All cube states can be solved using only one of these PLLs: J(a), J(b), R(a), R(b)

[blah]

Something bizarre happened today. I was constipating really bad in the bathroom, so to ease the pain, my mind wandered off into the realm of cubes And I arrived at this result (as the thread title says) without a cube in my hands and with a really bad constipation

I'm sure I'm not the first to have discovered this, but I haven't seen it documented anywhere yet, so anyway, my "hard"ship aside, here's the proof with (I think) the most common J perm:

(R U R' F' R U R' U' R' F R2 U' R' U' y)*3 = U (QED ) (actually, it's U y', or Dw)

Same goes for all the other PLLs I mentioned. The final rotation can be either y or y', it doesn't matter. Edit: Same goes for all 4 G perms too!

The important stuff:

Edit: To clarify that this is not a trivial result, the emphasis is on "All cube states can be solved using only one of these PLLs: J(a), J(b), R(a), R(b)". Also, to make it clearer, this is what I meant: All cube states can be solved using only one of these PLLs: J(a), J(b), R(a), R(b), and cube rotations, and nothing else - no setup moves, no AUFs. Clear enough now?

 

[blah]

An interesting result that follows is this:

Since (R U R' F' R U R' U' R' F R2 U' R' U' y)*3 = Dw
and (Dw')*3 = Dw,

=> (R U R' F' R U R' U' R' F R2 U' R' U' y)*3 = (Dw')*3
<=> R U R' F' R U R' U' R' F R2 U' R' U' y = Dw'

Which is obviously false

 

[Lord Voldemort]

Well, you can easily solve corners with just the J perms, and 90 degree turns are 3 J-Perms, so you don't need the R-Perms. I suppose if you can force a 90 degree turn, then you can use setups and Old Pochmann to solve the edges. (I think that's the easiest way to solve it, obviously you can solve it by doing 3 J perms to do a turn, but that would take 150+ J perms.)

 

[blah]

Well, you can easily solve corners with just the J perms, and 90 degree turns are 3 J-Perms, so you don't need the R-Perms. I suppose if you can force a 90 degree turn, then you can use setups and Old Pochmann to solve the edges. (I think that's the easiest way to solve it, obviously you can solve it by doing 3 J perms to do a turn, but that would take 150+ J perms.)

Dude dude dude no no no you got me WROOONG! I'll edit my first post to make it clearer.

 

[Stefan]

you don't need the R-Perms

You get 99 points for that observation (would be 100 but you're missing one).

And the G perms work as well.

 

[blah]

you don't need the R-Perms

You get 99 points for that observation (would be 100 but you're missing one).

I think Lord Voldemort missed my point when he made that statement. But I don't get what you're trying to say.

And the G perms work as well.

Ah, yes. Thanks for that But that's if you define a G-perm as a corner 2-cycle and an edge 4-cycle When most cubers say G-perm, they mean a corner 3-cycle and on edge 3-cycle. Just like when they say J-perms or R-perms, they mean a corner 2-cycle and an edge 2-cycle, not a corner 3-cycle and an edge 3-cycle.

 

[Stefan]

Clear enough now?

It was already clear before your clarification, both from the title and the proof. Though I admit I might've had an advantage cause I've seen this before.

you don't need the R-Perms

You get 99 points for that observation (would be 100 but you're missing one).

I think Lord Voldemort missed my point when he made that statement. But I don't get what you're trying to say.

Clear now?

 

[AvGalen]

Your J perm has a U' at the end. Please try the same with this one: L' U R U' L U2 R' U R U2' R'

 

[fanwuq]

Your J perm has a U' at the end. Please try the same with this one: L' U R U' L U2 R' U R U2' R'

Why? The permutation itself is exactly the same.

 

[blah]

you don't need the R-Perms

You get 99 points for that observation (would be 100 but you're missing one).

I think Lord Voldemort missed my point when he made that statement. But I don't get what you're trying to say.

Clear now?

Wow, I got myself Pochmann-owned!

Your J perm has a U' at the end. Please try the same with this one: L' U R U' L U2 R' U R U2' R'

That's funny, I actually used the inverse of this alg in my original post, but I edited it because I think most people will be more comfortable with the "standard" J.

 

[Stefan]

And the G perms work as well.

Ah, yes. Thanks for that But that's if you define a G perm as a corner 2-cycle and an edge 4-cycle

Well, I didn't even notice. I just copied and tested LU'RU2L'UR'F'B'U2FBU from speedcubing.com. Indeed it's 2+4 like you said. However... prove that no G suffices to solve the cube when it's 3+3. Maybe you just didn't find the proper way to utilize it.

A-perm works as well: (F'LF'R2FL'F'R2F2Uy')3. Yeah I know, non-standard. But still. Standard isn't always meaningful. I prefer R'U2RU2R'FRUR'U'R'F'R2 for R-perm, without the standard U' at the end.

 

[blah]

prove that no G suffices to solve the cube when it's 3+3. Maybe you just didn't find the proper way to utilize it.

3 (3+3)-Gs can do a U2. So... we can use (3+3)-Gs to solve any cube state in the last stage of Thistlethwaite's algorithm? Lemme try some more to see if I can get a quarter turn.

 

[Stefan]

Lemme try some more to see if I can get a quarter turn.

Good to see I was wrong.

 

[blah]

Lemme try some more to see if I can get a quarter turn.

Good to see I was wrong.

Double-owned I'll find a proof then.

Edit: Found

Spoiler

Odd vs even. Am I right?

 

[Stefan]

To clarify: I was wrong because you didn't see it immediately (I thought it would be too easy and not worth asking).

 

[RampageCuber]

T Perm (R U R' U' R F R2 U' R' U' R U R') *6 is a U2 also

 

[blah]

@Stefan: I did notice that the Rs and the Js are all corner 2-cycles even before you told me about the Gs, but it didn't occur to me that it had anything to do with the result; thought it was just a coincidence

 

[blah]

T Perm (R U R' U' R F R2 U' R' U' R U R') *6 is a U2 also

I've proven why an unlimited number of (3+3)-Gs and cube rotations can't produce a quarter turn. Now you prove why the same goes for T-perms

Hint: Same proof goes for E, F, H, N, V and Y.

 

[Stefan]

Edit: Found

Spoiler

Odd vs even.

Am I right?

I suspect you mean the right thing, but it's not clear what you mean.

 

[Stefan]

Btw, if you want a real challenge, then come up with a nice proof for Last Layer with 3 Faces.