4×4×4 parity error easiest solution?

[alkanova]

hi,I started with the four by four a couple of day ago but sometimes I get parity errors I am not able to solve it,I don't use algorithms I use the-pairing-then-solving-like-a-3×3×3 way. I solve it with a mixture of things and then use commutators to finish, the simplest pattern I got to is solving the whole cube then keeping one whole raw flipped and upside down too like this


1.can someone help me with how to correct it all at once if possible?
(remember I don't use algorithms so please don't use something too long or too complicated... if possible XD )

2.Moreover I saw something like avoiding such errors with counting or such can someone explain this for me, please?

 

[d4m4s74]

You'll need a pretty long alg
I use (Rr)2 B2 U2 (Ll) U2 (Rr)' U2 (Rr) U2 F2 (Rr) F2 (Ll)' B2 (Rr)2

and then you'll have to redo your LL

 

[nitrocan]

You have double parity there.

Hold the cube with orange on top and blue on front. Then do this:
Rw2 B2 U2 Lw U2 Rw' U2 Rw U2 F2 Rw F2 Lw' B2 Rw2

Then rotate the cube so that you will need to swap the UF and UB edges. Then do r2 U2 r2 Uw2 r2 u2

 

[d4m4s74]

You have double parity there.

Hold the cube with orange on top and blue on front. Then do this:
Rw2 B2 U2 Lw U2 Rw' U2 Rw U2 F2 Rw F2 Lw' B2 Rw2

Then rotate the cube so that you will need to swap the UF and UB edges. Then do r2 U2 r2 Uw2 r2 u2

or solve the OLL and PLL parity at the same time

(Rr)2 B2 (Rr)' U2 (Rr)' U2 B2 (Rr)' B2 (Rr) B2 (Rr)' B2 (Rr2) B2

 

[Stefan]

Or you could try to understand parity, simply do r and clean up with commutators.

 

[alkanova]

thanks
what about avoiding the errors?,can someone explain the way for me please? I read a little about it but didn't totally understand

 

[qqwref]

The easiest possible solution is this: just solve it just like the 3x3x2 cube (Domino), which is the same thing as the 3x3x3 except there is no middle layer and you can only turn R2, F2, L2, B2, U, D. One possible solution would be

[insert edge] [domino T-perm] [solve edges]
(R2 Uw R2 Uw' R2) (U [R2 Uw R2 Uw' R2 y' R2 Uw' R2 Uw R2 y] U2) ([R2 Uw2 Rw Uw2 R2] Uw' R2 Uw R2 Uw [R2 Uw2 Rw Uw2 R2] Uw R2 Uw' R2 Uw').

This is all combinations of the intuitive 3x3/Domino algorithms R2 U R2 U' R2 and R2 U' R2 U R2 (insert a corner into bottom layer) and (R2 U2)3 (swap UB and UF) so it should be straightforward to see what is going on.


About avoiding parity completely: there are a bunch of methods people have thought of which involve counting the 'cycles' of the edge pieces (look into 4x4 BLD if you want help with this), but they are way too complex and I don't think anyone has ever seriously used them. If you really want to avoid it, try switching to a method like rachmaninovian's which solves all of the edges and corners first, so that you can fix parity much faster (usually with something short like r2 U2 r U2 r2).

 

[alkanova]

thanks a lot ^^

 

[rjohnson_8ball]

I think the usual way (for those who are used to F2L, OLL, PLL) is to just do the F2L, but before doing anything with the final layer, at this time check to see if you have an odd number of paired edges to flip. If so, do the parity fix, then finish the final layer. For 4x4, the wide turns for L and R make the fix pretty fast.