# Intuitive PLL?



## Repsela (Jul 16, 2014)

Is there anyone who can perform intuitively the algorithm like PLL? I mean, is there a way of understand them like the commutators?


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## Lucas Garron (Jul 16, 2014)

For most of them, the fast algs don't have intuitive decompositions that would help your ability to memorize or perform them.
(Some, like the H perm and some Zs, are certainly different.)

For example, I know how to motivate why the standard 2-gen U-perm works, but I don't have a great reason why you should expect it to work before trying it. (As opposed to the optimal alg, which is pretty much obvious.)

See the PLL page on the wiki for some algs decomposed into commutators and conjugates.


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## guysensei1 (Jul 16, 2014)

Repsela said:


> Is there anyone who can perform intuitively the algorithm like PLL? I mean, is there a way of understand them like the commutators?



There are definitely ways to understand the PLLS. The 2 A perms are a conjugate with a commutator inside, the H perm is a conjugate, Y perm is a combination of 2 OLLs... etc


However, some PLLS are hard to understand intuitively. (Can any one explain how the 'standard' G perm Algs work?)


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## Stefan (Jul 16, 2014)

Not entirely sure what you mean, but... intuitive PLL? Sure. For example you can use L' R F2 R' L to _"take the UF edge out (and put the DF one in)"_ (with the F2 turn, after the corners got out of the way with L' R) and permute edges that way, for example:

(L' R F2 R' L) u2 (L' R F2 R' L) u (L' R F2 R' L) u (L' R F2 R' L)
or
(L' R F2 R' L) u (L' R F2 R' L) u' (L' R F2 R' L) u' (L' R F2 R' L) u' (L' R F2 R' L) u (L' R F2 R' L) u

Same for corners using R' D2 R. And do a quarter U in the beginning in case of parity.

I don't like to teach beginners (as it takes away their opportunity to solve on their own) but if I want to show someone that solving the cube is doable and understandable, this is how I show them. And before using turns, I tell them to really think of physically taking one piece out of the cube and putting it back in where it belongs (pushing out a piece, which then gets placed back in where it belongs, and so on).


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## Filipe Teixeira (Jul 16, 2014)

H perm:
swap UF-UB and DF-DB, do a U and swap UF-UB and DF-DB back and do a U'.
(M2 U2 M2) U (M2 U2 M2) U'

U perm
Sexy move to make two adjacent edges opposite, swap UF-UB and DF-DB, undo sexy move, swap UF-UB and DF-DB back.
(with cancellations):
(R U R' U') (M2 U2 M2) (U' R U' R') (M2 U2 M2 U2)

A perm
Finish with the sexy method.
z2 (R U R') D2 (R U' R') D (R U R') D (R U' R')


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## DGraciaRubik (Jul 16, 2014)

As Stefan told you with the edges, you could treat J Perms (L and J) as 3 pairs of edges and corners:

(r' D' r) U (r' D r) U (r' D' r) U2 (r' D R)

AS you can see r' D' r takes a block out and the U moves place the next block so that the block is replaced with the old block.


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## TDM (Jul 16, 2014)

I tried explaining most of my PLLs here, after the alg (fourth line) but there were a few (like R perms) that I couldn't explain.


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## AvGalen (Jul 23, 2014)

Corner-only or Edge-only PLL's are easy to do intuitively, for example with <R' D2 R> reposition U layer, repeat for corners and <M' U2 M> reposition U layer for edges. This would cover things like A,E,U,H,Z
Most Parity-PLL's (2 edges, 2 corners) are easy to understand as 2 OLL's or as a [L U R] alg (also known as playing with pairs). This would cover things like T, Y (2 OLL's), J, N and V (playing with pairs)
All G-Perms can be done as a A and U but I wouldn't know any intuitive way for this


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## porkynator (Jul 23, 2014)

J-perm as a pair commutator:
[R2: [D B2 D', Fw2]] = R2 D B2 D' Fw2 D B2 D' Fw2 R2

Another J-perm is explained here, under Strategy #5: Down-under, last example:
U2 L U L' U B2 D' R D B2
But it's a bit hard to understand imo


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