# Minimum number of cube rotations to solve f2l



## badmephisto (Feb 22, 2009)

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!


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## blah (Feb 22, 2009)

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?


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## Lucas Garron (Feb 22, 2009)

Why is this so hard?

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>


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## Escher (Feb 22, 2009)

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.


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## d_sprink (Feb 23, 2009)

*-*



Escher said:


> 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.


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## qqwref (Feb 23, 2009)

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?


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## Faz (Feb 23, 2009)

qq: I got 15 rotations for this - but it largely depends on the LL.


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## Lucas Garron (Feb 23, 2009)

fazrulz said:


> qq: I got 15 rotations for this - but it largely depends on the LL.


He's asking for a constant.


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## Waynilein (Feb 23, 2009)

With a method similar to ZZ-d, that number could probably be 5 or less.


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## teller (Feb 23, 2009)

qqwref said:


> 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?


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## Stefan (Feb 23, 2009)

teller said:


> Edges are either good or bad depending on that y, whether you fish them out of the middle layer or not, no?


No.

(consider the superflip)


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## teller (Feb 23, 2009)

StefanPochmann said:


> teller said:
> 
> 
> > Edges are either good or bad depending on that y, whether you fish them out of the middle layer or not, no?
> ...




Got it...looks like flipped edges are "bad" for all y.


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## Escher (Feb 23, 2009)

teller said:


> Got it...looks like flipped edges are "bad"_ for all_ y.



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.


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## teller (Feb 23, 2009)

Escher said:


> teller said:
> 
> 
> > Got it...looks like flipped edges are "bad"_ for all_ y.
> ...




For middle-layer edges using <R, U, L>? I don't see it...you can't unflip one without using F...?


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## Escher (Feb 23, 2009)

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.


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## badmephisto (Feb 24, 2009)

Oh my god I'm an idiot. I meant <R U L> of course! This is stupid. I'm sorry. (sorry Lucas)





qqwref said:


> 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.


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## Lucas Garron (Feb 24, 2009)

Changing the qhole question, eh?

<R,U,L> is also obviously one:

Define edge orientation via FUB.
Flip misoriented edges by moving from LL into F2L/vice versa.
y, all edges should be oriented for <U, R, L>


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## badmephisto (Feb 24, 2009)

Lucas Garron said:


> Changing the qhole question, eh?
> 
> <R,U,L> is also obviously one:
> 
> ...



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>


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## cuBerBruce (Feb 24, 2009)

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.


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## qqwref (Feb 24, 2009)

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).


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## tim (Feb 24, 2009)

Actually the question is "minimum number of cube rotations". So the answer is 0. "Maximum number of cube rotations" would be 1.


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## qqwref (Feb 24, 2009)

That's not what we mean when we say "minimum", tim... it's more technically some kind of min-max problem, since in this (and similar God's Algorithm type computations) we're concerned with the algorithm which produces the minimum possible value for the maximum number of required cube rotations (or turns or whatever) over all allowed scrambled positions.


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## badmephisto (Feb 25, 2009)

lol i wasn't aware that there were so many interpretations of my question! 
Naturally, the cross is solved. I'm referring to F2L as the stage of Fridrich method, not the actual first two layers -- a common use of the terminology.
And naturally I am only concerned with y/y'/y2 rotations... I'm trying to apply this stuff to an actual solve... have you seen someone use x rotations to insert some F2L pairs?

The question where you are constrained to only <R,U> is interesting, but seems much harder to answer. And the answer probably does not yield any interesting insights, because people solve slots one by one, and dont really plan ahead to optimize number of rotations... for good reasons too! so never mind.


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## Gparker (Feb 25, 2009)

Im a little lost on this forum so my ansewer might be completely different from everyone elses, please dont make fun of(xD)

i use d turns for f2l instead of cube rotations, and i actualyy do sometimes insert f2l pairs with an x to it


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## qqwref (Feb 25, 2009)

badmephisto said:


> lhave you seen someone use x rotations to insert some F2L pairs?



Cross on left? (But of course it's the same solve, just <U,D,R,x> instead of <L,R,U,y>.)


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## teller (Feb 25, 2009)

Thinking about this (and badmephisto's recent video) has changed the way I look at F2L during actual practice.

I used to spot an edge in the middle layer and not really notice whether or not it was flipped--I'd want to set up an easy case with its corner cubie which often required a cube rotation as I was ejecting it. To my dismay I would immediately need another cube rotation in order to execute the easy case. This was frustrating.

Now, if I see that it's flipped, I look for a way to pop it out without the rotation, even if its not my "preferred" setup with its corner, and voila! No more double-rotation!

For recognition, I've determined that if the edge's "outer" (not on the U face) stickers cannot match the L or R center pieces it means a cube rotation is required. This isn't too hard to recognize as an edge is either aligned with the red-orange axis or blue-green (for my color scheme).

So...thanks!


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