When to use F, R, U, R', U', F'

[Mik]

I know you can use this algorith in the OLL step of Fridrich method to solve the one that looks like a T, but what else is this for? I know you can use, the inverse of this algorithm to orientate pieces, but can you do the same with this algorithm. The way I learned you use the inverse if there are edge pieces at the 12/9 o'clock positions. Are there any specific cases like that for this algorithm?

 

[Jason Baum]

Yeah, that alg flips the UF and UB edges, so you can use this alg to orient two edges when the good edges are at UL and UR (I call it the "line case").

 

[AvGalen]

First you say that this case is for the T-OLL and then you ask if it orients pieces? OLL is all about orienting pieces.

The be more helpfull, the T-OLL and its inverse are used in 2 look OLL to orient the edges. (some corners are twisted to, but you don't care about that in 2 look OLL)

A nice variation is this F (RUR'U')(RUR'U')(RUR'U') F' which only changes the orientation of corners. It solves the same case as extended Sune, only from a different angle. This alg can be performed in about 1 second (14 moves!) and is about as fast as extended sune

 

[TimMc]

Why not use R' F' U' F U R to orient the edges in the LL instead?

I actually execute this as:
R' y x r' U' r U y' x' R

Case #1 - Two Oriented Edges near F and B (2 aedges already solved, one at FU and other at UB):
(R' y x r' U' r U y' x' R)

Case #2 - Two Oriented Edges near L and B (2 edges already solved, one at LU and other at UB):
(R' y x r' U' r U r' U' r U y' x' R)

Case #3 - No Oriented Edges (Case #1 + Case #2):
(R' y x r' U' r U y' x' R) (U2) (R' y x r' U' r U r' U' r U y' x' R)

EDIT: I'm not 100% sure that my x and y notation is correct. Keep in mind that when you apply x or y LRFBUD notation changes to new faces respectively.

y - 1/4th clockwise turn around the axis where U center and D center denotes the axis (move R face across to F face)
-> assume every face now has new LRFBUD notation after applying y
x - 1/4th clockwise turn around the axis where L center and R center denotes the axis (move F face up to U face)

It's still effectively the same as: R' F' U' F U R

I've just tried to transpose the algorithm into notation that's the same as how I actually execute R' F' U' F U R. If I hold the cube still while trying to execute this algorithm I find it very time consuming and difficult to do, so I use slight re-grips and turn two faces at a time with my right hand because it's faster for me than it is to use my left hand or F face turns.

Tim.

 

[TheBB]

If you do it twice, but do f,f' instead of F,F' in the second one, you also get an OLL.

Edit: That goes without switching F-f also, of course. I use both variations. (Two different cases).

 

[jazzthief81]

This algorithm can also be very useful for solving F2L cases where the edge is flipped. This means that the pair can't be solved without using F moves or without changing the orientation of the cube. Here are a few examples (do inverse to set up):

R U (F R U R' U' F') R'
R' U (F R U R' U' F') R
(F R U R' U' F') R' U' R
(F R U R' U' F') L U2 L'

There are numerous other examples. I think that for some cases they provide a very short and easy solution and I favour them over more intuitive solutions.

Lars

 

[Mik]

Alright then.
So basically I use the F, U, R, U', R', F' algorithm to set up a case where I can solve the OLL part (I don't know too many algorithms). I'm wondering if the inverse can used for the same purpose, as other algorithms I've tried often return the cube to its original state (before I use the algorithm but with F2L solved) within 2 or 3 uses of the algorithm.
Also, AvGalen, isn't that algorithm the same as this one (the one I currently use): R, U2, R', U', R, U, R', U', R, U', R' ?
TimMC, Maybe I'm doing something wrong but about 2 of those algorithms didn't quite work out for me.
Jazzthief81, none of yours worked for me...

 

[abbracadiabra]

Mik - one of the best ways to see what an alg will do to your cube is to apply it to an already solved cube. Then you can clearly see what moved where, and incorporate it as needed into your solves. If you try that you'll notice that the the inverse of F U R U' R' F' (i.e., F R U R' U' F) also orients two edges on the top layer. One orients two opposite edges, and the other orients two adjacent edges.

As for Jazzthief81's algs - they work perfectly well, and in fact are brilliant. He suggests that when solving the first two layers (F2L) you can move a flipped edge piece onto the top layer and solve it with a simple OLL alg, then put it back into the middle layer. I'm going to try to incorporate his suggestion, because I think he's right that in many cases it simplifies the F2L solution.