# 2-look PLL puzzle

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#### cuBerBruce

##### Member
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.

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.

#### Scigatt

##### Member
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.

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

#### cuBerBruce

##### Member
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.

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.
You're doing this on a computer, aren't you?
Yes, I used GAP.

#### Stefan

##### Member
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.

#### brunson

##### Member
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.

#### fanwuq

##### Member
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.
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

##### Member
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.

#### Zava

##### Member
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.

#### Stefan

##### Member
minimum is 5.
How did you manage to miss the whole discussion showing that it's possible with four?

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?

#### Stefan

##### Member
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?
A (y) A
No, not the H PLL, I meant H corners. As in, after spending an alg on solving the edges. So?

#### Stefan

##### Member
I think it is possible to avoid it to happen.
Nope, his method is really incomplete.

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