F R' F' R U S' R U' R' S

Solves a terrible dot OLL case. It's the inverse of the alg another dot case (as far as I know only teller west and I use this alg). I may actually switch to this

EDIT: I realised it's mirror is already on algdb but in a weird wide-turn version. Still cool
 
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I have an algorithm I found by messing around once again, this time it's the first WVLS case here: http://algdb.net/Set/WVLS

I have an alg from the back (if you do a y2): L F L' U' L U L F' L2

I actually found it by doing the mirror: R' F' R U R' U' R' F R2

Pretty fingertrick friendly and I haven't seen it on the site anywhere so I'm assuming it's a new alg. Let me know if you guys have seen it or not.
 
I have an algorithm I found by messing around once again, this time it's the first WVLS case here: http://algdb.net/Set/WVLS

I have an alg from the back (if you do a y2): L F L' U' L U L F' L2

I actually found it by doing the mirror: R' F' R U R' U' R' F R2

Pretty fingertrick friendly and I haven't seen it on the site anywhere so I'm assuming it's a new alg. Let me know if you guys have seen it or not.

That's pretty cool. It's nice to have an RUF Option of that WV case. You should post that one to algdb.net.
 
I think I've discovered a new algorithm for OLL case #23 (please correct me if this has already been found)

It takes the F2L pair on the right out, then takes the left out, then places them back in their correct positions.
However, the L layer remains so that the cross edge is on the front until the pair is reinserted.
This is the alg:

c8fad1e563338a1f42a5a8fb73ab1ddf.png


R U R' U' L' U R U' (M' x')
Note that the (M' x') at the end can be replaced with (R' L).

This is my first time submitting an algorithms, so please feel free to give advice :)
 
I think I've discovered a new algorithm for OLL case #23 (please correct me if this has already been found)

It takes the F2L pair on the right out, then takes the left out, then places them back in their correct positions.
However, the L layer remains so that the cross edge is on the front until the pair is reinserted.
This is the alg:

http://gyazo.com/c8fad1e563338a1f42a5a8fb73ab1ddf.png

R U R' U' L' U R U' (M' x')
Note that the (M' x') at the end can be replaced with (R' L).

This is my first time submitting an algorithms, so please feel free to give advice :)

Nice example of a commutator :) (notice it only moves 3 corners)
[R U R' U', L']

Also, here's a good place to look
https://www.speedsolving.com/wiki/index.php/OLL

your alg is the same as the first one here, except they did the M' first :)
 
Y perm from y2: [L' U' L U: R2] U2 (R2 U')3
Similar to R' U' R F2 R' U R d R2 U' R2 U' R2, but this one is from y2 instead of the normal AUF. It isn't as good as my normal y2 alg, but I thought I'd post it anyway.
 
Y perm from y2: [L' U' L U: R2] U2 (R2 U')3
Similar to R' U' R F2 R' U R d R2 U' R2 U' R2, but this one is from y2 instead of the normal AUF. It isn't as good as my normal y2 alg, but I thought I'd post it anyway.

I use that for OH. Learned it from Antoine's site (mirror tho). Now I'm curious: what's your 'normal' y2 alg?
 
F2L - To swap 2 edges in adjacent slots *and* flip them both:

Put both bad edges on the left side of the cube [if necessary using a setup AuF (as in u, not U) or an E-slice move] and then do a wide E perm:

x' (r U' r' D) (r U r' D') (r U r' D) (r U' r' D')

...this will leave both slots' corners in place. Just undo any AuF or E-type setup moves at the end.


I found this while playing with making regular algs into wide algs (probably not the first person to find it though). It is pretty fast, but is there perhaps a more efficient way to do that?
 
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