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

1. Originally Posted by 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.
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.

2. Originally Posted by Sa967St
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.

3. 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.

4. Originally Posted by Xishem
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.

Originally Posted by ~Phoenix Death~
Yeah right.

5. Originally Posted by miniGOINGS
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}".

6. 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. :\

7. Originally Posted by Stefan
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).

8. Originally Posted by riffz
their horrible misinterpretations of what you said
maybe he didn't say it right

lern2communicate

learn from sarah, she knows how to communicate

9. Originally Posted by crashdummy001
maybe he didn't say it right
lern2communicate
?
It was explained very well, some of you just didn't read it properly.

Originally Posted by crashdummy001
learn from sarah, she knows how to communicate
The example shouldn't really have been necessary. :/

10. 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.

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