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Help with commutator notation?

G2013

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I know how commutators and conjugates work, I can read commutator notation such as [R', U] [R2: [F', U'] [R, F] ] (U'), but I have some doubts...
¿Can that decomposition be done with any algorithm? ¿Even with an entire solve? If it is possible, ¿how do you do it?
Thanks for your time
 

suushiemaniac

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No, it can't be done with any algorithm, because most algorithms don't have the repeating patterns like setups or inverses.
The commutator notation [A , B] is read as A B A' B', but if there is no A' and/or B' in an alg, it thus can't be written in this notation.
 

whauk

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You should probably specify "that decomposition". If you give me any algorithm A1 A2 A3... An (with Ai being any turn) i can "decompose" that into [A1:A1]... [An:An], which is not what you are interested in, I guess?
 

G2013

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What I'm trying to say (english is not my language) is:
Suppose someone gives you an algorithm, like R U R' U' R' F R2 U' R' U' R U R' F', and tells you to write it in commutator notation. How would you do it?
I know that for example this 3-cycle R U R' D R U' R' D' can be written as [R U R', D], and I also know that the algorithm above (T perm) also can be written with that notation, but I don't know how to "translate" from one notation (standard) to the other (commuator notation). Is there a method? Or you just have to think?
 

Stefan

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Suppose someone gives you an algorithm, like R U R' U' R' F R2 U' R' U' R U R' F', and tells you to write it in commutator notation. How would you do it?
I know that for example this 3-cycle R U R' D R U' R' D' can be written as [R U R', D], and I also know that the algorithm above (T perm) also can be written with that notation

I doubt that.
 

Ranzha

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I doubt that.

In the style of the original post, [R, U] [R', F] [F [R: U']: U']. Only problem is the (F' F) in the middle.

Because commutators are (usually) formed with clearly distinguishable parts that are inverses of each other, and this T-perm alg doesn't have that property, I'd be very keen to believe there's no way to express that T-perm as one commutator.
 

Stefan

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Those styles are obviously meaningless (and btw shouldn't be called "commutator notation") because of what whauk had already said and because of the single move allowance. If his reply to whauk is supposed to be an answer, I can only interpret it as him changing his question to single pure commutator. And that's not possible for T perm because of parities (unless you cheat).
 
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G2013

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Those styles are obviously meaningless (and btw shouldn't be called "commutator notation") because of what whauk had already said and because of the single move allowance. If his reply to whauk is supposed to be an answer, I can only interpret it as him changing his question to single pure commutator. And that's not possible for T perm because of parities (unless you cheat).
What do you mean when saying (writing) "unless you cheat"?
T PLL: [R, U] [F: [F', R'] [R U' R': (U')] ]
From PLL page of the wiki
 

Stefan

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Cheating meaning [post=1000394]stuff like this[/post].

I don't know whether I ever tried rewriting algs "like this", I guess I don't see the purpose unless the alg was developed with conjugate(s)/commutator(s) in the first place or unintentionally happens to have a really meaningful structure. I don't see anything interesting in those F or T perm representations, only clutter and even canceling moves. But if there's something, someone please point it out :).

Anyway, since I don't do this, I don't know how to other than look and think, sorry. I guess this could be turned into a challenge like finding a method to rewrite an alg with conjugates/commutators, cancelling moves allowed, and the goal being something like using as few letters as possible.

T PLL: [R, U] [F: [F', R'] [R U' R': (U')] ]
From PLL page of the wiki

What's your point?
 
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Christopher Mowla

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In the style of the original post, [R, U] [R', F] [F [R: U']: U']. Only problem is the (F' F) in the middle.

Because commutators are (usually) formed with clearly distinguishable parts that are inverses of each other, and this T-perm alg doesn't have that property, I'd be very keen to believe there's no way to express that T-perm as one commutator.
When you mention writing this as one commutator, do you mean as [x,y](q), were q is a single quarter turn move or can q be a conjugate of a quarter turn? Or were you referring to another structure?

What I'm trying to say (English is not my language) is:
Suppose someone gives you an algorithm, like R U R' U' R' F R2 U' R' U' R U R' F', and tells you to write it in commutator notation. How would you do it?
I know that for example this 3-cycle R U R' D R U' R' D' can be written as [R U R', D], and I also know that the algorithm above (T perm) also can be written with that notation, but I don't know how to "translate" from one notation (standard) to the other (commuator notation). Is there a method? Or you just have to think?
You're English is very good. Just note that we don't start interrogative sentences with ¿ as is done in Spanish.

Anyway, I'm the one who made all of those 2-cycle PLL decompositions on the PLL wiki page. I didn't use a method, exactly. You basically just have to think. I'd say one of the most difficult algorithms to decompose into commutators and conjugates was the N-perm:
[L R: [R', U'] [U2: [U': [L', U] ] [R', U'] ] [U', L'] ] (U')

Since this was such a difficult one, I showed how I did it in the "Example Decomposition" spoiler in my forgotten thread where I originally posted all 2-cycle decompositions before I added them to the wiki page (and organized the 2-cycle PLLs to what it is today). (There is no content in that thread besides the example decomposition which is not now in the wiki.)

I know how commutators and conjugates work, I can read commutator notation such as [R', U] [R2: [F', U'] [R, F] ] (U'), but I have some doubts...
¿Can that decomposition be done with any algorithm? ¿Even with an entire solve? If it is possible, ¿how do you do it?
Thanks for your time
Even though I created all of those decompositions (it took me a couple of days to do all of the 2-cycle PLLs in the wiki), I don't find that most are particularly useful in understanding the algorithm. I'm sure some algorithms can be explained like this, as I explained J-Perms using my decompositions here.

I don't think it's possible to decompose every algorithm into a product of short (4 move) commutators as I did for the 2-cycle PLLs, but clearly we can write any algorithm, scramble algorithm, full solve, etc., using a product of commutators + a single quarter turn if there are an odd number of quarter turns (note however that I'm not referring to the 3x3x3 supercube.).

Below here are single commutators which are equivalent to the building blocks of all sequences (should we break up all sequences into individual quarter face turns, for example).

R U (by Stefan Pochmann)
[R, L R U2 F B' U' D']

R U' (my alg...using Stefan's R U idea)
[R, M' R U' M2 U R' D' F B' E' L E]

M2 (my new alg)
[U' D' S E S U D E', B F E' F2 E' F B']

U2 (by Oscar Roth Andersen)
[L R U2 R' L', U]

U D (my new alg)
[U, F2 R2 B2 S' D2 S]

L' R
[R, y2]

U2 D2
[U2, x2]
 

Christopher Mowla

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I don't understand. If you're not afraid to use rotations, why not use [M, y2] for M2?
I don't know what there is to not understand. I don't like using cube rotations, but I was lazy to make commutators for the other two cases, so I "cheated", as I didn't want to spend more time on that post than I already did...I was interrupted with the database error.

Why not [[L U' R: d2]:U2] (U')? Then just write it in an equivalent form without d2, and it shows how it actually works.
That's a great decomposition! As I have said, I created all of those decompositions in a few days, and thus if you or anyone else finds more useful ones, he or she may replace mine at anytime. (In addition, I was determined to write a decomposition for this algorithm which used commutators, and thus I probably could have used that conjugate one you found. However, I can see now that it's more important to have decompositions which help people understand algorithms more than it is to decompose them in a specific manner.)

I primarily posted my decompositions in the wiki to save a lot of questions like this about decomposing the parity PLLs, and apparently my plan has worked very well so far. (Lucas, I never intended them to be used for speed, if that's why you mentioned what you did earlier.)

It's interesting that you caught that typo considering that you specifically mentioned that you ignore most of my posts. I guess it's good that you caught me saying "you are" instead of "your" probably for the first time on the forums. Thanks. I guess you did your good deed for the day?

Thanks everyone, specially cmowla! Now I get it ;)
You're welcome. I did mark a few other 2-cycle PLLs to be "very hard" to decompose in my old thread where the example decomposition is (I apparently marked the other algorithms in PLL spoilers at the end of my post), so I guess you could quiz yourself by decomposing the "easier" ones and then trying to decompose the "harder" ones once you get more practice.
 
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