# New 3-Cycle variation?



## Cuber3 (Nov 5, 2009)

I was thinking of a better way to do 3-cycle method and then thought about how BH works (EDIT: In terms of cycing stickers). Instead of interchanging the pieces, you instead interchange the stickers. Why not do the same thing with 3-cycle? Just change your memory system so that you have a number/letter/whatever for each sticker and then just solve it that way. There are less things to remember (no need for a separate orientation step) and you don't need to worry about messing up the orientation when doing permutation.

Please tell me if something in this doesn't work or this method has been invented before and I haven't heard of it (in that case I will :fp myself alot).

EDIT:Yeah, I really totally fail. Also I wasn't trying to make a better method than BH, just a more efficient 3-cycle, but I guess I should have looked a bit more.


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## yoruichi (Nov 5, 2009)

interchangeing pieces and stickers are the same thing
u can look at a piece as 2 stickers so there are always 2 cycles
like UF -> UR is the same as UF -> UR and FU -> RU
so yea so 3 cycling is just 3 cyling so theres no invention


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## a small kitten (Nov 5, 2009)

This has been done for a couple of years already. ^ he uses this.


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## 4Chan (Nov 5, 2009)

Yeppers, that's kinda old.
My BLD method can do that too~
(ZBLD method)


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## Cuber3 (Nov 5, 2009)

Okay, :fp myself. But now I'm wondering why this isn't more popular than the regular 3-cycle method. 

And, I don't get how cycling stickers and cycling pieces is the same thing. Here is an example.

LU => UB => BD

If you do D2 F2 U' (Clockwise U Permutation) U F2 D2 then the pieces will be cycled but the orientation will stay the same. That is cycling pieces. But if you do b' L B U2 (Counter-clockwise U permutation) U2 B' L' b the orientation has changed (been fixed). Look at Badmephisto's video on commutators. Skip to around 5 minutes in and he is talking about the cycle of pieces/stickers.


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## Lucas Garron (Nov 5, 2009)

Cuber3 said:


> Okay, :fp myself. But now I'm wondering why this isn't more popular than the regular 3-cycle method.


[stefan-mode]But it *is*.[/stefan-mode]


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## yoruichi (Nov 5, 2009)

Cuber3 said:


> Okay, :fp myself. But now I'm wondering why this isn't more popular than the regular 3-cycle method.
> 
> And, I don't get how cycling stickers and cycling pieces is the same thing. Here is an example.
> 
> ...



lolz0rs that doesnt cycle LU UB and BD it cycles UL -> UB -> DB
they aint the same
first letter is the sticker ur talking about
of course that cycle is the same as LU -> BU -> BD
plus theres no such thing as orientation; u just define it under certain restrictions to make bld cubing harder


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## Escher (Nov 5, 2009)

Lucas Garron said:


> Cuber3 said:
> 
> 
> > Okay, :fp myself. But now I'm wondering why this isn't more popular than the regular 3-cycle method.
> ...



M2 is a 3 cycle method, amirite?

on-topic: I actually just tried a few edges-only BLD solves using EPLLs - it's a lot of fun


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## yoruichi (Nov 5, 2009)

nope uarntrite
M2 is a 2 cycle method


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## Escher (Nov 5, 2009)

yoruichi said:


> nope uarntrite
> M2 is a 2 cycle method



but RUR'U' M2 URU'R' M2 is a 3 cycle :confused:


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## yoruichi (Nov 5, 2009)

yea but u dont do M2 at the end


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## Lucas (Nov 5, 2009)

yoruichi said:


> nope uarntrite
> M2 is a 2 cycle method



M2 is "hidden commutators" and some other 3-cycle algoritms. Commutator is a 3-cycle.

I say that it is a hidden commutator because it has the form: (s) M2 (s)' (t) M2 (t)'.... the same as (s) M2 (s)' (t) M2 (t)' (s) (s)'...
And this the same as (s) (M2 ((s)' (t)) M2 ((t)' (s))) (s)'
The same as (s) (M2 ((s)' (t)) M2 ((s)' (t))') (s)'
The same as M2 ((s)' (t)) M2 ((s)' (t))', a commutator...


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## yoruichi (Nov 5, 2009)

even if its a 3 cycle its still a 2 cycle method
and by that i mean u only place one edge at a time


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## dbeyer (Nov 6, 2009)

Quote:
If you do D2 F2 U' (Clockwise U Permutation) U F2 D2 then the pieces will be cycled but the orientation will stay the same. That is cycling pieces. But if you do b' L B U2 (Counter-clockwise U permutation) U2 B' L' b the orientation has changed (been fixed). Look at Badmephisto's video on commutators. Skip to around 5 minutes in and he is talking about the cycle of pieces/stickers. 

lolz0rs that doesnt cycle LU UB and BD it cycles UL -> UB -> DB
End Quote:
Comment:
Wow ... that is very very in effective.
A) UL -> UB -> DB, the setup you used was not efficient.
a) U' B2 would suffice. U2 B2 ... even d' R2
- so now you want to do a 3-cycle on the U layer.
using a transposition from this alg
R' U R' U' R' U' R' U R U R2 (11 htm)
15 moves ... I can solve this in 8 moves (STM). I can solve it very fast with a 9 move (STM).

Again, UL -> UB -> DB
y' L2 D'M2D L2 D'M2u
Likewise ...
D2M2D' L2 DM2D' L2 D'
The pure case is
B2 D'S2D B2 D'S2D

This is just a very very basic example.


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