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

[Stefan]

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]

And?

 

[Hershey]

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

 

[Stefan]

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

 

[sa11297]

yeah that is pretty cool

 

[Xishem]

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]

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]

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~]

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

 

[Zubon]

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]

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]

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]

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]

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.

I always knew about this

Yeah right.

 

[Stefan]

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]

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]

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.

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]

their horrible misinterpretations of what you said

maybe he didn't say it right

lern2communicate

learn from sarah, she knows how to communicate

 

[Sa967St]

maybe he didn't say it right
lern2communicate

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

learn from sarah, she knows how to communicate

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

 

[Athefre]

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.