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 Post subject: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 7:31 am 
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what are the brackets in some algorithms for?

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 Post subject: Re: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 7:33 am 
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Help seperate it so it's easier to remember. Sometimes if you have to repeat something there will be a part within brackets and then a x2 or x3 or something after it.

For example, a T-Perm: (R U R' U') (R' F) (R2 U' R') (U' R U R' F')

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 Post subject: Re: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 11:18 am 
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I think he meant brackets like a [3,1] commutator, rather than parentheses ()

If not, how does the commutator notation work?

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 Post subject: Re: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 11:39 am 
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SEBUVER wrote:
Help seperate it so it's easier to remember. Sometimes if you have to repeat something there will be a part within brackets and then a x2 or x3 or something after it.

For example, a T-Perm: (R U R' U') (R' F) (R2 U' R') (U' R U R' F')

How British :D
We Americans know that those are actually called "parentheses". Brackets are "[ ]".

NType3 wrote:
I think he meant brackets like a [3,1] commutator, rather than parentheses ()

If not, how does the commutator notation work?


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 Post subject: Re: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 11:43 am 
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I'm still unsure what a [3,1] would do, though. It can't really flip a piece in the 3 position with one in the 1 position, can it?

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 Post subject: Re: what are brackets in algorithms for?
PostPosted: Sun Feb 13, 2011 4:09 pm 
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NType3 wrote:
I'm still unsure what a [3,1] would do, though. It can't really flip a piece in the 3 position with one in the 1 position, can it?
To write [3,1] is shorthand for saying you have an algorithm of the form X Y X' Y' where X is 3 moves and Y is 1 move.

As you probably know, one of the easiest ways to discover new routines is to find some sequence of moves that isolates a piece. We call that the X part. Once the piece is isolated you move it and put another piece in its place. We call that the Y part. Then if you undo the X part and then undo the Y part you will have created a 3-cycle of the pieces.

So for example on the Rubik's cube, do R' F R F'. That will isolate a corner in D. It just so happens that the sequence is also a [1,1] commutator because X = R' and Y = F.

But we'll call the whole 4-move sequence X.

If you then do D, and then undo X, and then do D' you will have a |X| = 4 and |Y| = 1 or [4,1] commutator. Written out it is [R', F, R, F', D, F, R', F', R, D'].

If you want to describe the length of the sequences you can call it [4,1] or you can get more specific and call it [[1,1],1].

One of the many benefits to commutators is that you don't have to memorize the whole long sequence, you can memorize the parts. Also, if you keep the same X but change the Y, you can vary what gets cycled without memorizing anything else. Replace D with D2 in the above sequence to see what I mean. Finally, if you are memorizing algorithms you rarely bother to memorize the inverse. With a commutator though, you just have to apply the sequence as [Y,X] rather than [X,Y] and it is the inverse. That way you don't think about doing something backwards but it ends up being backwards anyways.

The [4,1] I showed above can be shortened to [3,1]. I will leave that as an exercise for the reader.

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