[转帖]Chris 讲解的 commutator

warl0rds

正在学习彳亍法，竟然发现有这么好的帖子，收藏收藏，精辟哪！   推导公式原来这么有趣

harold_han

看英文的实在太累了呀！

skydragon7

.........楼主不厚道 英文也帮忙翻译下嘛

mylxc60

有点头疼啊  看得迷糊了

录

看不懂啊？哪位英文高手麻煩翻譯一下吧[em02]

彳亍

http://dbeyer.110mb.com/centers_commutators.txt

再来一个，今天丰收  ：）

I am trying to do the work that you promised so long ago =P
No offense just trying to help others and myself out.

Interchangeability -- Two pieces must be setup onto the same face.
Interchangeable -- To be on the same face of the Cube
There are three cubies per commutator that recieve a net effect:
Reference cubie -- The One that will be solved once a cycle of 2K+1 length cycle is completed.  (where K is an integer). [Ref]
Interchangeable Cubie -- This cubie and the reference cubie are interchangeable.  [Int]
Action Cubie --  This cubie is the lone cubie.  The swapping of the Interchangeables occurs here.  [Act]

The goal of this tutorial is to use the best setup and the correct commutator for the setup
A standard Commutator is composed of 4 parts
ABA'B'
Where A is comprised of three moves, ru'r' for example -- the first and last move must have a net change of 0 and may not be a double turn.
B is a turn of a one of the outermost slices parallel to the second move in part A
There are many cycles that can be created, by this subset:
rr'[U] -- r'r'[U]'
r'r[U] -- r''r[U]'
Any notation encased in brackets allows any turn to be used, such as U, U2, or U'

A Standard Commutator would solve the Cycle Cycles the 
ABA'B'
Ref -> Int -> Act
The Inverse  of a commutator
BAB'A'
Ref -> Act -> Int

Let Us analyze 3-cycles on big cube centers by this cycle LMN
ABA'B' would solve this cycle if L was the Ref, M was the Int, and N was the Act
LNM
BAB'A' would solve LNM under the same setup

Inverse Commutator is the first fundamental variation of the standard commutator

Variation 2:  Which Piece is made interchangeable with Ref?
Standard
Ref = L
Int = M
Act = N
Variation 2:
Ref = L
Int = N
Act = M
ABA'B' will cycle LNM
BAB'A' will cycle LMN
See the effects of changing the Int and Act:
*New notation [r|i,a]  Where r is the ref, i is the int, and a is the act*
[L|M,N]  and the Commutator ABA'B'
[L|M,N]' is the same setup with the Inverse Commutator BAB'A'

So, looking at this there are 3 Possible References, 2 Possible Interchangeables, and Two Commutators, at Total of 12 ways to cycle 3 centers.

Once Int has achieved interchangeability, we then Make Int the Ref, because the commutator is requires less setup by turning the reference face so that the Ref is in the location of the Int. 
If the cycle was [Ubl|Brd,Ldf]  The Commutator would be 
S;A;B -- Relates to Setup, Part A, Part B, this is performed as SABA'B'S'
r'; l'd'l; U'
What if the cycle was [Ubl|Brd,Ldb]?
r'U; rd'r'; d'
This moves the Ref into the location that Int was set to
That is the equivalent of 
r'L; l'd'L; U' or r'L2; lu'l'; U'
So Here are the 6 ways to cycle L->M->N
[L|M,N]
[L|N,M]'
[M|N,L]
[M|L,N]'
[N|L,M]
[N|M,L]'

Notice how this cycles just as a cycle does, it's just looking at the cycle from different points.  
Simply because L->M->N-> = M->N->L = N->L->M = N<-M<-L = M<-L<-N = L<-N<-M
Now this is an arbitrary explination, no actual examples.  Just theory, but writing this email helped me personally understand cycles too.  Now I also use another cycle.  

End of Guide
Personal Exploration
Let A contain the outer face turn and B contain the inner slice turn.
A: rU'r'
B: 
What a Commuator like this does is bringing two cubies from the centers into the another layer, the Outer Slice turn then  so that the one center cubie goes into the other location.  Then adjust that slice.  
Here is how you can visualize what the cycle will be.  Move a slice and turn an adjacent face counter or clockwise.  Now, which piece stays in the slice?  
The piece moved out of the slice is shot to the new face.  You adjust the the slice that the transposed cubie was in.  

With the commutator A: rU'r'; B:  
The Fur is moved out of the r slice.  The Fdr stays in the r slice and is shot to the Fur by A.  Then adjusting the u slice.  Then A'B' to complete the cycle.
Other commutators that work:
A:  rUr'; B:  [d]
A:  r'Ur; B:  
A:  r'U'r; B:  [d]
A:  r'Dr; B:  [d]
A:  r'D'r; B:  
A:  l'Ul; B:  
A:  lD'l'  B:  [d]

Now look at the pattern of the First two moves of A and B.
If the sign of A1 and A2 match then B must be the opposite slice of A2
If the sign of A1 and A2 don't match, then B must have the same slice of A2
A:  rU'r'  B:    A1's direction doesn't match  A2's direction so B be must be on the same half of the plane.  
A:  rUr'  B:  [d]  A1's and A2s direction match so B must be on the opposite halves of the plane.

If you like this concept, try and use them.  

There is my theoretical guide to center commutators and how to use them correctly.
In summary there are 12 commutators that you can make based on three cubies.  Millions of ways to set them up.  And 6 ways to cycle in one direction.

Later,
Daniel Beyer

tjhdd-a123

懂了，有点麻烦。

分析就要花很多时间/

彳亍

方法能解决所有三角循环，其中一些情况加setup moves ,应该不会超过两步，难点在于情况太繁杂，要熟练得花些工夫。

[此贴子已经被作者于2006-12-11 0:24:13编辑过]

彳亍

简单的说，循环的三个块需满足如下三个特征：
1、 A B 属于同一层（以 U 面为例），而 C 必须不在这一层，以 D 面为例

2、 A 能通过U 面的转动 到 B 位置，如 U U2 或者 U'，但不影响 C

3、C 能通过 （x y x）的方法与 A 或 B 交换位置，这个过程只能

   影响 U 面的某一个块。

那么， A → B →C 的循环就可以通过
     (过程2） (过程3） (过程2’） (过程3') 来实现。

以角块循环举例：

 UFR → LDF → UBL

1、UBL ， UFR 同在 U 面， LDF块在 D 面。
2、UBL ， UFR 可以通过U2 交换位置，不影响 LDF块。
3、UFR → LDF 可以通过 (R' D R)交换位置，只影响了 U 面的
   一个角块，就是 UFR
那么，UBL → UFR → LDF 的循环方法就是
   
    U2  (R' D R)  (U2)'  (R' D R)'  

    即 U2 R'D R U2 R'D' R