# 2-look PLL puzzle



## Scigatt (Mar 1, 2009)

As it is, 2-look PLL is usually done with 6 algs(I used J, Y, and edge perms). What I want to know is what is the fewest number of PLLs needed to 2-look that step, and to provide a concrete example with the fewest number of PLLs. (I know 5 is possible, and I suspect 4 is too.)


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## MistArts (Mar 1, 2009)

You can just use one J and one U.


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## nitrocan (Mar 1, 2009)

Or A and U.


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## Scigatt (Mar 1, 2009)

I said 2-look(as in max application of two algs.) not 2-stage.

Mistarts:One of the Gs will necessitate a 3 alg solve.
Nitrocan:Any of the Ns will need 4 algs.


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## teller (Mar 1, 2009)

Scigatt said:


> As it is, 2-look PLL is usually done with 6 algs(I used J, Y, and edge perms). What I want to know is what is the fewest number of PLLs needed to 2-look that step, and to provide a concrete example with the fewest number of PLLs. (I know 5 is possible, and I suspect 4 is too.)




I'm pretty sure you need 6. If you only have 5, there will be a case where it forces you to 3-look it.


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## Scigatt (Mar 1, 2009)

teller said:


> Scigatt said:
> 
> 
> > As it is, 2-look PLL is usually done with 6 algs(I used J, Y, and edge perms). What I want to know is what is the fewest number of PLLs needed to 2-look that step, and to provide a concrete example with the fewest number of PLLs. (I know 5 is possible, and I suspect 4 is too.)
> ...



You need 6 for 2-look if you have to do corners and edges separately. You can go lower if you integrate them.


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## Paul Wagner (Mar 1, 2009)

Okay, there are two possible permutations of corners, Opposite and Adjacent. No there is 4 possible permutations of the edge, U-C, U-B, Z, H so you need a maximum of six.


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## Unknown.soul (Mar 1, 2009)

Paul Wagner said:


> you need a *maximum* of six.



Minimum, not maximum.


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## Scigatt (Mar 1, 2009)

Paul Wagner said:


> Okay, there are two possible permutations of corners, Opposite and Adjacent. No there is 4 possible permutations of the edge, U-C, U-B, Z, H so you need a maximum of six.



You keep trying to separate corner and edge perm. You have to do them simultaneously to get lower than 6.

Edit: Also, why was this thread moved? This is a puzzle of theoretical interest, not a beginner-type question. Can someone move it back?


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## blade740 (Mar 1, 2009)

Just pointing out that you mean "2-alg" rather than "2-look" I could recognize a counterclockwise U in one look, but perform it as two clockwise Us.


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## Scigatt (Mar 1, 2009)

blade740 said:


> Just pointing out that you mean "2-alg" rather than "2-look" I could recognize a counterclockwise U in one look, but perform it as two clockwise Us.



I thought it was common knowledge that they were considered equivalent.


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## blade740 (Mar 1, 2009)

No, they're not. 2-look means you recognize twice. You can do a 2-look PLL with two algs (A/U) It's not the most efficient, though. More likely you want 2-ALG PLL.

As an example, look at compound OLL ( http://cube.garron.us/algs/compOLL/index.htm )

It is a 1-look, 2 alg OLL system.


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## Scigatt (Mar 1, 2009)

blade740 said:


> No, they're not. 2-look means you recognize twice. You can do a 2-look PLL with two algs (A/U) It's not the most efficient, though. More likely you want 2-ALG PLL.
> 
> As an example, look at compound OLL ( http://cube.garron.us/algs/compOLL/index.htm )
> 
> It is a 1-look, 2 alg OLL system.



Okay...make it 2-Alg.


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## Ellis (Mar 1, 2009)

Scigatt said:


> Okay...make it 2-Alg.



Then its six. T, Y, U, Ub, H, Z... could it really be less than that? You said 5 was possible, what 5?


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## Scigatt (Mar 1, 2009)

Ellis said:


> Scigatt said:
> 
> 
> > Okay...make it 2-Alg.
> ...



I'd rather you find out for yourself, but...



Spoiler



Js, Rs, and H


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## Ellis (Mar 2, 2009)

Scigatt said:


> Ellis said:
> 
> 
> > Scigatt said:
> ...



How would you solve an F-Perm with two of those?
Actually nm


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## cuBerBruce (Mar 2, 2009)

If my calculations are correct, if you select a set of two out of the 21 PLL algs, and execute up to two of those algs (from any angle), you can solve as many as 58 out of the 72 cases. (T-Perm and either U-Perm is one of the ways to do that.) I wouldn't be too surprised if having a 3rd alg in the set would allow you to solve all 72.


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## Scigatt (Mar 3, 2009)

Which 58 does it hit? Depending on where you start, there are ones that are a lot harder than others. 

Here's a classification scheme I devised for the PLLs. I hope it can help. It's not quite comprehensive yet, but it's a start.



Spoiler



Terminology:
C-E dual(or dual): The corners of one are isomorphic to the edges of the other, and vice-versa.
C: corners E: edges
s: solved(not neccesarily rel to other) a:adjacent swap equivalent. o: opposite swap equ.

Interactions(edges and corners)
a->a --> s, a or o. a->o or o->a --> a, o->o -->s

CsEs
Solved: Self-dual
H:Self-dual

CsEa
U(a): A(b) dual(from wiki)
U(b): A(a) dual

CsEo
Z:E dual

CaEs
A(a):U(b) dual
A(b):U(a) dual

CoEs
E:Z dual

CaEa
J(a):J(b) dual
J(b):J(a) dual
R(a):R(b) dual
R(b):R(a) dual
G(a):G(c) dual
G(b):G(d) dual
G(c):G(a) dual
G(d):G(b) dual

CaEo
T:Y dual
F:V dual

CoEa
Y:T dual
V:F dual

CoEo
N(a):N(b) Dual
N(b):N(a) Dual



Also, a possibility for 4


Spoiler



Js or Rs and a G-dual pair.


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## Kit Clement (Mar 3, 2009)

Corners: Acw or Accw, Y.
Edges: Ucw, Uccw, Z, H.

So six, unless you do something ridiculous like what Scigatt mentions.


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## JohnnyA (Mar 3, 2009)

For an efficient begginer speedsolving LL method, you would need three for corners and four for edges, so 7.

For fewest algs, I'm sure it could be less than 6.


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## cuBerBruce (Mar 4, 2009)

Scigatt said:


> Which 58 does it hit?



For the specific case I mentioned, where you use T-Perm and one U-Perm, all cases can be solved in two alg executions except the F-Perms, N-Perms, V-Perms and Y-Perms.



Spoiler



Using a set of three PLL algs, I found that the most you can solve (with no more than 2 alg executions) is 71 out of 72 cases. (I'm basing the PLL names from the speedsolving.com Wiki page.) For example, with Ga, Ab, and Jb; you can solve all cases in 2 alg executions except Nb. So Ga, Ab, Jb, and Nb will solve all. *So it appears to me that 4 is the smallest number of PLL algs that can be used to solve any PLL with at most 2 alg executions.*

I calculated 16 cases where 71 out of 72 cases can be solved. Pick any G-Perm, any U- or A-Perm, and then some J- or R-Perm will allow you to reach 71 out of 72 cases with no more than 2 alg executions.


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## Scigatt (Mar 4, 2009)

cuBerBruce said:


> Scigatt said:
> 
> 
> > Which 58 does it hit?
> ...



You're doing this on a computer, aren't you?


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## cuBerBruce (Mar 4, 2009)

Scigatt said:


> cuBerBruce said:
> 
> 
> > Scigatt said:
> ...



Yes, I used GAP.


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## Stefan (Mar 4, 2009)

kippy33 said:


> So six, unless you do *something ridiculous like what Scigatt mentions.*



You mean the kind of thing this thread is actually about? Which was btw started by Scigatt?

You people saying "6": You're completely missing the point and should just ignore the "Puzzle Theory" subforum.


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## brunson (Mar 4, 2009)

blade740 said:


> No, they're not. 2-look means you recognize twice. You can do a 2-look PLL with two algs (A/U) It's not the most efficient, though. More likely you want 2-ALG PLL.
> 
> As an example, look at compound OLL ( http://cube.garron.us/algs/compOLL/index.htm )
> 
> It is a 1-look, 2 alg OLL system.


I think that's a specious differentiation. 

In the common vernacular, "look" implies recognizing the case and applying the appropriate algorithm. If you see a ccwA and perform cwA*2, then you've simply used a sub-optimal algorithm to solve the case. 

The standard Y perm is actually a combination of 2 OLLs (37+33), as is the standard T perm (same OLLs, reversed) with a couple of moves cancelled out. By your definition each of those could be definited as two algs.


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## fanwuq (Mar 4, 2009)

brunson said:


> blade740 said:
> 
> 
> > No, they're not. 2-look means you recognize twice. You can do a 2-look PLL with two algs (A/U) It's not the most efficient, though. More likely you want 2-ALG PLL.
> ...



It's a bit different idea because he is speaking about combining 2 PLLs and the Y perm is 2 OLLs. In that case, you might as well as say the Y perm is made up of 17 single turn algs.

If you use only one A perm and one U perm, it can be 1 look and as many as 2 of the same A and 2 of the same U.

The question here should actually be applying how many distinct permutations of the LL is sufficient to reach all PLL cases. Treat the permutation as the function, not the manual execution of the alg.

Edit:
The title of the thread is actually proper. It is 2 PLL, not 2 alg or 2 look.


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## brunson (Mar 5, 2009)

I think my point may have been more clear if I'd quoted his first posting instead of the second. Performing ccwU twice to solve a cwU should be considered a different algorithm from ccwU, there are just shorter cwU algs available.

Maybe it's a moot point.


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## Zava (Mar 8, 2009)

minimum is 5.
you start with edges, you only need a T and a J perm for adjacent and diagonal swap, after that, you need A, A, E.


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## Stefan (Mar 8, 2009)

Zava said:


> minimum is 5.


How did you manage to miss the whole discussion showing that it's possible with four?



Zava said:


> you start with edges, you only need a T and a J perm for adjacent and diagonal swap, after that, you need A, A, E.


And how do you solve H corners?


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## Kenneth (Mar 9, 2009)

StefanPochmann said:


> Zava said:
> 
> 
> > you start with edges, you only need a T and a J perm for adjacent and diagonal swap, after that, you need A, A, E.
> ...



A (y) A

It's the basic begginners method Stefan 

Yea, I know, ppl like us newer was that low


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## Stefan (Mar 9, 2009)

Kenneth said:


> StefanPochmann said:
> 
> 
> > Zava said:
> ...


No, not the H PLL, I meant H *corners*. As in, after spending an alg on solving the edges. So?


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## Kenneth (Mar 9, 2009)

Yes, I understand that, but I think it is possible to avoid it to happen.


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## Stefan (Mar 9, 2009)

Kenneth said:


> I think it is possible to avoid it to happen.


Nope, his method is really incomplete.


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