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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.)

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.)

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.)

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.

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.

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.

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.

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.

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