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Isaac Paurus

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Aug 9, 2012
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457
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your house.
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.
 

omer

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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.
 

Isaac Paurus

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Aug 9, 2012
Messages
457
Location
your house.
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.
 
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elliotsherrow

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Feb 7, 2012
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USA
WCA
2013SHER01
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!
 

sneaklyfox

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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

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Dec 1, 2012
Messages
205
Location
Israel
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...
 

Christopher Mowla

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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).

Case 1
jIejYZDlPEI3j_e.jpg

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
jzSCzmGU88YF3_e.jpg

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
j5ckKAPaOgNzl_e.jpg

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)
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.
 
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