Last Layer with 3 Faces

[HASH-CUBE]

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??

 

[irontwig]

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/

 

[Stefan]

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.

 

[HASH-CUBE]

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.

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

 

[rokicki]

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?

 

[Stefan]

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?

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.

 

[nitrocan]

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?

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.

 

[Stefan]

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?

then the OLL algs for the Last Layer.

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

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

Where's the proof?

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.

 

[nitrocan]

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.

 

[Stefan]

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.

 

[nitrocan]

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.

 

[tikva]

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.

 

[Lofty]

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.

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.

 

[nitrocan]

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

 

[trying-to-speedcube...]

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>

 

[rokicki]

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.

 

[blah]

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.

 

[blah]

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?

 

[Stefan]

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?

Consider it a joke.

 

[blah]

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?

Consider it a joke.

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.