A Collection of Algorithms

[Lid]

r U R' U' r' F R2 U' R' U' R U R' F' (not found in speedsolving oll’s page)

I think I'll use it as OLLCP.

There is a reason why it's not on the wiki, it's way too long compared to: x' (R U' R) U2 (L' U L) U2 R2 x (9 vs 14 htm).

Mirror/inverses:
x' (L' U L') U2 (R U' R') U2 L2 x
x' R2 U2 (L' U' L) U2 (R' U R') x
x' L2 U2 (R U R') U2 (L U' L) x

[sa11297]

M' U M U2 M' U M. Try this.

already known common oll. It is pretty fast, but I have switched to another alg.

[mati1242]

www.boca.bee.pl

Site is already opened so you can search any algorithms you want ! ;]

[sa11297]

www.boca.bee.pl

Site is already opened so you can search any algorithms you want ! ;]

I have never seen this site, but it looks really nice. I will search around it and tell you what I think.
EDIT: this is really nice. Maybe you can add pyraminx, sq 1 and megaminx?

[sneaklyfox]

www.boca.bee.pl

Site is already opened so you can search any algorithms you want ! ;]

Nice site! I like the the voting thing.

[Cubenovice]

www.boca.bee.pl

Just my luck:

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[mati1242]

sa11297 - I'm not owner of the site so I can't add any things to it but i really recommend it to anyone who wants to find specified algorithms for 2x2 and 3x3.

Cubenovice- I don't know why you having this message... I don't have any troubles while browsing this site and I'm sure in 90% that the site is clear and safe.


EnjoY~!

[Robert-Y]

I found this nice COLL/ZBLL alg by hand a few minutes ago:

(y') r U R' U' r' F R2 x' U R' D' R U' R' x

[cubeone]

This is a j-perm that I found in a cubing book that I have but is not in the ss wiki.

F2' L' U' r U2' l' U R' U' R2 (x2)


Hopefully there aren't any copyright issues because of the fact that it was from a book. (By the way, in case you were wondering, the book is Speedsolving The Cube by Dan Harris.)

[cmowla]

This is a j-perm that I found in a cubing book that I have but is not in the ss wiki.

F2' L' U' r U2' l' U R' U' R2 (x2)


Hopefully there aren't any copyright issues because of the fact that it was from a book. (By the way, in case you were wondering, the book is Speedsolving The Cube by Dan Harris.)

It's in the wiki, just not in that exact form:

(y x') R2 u' R' u R2 (x' y') R' U R' U' R2 (x2)
and
(y2 x) U2 r' U' r U2 l' U R' U' l2

are equivalents.

All are equal to the decomposition: [B2 R': (U')] [L': [D, L'] ] .