# 2 4x4 PLL parity's



## martijn_cube (Mar 1, 2008)

i have two 4x4 PLL parity's i can't solve. and i can't find them in the dan harris Doc. the first one i had at first, then i screwed-up and had to solve the cube from half the F3L, and then got parity 2. so no i'm stuck 











tnx


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## Johannes91 (Mar 1, 2008)

What do you mean by "parity"? Seems even in both cases to me. But anyway, here's how I'd solve them:

[[F' U' B' : u2], U2] = F' U' B' u2 B U F U2 F' U' B' u2 B U F U2
[r l : [[R' L' : d2], U]] = r l R' L' d2 R L U R' L' d2 R L U' r' l'


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## martijn_cube (Mar 1, 2008)

well i could not really recon a normal 3x3 pll, so i thought it was a parity then?
can't seem to get F' U' B' u2 B U F U2 F' U' B' u2 B U F U2 to work for the first one.


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## Johannes91 (Mar 1, 2008)

martijn_cube said:


> well i could not really recon a normal 3x3 pll


Because ... it's a 4x4? In the reduction step, before solving it as a 3x3, you should pair all edges.



martijn_cube said:


> so i thought it was a parity then?


What some people call "PLL-parity" means that two edge-groups are swapped. If you want to think that it's a 3x3, it looks like 2 edges are swapped, which is not possible on a 3x3.


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## martijn_cube (Mar 1, 2008)

Johannes91 said:


> martijn_cube said:
> 
> 
> > well i could not really recon a normal 3x3 pll
> ...



that's the problem i thinks this time. normally i solve it te easy way and pair up all the edges first. but this time i solved in completely like a 3x3 from the beginning. so ceters - cross - f3l - OLL - PLL


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## masterofthebass (Mar 1, 2008)

martijn, you are going around this by combining direct solving with reduction. If you use this method, then there's no possibility that you will get OLLs or PLL without doing some commutators to fix the LL.


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## Inusagi (Mar 1, 2008)

You should have done them before the 3x3x3 part. It's step 2. You'll find them in Dan Harris site.


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## martijn_cube (Mar 1, 2008)

yeah i know how to do the normal beginners paring solve, and then end it with 3x3. but this time i wanted to solve it in a different way, so i solved in entirely like an 3x3. and ik works very good until the PLL  now i have a very odd PLL. and i understand now why, but there must be some alg for this right?


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## Swordsman Kirby (Mar 1, 2008)

Reduction + Direct Solving?

Fail.

Second case can be done like:

[r' b' D b D2 r, U2]

I beat Johannes for that one.


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## martijn_cube (Mar 1, 2008)

Swordsman Kirby said:


> Reduction + Direct Solving?
> 
> Fail.
> 
> ...



can you type the full alg for that one? i'm not really familliar with commutators. that is one right?
and small r is only the middel-right slice right? not rR.
my 4x4 is now in postition 1. do you have an alg for that one too?


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## Johannes91 (Mar 1, 2008)

Swordsman Kirby said:


> Second case can be done like:
> 
> [r' b' D b D2 r, U2]
> 
> I beat Johannes for that one.


That's the same as my first, he edited the post to include pictures instead of links and changed the order.

I still think F' U' B' u2 B U F is more elegant than r' b' D b D2 r. 

Edit:


martijn_cube said:


> can't seem to get F' U' B' u2 B U F U2 F' U' B' u2 B U F U2 to work for the first one.


It works for the second one.


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## martijn_cube (Mar 1, 2008)

ow ok, i changed the order yeah.
but do you know an alg for the first one? or a way for me to find it somewhere?


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## Johannes91 (Mar 1, 2008)

Doesn't the second one in my first post work?

r l R' L' d2 R L U R' L' d2 R L U' r' l'


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## martijn_cube (Mar 1, 2008)

sorry, i kinda missed that one  but it works  it's solved now.
but how do you find these algs? because there are not very common PLL's right?


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## Swordsman Kirby (Mar 1, 2008)

Well, in the two main approaches, reduction to 3x3 and direct solving...

Reduction: Those cases never come up.
Direct Solving: If you're good you could avoid those cases 99.99% of the time. (Left the .01% for that being the starting PLL case)


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## martijn_cube (Mar 1, 2008)

ok, so i think i can better use reduction, then direct solving. or find a different way of direct solving.
but about direct solving. we found a fridrich way for the 3x3. but shouldn't there be a direct solve solution like fridrich for the 4x4 and 5x5 etc, do you think?


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## masterofthebass (Mar 1, 2008)

there is... It's called K4, for the 4x4. It's a direct solving method that has been optimized for the 4x4. You can also expand it to the 5x5, but it's a little more extensive.


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## martijn_cube (Mar 1, 2008)

ow ok. great. that's very interesting. do you have a good link? what time do they get with the K4?

edit: i watched a couple k4 tutorials on youtube, and it's almost the same idea as i had on this topic. start with centers, then cross, then F3L then OLL and PLL. so thats just like a normal 3x3, but with a little pairing.


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## Swordsman Kirby (Mar 2, 2008)

martijn_cube said:


> ok, so i think i can better use reduction, then direct solving. or find a different way of direct solving.
> but about direct solving. we found a fridrich way for the 3x3. but shouldn't there be a direct solve solution like fridrich for the 4x4 and 5x5 etc, do you think?



You COULD use just reduction, like most people.


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## Kenneth (Mar 2, 2008)

First case : 2x(Rr2 F2 Rr2 U)

Or: Rr2 F2 Rr2 U Rr2 F2 Rr2 U (if the multiplyer was not understanable =)

I actually use that alg 

I got a huge load of algs to fix edges when direct solving big cubes. If you like you can go to the Y-group and look for a thread "3 step ELL for 4x4x4" or something like that. there I describe my LL method and also list algs for all possible cases.


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## martijn_cube (Mar 2, 2008)

wich yahoo group do i have to look for. i once found a good one, but can't remember wich one it was.
and that's a very nice alg. very easy. 
only, if i do the alg backwards, i'll get a somewhat different PLL.
the blue and green cubie are right, but the purple is one cubie down, and the red is one up. so now they are all on the same spot of one side. so that's a bit different.


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## Kenneth (Mar 2, 2008)

http://games.groups.yahoo.com/group/speedsolvingrubikscube/message/36777


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## martijn_cube (Mar 2, 2008)

thanks.


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