# Aa+Ua = Ab+Ub = F+Y = T+V = {Ga,Gb,Gc,Gd}



## Stefan (Apr 29, 2011)

As a side product of something, I found out that doing Aa and Ua (or Ua first, or the other combinations in the title) exactly produces the four G perms (which one depends on the AUF between the two perms). And these are all combinations that do this.


----------



## Xishem (Apr 29, 2011)

And?


----------



## Hershey (Apr 29, 2011)

Man, even when I am sarcastic I don't sound as serious (as this post sounds).


----------



## Stefan (Apr 29, 2011)

And I found it neat but didn't want to spam the theory forum.


----------



## sa11297 (Apr 29, 2011)

yeah that is pretty cool


----------



## Xishem (Apr 29, 2011)

I just always thought this was obvious. I always saw G perms as a 3-cycle of corners and a 3-cycle of edges, because all that was solved were the 2 adjacent pieces.


----------



## Stefan (Apr 29, 2011)

Xishem said:


> I just always thought this was obvious. I always saw G perms as a 3-cycle of corners and a 3-cycle of edges, because all that was solved were the 2 adjacent pieces.


 
Of course it's obvious that each G can be done as *some* A and *some* U. But I don't find it obvious that all four can be done with Aa+Ua alone, without using Ab or Ub. Same for Ab+Ub. And I have no idea how it was obvious to you that F+Y and T+V can do this as well, and that there are no other pairs that do this. Pretty impressive if all this was obvious to you.


----------



## Xishem (Apr 29, 2011)

I think it's only obvious that it can be produced with only Aa+Ua or Ab+Ub, as the edge permutation and corner permutations are completely separate of each other.

And you're right, the other combinations were not obvious to me, but the A+U seemed to be the main point of your post.

Edit: Edge permutation and corner permutation aren't separate of each other. Sorry. What I meant was that 3-cycles of edges and 3-cycles of corners are separate of each other.


----------



## ~Phoenix Death~ (Apr 29, 2011)

I always knew about this, but this is good to know.


----------



## Zubon (Apr 29, 2011)

I don't get it. By looking at the diagrams of the G-perms, anyone can see that they are combinations of various A and U perms.

Did you make this thread simply to say that you are surprised that you didn't notice this before? Or is there something else I am missing?


----------



## Sa967St (Apr 29, 2011)

Xishem said:


> I just always thought this was obvious. I always saw G perms as a 3-cycle of corners and a 3-cycle of edges, because all that was solved were the 2 adjacent pieces.


That's not what the title means.

Get four cubes. On one of them do Aa, followed by Ua. On another do Aa followed by y Ua. On another do Aa followed by y2 Ua. On the last one do Aa followed by y' Ua. You will have all 4 different G perms in front of you.

Same goes with Ab+Ub, F+Y, T+V instead of Aa+Ua.


----------



## Xishem (Apr 29, 2011)

Sa967St said:


> That's not what the title means.
> 
> Get four cubes. On one of them do Aa, followed by Ua. On another do Aa followed by y Ua. On another do Aa followed by y2 Ua. On the last one do Aa followed by y' Ua. You will have all 4 different G perms in front of you.
> 
> Same goes with Ab+Ub, F+Y, T+V instead of Aa+Ua.


Ah. I see. This makes more sense.


----------



## miniGOINGS (Apr 29, 2011)

So an Aa followed by an Ua (from ANY angle) will always produce one of the Gs? And same goes with the other combos there? Interesting.

EDIT: So the angle that the second algorithm is applied determines which G? And each angle corresponds to a different G? Woah. Makes sense now.


----------



## Stefan (Apr 29, 2011)

Xishem said:


> I think it's only obvious that it can be produced with only Aa+Ua or Ab+Ub, as the [...] 3-cycles of edges and 3-cycles of corners are separate of each other.



I disagree. Just looking at the PLL diagrams, the Gs look like this:

Ga = Ua + Aa
Gb = Ub + Ab
Gc = Ub + Ab
Gd = Ua + Aa

Two are Aa+Ua, two are Ab+Ub. Not all four Aa+Ua, or all four Ab+Ub.



~Phoenix Death~ said:


> I always knew about this


 
Yeah right.


----------



## Stefan (Apr 29, 2011)

miniGOINGS said:


> So an Aa followed by an Ua (from ANY angle) will always produce one of the Gs?


 
Yes, but not only that. Also the other direction. Not only is any Aa+Ua some G, but also any G is some Aa+Ua. All four Gs are generated. That's why I wrote _"*exactly* produces the four G perms"_ and used "= {Ga,Gb,Gc,Gd}" rather than "⊂ {Ga,Gb,Gc,Gd}".


----------



## riffz (Apr 29, 2011)

Pretty kewl.

I thought given your reputation people wouldn't assume their horrible misinterpretations of what you said were indeed what you meant. Guess not. :\


----------



## Xishem (Apr 29, 2011)

Stefan said:


> I disagree. Just looking at the PLL diagrams, the Gs look like this:
> 
> Ga = Ua + Aa
> Gb = Ub + Ab
> ...


True, but I still think it can be fairly easily implied that if you apply a 3-cycle of edges and then a 3-cycle of corners that it can only produce a G perm, and since each 3-cycle of corners is being applied to a different orientation of edges, (I think) that each PLL essentially _has_ to be a different PLL. Therefore, if it can only produce a G perm, and each of the final cases has to be different, only the 4 G perms could be produced. It's not 100% obvious, but I think it can be somewhat implied (at least with the Us and As).


----------



## crashdummy001 (Apr 29, 2011)

riffz said:


> their horrible misinterpretations of what you said


 
maybe he didn't say it right

lern2communicate

learn from sarah, she knows how to communicate


----------



## Sa967St (Apr 29, 2011)

crashdummy001 said:


> maybe he didn't say it right
> lern2communicate


?
It was explained very well, some of you just didn't read it properly.



crashdummy001 said:


> learn from sarah, she knows how to communicate


The example shouldn't really have been necessary. :/


----------



## Athefre (Apr 29, 2011)

Thom has a nice "Random Cubing Discussion" topic you could have posted this in without having people hatin on you. It probably wouldn't have gotten as much attention though.


----------



## crashdummy001 (Apr 29, 2011)

Sa967St said:


> It was explained very well, *most *of you just didn't read it properly.


 
fixed

if most people (83% up to your post) don't understand, then it wasn't conveyed properly. end of story.


----------



## Sa967St (Apr 29, 2011)

crashdummy001 said:


> fixed
> 
> if most people don't understand, then it wasn't conveyed properly. end of story.


 
False. It means you need to lrntonotation.


----------



## Stefan (Apr 29, 2011)

Xishem said:


> if you apply a 3-cycle of edges and then a 3-cycle of corners that it can only produce a G perm



Does this look like a G perm to you?
(R' U R' U' R' U' R' U R U R2) (R' F R' B2 R F' R' B2 R2)

Or this?
(R' U R' U' R' U' R' U R U R2) y2 (R' F R' B2 R F' R' B2 R2)



Athefre said:


> Thom has a nice "Random Cubing Discussion" topic you could have posted this in without having people hatin on you.



Ah yeah, that might've been a good place. I've noticed that thread but never really looked into it...


----------



## TiLiMayor (Apr 29, 2011)

So this is the side product of what? It may be found a little more interesting than this curious not sarcasm.


----------



## Xishem (Apr 29, 2011)

Stefan said:


> Does this look like a G perm to you?
> (R' U R' U' R' U' R' U R U R2) (R' F R' B2 R F' R' B2 R2)
> 
> Or this?
> (R' U R' U' R' U' R' U R U R2) y2 (R' F R' B2 R F' R' B2 R2)


Touché. I never took into account the fact that Ja, Jb, Ra, and Rb perms also have AUFs where only one corner and one edge is solved. Are there any more?

I guess only ignorance makes things "obvious".


----------



## Sa967St (Apr 29, 2011)

TiLiMayor said:


> So this is the side product of what? It may be found a little more interesting than this curious not sarcasm.


http://www.speedsolving.com/forum/s...PLL-time-attack-orders-return-to-solved-state


----------



## Stefan (Apr 29, 2011)

TiLiMayor said:


> So this is the side product of what?



Side product of computing combinations of two PLLs. My results and program:
http://www.stefan-pochmann.info/misc/PLL_combinations.html - any other interesting things in this table?
http://www.stefan-pochmann.info/misc/PLL_combinations.txt

That in turn was my side product of attempting to count PLL attacks returning to solved.



Xishem said:


> I never took into account the fact that Ja, Jb, Ra, and Rb perms also have AUFs where only one corner and one edge is solved. Are there any more?



That's a good way to put it, I like that. And no, those are all.


----------



## Lucas Garron (Apr 29, 2011)

This *is* cool, and I think perfectly justified for a new thread. In fact, this should probably actually be in Puzzle Theory; it deserves to be there more so than the many threads people regularly post there (I regularly move them out; I don't even wanna know what a mess that subforum would be if I didn't moderate it constantly), although that wouldn't really free it from posts that detract from the discussion.

Ua and Aa I would have believed, but F+Y and T+V I wouldn't have suspected (but note that the latter two obviously imply each other's possibility by corner-edge duality).


----------



## Stefan (Apr 29, 2011)

Feel free to move it. By now with the discussion, including your very cool duality observation (*), I agree it's more theory than anything else. When I started it, I had no idea where it would go or whether it would get replies at all.

(*) In case someone doesn't know what that duality means: T is like Y, and V is like F, only with roles of edges and corners switched (describe how T looks like, replace "edge" with "corner" and "corner" with "edge", and you should have a description of Y). Or imagine you could also do eighth turns rather than just quarter turns, so you'd actually have edges and corners switch places in the cube.

Although, Lucas... while I can clearly see that duality between F+Y and V+T, I must admit I don't that easily see the duality between the G perms (yeah I just don't see it, they're too complicated... but I understand it must be there )


----------



## Tim Major (Apr 29, 2011)

Arnaud showed me this. I asked why he used A-perm for 2-look PLL, he said he only did with the G-perms, because then he always knows which EPLL he'd have.

Edit: This was only related to G-perms, didn't know about the other two


----------



## Stefan (Apr 29, 2011)

Stefan said:


> I don't that easily see the duality between the G perms (yeah I just don't see it, they're too complicated... but I understand it must be there )


 
Had to paint it:






Now I see Ga and Gc are duals, as are Gb and Gd.


----------



## toastman (Apr 29, 2011)

I learnt my 2LOLL by using T-Perm+U-Perm for basically everything. 
Next I learnt H Perm (as it's so easy to memo), and Z Perm (again, easy to memo).
Then Y-Perm (gives you Y, and the 2 N-Perms).

A-Perms I learnt quite late. x R' U' R always confused the heck out of me, I'd have to stop and think for 5 seconds and mis-execute. In fact, I've been "properly" speed-cubing for 5 months now and only in the last 2 weeks am I really starting to nail my A-perms fast and consistent. 

Even then I still sort-of have to think "Block on the left side, headlights on top-or-bottom, rotate-cube-toward-headlights, move-r-face-the-other-diretion, fingertrick.."

As a result I'm either going to find BH-commutators very, very easy or very, very hard.


----------



## Lucas Garron (Apr 29, 2011)

Stefan said:


> Now I see Ga and Gc are duals, as are Gb and Gd.


Yeah, that's exactly the image I had in mind, thanks for explaining things to everybody. If you draw each arrow the same (instead of different colors), it becomes more obvious: a G-perm is two three-cycles in opposite directions, fully interleaved (none leaves a hole in the other).
T/Y/V/F are a bit funny because they're all self-inverses and self-mirrors, though their combinations are asymmetric – making twice the possible algs.

So, the question is: Is there an intuitive reason the two 3-cycles produced by T+V (or F+Y) should interleave completely?


----------



## riffz (Apr 29, 2011)

crashdummy001 said:


> maybe he didn't say it right
> 
> lern2communicate
> 
> learn from sarah, she knows how to communicate


 
Learn to read. I understood it.



crashdummy001 said:


> if most people (83% up to your post) don't understand, then it wasn't conveyed properly. end of story.


 
Are you serious?


----------



## Cubenovice (Apr 29, 2011)

Tim Major said:


> Arnaud showed me this. I asked why he used A-perm for 2-look PLL, he said he only did with the G-perms, because then he always knows which EPLL he'd have.
> Edit: This was only related to G-perms, didn't know about the other two



Isn't it quite common to use A-perm for 2-look PLL?
I already realised A-perming a G-perm always results in U-perm but didn't notice the a-a and b-b relation.
I also used to AUF the G-perm with headlights in the back.

Seems I can skip that AUF now.

Cool stuff!

Idea: got no cube with me so I cannot check right now:
Could you start with the G in such an orientation that you can cancel the R2 at the end of A with the R2 beginning of U?


----------



## Cubenovice (Apr 29, 2011)

toastman said:


> A-Perms I learnt quite late. x R' U' R always confused the heck out of me, I'd have to stop and think for 5 seconds and mis-execute. Even then I still sort-of have to think "Block on the left side, headlights on top-or-bottom, rotate-cube-toward-headlights, move-r-face-the-other-diretion, fingertrick.."



Try to think of x R’ U R’ as l U R’ 
I actually think along the lines of “pulling U R’ ‘over’ the cube”


----------



## Stefan (Apr 29, 2011)

Lucas Garron said:


> If you draw each arrow the same (instead of different colors), it becomes more obvious:



Bawww... but I did that on purpose... note the colors do *not* stick to the type of piece but to which cycle it is relative to the non-moving pieces. Pink is the cycle counterclockwise-next from the dots, black the clockwise-next. For Ga and Gb, pink is edges. For Gc and Gd, pink is corners. For me that makes it *easier* to see the relation.



Lucas Garron said:


> So, the question is: Is there an intuitive reason the two U-perms produced by T+V and F+Y should interleave?



Two U-perms? Are you comparing T+V and F+Y? I don't get it.



Cubenovice said:


> Idea: *got no cube with me* so I cannot check right now:
> Could you start with the G in such an orientation that you can cancel the R2 at the end of A with the R2 beginning of U?



http://alg.garron.us/
http://thearufam.brinkster.net/cube/wrapplet.asp
[noparse][/noparse]


----------



## Lucas Garron (Apr 29, 2011)

Stefan said:


> Two U-perms? Are you comparing T+V and F+Y? I don't get it.


Uh, *two three-cycles. In either combination, since they're the same.


----------



## Cubenovice (Apr 29, 2011)

Stefan said:


> http://alg.garron.us/
> http://thearufam.brinkster.net/cube/wrapplet.asp
> [noparse][/noparse]


 
You expect me to know algs by heart?

You are giving me too much credit...


----------



## DavidWoner (Apr 29, 2011)

Lucas Garron said:


> So, the question is: Is there an intuitive reason the two 3-cycles produced by T+V (or F+Y) should interleave completely?



Well I view the defining trait of an interleaved corner and edge cycle as the 1x1x2 block that is preserved. Working with F+Y(proving T+V as well is obviously not needed), two of the angles are obvious, as a portion of the 1x1x3 block is preserved. For Fperm(bar in front) y Yperm(standard) and the y2 one, there only one corner that moves to the position opposite its origin, and one edge as well. If you think about the N-perms, it's easy to see how this will always result in the formation of a 1x1x2 block.


----------



## Forte (Apr 29, 2011)

I thought this was cool ):


----------



## Lucas Garron (Apr 29, 2011)

Cubenovice said:


> You expect me to know algs by heart?
> 
> You are giving me too much credit...


Yes, we do. Either that or you look them up: http://www.speedsolving.com/wiki/index.php/PLL

I find it disrespectful that you expect us to give you "less credit" (i.e. time and forum space for special treatment) while we're having a serious discussion.



DavidWoner said:


> Well I view the defining trait of an interleaved corner and edge cycle as the 1x1x2 block that is preserved. Working with F+Y(proving T+V as well is obviously not needed), two of the angles are obvious, as a portion of the 1x1x3 block is preserved. For Fperm(bar in front) y Yperm(standard) and the y2 one, there only one corner that moves to the position opposite its origin, and one edge as well. If you think about the N-perms, it's easy to see how this will always result in the formation of a 1x1x2 block.


Well, the two angles are obvious. I see what you mean about the N-perm, having the edge and corner each move to the opposite side makes sense. But the fact that the two types of overlap produce exactly all the G-perms feels like a truth I have to "get used to" rather than something that just falls out of some intuitive observation.


----------



## Cubenovice (Apr 29, 2011)

OK, perhaps I should have just mentioned "too busy at work to check via online resources" instead of trying to be funny.

I meant no disrespect meant at all.


----------



## DavidWoner (Apr 29, 2011)

Lucas Garron said:


> Well, the two angles are obvious. I see what you mean about the N-perm, having the edge and corner each move to the opposite side makes sense. But the fact that the two types of overlap produce exactly all the G-perms feels like a truth I have to "get used to" rather than something that just falls out of some intuitive observation.


 
I totally agree- it makes sense if you sit down and look at it, but it's certainly not obvious.


----------



## Cool Frog (Apr 30, 2011)

Stefan said:


> Does this look like a G perm to you?
> (R' U R' U' R' U' R' U R U R2) (R' F R' B2 R F' R' B2 R2)
> 
> Or this?
> (R' U R' U' R' U' R' U R U R2) y2 (R' F R' B2 R F' R' B2 R2)


 
This thread destroyed my mind. Cubing has been contributing to some major headaches lately (I think they feel good)
Ah yeah, that might've been a good place. I've noticed that thread but never really looked into it...


----------



## TiLiMayor (Apr 30, 2011)

Stefan said:


> Side product of computing combinations of two PLLs. My results and program:
> http://www.stefan-pochmann.info/misc/PLL_combinations.html - any other interesting things in this table?
> http://www.stefan-pochmann.info/misc/PLL_combinations.txt
> 
> That in turn was my side product of attempting to count PLL attacks returning to solved.


 
I loved the table, now I have a complete way of teaching 4LLL, not only interesting but this is just perfect for me, thank you, ye.


----------



## Stefan (Apr 30, 2011)

DavidWoner said:


> Well I view the defining trait of an interleaved corner and edge cycle as the 1x1x2 block that is preserved. Working with F+Y(proving T+V as well is obviously not needed), two of the angles are obvious, as a portion of the 1x1x3 block is preserved. For Fperm(bar in front) y Yperm(standard) and the y2 one, there's *only *one corner that moves to the position opposite its origin, and one edge as well. If you think about the N-perms, it's easy to see how this will always result in the formation of a 1x1x2 block.



I'd like to emphasize that "only" there a bit more, because the existence of a 1x1x2 block alone isn't enough. It's also needed that all other pieces are moved. With exactly one corner and one edge moving opposite their origin, append a U2 and you have exactly that, those two pieces are back home and all others are not.

Now with exactly three edges and three corners moved, it's clear you have two 3-cycles (no other way to move three pieces), but it could also be a J-perm which also has a solved 1x1x2 block and two 3-cycles (in the same direction, which is why it's not a G).



TiLiMayor said:


> I loved the table, now I have a complete way of teaching 4LLL


 
Hmm, how is that? What did it give you for that that you were missing before?


----------



## oll+phase+sync (May 10, 2011)

Lucas Garron said:


> So, the question is: Is there an intuitive reason the two 3-cycles produced by T+V (or F+Y) should interleave completely?


 
For me it was obvious T+V would work, the moment I understood why T+U don't work.


----------



## Lucas Garron (May 10, 2011)

oll+phase+sync said:


> For me it was obvious T+V would work, the moment I understood why T+U don't work.


Hmm, I think I see it now, but only from thinking about the opposite-swapping adjacent-swapping parts of the algs carefully. (They're interleaved duals, done in opposite order – leading to each possible direction.)

Are you sure you understood exactly why the T+V should work, exactly with that interleaving, and exactly with those opposite directions of permutations?


----------



## oll+phase+sync (May 11, 2011)

Lucas Garron said:


> They're interleaved duals, done in opposite order – leading to each possible direction.
> Are you sure you understood


 I don't understand your words but my simple thoughts are:
1. Thought: If Gx = T + Uy then Mirror(Gx) = T + Mirror(Uy) thats why T and U don't work.
2. Thought: Ga + T produces a 2x2x1 Block and Mirror(Ga) + T also produces also 2x2x1 block both can be solved by V not by Y

So a this Point I knew it can only be V, even so a reasoning for the other two Gs is missing.


----------



## Stefan (May 11, 2011)

oll+phase+sync said:


> 1. Thought: If Gx = T + Uy then Mirror(Gx) = T + Mirror(Uy) thats why T and U don't work.


 
I don't quite understand this. Similarly I could say this:
_If Gx = T + Uy then *Inverse*(Gx) = T + *Inverse*(Uy)_

But that is false:

Gb = T + Ua
*True*: Inverse(Gb) = T + Ua
*False*: Inverse(Gb) = T + Inverse(Ua)




oll+phase+sync said:


> 2. Thought: Ga + T produces a 2x2x1 Block and Mirror(Ga) + T also produces also 2x2x1 block both can be solved by V not by Y


 
Not necessarily. Whether you get a 2x2x1 block depends on the AUF between the algs.
And what about all the other PLLs?


----------



## oll+phase+sync (May 12, 2011)

Stefan said:


> I don't quite understand this. Similarly I could say this:
> _If Gx = T + Uy then *Inverse*(Gx) = T + *Inverse*(Uy)_
> 
> But that is false:
> ...


 
That means T + Ua = Ub + T (+ may contain AUF) ? (no cube at hand)

I used mirroring for my argument because that much easier to predict. When I have Ga / Gc (these two I call mirrors of each other) I can use a T to extend the Headlights to an 3x1x1 block in one case I need a Ua in the other Ub to solve the cube.

For Gb/Gd the 3x1x1 block is build opposite the Headlights, same argument but harder to see.



Stefan said:


> Not necessarily. Whether you get a 2x2x1 block depends on the AUF between the algs.


There is no AUF freedom when I want to solve a G by doing T + V the T must alway break the Headlight. Since V and Y are the only Diagonal PLLs wich do an Edge 3-cycle there is no need to check other PLLs.

EDIT
An obvious argument, why e.g. T + J cannot produce 4 Gs is that T+J sometimes solves all corners and sometimes produce a diagonl PLL

Question:
Aa + Ab = {solved,Aa,Ab,E}
Ua + Ub = {solved, Ua, Ub, Z}

That looks like alwas getting exact 4 Results, (cause there are just 4 AUF cases it can't be more )

But can it be less for essencial, different input algs?


----------



## nerd (May 21, 2011)

that is actually really cool me and my friends make up things like this but not this crazy


----------



## ct5010 (May 27, 2011)

I solved G-Perms with 2 Look PLL just like this!! Just 1 Edge and 1 Corner Cycle.


----------



## Stefan (May 27, 2011)

ct5010 said:


> I solved G-Perms with 2 Look PLL just like this!! Just 1 Edge and 1 Corner Cycle.


 
Which edge-cycle and which corner-cycle?


----------



## vcuber13 (May 27, 2011)

im sure hes meaning 2lpll


----------



## uberCuber (May 27, 2011)

vcuber13 said:


> im sure hes meaning 2lpll


 
I would hope so since it says "2 Look PLL" in his post.


----------



## vcuber13 (May 28, 2011)

apparently, i fail at reading


----------



## ct5010 (May 28, 2011)

Stefan said:


> Which edge-cycle and which corner-cycle?



A and U Perm. (Yeah I know they are actually 3-cycles.)


----------



## Rpotts (May 28, 2011)

ct - you are missing the point of the thread.


----------



## Stefan (May 28, 2011)

ct5010 said:


> A and U Perm.


 
Which A perm and which U perm?


----------



## uberCuber (May 28, 2011)

Stefan said:


> Which A perm and which U perm?


 
when I do lol beginner LBL + 4LLL solves, I always solve G perms with Aa and Ua


----------



## Cubenovice (May 28, 2011)

Same here (since I have forgotten my G's...)
G-perm headlights in back: right hand A-perm gives righthand U (ACW)


----------



## StachuK1992 (Aug 18, 2011)

I think the following alg combinations allow for PLL to be solved in 2 algs by memorizing only 3 algs:
Algs used: 1 9 10
Algs used: 2 5 10
Algs used: 2 5 11
Algs used: 2 11 13
Algs used: 5 12 19
Algs used: 5 19 20
Algs used: 6 7 11
Algs used: 6 7 21
Algs used: 6 12 19
Algs used: 6 19 20
Algs used: 7 13 15
Algs used: 10 11 12
Algs used: 10 11 17
Algs used: 11 12 15
Algs used: 11 17 18

I think this is the most relevant place to put this.
FYI this is very relevant to the OLLCP (hax) thread.
If someone could confirm that these results are correct, I have further plans...

Relevant:
algs = ["",
"R' U R' U' R' U' R' U R U R2", #This is "1"
"R U' R U R U R U' R' U' R2",
"R2 U2 R U2 R2 U2 R2 U2 R U2 R2",
"R2 U R2 U' R2 F2 R2 U' F2 U R2 F2",
"R U' R F2 R' U R' U' R2 F2 R2",
"R2 F2 R2 U R U' R F2 R' U R'",
"R U R' U R' U' R F' R U R' U' R' F R2 U' R2 U R",
"F' U2 R' U F U' F' U' R U' F",
"R U R' F' R U R' U' R' F R2 U' R'",
"R' U2 R U2 R' F R U R' U' R' F' R2",
"F' R' U' R F' R' U F' U' F' U F R F2",
"F2 R2 U' R' U' R F2 R' U R F2 U R2 F2",
"F2 R2 U' F2 R' U' R F2 R' U R U R2 F2",
"R' U' R2 U' R2 U' R U2 R2 F U R' U' R F' R",
"R' F R' U R U' F' R2 U2 R' U R2 U R2 U R",
"F' R U R' U' R' F R2 F U' R' U' R U F' R'",
"R' U R' F R F' R U' R' F' U F R U R' U' R",
"R U R' U' R' F R2 U' R' U' R U R' F'",
"F R U' R' U' R U R' F' R U R' U' R' F R F'",
"F' U2 F' U' R' F' R2 U' R' U R' F R U' F",
"R U' R U F R F2 U F U' F R' F' R'"]


----------



## Godmil (Aug 18, 2011)

what do those numbers relate to?


----------



## StachuK1992 (Aug 18, 2011)

Oh, I'm so sorry, editing post.


----------



## Stefan (Aug 25, 2011)

StachuK1992 said:


> I think the following alg combinations allow for PLL to be solved in 2 algs by memorizing only 3 algs



From a previous discussion:



cuBerBruce said:


> Using a set of three PLL algs, I found that the most you can solve (with no more than 2 alg executions) is 71 out of 72 cases.


----------

