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

[Forte]

I thought this was cool ):

 

[Lucas Garron]

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.

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]

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]

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]

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]

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]

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).

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]

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]

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]

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]

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)

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]

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)

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.

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]

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

 

[ct5010]

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

 

[Stefan]

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]

im sure hes meaning 2lpll

 

[uberCuber]

im sure hes meaning 2lpll

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

 

[vcuber13]

apparently, i fail at reading

 

[ct5010]

Which edge-cycle and which corner-cycle?

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

 

[Rpotts]

ct - you are missing the point of the thread.