[转帖]Chris 讲解的 commutator

彳亍

转自blindfoldsolving-rubiks-cube@yahoogroups.com

I use a three cycle approach, and I don't use algs at all for the 4x4
or 5x5 centers, I use the following intuitive process to come up with
commutators:

Setup:
------
1) You must have two pieces on the same slice, and one piece off that
slice. These are the "slice pieces", and they are on the "main slice."

1a) The two pieces on the main slice must be "interchangeable" (as per
Doug Li), meaning that one piece can be placed into the position of
the other with just a turn of that slice.

2) The piece not on the main slice is called the "lone" piece. This
piece must easily be able to replace one of the slice pieces without
affecting any other pieces on the main slice.

The Commutator:
---------------
Step 1) Place the lone piece into the main slice such that it replaces
one of the slice pieces. The location in the main slice where you
just replaced that piece is now called the "action" spot. It is
literally where you will do all the work for this commutator.

Step 2) Interchange the other slice piece "the one you have not
affected yet" into the action spot by turning the main slice.

Step 3) Undo the move you did in step 1

Step 4) Undo the main slice turn you did in step 2

And that's it.

Ok here's an example.

The cycle is dfL->frU->blU->dfL and is just a three cycle.

I literally think through all of the steps I listed above each time I
encounter a three cycle.

Setup:
------
1) My "main slice" is the U layer. My "slice pieces" are frU and blU.

1a) I can interchange blU to the frU spot by doing the slice turn U2.
I can interchange frU to the blU spot by doing U2.

2) The dfL piece is the "lone" center. This is because it is not on
the U slice. I can place the dfL piece, the lone center, into the blU
spot by doing l' d' l. I can place the dfL piece into the frU spot by
doing r' d r. Either move does not affect any other piece in the U
layer, so it is my choice as to which to use during the solve.

The Commutator:
---------------
Step 1) I have to place the lone center into the main slice, such
that I replace one of the slice centers without affecting any other
piece in the U layer. I can do this in one of two ways. I can
replace blU with the move l' d' l or I can replace frU with the move
r' d r. I choose r' d r. The reason is because the dfL center is
supposed to go to the frU center in this cycle.

At this step I perform the moves r' d r.

Step 2) I have now replaced the frU center with the lone center. Now
the frU location on the cube becomes the "action" spot. I now have to
place the other main slice center into the action spot. I can do this
with the move U2. This moves the blU center to frU, and into the
action spot.

Step 3) I now undo the move I did in step 1. This moves the piece
that was originally in the frU location back into the action spot.
(Step 1 sort of moves it into limbo, and Step 3 brings it back). So
at this step do r' d' r.

Step 4) I now undo the main slice turn I did, the U2, with its inverse
of U2.

--------------

Final result: I performed the algorithm r' d r U2 r' d' r U2 for the
three cycle dfL->frU->blU->dfL.

------------------------------------------

I do not have any algs memorized for either of the centers or edges
step for either of the 4x4x4 or 5x5x5. I actually go through this
process every time. Some types of commutators come up more frequently
though, so in a sense I do have them memorized, but not as algs. I
still think of them in terms of the steps above, I just sort of group
many steps together if it is a very common case.

If your current three cycle on your centers is not setup correctly,
then you will need to do a setup turn, or maybe even two.

Here is an example.

Harder three cycle:
-------------------
dfL->drF->dbR->dfL

This is a three cycle, but let's look at the setup.

Setup:
------
1) My main slice is the inner d layer. However all three pieces are
now "slice" pieces, and I have no "lone" piece.

1a) Any of the three pieces can be interchanged with any other with a
d turn, so this part is ok.

2) I have no lone piece, this is a problem.

Setup Fix:
----------
I have to fix the current state before I can go on, because I haven't
even satisfied the setup part. There's no way I can do a "freestyle"
commutator from this point. By freestyle commutator I mean the steps
I have been describing here.

To fix this situation I need to look at my current setup. The only
thing wrong is that I have no lone piece. Everything else is ready to
start. In that case I just need to turn one of my three d layer
centers off the d layer. I choose to do this with the move f.

Move performed at this step: f

I have now placed the dfL piece on the U layer, at lfU. So my cycle
is now lfU->drF->dbR->lfU.

In this new three cycle I have to look at my setup:

New setup:
----------
1) My main slice is the inner d layer. My slice centers are the drF
and dbR centers.

1a) I can interchange the slice centers with either a d or d' move.

2) My lone center is at flU

The Commutator:
---------------
1) I need to place the flU piece into the d layer such that I replace
one of the slice centers, and also such that I don't affect any other
piece on the d slice. I can do this with either r U' r' or b U2 b'.
I choose to do this via r U' r'. This is because the lfU piece is
supposed to go to the drF spot in this cycle.

Move performed at this step: r U' r'

2) I have replaced the drF spot, so it becomes my action spot. I now
need to move the other slice center into the action spot. I do this
with the turn d'.

Move performed at this step: d'

3) I now undo the alg I did in step 1 with r U r'. This brings the
original piece that was at drF and places it in the action spot. It
brings it back from being in limbo so to speak.

4) I now undo the main slice turn with d.

Don't forgot to undo the setup turn you did once you finish the
commutator. My setup turn was f, so I now undo with f'

---------

Final result:

I performed the alg f r U' r' d' r U r' d f' to solve the harder three
cycle dfL->drF->dbR->dfL.

This isn't really an alg though, notice how it was all very intuitive.
You can understand the reasoning and the purpose behind every move.
In fact you are literally coming up with this entire commutator on the
fly. This does not require memorizing anything more than the process
I've listed with each step and what you need to do at each one.

-----------------------------------------

I hope this helps. I know it's a lot to read, but that is exactly
what I do for both the centers and edges. The only "algs" you have to
learn are how to insert the lone edge into the main slice without
affecting any other piece on that slice. That's the only "alg" part,
and all those algs are 100% intuitive and all only 3 moves long.

If you want to solve using a three cycle approach I give this one my
highest recommendation. My record solving phase on a successful 4x4x4
blindfolded solve is about 3 minutes 22 seconds. My average is
probably 3:55-4:05 right now, and I am still improving. I think I can
get my average sub-4 all the time with more practice and more
experience with these commutators.

Let me know if this helps, but that's exactly how I approach the
centers and edges of both the 4x4x4 and 5x5x5 cubes.

Chris

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
 

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]

tjhdd-a123

懂了，有点麻烦。

分析就要花很多时间/

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

录

看不懂啊？哪位英文高手麻煩翻譯一下吧[em02]

mylxc60

有点头疼啊  看得迷糊了

skydragon7

.........楼主不厚道 英文也帮忙翻译下嘛