4x4 Parity Algorithms

[kunparekh18]

I don't think that's how it should be solved. First all centers, then all edges, then 3x3 stage.

 

[Username]

I don't think that's how it should be solved. First all centers, then all edges, then 3x3 stage.

Unless you're using K4 or any other method

 

[kunparekh18]

Unless you're using K4 or any other method

I should have specified. I meant reduction.

 

[CubezUBR]

ok, i kind of made up that method because it was what i saw on bob burtons cubewhiz.com. what method do you recomend then?

 

[Username]

ok, i kind of made up that method because it was what i saw on bob burtons cubewhiz.com. what method do you recomend then?

Yau

 

[CubezUBR]

I'm curious, what do you average with that?

about 2 mins 30 secs. i had a 4x4 for 2 days now

 

[Gordon]

That is fast for 2 days...

 

[qqwref]

This is not PLL parity, but actually OLL parity - you can see that you only have to switch two edge pieces. The fact that all the yellow stickers on top is actually irrelevant.

PS: Your method is fine, feel free to keep using it if you like it.

 

[Kirjava]

This is not PLL parity, but actually OLL parity

it hasn't been reduced to 3x3x3 - this is the edge permutation parity, of which OLL parity is an example.

 

[qqwref]

That's just a difference in terminology. I like to call all wing permutation parities "OLL parity" because it's short without being ambiguous. (Nobody could think I'm talking about PLL parity, which is the important thing to me.)

 

[Kirjava]

That's just a difference in terminology. I like to call all wing permutation parities "OLL parity" because it's short without being ambiguous.

Just using 'parity' is non-ambiguous in a non-reduction context and is shorter.

OLL Parity could mean a specific case in this context, and is ambiguous.

 

[Christopher Mowla]

Besides

R B (r U2 r U2 F2 r F2 l' U2 l U2 r2) B' R'

, if you want an alternate algorithm, this might be the best I have come up with for speed:
x r U' R' U' r U R U M' U2 r U2 r' U2 l U2 r2 x'

For a large list of algorithms for this case (well the mirror of your specific case), see this.

EDIT:
I just noticed you can insert an R move which could make this faster without adding moves (when applying to the 4x4x4).
x r U' R' U' r U R U (M' R) U2 r U2 Rw' U2 l U2 r2 x'

 

[Christopher Mowla]

Besides, if you want an alternate algorithm, this might be the best I have come up with for speed:
x r U' R' U' r U R U M' U2 r U2 r' U2 l U2 r2 x'

I just found these three which are related algorithms.
x r' U2 r' U2 M U' R' U' r' U R U r U2 l' U2 r2 x'
Lw' U2 r' U2 (M R') U' R U' r' U R' U r U2 l' U2 Rw2 x'
x Rw R U' R U' r U R' U M' U2 r U2 r' U2 l U2 Rw2 x'

I think the first alg listed in this post is even faster than the alg in my previous post (maybe the inverse is faster).