# How-To Make Your Own OLLs



## Lotsofsloths (Jan 13, 2008)

Just thought of this idea 
Yes I know this is ALOT more moves that premade algorithms, but these are algorithms the user is comfertable with, and can be made in a jiffy.

*Step 1; Making the algorithm: *Put your last layer color on the U face and experiment with the F2l, by this I mean seeing how you can take out an 
F2l pair, and reinsert it. BUT: You cannot just apply the inverse 

Example: R U' R' U F' U2 F U' R' U2 R2 U R2 U R

*Step 2; What case is the algorithm used for?: *Now, you are seeing which of the 57 cases your unique algorithm falls under. Repeat your algorithm until it solves itself. You have to remember the case that it solves.
Say you have this case: |try1|try2|try3|try4|solved|, you would have to remember the case of try4.
HINT: goto this site, http://solvethecube.110mb.com/tools.html#order,type in your algorithm, and find out how many times it takes to solve itself.

Example: My algorithm: R U' R' U F' U2 F U' R' U2 R2 U R2 U R, takes 12 times to get the cube back to its original state. This means the after the 11th time I execute that algorithm, thats the last layer orientation it will solve.

*Step 3; Using you algorithm!: *Now that you've developed your algorithm, whenever you run into that case, preform you algorithm 

Example: Whenever I run into this OLL, I know I have to my algorithm(R U' R' U F' U2 F U' R' U2 R2 U R2 U R).







Hope I helped 
- Daniel Saha


----------



## alexc (Jan 13, 2008)

That's cool, except as you said, it's not the optimal solution all the time.


----------



## Lotsofsloths (Jan 13, 2008)

Yea, its random 
But they sure are easy to remember!
Especially since you made them, they come more natrally.


----------



## badmephisto (Jan 13, 2008)

in fact, all you have to do to find out which case it is is to apply the inverse of that algorithm on the solved cube
alg: R U' R' U F' U2 F U' R' U2 R2 U R2 U R
inverse: R' U' R2 U' R2 U2 R U F' U2 F U' R U R' (read backwards and negate all)

this is in fact how they got most of their algorithms for OLL back while the computers couldn't do it. In fact, a friend of mine who derived the solution to the cube himself used this technique to derive all of his own OLL's


----------



## Lotsofsloths (Jan 13, 2008)

!!
All of his own OLLs!
Thats awesome, I'm gonna try that!
Also, most OLLs are mirrors and inverses so it should work(making one algorithm into 3 )


----------



## Leo (Jan 13, 2008)

I've been using this for awhile, always easier to remember algs you find you yourself  I also find out which of my PLLs are made of two OLLs and find out which case they belong two. I've gotten OLLs off of an E perm, Y perm, and..something else?


----------



## joey (Jan 13, 2008)

T-perm, but thats just the same as the Y-perm


----------



## badmephisto (Jan 13, 2008)

i have an intuitive way to do 3cycle of edges, it involves taking out FR pair, then taking out FL pair, then inserting the FR pair back, and then inserting the FL pair back. Result = 3cycle. Do it the other way to get the other 3cycle.

so this technique is good even for PLL's...


----------



## Lotsofsloths (Jan 13, 2008)

http://www.speedsolving.com/showthread.php?t=2201

Same thing happened to me!!


----------



## doubleyou (Jan 14, 2008)

this is also a good way of thinking about it when learning any new algorithm that you find on the net.

watch how the alg handles F2L pairs. 

when you get aware of how it works, execution is also optimized because you are so used to handle F2L cases


----------



## Lotsofsloths (Jan 14, 2008)

Yea, thats how I remeber my last layer algorithms..


----------



## soccerking813 (Jan 29, 2009)

Wow, this is really interesting.
I have often wondered exacctly how algoithms were made (without computers).


----------



## IamWEB (Jan 29, 2009)

soccerking813 said:


> Wow, this is really interesting.
> I have often wondered exacctly how algoithms were made (without computers).



Jessica (Fridrich) used the commutator principle to invent her algorithms.


----------



## soccerking813 (Jan 29, 2009)

And a commutator is like R' D' R D, correct?


----------



## trying-to-speedcube... (Jan 29, 2009)

soccerking813 said:


> And a commutator is like R' D' R D, correct?



A commutator is any algorithm in the form of ABA'B'.

It does something to a piece (RUR'U'RUR'), like flipping a corner.
Then there is a move (D) to put another piece in that place.
Then the first move is undone (RU'R'URU'R') to solve all pieces and flip the corner you put in that place.
Finally, you undo your move (D') to put the piece back.

This commutator flipped the corners FRD and BLD.


----------



## d4m4s74 (Jan 29, 2009)

don't you mean FRD and FLD?


----------



## trying-to-speedcube... (Jan 29, 2009)

d4m4s74 said:


> don't you mean FRD and FLD?



true

blablablablablablabla


----------



## d4m4s74 (Jan 29, 2009)

trying-to-speedcube... said:


> d4m4s74 said:
> 
> 
> > don't you mean FRD and FLD?
> ...



BLD could also be a pun

just change D into D2


----------



## soccerking813 (Feb 19, 2009)

I found a really nice U-perm for PLL using this.
R U R' U' L' U' L U2 R U' R' U' L' U L. (Counter-clockwise)
And for clockwise it is the same thing pretty much:
L' U' L U R U R' U2' L' U L U R U R'
Not good in move count, but easy to execute and remember if you look at what is happening.


----------



## Escher (Feb 20, 2009)

soccerking813 said:


> I found a really nice U-perm for PLL using this.
> R U R' U' L' U' L U2 R U' R' U' L' U L. (Counter-clockwise)
> And for clockwise it is the same thing pretty much:
> L' U' L U R U R' U2' L' U L U R U R'
> Not good in move count, but easy to execute and remember if you look at what is happening.



thats what badmephisto meant earlier...

i found those a while ago - I don't like them for 3x3, but i use them for megaminx : if you leave a gap between the pairs you take out, then you get U perms, and if you don't, you get A perms


----------



## trying-to-speedcube... (Feb 20, 2009)

d4m4s74 said:


> trying-to-speedcube... said:
> 
> 
> > d4m4s74 said:
> ...



Yes, that's what I thought first. But then I changed D2 into D so people could more easily see that D' is the inverse of D. D2-->D2 isn't as clear. But I forgot to change the corners that were flipped.


----------

