The One-Answer Question Thread [closed]

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cmhardw said:
miniGOINGS said:
You just gained a whole lot of respect from me. The reason for my question, I was wondering if I should learn PLL so that if I ever got a corner orientation skip, I would have a 1LLL solve with ZZ. Thanks.
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If you think it's worth it, 1/27 chance is not too terrible. Plus full PLL is only 21 algs, which is not too bad considering the chance of being able to use one of those cases.

To give you an idea, if you do an average of 5 in a competition you have a
1-(26/27)^5 = 17% chance approximately of getting at least one corner orientation skip in your average.

Chris
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Wow, ok.
 
I am planning to order a type f ii and a type d ii along with some other cubes, do you think I should order an old type A core to put in the d ii as some people suggest, or just leave the original hardware in there? Also, is buying a new storebought, (because my original broke) and ordering some cube4you screws and a core and put in some random centers a good idea?
 
In a place called Rubikland:cool:. A place where all cubers meet. A big ass fight to be tough to solve the cube faster than anyone.:p So far 1 cuber can beat all of you that easily. With the new features choose your cube. C4U, DIY, Rubik's, Diansheng, or even those crappy cube at dollar store:cool:. Choose your player:). Then you choose your opponent:D. This battle is fill with (swine flu):eek: cubes that you labor for
the rest of your life:(. Don't just leave yet, those cube will just push you down from cubeland to noobland :eek:(since your not fast enough to speedcube the cube.).

Are you ready for the challenge, or you be willing to become a noob?:mad:
 
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Been going through my head for a while now: how are there only 493 ZBLL cases (1 is solved)? I get 12 edge permutations per COLL, 40 COLLs, and 21 PLLs. 12 * 40 + 21 = 501.

I've thought of any ways edge cycles could overlap, and I got a reduction on 2 H sets with U2. That eliminates 1 U * 4 positions + 1 Z per case, or 10 algs. But that would yield 491 algs.

Also, for those cases, you would have to recognize that a U2 is needed, making it, in reality, another case.

Why are there only 493 ZBLL cases?
 
eastamazonantidote said:
Been going through my head for a while now: how are there only 493 ZBLL cases (1 is solved)? I get 12 edge permutations per COLL, 40 COLLs, and 21 PLLs. 12 * 40 + 21 = 501.

I've thought of any ways edge cycles could overlap, and I got a reduction on 2 H sets with U2. That eliminates 1 U * 4 positions + 1 Z per case, or 10 algs. But that would yield 491 algs.

Also, for those cases, you would have to recognize that a U2 is needed, making it, in reality, another case.

Why are there only 493 ZBLL cases?
Click to expand...

I'm pretty sure Cubes=Life was saying he has some tricks that cuts down on the cases, also Mirrors and inverses.
 
For ZZ, in which order should I locate bad edges for EOline? U layer, D layer, E layer? F, S, B? M, L, R?
 
Got one for you.

Is there a sensible algorithm for an F permutation that does not require any axis rotations? IE no X, Y, or Z?

Kinda random, but I can't find one...
 
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