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I was recently trying to find the absolute upper bound on the number of cube rotations needed to solve f2l. For a given cube rotation you are constrained to use only <R,U,L> (and D if you want to... I dont think it changed anything) The problem seems to be pretty hard, and my current guess is 2, for extremely long reasons.
Has anyone ever looked at this problem? Can anyone guess the answer?

EDIT: Changed <F, U, L> to <R, U, L>, of course, I'm sorry!

If you can do <F,U,L>, then you should force your other hand to do the same as well. And I think F2L can be solved with zero cube rotations using <F,U,L,R>. In fact, I'm pretty sure, though I don't have solid proof. I'll work on it when I'm free I guess.

If your right hand really is so weak it can't do a thing (like my left), then it'd be better if you did cross on right, and solve the cube using <L,U,Rw>.

Okay, I didn't answer any part of your question did I?

Two for <F,U,L> - calling the original FLBR:1234 -

Make sure 34 pieces are not in FL, y2

Place 34 at FL, y2

Solve other three pairs.

Zero for <F,U,L,R> (3-gen, and F for EO)

EDIT:
Ignore that, it's obviously one for <F,U,L>, even if you don't get to choose starting orientation:
Only two F2L pieces are hidden at BR. One F2L pair of BL, FL, and FR must be slit over F,U,R. Place it, and rotate it to BR. Solve the others with <F,U,L>

I think its one, using RUL, providing that you don't mind using one or two moves to affect the position of one or two bad edges... I'm not sure if this is correct, its completely a wild guess.

I think its one, using RUL, providing that you don't mind using one or two moves to affect the position of one or two bad edges... I'm not sure if this is correct, its completely a wild guess.

I think you're right. You get all the ones possible with one rotation, then simply rotate the cube c-cw or cw, either works, and get the others... unless you accidentally mess up some of the first ones, then you gotta move it back.

I'm going to agree with everyone else and say <F,U,L> is silly. Also <R,U,L> takes at most one cube rotation (since all you have to do is solve all the "good" edges first and then put the other edges on U and do a y to "flip" them), but if you do that you're going to waste a lot of moves.

Here's an interesting question: if you are only allowed to use <R,U> turns and rotations, what's the fewest number of rotations that you can always solve a cube in?

Also <R,U,L> takes at most one cube rotation (since all you have to do is solve all the "good" edges first and then put the other edges on U and do a y to "flip" them), but if you do that you're going to waste a lot of moves.

For <R,U,L> I don't see why you would have to place them in U. Edges are either good or bad depending on that y, whether you fish them out of the middle layer or not, no?

wrong again Me and Stefan already had this discussion (well, more of a 'think again... think again... no... hmm, thats interesting, but no... fine this is it.') on the 'ZZ cubers' thread.

wrong again Me and Stefan already had this discussion (well, more of a 'think again... think again... no... hmm, thats interesting, but no... fine this is it.') on the 'ZZ cubers' thread.

i meant what lucas garron said earlier concerning RUL. you said 'for all y' which confused me. It seemed in your original post like you meant 'when doing RUL and cube rotations, a bad edge is always a bad edge', which is incorrect.

EDIT
in fact, qqwref's post is very clear. you are just wrong in that first post of yours.

Also <R,U,L> takes at most one cube rotation (since all you have to do is solve all the "good" edges first and then put the other edges on U and do a y to "flip" them), but if you do that you're going to waste a lot of moves.

I don't think thats right. I was not able to solve the case where all edges are flipped in 1 rotation only... I only tried once though so I'm not too confident about that, ill try again tonight.

yea you're right. That line of though was the first thing I thought of, but then I couldn't figure out how to solve that case with 4 wrongly flipped edges, and that made me doubt it. And I just tried now and it works without any problems... so my bad, i guess i was really tired or otherwise confused :s

but its not exactly 'obvious' to me... and I don't know exactly what you mean by defining orientation via FUB. I should really look more into this stuff

now i'm really wondering about the number needed for <R, U>

I think the problem could have been stated more clearly. For instance, is it to be assumed cross edges are initially solved? (Since badmephisto now is saying he was concerned about the case '4 wrongly flipped edges' rather than 8 wrongly flipped edges, it seems to me he assumes the cross pieces are considered to be solved in the initial state.) Is it assumed only { y, y', y2 } cube rotations are allowed, or any of the 23 possibilities for cube rotations?

It appears Lucas assumes cross edges are initially solved. If you don't assume that, then one cube rotation is not sufficient for the <F,U,L> problem (yes I mean F, not R). For example, Pons Asinorum (U2 D2 L2 R2 F2 B2) can not be solved in <F,U,L> with only one cube rotation. To solve in <F,U,L> with only one cube rotation, you need to solve a 2x2x2 block, and then use the cube rotation to move the block to DRB corner. But from Pons Asinorum, at least one edge that you'll need for building some 2x2x2 block will be trapped in the DRB corner.

The question was relating to the Fridrich users' conception of F2L, that is, the cross pieces are already solved and you just have to solve the remaining pieces in the lower two layers. Only y/y'/y2 rotations are considered but it doesn't matter because allowing other rotations would not reduce the optimal number (that is, one rotation).