# my 2x2 method



## skinnyandweak (Jan 16, 2008)

well i think lots of people have already thought of this, but i just wanted to share and see what you guys think.

since i don't want to learn all guimond cases for the first step, i decided to do something else.

first, i would solve one face with oppisite colors. like 3 green and 1 blue, 2 red and 2 orange, and stuff like that. then, i would go on to the top layer and solve one of the 7 OLL cases, but they are not one color, but mixed with the 2 oppisite colors you started with. after that, i would do the guimond part where you make both faces the same colors, and same with the last part, permuting both layers.

so all i basicly did was just do the beginning kinda like fridich, and the rest like guimond. what do you guys think? is this good? bad? it's just temporary though =P


----------



## Erik (Jan 16, 2008)

It's a possible method, I use it sometimes when the opposite layer face is already made and nothing else is nice, but it's not very original see: http://erikku.er.funpic.org/rubik/OFOTA-1.htm only you do the orientation and seperation in seperate steps....


----------



## skinnyandweak (Jan 16, 2008)

link not working =P

and yeah, it's not much different from guimond, and basically the same idea. only the first part where you get the the bottom and top to be oppisite colors is different. it's just easier for me, because i don't really have to learn anything but the permutation of last 2 layers. doesn't work too bad =P.

edit : and can someone help me with making short algs to solve the top? since it doesn't have to be all the same color, just oppisites. like for the triple sune (i think that's what it's called), with the cross and 2 pairs of corners pointing oppisite directions, i found out that R2 U2 R' would solve it. very fast. but i'm not a great cuber myself, so i can't think of other algs to solve the other cases faster. so if anyone can help, that'll be great.


----------



## Swordsman Kirby (Jan 17, 2008)

"all guimond cases"?

There's only two more than Ortega...


----------



## masterofthebass (Jan 17, 2008)

Yes, but Tim, the Ortega 'cases' are just OLL that most people know. The issue with people learning ortega, would be learning the last PBL algorithms that both guimond and Ortega share. Also, guimond cases are somewhat "foreign" to people just starting to learn 2x2, IMHO.


----------



## Kenneth (Jan 17, 2008)

Because you have mixed colours in the faces you do not have to preserve those = you can mix the opposite faces while solving the OLL. Ortega or 3x3x3 OLL's preserves faces and that makes the algs longer than they have to be. But there are shorter ones.

pi : F2 U F2 l' u2 l
H : F2 U2 F/F'
S : F2 U' F U' F/F'
-S: F2 U F' U F/F'
L : R U2 R' F' U' F/F'
T : F R U2 R' F/F'
U : F U2 F U2 F/F'

From my BCE tread, last turn F/F' can be done in any direction, that way you can start separation while doing the last turns of the OLL. "F" and not "R" because I'm solving F-side corners if the ones at BD are solved so the algs preserves DBR and DBL, transpose the algs to R if you prefer that. If you do not care about preserving DB, then it may be even shorter algs than these =)


----------



## Erik (Jan 17, 2008)

sorry but I don't like those algorithms at all Kenneth  all those F moves????:S
If you want to learn ortega you'll have to learn 3 extra algorithms of which one is only 3 half turns. If you want to learn guimodn you have to learn about 15 new algorithms of an average of 4 turns, I've had to learn worse things than that


----------



## skinnyandweak (Jan 17, 2008)

can't all those F turns be translated into R turns? and i think the way i do it is pretty good for not learning anything but PBL algs. it's easy, and getting the first face usually takes about 2-3 moves. so i guess it's comparable to guimond. i never had to do more than 3 moves to get the first face. the only con is the OLL. but i still like it =P


----------



## Kenneth (Jan 17, 2008)

Think I wrote "transpose the algs to R if you prefer that."

It's easy, for:

H : F2 U2 F/F'

do this:

H : R2 U2 R/R'

But ok, here's all of them:

pi : R2 U R2 F' U2 F
H : R2 U2 R/R'
S : R2 U' R U' R/R'
-S: R2 U R' U R/R'
L : R U2 R' F' U' F/F'
T : F R U2 R' F/F'
U : R U2 R U2 R/R'

For T and L you have to stick with the originals, there are already both R and F in them.


----------



## Erik (Jan 17, 2008)

good algorithms then Kenneth


----------



## skinnyandweak (Jan 17, 2008)

which cases are which? i don't know which one solves which OLL =P. sorry, i'm just new to cubing


----------



## Kenneth (Jan 17, 2008)

Do the alg backwards and it sets up the case.

Example: R U' F' <--> F U R' = backwards order of turns and inverse rotation.


----------



## skinnyandweak (Jan 18, 2008)

okay i got it. those are some pretty good algs. except i can't get the T to work, i just end up with L . maybe something is wrong?


----------



## Kenneth (Jan 18, 2008)

skinnyandweak said:


> okay i got it. those are some pretty good algs. except i can't get the T to work, i just end up with L . maybe something is wrong?



Hmm. that's the alg I have written down, must be an error from start I think =)

Have not got the time to fix it right now.

But you can try to find one by yourself, it's not hard because of the shortness of the algos, just do some turns and see what's happening.


----------

