# Commutator notation extension - [A,B,C]

Discussion in 'General Speedcubing Discussion' started by Kirjava, Mar 23, 2012.

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1. ### KirjavaColourful

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Mar 26, 2006
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snkenjoi
While some extensions to the notation have been proposed in the past, this is a common group of algorithms and the notation is already somewhat being used.

I propose that [A,B,C] = [A,B] [B,C].

After expanding, a cancellation in the middle gives A B A' C B' C'.

For example, [l', RUR'U', r] is a common 4x4x4 last layer algorithm.

This notation can be extended further -

[A,B,C,D] = [A,B] [B,C] [C,D] = A B A' C B' D C' D'

but I am yet to find a use for it

Finally, just as [A,B]' = [B,A], [A,B,C]' = [C,B,A].

2. ### ben1996123Banned

May 17, 2009
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Nice, this will make it easier to write some stuff (probably, I don't know much about commutators).

It reminds me of the chain rule.

3. ### riffzMember

Oct 3, 2008
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I like it. Probably wouldn't use it too often but being able to avoid typing out an entire cyclic shift is good enough for me

[F R', U2, R' F]

4. ### aronpmMember

Sep 9, 2009
Ooh, that is a good use for this. I never thought of it before. I've only used it for 2*3 cycles and 5 cycles

5. ### KirjavaColourful

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snkenjoi
btw

[R, U, U2, R]

828
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Sep 17, 2009
New Orleans, LA
4EverTrying
I haven't seen that many algorithms that contain a product of two (or more) non-conjugated (or non shifted) commutators. Moreover, I have seen that algorithms for 2 3-cycle cases are just as brief (or even briefer) if handled with one commutator instead of 2. I think it would be much more useful to establish a notation extension that illustrates cyclic shifts rather than a product of two commutators.

7. ### MostEdMember

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I though it was super duper complex!

8. ### KirjavaColourful

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Are you sure? K4 makes heavy use of them.

I mentioned this yesterday, but didn't see much use in it. Can you elaborate?

Do you have anything in mind? I think a single delimiter to denote the location of the shift could be a solution.

828
3
Sep 17, 2009
New Orleans, LA
4EverTrying
Well, the only real use I would see is showing the decomposition of an algorithm more effectively (even permutation or odd permutation). (Sure there are most likely move cancellations with these, but is it not so with the product of commutators? If the two commutators affect only one piece type each and there are no move cancellations between them, then they should be seen as two different algorithms, not one. Of course, I'm not saying you ever said this, but just for sake of argument).

I think a single delimiter is great too. No I don't have anything in mind to suggest, because honestly, I can't recall a single time when my suggestions were considered. I just like to post my opinion which usually gives people in the thread something to oppose...in doing so, "the people find direction." If you all do decide to do this (which I highly doubt, because the notations which are currently standard "shall not be modified" ), then I'm sure you all will agree on some symbol to use as a delimiter (I'm not going to waste my time...again).

10. ### KirjavaColourful

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snkenjoi
When you intend to use it, just write that "|" (or whatever) denotes a cyclic shift. Maybe it'll catch on.

828
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Sep 17, 2009
New Orleans, LA
4EverTrying
Whatever happened with this?

12. ### qqwrefMember

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Nobody used it enough to make it become a standard, and it was confusing when people did use it.

13. ### KirjavaColourful

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snkenjoi
When did this happen?

14. ### qqwrefMember

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I think I saw it once or twice in a random cubing thread or something. Not sure.

15. ### KirjavaColourful

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snkenjoi
that was said with the least amount of conviction ever

16. ### qqwrefMember

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qqwref2
This topic was 9 months ago, and you expect me to remember details of where I saw an obscure notation on this forum?

17. ### KirjavaColourful

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no, I think you added the part about confusion without really thinking to add weight to your statement, then when pressed for an example came up with "I think I saw it once or twice in a random cubing thread or something. Not sure."

18. ### qqwrefMember

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