# Last Layer with 3 Faces



## HASH-CUBE (May 7, 2009)

okay, so i was just scrambling my cube with the 3 faces (F R and U) and solve it using only (F R and U) and it's possible for me.

the thing is i thought i will scramble the cube with all the faces, and then to create 2x2x2 block, then keep that block in BLD (Back Left Down) position, then try to solve the rest using only the 3 Faces (F R and U), and i did it, so i tried it out alot of times, and i success, now i tried with PLLs, yep, i can solve any PLL with only F R U, even if they were created using other than these faces.

so the thing is, PLLs, OLLs, can be solved using only F R and U, so this means i can solve the Last layer with only F R U.

that's what i came up with, just want to view it here 

how about you guys, what do you think of making algorthms to solve OLLs and PLLs with F R U only??


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## irontwig (May 7, 2009)

HASH-CUBE said:


> how about you guys, what do you think of making algorthms to solve OLLs and PLLs with F R U only??



http://www.ai.univ-paris8.fr/~bh/cube/


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## Stefan (May 7, 2009)

HASH-CUBE said:


> what do you think of making algorthms to solve OLLs and PLLs with F R U only??


Why? As a sort of riddle, only allowing FRU? Or for regular solving because you think FRU algs are often fastest? Or for another reason?

When I search for algs for certain effects, I often use ACube and restrict it to only RU or FRU. I think many people do. And yes, after the LBD 2x2x2 block you can always solve the rest of the cube with FRU moves.


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## HASH-CUBE (May 7, 2009)

StefanPochmann said:


> HASH-CUBE said:
> 
> 
> > what do you think of making algorthms to solve OLLs and PLLs with F R U only??
> ...



thanks stefan, i was thinking that algorithms like having only FRU will be faster, just wondering


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## rokicki (May 7, 2009)

StefanPochmann said:


> And yes, after the LBD 2x2x2 block you can always solve the rest of the cube with FRU moves.



Such an assertion always begs a proof. What's your most concise proof of this, Stefan?


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## Stefan (May 7, 2009)

rokicki said:


> StefanPochmann said:
> 
> 
> > And yes, after the LBD 2x2x2 block you can always solve the rest of the cube with FRU moves.
> ...


Um, ok... you caught me. Alright, I'll try to prove it. Guess I'll try to find an algorithm that emulates an L turn by using only <F,R,U>.

And hey, didn't know you're here! And apparently you've somewhat been actively reading.


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## nitrocan (May 7, 2009)

rokicki said:


> StefanPochmann said:
> 
> 
> > And yes, after the LBD 2x2x2 block you can always solve the rest of the cube with FRU moves.
> ...



It seems obvious, but you should know that after the LBD 2x2x2, with using F R and U, you can orient all the edges in every possible way. 

F U' R' F2 for the LF
R U F for the RF
R' U F R2 for the RB
then the OLL algs for the Last Layer.

The PLL can be just done with the U perm and T perm combined, every PLL can be done in F R and U.


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## Stefan (May 7, 2009)

nitrocan said:


> after the LBD 2x2x2, with using F R and U, you can orient all the edges in every possible way.
> 
> F U' R' F2 for the LF
> R U F for the RF
> R' U F R2 for the RB


Yeah, well... it's not enough to show that you can do these individual things. You need to prove that you can also reach them all at the same time. Since each of those algs shuffles a lot of pieces around in addition to flipping one edge, this doesn't prove much at all. Plus, what about the other five F2L pieces?



nitrocan said:


> then the OLL algs for the Last Layer.


And where's the proof that you can do that in <F,R,U>?



nitrocan said:


> The PLL can be just done with the U perm and T perm combined


Where's the proof?



nitrocan said:


> every PLL can be done in F R and U.


I assume there's a "so" missing at the beginning here? Otherwise I request proof.

So yeah... I believe this really isn't quite trivial to do right and nicely. Which is why I omitted proof in my original statement.


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## nitrocan (May 7, 2009)

OLL: Orient the edges using F R U R' U' F', then orient the corners using R U R' U R U2 R'. (I don't think there's much need to elaborate this)

PLL: Do the corners with the T perm (R U R' U' ...). If two that are adjacent that need to be swapped, AUF so that they are on the L face, then do the algorithm. If they are diagonal, do the Y perm (F R U' ...) Then permute the edges using R and U.


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## Stefan (May 7, 2009)

Alright, I'd say this might count as "proof" for OLL and PLL. But what about F2L?

An idea I have for F2L:


Spoiler



Being able to do OLL and PLL means we can do everything with the U layer that we want (except violating the usual edge orientation, corner orientation, and overall permutation invariants of the cube). Thanks to symmetry, we can also do the same with the F layer and the R layer. So after the DBL 2x2x2, solve the FDL corner by first getting it to the F layer (if it's not already there, do something appropriate with the U or R layer) and then do something appropriate with the F layer to solve the FDL corner. In a similar way, solve the remaining F2L pieces. The "appropriate something" is always to bring the current piece closer to its goal place while keeping the previously solved pieces solved.



Other (unfinished) ideas:

- The corners of this restricted 3x3x3 can be solved because they're just an unrestricted 2x2x2. Now I'd have to prove I can solve the edges from there.

- If I prove that I can solve the edges, then proving I can afterwards solve the corners could be done using my old blindsolving method.

How I convince myself that it is true is that I have plenty of freedom for setup moves and that I can use a Tperm for swapping two edges and two corners, which enables me to do everything. Sadly this is far from being a proof.


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## nitrocan (May 7, 2009)

Narrowing it a little further:


Spoiler



All the positions that have the edges oriented correctly, can be solved using 2-gen algorithms. (R U for DRF, F U for DLF and R U again for DRB, see Petrus Method) So it's required to find out the solutions for the positions that don't have the edges oriented correctly.


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## tikva (May 7, 2009)

nitrocan said:


> Narrowing it a little further:
> All the positions that have the edges oriented correctly, can be solved using 2-gen algorithms. (R U for DRF, F U for DLF and R U again for DRB, see Petrus Method)



That's wrong. Even with oriented edges you can't solve PLL A by turning only two adjacent faces.


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## Lofty (May 7, 2009)

tikva said:


> nitrocan said:
> 
> 
> > Narrowing it a little further:
> ...



Well since Stefan already showed that the corners can be solved so I think he is assuming that that is already done. If one orients all the edges and solves the corners the remaining F2L edges can be inserted in 2-gen. Also if we want a very inefficient proof that we can orient the edges with <R,U,F> we can insert the edge into the FR spot with <R,U> and remove it with <F,U> and its orientation will be changed.


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## nitrocan (May 7, 2009)

I'm sorry I meant the F2L, not the LL.


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## trying-to-speedcube... (May 7, 2009)

I know a 2-gen A-perm ^^ (Spoiler below)

R2 z' R2 z R z R z' R' z' R2 z R z R' z' R

Anyway, I suck at real proof, but each LL can be solved with <F, R, U>, just because you can orient edges with F R U R' U' F', orient corners with <R, U>, permute corners with T-perm, and permute edges with <R, U>


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## rokicki (May 9, 2009)

*Quick proof*

I think if we can find a 3-cycle on edges and a double edge twist, both of which leave the corners alone, we are done.

But being too lazy to do that, I just asked gap, following along the lines of this:

http://www.gap-system.org/Doc/Examples/rubik.html

So using only the three generators:

cube3 := Group(
( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),
( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),
(17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11)) ;

<permutation group with 3 generators>

Size(cube3) ;

170659735142400

Factorial(7) * 3^7 * Factorial(9) * 2^9 / 12 ;

170659735142400

Those look sufficiently close, so Stefan was right.

As if anyone doubted it.


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## blah (Jul 15, 2009)

I haven't read any other post other than the OP's, so I dunno if this has been suggested before, but how about this:

Old Pochmann (QED)

Edit: Finished reading. Apparently Stefan himself has suggested this. I don't see what's wrong with it. Or does anyone require an elaborate proof on how this works? I _have_ thought about solving every single piece before making this post, it's not just a rash unsubstantiated conclusion.


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## blah (Jul 15, 2009)

StefanPochmann said:


> Guess I'll try to find an algorithm that emulates an L turn by using only <F,R,U>.



I thought this was trivially impossible?


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## Stefan (Jul 15, 2009)

blah said:


> StefanPochmann said:
> 
> 
> > Guess I'll try to find an algorithm that emulates an L turn by using only <F,R,U>.
> ...


Consider it a joke.


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## blah (Jul 15, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > StefanPochmann said:
> ...



So is my proof good enough? Or would you like a thorough explanation? That's the first (and only) thing that came to my head before I read all the other posts.


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## Stefan (Jul 15, 2009)

blah said:


> So is my proof good enough?


Not by a long shot.


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## blah (Jul 16, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > So is my proof good enough?
> ...


How about now?

Setup moves for edges (HTM optimal):
UB: R2 U' R2
BU: U F' R' F U2 F2 U
UL: 
LU: U2 R' U F' U
UF: R2 U R2
FU: R F R U2 R2
FL: U' F2 U' R U2
LF: U' F U
FD: U2 F' R F U2
DF: U' F2 U
FR: U2 R U2
RF: U' F' U
DR: U2 R2 U2
RD: U2 R U F' U
BR: U2 R' U2
RB: U2 R2 U F' U

Setup moves for corners (HTM optimal):
URB: U F U'
RBU: R2
BUR: R' F
UFR: F
FRU: U F2 U'
RUF: R'
ULF: F R'
LFU: F2
FUL: U' R' U
DFL: F'
FLD: F' R F
LDF: F2 R'
DRF: F' R'
RFD:
FDR: R F
DBR: R2 F
BRD: R F' R'
RDB: R

Parity: U R U' R F2 U R U R U' R' U' F2 R2


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## Stefan (Jul 16, 2009)

Okay, that does convince me (for me with my knowledge of how old pochmann works, it's alright as proof). Although I must admit I only checked the first few setups. Checking all the remaining ones would've been tedious. Not as much as you finding them, of course. Anyway, this is exactly what makes it an unelegant proof. It's not like _"three Rperms give me a U turn, so the cube can be solved with Rperms alone"_, you know? Well, you can shorten it by saying you can solve corners (and thereby also take care of parity) because their movement is unrestricted. But the edges part would still be tedious and unelegant.


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## blah (Jul 16, 2009)

StefanPochmann said:


> Okay, that does convince me (for me with my knowledge of how old pochmann works, it's alright as proof). Although I must admit I only checked the first few setups. Checking all the remaining ones would've been tedious. Not as much as you finding them, of course. Anyway, this is exactly what makes it an unelegant proof. It's not like _"three Rperms give me a U turn, so the cube can be solved with Rperms alone"_, you know? Well, you can shorten it by saying you can solve corners (and thereby also take care of parity) because their movement is unrestricted. But the edges part would still be tedious and unelegant.



Yeah I didn't like my proof because of the elegance issue too 

I've had another idea though, with a missing part I haven't had the time to think about.
1. Missing part: Solve FL, FD, RD, RB
2. ELS
3. U/H/Z-perms to solve U edges
4. Commutators for corners (haven't thought about proving this yet, but my gut tells me there's a trivial proof)

That said, this isn't much more elegant either, if at all


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## mrCage (Aug 2, 2009)

Edge 2-flip:
F' U F2 U' R' F' U' R2 U R' F2 R2 F2 R2

Corner 2-twist: (2 mirrored sunes, sort of):
R' U' R U' R' U2 R U2 - R U R' U R U2 R' U2

Corner 3-cycle:
R' F U2 F' R F R' U2 R F'

Edge 3-cycle:
R U R U R U' R' U' R' U'

Apart from the edge 2-flip it's all basic knowledge 

Per


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## Stefan (Aug 2, 2009)

Per, what's that post supposed to be?


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## mrCage (Aug 3, 2009)

StefanPochmann said:


> Per, what's that post supposed to be?


 
All the required building blocks for a <U,R,F> solving strategy

On the other side once <U,R,F> solving strategy has been shown, one can quite trivially extend to another version of solving without D turns entirely.

Per

And while we are at it show constructively how to do 4x4x4 solving in the following subgroups:

<U,u,R,r>, <U,u, F,f,R,r> and <U,u,F,f,R,r,B,b,L,l>


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## TMOY (Aug 3, 2009)

Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...


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## mrCage (Aug 3, 2009)

TMOY said:


> Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
> In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...


 
The last i suggested was a joke (trivial). I'm still serious about <UuRr> and <UuRrFf> though

Per


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## mrCage (Aug 3, 2009)

TMOY said:


> Last one is easy: Rrl'L' performs an x rotation, you can then get D and d moves by Rrl'L' F R'r'lL and Rrl'L' f R'r'lL.
> In fact it is possible to show that every solvable position on the 4^3 can be reached using only U, Uw, Lw and Rw moves. I'll let you figuer out how...


 
Well, only need to show that R and L are possible. How to do the rest is quite trivial Gimme some time on this one

Per


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## d_sprink (Aug 10, 2009)

I did fused-cube solves for fun, and if you know all the right-handed algs that use F,R, and U only, you can mirror them with a U' , rotate the cube so that R becomes F, and mirror it with your new F U and L. Worst comes to worst, do a 4L-LL, but it's not usually necessary. It can be done very easily.

Edit: Reading some more posts, it's funny how far this has drifted from the original subject. Just thought it was worth mentioning.


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## calekewbs (Aug 10, 2009)

well, you can orient 2 edges with F R and U, so edge orientation is possible. You can orient all corners with 2-gen, so OLL is possible.

You can permute edges with 2-gen algorithms so Edge perm is possible, and you can swap two corners with T perm (which is only U R and F) so LL with only U R and F is 100 percent possible.


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