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How To: Commutators for Big Cubes

Mike Hughey

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Sorry, Kenneth - for some reason that I don't fully understand, I completely failed to see your post. Yes, your algorithm works fine for this. And it's pretty easy to execute, with all the R's and U's. But it's actually one move more than mine. :)

But that does look generally useful; I'm going to have to play with it a while and see if I can get where I really understand it. Thanks!
 

Kenneth

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U R' stores the Ubr at Fur, U moves the one that initialy was at Ulf to Ubr so a r2 can swap that one with Frd, then the piece at Ubr is chanced to the stored one using U R U'. Then the second swap follows (r2) and the rest is restoration, first the last exchange using U R' U' and then the storage is restored using U' R U' (the two following U' are of course merged to a U2).

Hope I got it right, have no cube here right now :)
 
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mrCage

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Hi :)

The way i prefer to explain (most common) commutators (for 3-cycles at least) is to show how it works on a "cutplane". The 2 different kinds of cutplanes available are outer and inner layers. Inner layers are of course not functioning for corner cycles. Cycling only 3 centers on a 3x3x3 cube is not possible.

The idea of almost all my commutators is to do the following:

1 - Replace a single cubie on a cutplane (A- commonly 3 or 4 turns)

2 - Rotate the cutplane (B - always a single turn!!)

3 - Return the cubie removed from the cutplane back into its new position on the cutplane (A' - inverse of A. To invert a sequence start from the back and move to the front. Invert every turn as you go along. The inverse of UR is NOT RU but R'U'.).

4 - Undo the cutplane rotation.

Easy example on corners (any size cube:
1 - R' D R (A-part. Replaces one corner on the cutplane)
2 - U (B-part, the cutplane rotation)
3 - R' D' R (inverse of A. Returns the corner removed from cutplane back into a new position on the cutplane)
4 - U' (inverse of B-part)

This becomes R' D R U R' D' R U' (indeed a 3-cycle)

(Easy example on edges:
1 - r' D r (A-part)
2 - U (B-part, the cutplane rotation)
3 - r' D' r (inverse of A)
4 - U' (inverse of B-part)

This becomes r' D r U r' D' r U'

Note the intended similarity between the 2 examples.)

This whole thing becomes ABA'B' where the total length becomes length of A times two plus 2 (since length of B is 1!!)

Now for bigger cubes edges belong to different ORBITALS.
On a 4x4x4 cube UFr, URb, UBl and ULf all belong to the same ORBITAL as those positions are only a single turn of the cutplane apart.
UFl, URf, UBr and ULb belong to the same (but different) orbital

What to do if you want to cycle 3 edges of different ORBITALS?
Simple answer: position the edges into same orbital first.

[UNFINISHED POST - lunch !!!!]

[edit - i will finish this post some other time, probably completely rewrite it by the look of it !!]

- Per
 
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dbeyer

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Here is a special case that Chris and I have talked about for our commutator system.

Its actually rather interesting.
For wings, it's what I'd call an anti-slice commutator.

You have 3 pieces which all have no interchangeability.

Such as two pieces on the same face-plane, but different orbitals, and then the 3rd is on a parallel slice-plane. Its on the same face-plane as one of the first two but again on a different orbital.

URb -> FUl -> LFu

I don't know how many of you remember Fridrich's notation where she used slice and anti slice notation. Well this is where I'm getting it from.

The Setup and the First move of the insertion are anti slice turns which I think make for nice commutators mostly.

Now for that case.

URb -> FUl -> LFu
Lets invert it.
URb -> LFu -> FUl
I can so you the SAB commutator rather than the SBA commutator.

L' ] R'FR b' R'F'R b [ L

Another nice case:

These are called Ferris Wheels, just a nickname that Chris gave to them. It makes sense, I suppose.

These are cases where the S and B lead to a cancelation.

Basically: You'll see that two pieces are interchangeable. Such as the ULf and FLd. There is no 3 move insertion from either stationary point (where the cubies currently are) Yet you'll see somewhere on that face plane you can make an make a 3 move insertion.

So you move the first piece there, insert, move the second piece there, undo the insert, and then restore the main slice.

Like I said the ULf and the FLd, now the FUl.

I look and I see that doing l' moves the FUl to the UBl
I also see that doing U moves one of my main slice pieces to the UBl. So maybe I can do an insertion of Ul'U' and interchange?

No, it wouldn't work because the lone edge is distrubed by the turning of the U layer.

Now, I could move both pieces to say the DLb.
So.

Setup: L
Insertion: DlD'
Interchange: L

L ] DlD' L Dl'D' L' [ L'
=
L DlD' L Dl'D' L2'
=
L DlD' L Dl'D' L2
Now if you did it the last way with that cancelation with a clockwise L2, you'd see the L slice making a complete 360 degrees rotation. Which is what a ferris wheel does, which I guess is why Chris named it like such.
 

Bomber

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My brain is confuzzled!!!!

And you had to revive a thread that was last posted in april just to say that?

Good god.

And you had to post that didn't you? Are you Admin? Or even a Moderator? No.
Read this > http://www.speedsolving.com/forum/showthread.php?p=121233#post121233

I wasn't here 8 months ago so I obviously wasn't here to read it. I am going to read the thread and maybe post something relevant, I think it gives a lot more variety unlike the utter drivel you have just posted that is so repetitive on this forum.
Now I will read, and hopefully enjoy this topic.
 
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