emolover
Legend
That is a fake. It is just a 3x3 with cubies glues to the top and bottom sides. Don't even try wasting your money on that piece of $hi!. I would sigguest getting a cube4you version because it is the real thing.
The One-Answer Question Thread [closed]
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That is a fake. It is just a 3x3 with cubies glues to the top and bottom sides. Don't even try wasting your money on that piece of $hi!. I would sigguest getting a cube4you version because it is the real thing.
Phlippieskezer
Veteran
Bump. I mean slice turns (i.e. r, l) counting as one move, by the way.
Oh, and FYI, I asked if any one knows. "No" could be a valid answer to this question.
Edit: If possible, how about OBLBL's standard movecount as well?
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hic0057
Veteran
K4 is 120 moves on average.
JasonK
Premium Member
And where do you get this number from?
hic0057
Veteran
Godmil
Premium Member
'No' would only be a true answer if you knew that nobody knew. In which case it would also be a very informative answer.
MostEd
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Actually not, I was talking to kir the other day, and I and he made this wiki article look like something, then he cleaned it up and yeah its not fake
iseldoff
Regular
Damm! Ayway I have a c4u 3x3x7 but I was pretty unhappy with that seeing as how it doesn't Shape shift. Also I can't even solve it because all the tutorials on you tube require you to know how to solve the cube4you 3x3x5 first and I don't have it.so I can't figure it out. I have the same problem with my square 2 so if anyone wants to trade a squaire 1 for a never solved square 2 I'm down! Just message me and we will figure it out.
I ordered the fake silver one anyway so instead of ruining a good cube to make a extended one and go threw all that work I just got this but anyway
Does anyone know were I can get a fully proportional fully functional 3x3x5 or 3x3x7 for a decent price?
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Phlippieskezer
Veteran
'No' would only be a true answer if you knew that nobody knew. In which case it would also be a very informative answer.
Well, there are some variables to consider. i.e. How decent is the cuber at commutators (K4)? Which edge pairing technique is used?
Will these make major differences in movecount? (Obviously don't assume the cuber is a complete nub at commutators, because that would make everything fly through the roof >.>)
I don't want "no" as an answer, but for all I know, it's the true one.
Also - should have checked the wiki before asking... My bad. But thanks for the info, hic0057.
Jaycee
Veteran
What causes OLL and PLL parity on a 4x4? I think OLL parity has something to do with the number of turns when you're doing Reduction or something like that, but I could be wrong.
5BLD
Legend
The fact that there are two pieces in each dedge.
Tao Yu
Veteran
Its the number of inner layer turns. If it is odd the an odd parity will happen.
OLL parity is the swap of two edges
Do a 2R on the 4x4.
If you ignore the centres, you see that a 4-cycle of edges ha been done.
A 4 cycle can be made into 2 cycle and therefore OLL parity.
The centres can all be solved without effecting the edges
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Jaycee
Veteran
Thanks, I understand OLL parity now. What about PLL parity?
Tao Yu
Veteran
Do R2u2R2u2R2u2
The centres can be then solved
I made a video about a different way to solve OLL and PLL parity.
If you want to really want understand how parity works though, check out 4EverTrying's channel. He knows more about parity than anyone else on earth probably.
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Phlippieskezer
Veteran
"PLL Parity" technically isn't parity (AFAIK). You're switching edges in an "X" permutation (assuming they're opposite). It's a 2-2 cycle.
Edit:
Sorry, I don't understand this statement. You're turning a 4-cycle into a 2-cycle?
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Jaycee
Veteran
@Tao I'm confused. How is that the cause of PLL parity? :/ I feel stupid >.<
EDIT : Oh, I get it. Thanks Phlipp and Tao!
EDIT : Oh, I get it. Thanks Phlipp and Tao!
5BLD
Legend
You are transposing them in a sort of X shape as Phlippieskezer said. The cause of PLL parity in 4x4 and not in 3x3 is due to inner slices.
Tao Yu
Veteran
Here is a commutator PLL parity
u'[R'FRF'RU'R':2U 2D]U 2D'
Basically you are using the 2 flip at the end of reduction to swap one of the pairs of pieces in the x perm.
Then you are bringing in the other two pieces of the x perm into the flip position, and you do the flip backwards, so it flips the two pieces, and it solves the rest of the cube.
Then AUF(ADF as well, come to think of it)
Sorry if you can't understand me or don't understand commutators
u'[R'FRF'RU'R':2U 2D]U 2D'
Basically you are using the 2 flip at the end of reduction to swap one of the pairs of pieces in the x perm.
Then you are bringing in the other two pieces of the x perm into the flip position, and you do the flip backwards, so it flips the two pieces, and it solves the rest of the cube.
Then AUF(ADF as well, come to think of it)
Sorry if you can't understand me or don't understand commutators
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uberCuber
Legend
What he means is that a 4-cycle can be turned into a 2-cycle by just doing a 3-cycle (which doesn't change parity).
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