PLL's little brother: NEO-PLL

[FatBoyXPC]

If you want to avoid no edges oriented OLL's, you can force two edges to be oriented. For the simple RU'R' case, you can do the sledgehammer isntead (R'FRF'), and for the RUR' case, you can orient all 4 edges with R U' R2 F R F' U2 R' F R F'. That's very similar to OLL 17 (17 has R U2 instead of R U'), and it orients all edges. I posted in that thread I was going to find algs to orient 2 edges (in the event you had 2 edges already oriented) so you didn't have to do the F RUR'U' F' move, and especially f RUR'U' f' back to back with no edges oriented, but for all the RU'R' cases I found it was easier on move count to just sledgehammer it (depending on the two edges already oriented) or just solve it and F/f RUR'U' f'/F'.

 

[joey]

There are certain cases I don't like, especially some no edges oriented OLLs. Doing for example a sune and then NEOPLL seems a lot faster then.

This post made me understand.

 

[Toad]

If you want to avoid no edges oriented OLL's, you can force two edges to be oriented. For the simple RU'R' case, you can do the sledgehammer isntead (R'FRF'), and for the RUR' case, you can orient all 4 edges with R U' R2 F R F' U2 R' F R F'. That's very similar to OLL 17 (17 has R U2 instead of R U'), and it orients all edges. I posted in that thread I was going to find algs to orient 2 edges (in the event you had 2 edges already oriented) so you didn't have to do the F RUR'U' F' move, and especially f RUR'U' f' back to back with no edges oriented, but for all the RU'R' cases I found it was easier on move count to just sledgehammer it (depending on the two edges already oriented) or just solve it and F/f RUR'U' f'/F'.

But some two edges oriented algs are horrid...

eg this:

The OLL for it is awful but you could do:
y' F (U R U' R') F'
And then use NEOPLL...

It's cases like this that make it worth it.

 

[FatBoyXPC]

Well I'm still on 2LOLL so I only know the 7 required algs or whatever, plus a few dot ones (which I don't need anymore heh, due to me finding those algs). I just looked that up on the wiki and I'd probably just do F R U R' U' F' then corresponding COLL.

 

[Toad]

Lol why comment when you're only on 2LOLL...?

I was wondering why your post made no real sense

 

[trying-to-speedcube...]

OLLs I might use this on when I master NEO-PLL:

1, 2, 3, 4, 9, 10, 11, 12, 17, 18, 19, 20, 30, 34, 35, 41, 42, 50

That's 18/57, so in theory I could use NEO-PLL with an advantage about 1/3 of the time.

 

[Toad]

Go to sleep now Maarten...

 

[nlCuber22]

The OLL for it is awful

(y) R' F R F' R' F R F' R U R' U' R U R'
Just sayin'.

 

[trying-to-speedcube...]

Oh my god that algorithm is sooooo fast not

@Mark: But I don't want to @{

 

[joey]

maarten: this is your dad, using joey's account. Go to bed now please.

 

[Toad]

And no Maarten we don't care if there's a noisy cafe in your road, just sleep.

 

[trying-to-speedcube...]

BUT NOT WANT SLEEP DO @{

 

[Escher]

BUT NOT WANT SLEEP DO @{

@{

 

[rjohnson_8ball]

Seems like a good idea to me. Am I right to think the number of PLLs with 4 edges bad is equal to the number of PLLs with 4 edges good? If so, memorizing twice the number of PLLs is not too bad.

 

[Toad]

Seems like a good idea to me. Am I right to think the number of PLLs with 4 edges bad is equal to the number of PLLs with 4 edges good? If so, memorizing twice the number of PLLs is not too bad.

Yes it is exactly the same since orientation and permutation are not related on a cube

 

[Cride5]

Seems like a good idea to me. Am I right to think the number of PLLs with 4 edges bad is equal to the number of PLLs with 4 edges good? If so, memorizing twice the number of PLLs is not too bad.

Yup its the same number, but don't forget ... if you use this then there are some OLLs which you may as well forget

EDIT, ninja'd by a ninja-toad!

 

[Toad]

Seems like a good idea to me. Am I right to think the number of PLLs with 4 edges bad is equal to the number of PLLs with 4 edges good? If so, memorizing twice the number of PLLs is not too bad.

Yup its the same number, but don't forget ... if you use this then there are some OLLs which you may as well forget

Not forget, just learn new algs for.

And what I said is only strictly true because this OLL has the same symmetries as OLL solved (full symmetry ).

 

[trying-to-speedcube...]

Goodnight.

 

[Toad]

Goodnight.

Finally.

 

[Cride5]

Seems like a good idea to me. Am I right to think the number of PLLs with 4 edges bad is equal to the number of PLLs with 4 edges good? If so, memorizing twice the number of PLLs is not too bad.

Yup its the same number, but don't forget ... if you use this then there are some OLLs which you may as well forget

Not forget, just learn new algs for.

And what I said is only strictly true because this OLL has the same symmetries as OLL solved (full symmetry ).

Hmm, are you sure? I was thinking that theoretically, you only need to know about 1/2 of your OLL's if you're using NEOPLL. You certainly don't need any of the no-edge flipped cases, because you can just use OCLLs for them.

For cases with two flipped edges, it seems like only half the number need to be known. For example imagine sune with a vertical bar:

Well if you know that, then you don't need to know this:

because applying an alg for the first case on the second one will give you an all-edge flip OLL. The same story is true of all the 'bar' cases.

Similarly, take this adjacent edge-flip sune:

And this one.

Applying the alg for one of these cases to the other gives an all-edge flip case, meaning you only need to know the alg for one of these two cases. And similarly for all (I think) adjacent edge flip OLLs.
The only set which you need to know all algs for is the all-corners oriented OLLs, and one of them is a NEOPLL OLL-skip!


If this is indeed the case, then it means that using NEOPLL+PLL can involve less algs than straight OLL/PLL