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



## blah (Jul 15, 2009)

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 (Jul 15, 2009)

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 (Jul 15, 2009)

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 (Jul 15, 2009)

Lord Voldemort said:


> 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 (Jul 15, 2009)

Lord Voldemort said:


> 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 (Jul 15, 2009)

StefanPochmann said:


> Lord Voldemort said:
> 
> 
> > you don't need the R-Perms
> ...


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



StefanPochmann said:


> 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 (Jul 15, 2009)

blah said:


> 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.



blah said:


> StefanPochmann said:
> 
> 
> > Lord Voldemort said:
> ...


Clear now?


----------



## AvGalen (Jul 15, 2009)

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 (Jul 15, 2009)

AvGalen said:


> 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 (Jul 15, 2009)

StefanPochmann said:


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


Wow, I got myself Pochmann-owned! 



AvGalen said:


> 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 (Jul 15, 2009)

blah said:


> StefanPochmann said:
> 
> 
> > And the G perms work as well.
> ...


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 (Jul 15, 2009)

StefanPochmann said:


> 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 (Jul 15, 2009)

blah said:


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


Good to see I was wrong.


----------



## blah (Jul 15, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > Lemme try some more to see if I can get a quarter turn.
> ...


Double-owned  I'll find a proof then.

Edit: Found 


Spoiler



Odd vs even. Am I right?


----------



## Stefan (Jul 15, 2009)

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


----------



## RampageCuber (Jul 15, 2009)

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


----------



## blah (Jul 15, 2009)

@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 (Jul 15, 2009)

RampageCuber said:


> 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 (Jul 15, 2009)

blah said:


> Edit: Found
> 
> 
> Spoiler
> ...


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


----------



## Stefan (Jul 15, 2009)

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


----------



## blah (Jul 15, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > Edit: Found
> ...


Permutation parity.

To clarify:


Spoiler



Even parity is achievable by performing a finite number of an algorithm that achieves odd parity.
Odd parity is not achievable by performing any number of an algorithm that achieves even parity.

The simplest example is this:
Alg that achieves odd parity: U
Alg that achieves even parity: U2


----------



## cuBerBruce (Jul 15, 2009)

blah said:


> @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



This J has a corner 3-cycle: L R U2 R' U' R U2 L' U R'

Any PLL can have either even or odd parity in the corners, depending upon the AUF case you use.


----------



## blah (Jul 15, 2009)

I think you missed this:


blah said:


> StefanPochmann said:
> 
> 
> > And the G perms work as well.
> ...


----------



## Stefan (Jul 15, 2009)

blah said:


> Odd parity is not achievable by performing an infinite number of an algorithm that achieves even parity.


Please write "any number" rather than "an infinite number". I don't even have time to perform an infinite number at all. 



blah said:


> Alg that achieves odd parity: U


Sorry, I'd say that's even parity.


----------



## blah (Jul 15, 2009)

StefanPochmann said:


> blah said:
> 
> 
> > Odd parity is not achievable by performing an infinite number of an algorithm that achieves even parity.
> ...


Sorry, bad choice of words 



StefanPochmann said:


> blah said:
> 
> 
> > Alg that achieves odd parity: U
> ...


Okay okay, so I phrased it poorly  Odd parity for corners and for edges. Sum of parities is even. Was this the problem in my original post?


----------



## cuBerBruce (Jul 15, 2009)

blah said:


> I think you missed this:
> 
> 
> blah said:
> ...



And I would basically disagree with that comment...

Most speedsolvers would would say something like "I had a J-Perm" regardless of which angle case and which AUF case they actually had.

In BLD solving, yes, it's important to realize (and usually clear in such discussions) which AUF case and which angle case is meant.


----------



## blah (Jul 15, 2009)

cuBerBruce said:


> Most speedsolvers would would say something like "I had a J-Perm" regardless of which angle case and which AUF case they actually had.



I meant it in a different sense.

The reason it's called a J-perm is because in a PLL diagram, the arrows look like a J. When was the last time you saw a J-perm PLL diagram with a corner 3-cycle and an edge 3-cycle?

When people learn algorithms for the J-perm, they learn one that swaps 2 corners and 2 edges, not one that cycles 3 and 3. Likewise, when people learn algorithms for the U-perm, they learn one that cycles 3 edges, not one that cycles 4 corners and swaps 2 edges.


----------



## brunson (Jul 15, 2009)

blah said:


> I was constipating really bad in the bathroom, so to ease the pain, my mind wandered off into the realm of cubes


Familiar with the concept of TMI?


----------



## cuBerBruce (Jul 15, 2009)

blah said:


> cuBerBruce said:
> 
> 
> > Most speedsolvers would would say something like "I had a J-Perm" regardless of which angle case and which AUF case they actually had.
> ...



No, when people learn PLLs, they often learn one that needs an AUF move to make it correspond the canonical arrows diagram. These algs are usually written with an additional U, U' or U2 move in parentheses to indicate it's for a different AUF case. Often, there may also be a parenthesized y, y' or y2 at the beginning to indicate it's for a different angle case.


----------



## StachuK1992 (Jul 15, 2009)

brunson said:


> blah said:
> 
> 
> > I was constipating really bad in the bathroom, so to ease the pain, my mind wandered off into the realm of cubes
> ...


+1



cuBerBruce said:


> No, when people learn PLLs, they often learn one that needs an AUF move to make it correspond the canonical arrows diagram. These algs are usually written with an additional U, U' or U2 move in parentheses to indicate it's for a different AUF case. Often, there may also be a parenthesized y, y' or y2 at the beginning to indicate it's for a different angle case.


Much like the 'standard' E perm


----------



## Lord Voldemort (Jul 15, 2009)

blah said:


> Now you prove why the same goes for T-perms



Well, I don't know how lay out a formal proof of this, but T-Perm has an only an opposite swap of edges, so there are some cubes (most) that won't be solvable because they need an adjacent edge swap, which you can't do with pure opposite swaps (again, I don't know how to prove this formally). Same goes for any other PLL that swaps two opposite pieces. 

Someone correct me if I'm wrong.


----------



## blah (Jul 15, 2009)

brunson said:


> blah said:
> 
> 
> > I was constipating really bad in the bathroom, so to ease the pain, my mind wandered off into the realm of cubes
> ...



Wikipedia gave me a whole list of stuff with this acronym. Only Transmarginal Inhibition seems to fit the context  Too Much Information arguably fits the context of this thread too 



Wikipedia said:


> Patients who have reached this shutdown point often become socially dysfunctional or develop one of several personality disorders.


I don't consider cubing a "social dysfunction" or a "personality disorder"  More like - *thinks hard* - a disease


----------



## Tim Reynolds (Jul 15, 2009)

Lord Voldemort said:


> blah said:
> 
> 
> > Now you prove why the same goes for T-perms
> ...



Pieces in the M slice stay in the M slice.
Pieces in the E slice stay in the E slice.
Pieces in the S slice stay in the S slice.

Good enough?


----------



## blah (Jul 15, 2009)

Tim Reynolds said:


> Lord Voldemort said:
> 
> 
> > blah said:
> ...



For me? Yep  I don't remember who coined the term, but these are known as "circuits".


----------



## Lucas Garron (Jul 15, 2009)

blah said:


> Tim Reynolds said:
> 
> 
> > Lord Voldemort said:
> ...


Orbits? That's the group theory term, and it means what you're probably thinking of.


----------



## blah (Jul 16, 2009)

Lucas Garron said:


> blah said:
> 
> 
> > For me? Yep  I don't remember who coined the term, but these are known as "circuits".
> ...


Ryan Heise's Human Thistlethwaite Algorithm page (I _think_ this is where I got the term):

_Put cube into <L2,R2,F2,B2,U2,D2> group, named G3 ... *some text* ... In G3, each corner can only move between one of 4 positions, which I call a circuit._


----------

