"Extended" Commutators

[JBCM627]

(continued from another thread)

So commutators are usually defined as A B A' B'.

I was wondering about the idea of extending the individual blocks of moves, A and B, even further. Dealing with just one of them for now, let A be two sets of moves: A = X Y. A' = Y' X'. (This can be seen when doing A A' = (X Y)(Y' X') = no net effect.) So now replacing for A, you have a commutator defined as:
(X Y) B (Y' X') B'.

This can probably be more useful than my first/only current idea for it, which is:
X = (R' D2 R) = X'
Y = (F D2 F') = Y'
B = U
Which gives you the algorithm:
R' D2 R F D2 F' U F D2 F' R' D2 R U'
Which will do... well, you'll see . And of course, mirrors and inverses do similar things.

Has anyone else toyed with this before, or can find more uses for it?

 

[Lotsofsloths]

Post more examples, this time affecting the permutation xD

 

[MistArts]

Well, I think extended commutators are more suitable for 5 cycles.

 

[JBCM627]

A 5-cycle can be just 2 commutators, which isn't what I am talking about necessarily. This is still just 1 commutator, although probably does have the potential for simplified 5-cycles.

But ok: this affects permutation (although probably isn't too great, unless you really know what you are doing)

Let:
X = (R' D2 R)
Y = U
Z = (F D2 F')

And is of the form: A B A' B'
Where A = (X Y Z), and B = (Y')

commutator: (X Y Z) (Y') (Z' Y' X') (Y)

Algorithm: R' D2 R U F D2 F' U' F D2 F' U' R' D2 R U

(I think this post is typo-free)


----
Edit: Here's another one...

X = R' D' R
Y = U
Z = F D F'

Do: (X Z)(Y)(Z' X')(Y')
Algorithm: R' D' R F D F' U F D' F' R' D R U'

And just change Z to F D' F', and you will get another one that can be solved without use of the "extended" commutators

 

[MistArts]

A 5-cycle is just 2 commutators, which isn't what I am talking about necessarily. This is still just 1 commutator.

But ok: this affects permutation (although probably isn't too great, unless you really know what you are doing)

Let:
X = (R' D2 R)
Y = U
Z = (F D2 F')

And is of the form: A B A' B'
Where A = (X Y Z), and B = (Y')

commutator: (X Y Z) (Y') (Z' Y' X') (Y)

Algorithm: R' D2 R U F D2 F' U' F D2 F' U' R' D2 R U

(I think this post is typo-free)


----
Edit: Here's another one...

X = R' D' R
Y = U
Z = F D F'

Do: (X Z)(Y)(Z' X')(Y')
Algorithm: R' D' R F D F' U F D' F' R' D R U'

And just change Z to F D' F', and you will get one that can be solved without use of the "extended" commutators

This is acutally two commutators too. Except that it has 2 piece from the first cycle in the second. Making it a double swap.

 

[JBCM627]

This is acutally two commutators too. Except that it has 2 piece from the first cycle in the second. Making it a double swap.

Yeah I changed my wording of that just before you replied I think. I wouldn't consider it 2 commutators, as it is still of the form A B A' B'... while (A B A' B') (C D C' D') would be 2 commutators. I guess it depends on the specific A, B, C, and D.

 

[Lucas Garron]

I don't really care much about what you've done, since your X Y is just a split of A, and though it is comprised of two parts that would each commute well on their own, it's just a single comm.
I don't want to discount the idea, but can you explain how this gives us some new interpretation or understanding? Is it that you are more intuitively generating a commutator?

(X Y Z) (Y') (Z' Y' X') (Y), by the way, is:
(X Y (Z Y' Z' Y') X' Y) Comm-in-a-comm. Can be neat, but unless we can make the internal comm generally nicer in some brilliant way, it's rather just obfuscating.

 

[joey]

If A = X Z
A' = Z' X'.
So these are still commutators.

----
Edit: Here's another one...

X = R' D' R
Y = U
Z = F D F'

Do: (X Z)(Y)(Z' X')(Y')
Algorithm: R' D' R F D F' U F D' F' R' D R U'

Or you can do
R' F' R B2 R' F R B2
Yours is 14 moves, mine is 8.

 

[mrCage]

Hi

Here's another "extended" commutator

P=R'D2R
Q= B'U2B

Now do PQP'Q' = R'D2R B'U2B R'D2R B'U2B = (R'D2R B'U2B)*2 [1]

Then consider BR'D2RB' U2 BR'D2RB' U2 = (BR'D2RB' U2)*2 [2]

What we have done is a cyclical shift. This is not really an "extended" commutator. This latter splits simply into PQ P'Q' like so:

P=BR'D2RB'
Q=U2

Also look at this: R' D R2 F' R' F U2 F' R F R2 D' R U2 [3], also a commutator with

P =R' D R2 F' R' F
Q=U2

[2] and [3] have the same effect. [2] is shorter than [3]. I like to call [2] an overlapping commutator. The reason for this should be apparent. Overlapping commuttaors are often shorter than non-overlapping equivalents.

Finally, is R' F R F' an "intended" commutator??

- Per

(edit)

Challenge:
How can [3] be done shorter, modifying [2] ??

(/edit)

 

[mrCage]

(continued from another thread)

So commutators are usually defined as A B A' B'.

I was wondering about the idea of extending the individual blocks of moves, A and B, even further. Dealing with just one of them for now, let A be two sets of moves: A = X Y. A' = Y' X'. (This can be seen when doing A A' = (X Y)(Y' X') = no net effect.) So now replacing for A, you have a commutator defined as:
(X Y) B (Y' X') B'.

This can probably be more useful than my first/only current idea for it, which is:
X = (R' D2 R) = X'
Y = (F D2 F') = Y'
B = U
Which gives you the algorithm:
R' D2 R F D2 F' U F D2 F' R' D2 R U'
Which will do... well, you'll see . And of course, mirrors and inverses do similar things.

Has anyone else toyed with this before, or can find more uses for it?

Hi

Better to avoid using X and Y to denote sequences as X and Y has other meaning in the standard notation for cubes :rolleyes:

- Per

 

[mrCage]

Hi

Just came to think about this cute little commutator. This one also has some overlap and the result is 2 2-cycles.

P= R L' D L R'
Q = U

Does anyone notice anything peculiar about those 2-cycles? What is the order of the permutation??

- Per

 

[JBCM627]

I don't really care much about what you've done, since your X Y is just a split of A, and though it is comprised of two parts that would each commute well on their own, it's just a single comm.
I don't want to discount the idea, but can you explain how this gives us some new interpretation or understanding? Is it that you are more intuitively generating a commutator?

(X Y Z) (Y') (Z' Y' X') (Y), by the way, is:
(X Y (Z Y' Z' Y') X' Y) Comm-in-a-comm. Can be neat, but unless we can make the internal comm generally nicer in some brilliant way, it's rather just obfuscating.

It isn't really a new interpretation, nor do I particularly find it more intuitive I guess. The (useful) point is just to provide an understanding of what happens when you split A (or B) up into more algorithms. It is supposed to still be a single commutator.

It can also be thought of as 2 commutators (as MistArts said). It is namely a special case that has cancelations, although it is more confusing for me to think of it like this, as you have to involve the cancelations. Consider a case of the form: (A B A' B')(B C B' C'), which will give you (A B A')(C B' C'). This itself is interesting, but with an appropriate choice of algorithms A-C, it can be convienient to move A to the end, resulting in B (A' C) B' (C' A) = B (A' C) B' (A' C)'.

As for the comm-in-a-comm, that is only possible because I used Y twice, I think.

Or you can do
R' F' R B2 R' F R B2
Yours is 14 moves, mine is 8.

Yeah, that one was a bad example. The only particuarly useful one imo that I mentioned (or could even find) was the orientation one (R' D2 R F D2 F' U F D2 F' R' D2 R U'), which may not even be that great considering that there are faster ways to do that. I guess the parity-fixing one is ok.