A Collection of Algorithms

[Isaac Paurus]

Yeah I'm matt. You're the one with the glasses, correct? And no there are four.

Gd (block on left; headlights back (if block facing you)): R U R' (y') R2 u' R U' R' U R' u R2
Gc (block on left; headlights on right (if block facing you)): R2' u' R U' R U R' u R2 f R' f'
Gb (block on right; headlights back (if block facing you)): F' U' F R2 u R' U R U' R u' R2'
Ga (block on right; headlights left (if block facing you)): R2' u R' U R' U' R u' R2 F' U F

haha, yeah, im the one with the glasses. but i was saying the one i had was none of those. if i come across it again, i will check it out more intelligently.

 

[ottozing]

Weird G perms I found.

R2' F2 R' U R' F' R U R' U' R' F R2 U' R F2' R2'
R2' F2 R' U R2' F' R U R U' R' F R U' R F2' R2'

 

[Ranzha]

Gesus Fish: (R2' S2) (U l2' U' l2') (u R2 U' r2' F2)
Gewfro: (R U R' F') (R U R' U) (R U' R' U') (R' F R2 U R' U') (R U' R')

 

[omer]

haha, yeah, im the one with the glasses. but i was saying the one i had was none of those. if i come across it again, i will check it out more intelligently.

Maybe you're confusing it with one of the r perms, r perms also have headlights and a block.

 

[kirtpro]

R2' F2 R' U R2' F' R U R U' R' F R U' R F2' R2'

Cool thanks, imma steal this one off ya

update:
lol turned out not to work too well for me

 

[Isaac Paurus]

Weird G perms I found.

R2' F2 R' U R' F' R U R' U' R' F R2 U' R F2' R2'
R2' F2 R' U R2' F' R U R U' R' F R U' R F2' R2'

Thanks that top one is the one that makes the case that I was talking about and the bottom one solved it. Thanks

Maybe you're confusing it with one of the r perms, r perms also have headlights and a block.

No, the R perms blocks are always connected to the headlights, like in a T perm. The one I saw didn't connect with the headlights.

 

[elliotsherrow]

New OLL, similar to T perm

So I accidentally found out about this really fast algorithm. It's algorithm number 28 on speedsolving's oll page. It's the one where all is oriented but two adjacent edges. We all know how fast the T perm is. Personally it's one of my favorite algorithms. I just repeat it when I'm bored. But I discovered that r U R' U' r' F R2 U' R' U' R U R' F' is basically the exact same as the t perm, besides instead of R's you'll find r's. This is an extremely fast and easy to remember algorithm for anybody. I couldn't find it anywhere on the internet so I thought I'd give it to anybody who wants to use it!

 

[Sa967St]

It's a really nice alg for that OLLCP case.
It's cool that you discovered it, but it's not new.

 

[uniacto]

It's algorithm number 28 on speedsolving's oll page.

I couldn't find it anywhere on the internet so I thought I'd give it to anybody who wants to use it!

what.

Anyway, it seems alright, I'll probably try to do it better when I get onto my OLL's.

 

[elliotsherrow]

Well I figured as much. But I tried to find it and I couldn't anywhere. Do you use it?

 

[Sa967St]

Well I figured as much. But I tried to find it and I couldn't anywhere. Do you use it?

I use M' U M U2 M' U M for that OLL case.

For that OLLCP case I would use l' U R' U2 L U' L' U2 R2 x' which is similar to an A perm, but your alg is probably better.

 

[sneaklyfox]

So I accidentally found out about this really fast algorithm. It's algorithm number 28 on speedsolving's oll page. It's the one where all is oriented but two adjacent edges. We all know how fast the T perm is. Personally it's one of my favorite algorithms. I just repeat it when I'm bored. But I discovered that r U R' U' r' F R2 U' R' U' R U R' F' is basically the exact same as the t perm, besides instead of R's you'll find r's. This is an extremely fast and easy to remember algorithm for anybody. I couldn't find it anywhere on the internet so I thought I'd give it to anybody who wants to use it!

That's cool, but r U R' U' r' R U R U' R' is just faster and shorter if you want something similar.

 

[omer]

Can someone give me a good algorithm for case 23 OLL? (All Pieces oriented except 2 adjacent corners which point to the same direction, the headlights\superman case)
All the algs on the wiki are slow for me and I can't finger trick them...

 

[ThomasJE]

Can someone give me a good algorithm for case 23 OLL? (All Pieces oriented except 2 adjacent corners which point to the same direction, the headlights\superman case)
All the algs on the wiki are slow for me and I can't finger trick them...

R U R' U R U2 R2 U' R U' R' U2 R

 

[omer]

R U R' U R U2 R2 U' R U' R' U2 R

That's a nice one thanks, still pretty difficult for me though... I'll try to practice it more.

 

[ThomasJE]

That's a nice one thanks, still pretty difficult for me though... I'll try to practice it more.

It's basically Sune, than Anti-sune.

 

[omer]

It's basically Sune, than Anti-sune.

How could I not notice that? Looking at it this way is much easier

 

[Benjamins]

funny optimal T-perm without regrips

R2 D' F2 D R2 B2 D r2' U' F2

 

[Brest]

<RobertY> l U R' D R U R' D' F r U' r' R U2 l'
<Nibblr> inverse: l U2 R' r U r' F' D R U' R' D' R U' l'

 

[Christopher Mowla]

Here is a collection of algorithms for all "oriented" 1 2-cycle and 1 3-cycle last layer odd parity algorithms for the 4x4x4. (They can be applied to the nxnxn with some minor adjustments, if they cannot be directly transferred already). These cases arise in the K4 Method (they occur very often if you choose to "orient" all wings first and then permute them). To generate algorithms for all possible cases from the ones listed here, you take the inverse, take the mirror, and take the mirror and inverse of any of the algorithms listed under each case. (Of course you can also rotate them about y as well).

The * algorithms are ones that I got from CubeExplorer (and I conjugated them with two moves). The rest of the algs I created by hand. Note that it was very difficult for me to make these by hand because the "oriented" cases are so "picky" as to where the wing edges need to be located that for every "oriented" algorithm you see, I probably found 5 or more "unoriented" algorithms (not to mention that all wing edges must be in the same face). It took several attempts to come up with a method that would also produce relatively (and reasonably) brief algorithms, as well as algorithms which were not a total disaster executing. Of course, some of the algorithms listed might not be so pretty, and some are quite longer than I wished.

I haven't seen the 2-gen algorithms that was generated for the 4x4 LL scrambler for the K4 Method, so I don't know how to compare the length of my algorithms, but clearly we can see that most of the algorithms listed are fewer slice quarter turns as well as qtm than if we were to grab and conjugate the optimal algorithm that CubeExplorer can generate (some of the algorithms are quite low in ftm as well).

Spoiler: In 3 Edges

Case 1

y' x M l F2 Lw U2 l U2 Lw' B l' U2 l U2 F2 B' M' l' x' y (22,17)
M2 U' x' l2 U2 F2 l F2 U2 Lw2 U2 l' U2 Lw2 F2 l' x U' M2 (29,17)*
y' Lw F2 l U2 l U2 r' U M' l' U2 l U2 M U' M' F2 Lw' y (24,18) or y' Rw U2 Lw x' U2 l U2 r' U M' l' U2 l U2 M U' (Lw' r) F2 Lw' y (24,18)
y' F Lw2 F' r B2 l' B2 r' R B U2 l' U2 l B' R' F Lw2 F' y (25,19)
y' R Lw F2 Lw U' R' U' l U2 l' U' R U Lw' U2 l U2 F2 Lw' R' y (25,20)
r U2 l2 U M' U' Rw' U2 l' U2 Rw U M l U2 l' U l2 U2 r' (27,20)
y' R2 Lw F2 Lw U' R2 U' l U2 l' U' R2 U Lw' U2 l U2 F2 Lw' R2 y (29,20)

Case 2

y x M l F2 B' U2 l' U2 l B Lw U2 l' U2 Lw' F2 M' l' x' y' (22,17)
M2 U' x' l F2 Lw2 U2 l U2 Lw2 U2 F2 l' F2 U2 l2 x U' M2 (29,17)*
y Lw F2 U2 l' U2 Lw U' R U l U2 l' U R' U Lw' F2 Lw' y' (23,18)
y l F2 U2 l' U2 Lw U L' U' l U2 l' U' L U' Lw' F2 l' y' (23,18)
y x U l2 U' (Rw' l) B' l' U2 l U2 B Rw B2 l B2 l' U l2 U' x' y' (24,18)
y l F2 M U' M' U2 l' U2 M l U r U2 l' U2 l' F2 l' y' (24,18)
r U2 l2 U l U2 M' l' U Rw' U2 l U2 Rw U' M U l2 U2 r' (27,20)

Case 3

M2 U' r2 U2 F2 r U2 r2 U2 F2 r F2 r2 F2 U2 r' U M2 (31,18)*
y R B Rw' S2 B' U2 r U2 r' B Rw' U2 r U2 Rw S2 Rw B' R' y' (24,19)
y (r2 F2 U2 r2 U2 F2 r2) (x r2 U2 l' U2 l F2 r' F2 U' R' U' r' U R U r' x') y'
= y r2 F2 U2 r2 U2 x l' U2 l F2 r' F2 U' R' U' r' U R U r' x' y' (27,19)
R' U' Rw U2 l2 U l U2 M' l' U Rw' U2 l U2 Rw U' M U l2 U2 Rw' U R (31,24)
R2 U Lw U2 Rw2 U r U2 M r' U Lw' U2 r U2 Lw U' M' U Rw2 U2 Lw' U' R2 (33,24)

Spoiler: In 4 Edges

Case 1

B2 U l Rw U' r F2 r' F2 U l' U2 r' U2 Rw' U' B2 (23,17)
B2 U l Rw U' Rw F2 r' F2 R' U Rw' F2 r' F2 l' U' B2 (24,18)
B2 U l Rw U' R2 r F2 r' F2 R2 U Rw' F2 r' F2 l' U' B2 (27,19)
y F2 r' U' D2 l' U2 M l U Rw U2 l' U2 Rw' U' M' D2 U' r F2 y' (25,20)

Case 2

F2 l U' D2 r U2 R r' B Rw' U2 r U2 Rw B' R' U' D2 l' F2 (25,20)
F2 l U' D2 r U2 M r' U Lw' U2 r U2 Lw U' M' D2 U' l' F2 (25,20)

Case 3

F2 r (Uw d') x r' U R' U r' U' R U' r' U2 r' U2 r' x' (Uw' d) r' F2 (23,19)
x U2 B Rw2 U R U' r U2 r' U' R' U' Rw' U2 r' U2 Rw' B' U2 x' (25,19)
x U2 B' r U2 r U2 Rw U' L U' r U2 r' U' L' U Rw' r' B U2 x' (25,20)

EDIT: Also notice that some algorithms in cases 1 and 2 (in 3 edges and in 4 edges) are very similar...just inverting the outer layer quarter turns might be enough to generate an algorithm for case 2 from an algorithm for case 1, and viceversa.