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2-look PLL puzzle

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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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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

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#5
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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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.
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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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

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#10
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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#17
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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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.

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
Js or Rs and a G-dual pair.
 
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#20
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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