# Funny OLL algorithm & nice patterns

#### Joël

##### Member
Hey guys and girls ,

I recently found a new OLL algorithms... Actually, it's a commutator that was shown to me by Per, but I optimized it for my hands, and found out that it can be done quite fast:

z' (U' R U2 r' U) L2 (U' r U2 R' U) L2 z

The last L2 is just to make it a commutator... You don't need it for speedcubing. The algorithm has order 2, so you can see which case it is by just executing it.

You can also make a very nice pattern with this algorithm. Start with a solved cube and do the following moves:

(U' R U2 r' U) L2 (U' r U2 R' U) L2
z2 x (U' R U2 r' U) L2 (U' r U2 R' U) L2
z' U M2 U2 M2 U

I also found the logic behind another interesting pattern recently... It's called the 'python'. I always thought it looked too complicated, but the idea ia actually very simple, and the pattern looks pretty cool:

B2 M'U M'U M'U M'U2 M'U M'UM'U M' B2
R (U2 M2 D2 M2) R'

The first part just flips the FU and BD edges. The result should be a 'snake' pattern on your cube .

- Jo?l.

#### pjk

Staff member
Thats pretty cool.

#### Joël

##### Member
Originally posted by PJK@Mar 16 2006, 06:44 PM
Thats pretty cool.
Yes.. Patrick.

It's also very interesting to find out the structure of these algorithms.

z' (U' R U2 r' U) L2 (U' r U2 R' U) L2 z

You see, the first part flips one 'column' of three pieces in the L layer. Then, after L2, the seconds part is just the inverse of the first bit. That flips the other column, and brings back the rest of the cube!

Algorithms like this are called commutators. You can read all about it in my commutator tutorial.

#### Alexander

##### Member
cube in a cube
easy to remember:
U' L2 D' (R' F) *3 D L2 U' R2 B2 U2 (15,20)

kinda commutator style

#### Joël

##### Member
Yes, Alexander,

I memorised that a loong time ago! When I tried making a cube in a cube for speed at the Worlds, Per said I was cheating!!

- Jo?l

#### AbelBrata

##### Member
Here's a pattern accidentally found by my friend...
I tried to find the moves, since he couldn't recreate it

On a solved cube:
x LR U2 L'R' - F'B' U2 FB x'
and then
R2 Dw2 - M E M' E'

The first is H-permutation.