What is the minimum number of algorithms needed for 2 look last layer?

[Lars Petrus]

Thanks! I added those.

That set does miss one position: http://birdflu.lar5.com/?pos=Cd4a (Setup: B' U2 B U' F' U B' U' F2 U' B U F')

Still, that easily beats my 14 alg set by adding any alg that covers that position. The fact that I didn't have 10 of those 12 in the system shows the deficiencies in my small set search. I've mostly focused on finding fast sets, not small, and it shows.

How did you find this one?

 

[TDM]

The smallest set for all 494 LL positions with edges oriented is 14 algs: http://birdflu.lar5.com/alg_sets/4/algs.

Wow. What I love about these are they're all nice cases. Most are standard COLLs/easy ZBLLs.
It's unusual how only J93 and J417 affect EO though. Do you know which cases they help solve?

I made a table with speed-optimal algs, though I doubt many people will be able to use it for speedsolving.

 

[Lars Petrus]

If I remove J93/J417, only http://birdflu.lar5.com/?pos=Fo4A becomes unsolved, so that's why it's there. Kinda silly to have EO flipping algs in an EO set, but that's what dumb software will do.

When I made the "tiny" sets, only <= 10 moves algs were combinable, so that's why all the algs are short

If you want to "upload"/play around with that 14 alg set, I'll add the missing algs.

 

[xyzzy]

How did you find this one?

I wrote a script a while back to calculate the coverage of any set and ran it for a few hours (it was really slow) on random sets of 6 cases. Strange that it missed one case though; maybe there's a bug in my code, or maybe I copied the results wrongly back when I ran it. I do have two other 6-alg sets that cover all but one case.

Edit: Oh, actually the code was correct, but I looked up the wrong case in Birdflu. Replacing Ff4h / K1880 with Fl4F / K701 fixed it—full coverage with 12 algs!

 

[Niko Lopez]

what is the minumum number of algorithms needed such that for any last layer case, you can apply 2 algorithms and solve the cube?

it's probably some really silly thing that would be silly to recognize but it seems like an interesting question.

what about 3 look?

I can't really think of a way of finding such a method so i'll let the puzzle theory people find it instead

So I'm thinking (I'm probably wrong) that if you want to be able to do 2LLL every time using the minimum amount of algs, you will have to learn winters variation (I don't know if that counts as a LL alg but for now I'm not counting it) then you will have to learn 3 olls and 21 plls so a total of 24 algs, if you count winters variation, then it would be a lot higher, in this case you would want to learn how to do intuitive edge control (which is no Algs or if you wanted to learn the alg version you can learn VHS) then learn the 7 cross oll cases and the 21 pll cases which comes out to 28.

 

[Cale S]

So I'm thinking (I'm probably wrong) that if you want to be able to do 2LLL every time using the minimum amount of algs, you will have to learn winters variation (I don't know if that counts as a LL alg but for now I'm not counting it) then you will have to learn 3 olls and 21 plls so a total of 24 algs, if you count winters variation, then it would be a lot higher, in this case you would want to learn how to do intuitive edge control (which is no Algs or if you wanted to learn the alg version you can learn VHS) then learn the 7 cross oll cases and the 21 pll cases which comes out to 28.

This question is about full LL, so you can't influence it with WV

 

[Lucy Griffiths]

I think to understand the math here - I think we should understand the basics, called Fundamental counting principle. Once you get that, you can look at it as a path counting problem.

 

[Abram Lookadoo]

i have found a way to solve this using 45 algorithms

preform M2 S2
look 1: use adf and auf then orient all the corners and edges (31 algs)
look 2: use adf and auf then permutate all the corners and edges (14 algs) then auf all the side permutation with the centers, and adf all the corners permutation with the E side pieces
preform M2 S2

for 3 look, i found a way that uses 22 algorithms

preform M2 S2
look 1: use auf and adf then solve edge orientation and corner permutation (11 algs)
look 2: use auf then solve corner orientation (7 algs)
look 3: use adf then solve edge permutation (4 algs) then adf the bottom corners to match the E edges, and auf the top edges to match the centers
preform M2 S2

this is effective enough to turn 1 look into 1289 algorithms

 

[xyzzy]

i have found a way to solve this using 45 algorithms

preform M2 S2
look 1: use adf and auf then orient all the corners and edges (31 algs)
look 2: use adf and auf then permutate all the corners and edges (14 algs) then auf all the side permutation with the centers, and adf all the corners permutation with the E side pieces
preform M2 S2

this is effective enough to turn 1 look into 1289 algorithms

Very interesting ideas (using setup moves so you can reduce by both AUF and ADF is pretty cool), but (i) use the edit button instead of multiple posts and (ii) don't post the same thing in multiple threads.