is this consider CHEATING...???
It would be if it worked.
I still call it cheating and I agree that this doesn't work. If you do OLL-parity the centers don't change position like that

can 4x4 parity be avoided by using a different way to pair up the last 2 edges?
Sure. It's just not exactly easy to recognize.
Assuming normal reduction (centers, edges, 3x3x3, parities) you can see if you have OLL-Parity after centers and PLL-Parity after edges. That also means that you can solve/avoid those parities at the end of those stages.

But doing that is only useful for 4x4x4 FMC, not for speedsolving
 
Sure. It's just not exactly easy to recognize.
Assuming normal reduction (centers, edges, 3x3x3, parities) you can see if you have OLL-Parity after centers and PLL-Parity after edges. That also means that you can solve/avoid those parities at the end of those stages.

But doing that is only useful for 4x4x4 FMC, not for speedsolving

ok thanks. :D
 
Help with the 4x4 speed solve

I currently am slow at solving the 4x4 averaging around 6 minutes. I solve the centers, do advanced edge pairing, and the solve like a 3x3. Is there a faster method?

Centers: around 1:30 min.

Edges: around 2:30 min.

3x3: around 2:00 min

Please give me any suggestions on how to get faster.
 
1. Get a better 4x4 because your 3x3 phase is just speed and really depends on corner cutting. Your 4x4 times also depends on your 3x3 times.

2. Look ahead for pieces you want.

3. PRACTICE!!
 
@sir pochmann and @sir avgalen...

thanks for the heads up... i know it is... just wondering cuz no rules has been broken... i might as well cheat myself...

never encounter parity with my old cube... well i will try solving it again more to see if i encounter parity... but since then not yet encounter one... it is on my collection now...

thanks a lot... hehehehehhe... :D:D:D:D
 
There's no parity for the 2x2x2.

By the way, did I say something wrong there, since I have been getting interesting responses and couldn't quite get what's wrong.
 
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I'll explain this now, hopefully I'm correct.

My definition of "parity" is when there is an odd number of swaps necessary to solve a cube (occurs on 2x2, 4x4, Square-1, etc)
 
What if we left out the last 2 or 3 edges (i.e. leave them incorrectly paired), solved the 3x3 normally and then solve the last 2 or 3 edges?

This could be done, although unfortunately if you have parity you will of course still have parity. And if you don't have parity, fixing the edges will be slower... so it's probably not worth it.
 
You'd avoid parities in the sense that you don't have to put any extra work into solving them. (Remember the typical way to solve normal OLL parity is basically a domino 2-corner-cycle.) I think you'd have a hard time reducing it into a domino, though.
 
What if we reduced the 4x4 to a Rubik's domino and fixed it? Would we somehow avoid parities?
What do you mean with "fix it"?

Well, I assume what you mean is to reduce to a domino just like you reduce to a 3x3x3, and then try to solve it. While in the 3x3x3 version you can come across the "OLL parity" and "PLL parity" and thus your "3x3x3" is not necessarily solvable as a 3x3x3, the domino version doesn't have this issue and you can always solve your "domino" just like a domino. So if that's what you meant, the answer is "yes", and I'd like *you* to explain why.
 
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If you want to cheat, count the number of slice turns you make in a normal 4x4 scramble. If that number is:
1) Even
- Solve centers with an even number of slice turns. (even+even=even)
2) Odd
- Solve centers with an odd number of slice turns. (odd+odd=even)

Double turns count as 2.

You can do it without cheating :
You can determine during the permutation if the permutation of the edges is odd or even by counting the number of cycles of the edge permutation (an edge at its place counts as one cycle). Even number of cycles = even permutation.

Then you solve the centers by the way you describe. Well, you only need to solve 3 centers (or 2 not-opposite centers) as the others will be solved using an even number of slice turns (except if you are solving centers strangely !).
Pairing up edges won't mess this up.

The only problem is that it takes a very long time (about 45 sec for me).
 
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I just solved my first 4x4x4 today, so the first time I came across this issue. I decided to try solving without algorithms, just using Heise's cube principals. I managed to find the OLL parity case by counting the number of 'bad' edges (at the end of the edge pairing stage) in a similar manner to the beginning of ZZ. If its odd then a centre slice quarter turn, followed by only an even number of centre slice turns fixes this prob. I had a lot more problems with PLL parity as I was only able to detect it near the end of the solve. :(
 
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@ Stefan: (Yes, I meant solved, sorry I've used the word "fix" as in "repair" ever since I started speedcubing, I'll try and stop using it).

I'll try and answer your question (which I assume you probably have the answer to :)):

1. You can't orient pieces on the rubik's domino
2. You can always permute 2 of the same type of pieces e.g. R2 U2 R2 U2 R2 U2.

I think my idea is not a good one because as Michael basically said, you would have a hard time reducing it to a rubik's domino.
 
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Thanks :)

Ah it seems to me as though there would be too much effort involved to avoid parities; at the moment, I don't think there is a better solution than just solving the parities during the LL stage.

Btw, when is it okay to use the word "fix"? Is it only okay when you talk about parities?
 
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