言：

现在方法变了嘛？/不是先还原棱块。是先还原的角块。

当然角块的还原次数是偶数次，不用做，

但是奇数次的话不是还要做一个奇偶校验。?????

或者最后来直接交换UL和UB棱   ??
  但是不可能出现这种情况。？？？？？？？

先角先棱都没所谓的。奇偶校验肯定有，有可能需要影响一对棱

warl0rds

06年9月彳亍 已经开始研究位置方向一块还原？？  元老

老沙

原帖由 warl0rds 于 2009-5-9 21:51 发表 06年9月彳亍 已经开始研究位置方向一块还原？？  元老

yh
盲拧的发展历史。。。。。。。。。。。。。

小哲哲

老帖都顶出来了，不过也要想前吃醋前辈致敬啊@@