Avoiding parity errors in 4x4x4?

[ZamHalen]

Okay I'm going to give my original idea now even though i didn't like it because it involves partial cage(which I know very little about).The first step would be building a 3x3x4 block(getting only the centers and edges required)then the edges would be solved then put in their solved positions.the last part would be finishing the centers like in cage method (which I know almost nothing about).Then solving the rest 3x3x3.

Spoiler

predicted answer: it won't work you noob you know nothing about
4x4x4...blahblablah

 

[TMOY]

It doesn't work, sorry. You can avoid parity issues with pure cage because before you solve the last edges one of the slices is still entirely free; as soon as you have solved two adjacent centers this is not true anymore.

 

[whitelynx_nivla]

I believe there is, but I dont think that methode has been discovered for now. Will just have to experiment it with yourself.

My theory is (just a theory) there's might be some clue from the corner's position. Because those are the one that you want to put in the right direction when assembling the cubes....But then again..maybe I was wrong

 

[joey]

I believe there is, but I dont think that methode has been discovered for now. Will just have to experiment it with yourself.

My theory is (just a theory) there's might be some clue from the corner's position. Because those are the one that you want to put in the right direction when assembling the cubes....But then again..maybe I was wrong

>_>

 

[SuperStrawberry]

Unless you want to cheat then there is no way to avoid them, that's way they don't allow more than one logo on any cube in the WCA.

How does more than one logo help you cheat?

 

[SenileGenXer]

You need to know the algs.

You can avoid the edge pairing parity error - the first of the three kinds of parity error on a 4x4x4. Your not saving yourself that much time and the alg isn't that hard.

You avoid the edge pairing parity error by counting down edge pairs and watching exactly how many your solving. No edge pairs solved means that there are 12 pairs yet to do. Count down until you have just 3 pairs left to do and you don't have a parity problem. 3 pairs left to do can all solve into each other. Lose count or go to fast and have 2 pairs left to do; that is the first parity case.

I have a video I'm contemplating uploading about this, it's got some bad camera work and the worst example I could possibly get where at 5 pairs left to go I run into 3 pairs that could all solve eachother leaving me with the parity problem. It would have been faster to take those 3 then do the alg.

All three parity cases for the 4x4x4 seem to be made and locked in during the edge pairing process. If you could find the right rules for edge pairing it might be possible to avoid all of them but the OLL and PLL problems are tuff to spot at this stage.

 

[cubernya]

Oll is set during centres, pairing "parity" during edges, and Pll "parity" (quotes cause it's an X perm) durig edges(?)

 

[SenileGenXer]

In simplification when your solving edge-pairs you are splitting and reconnecting centers as well. Does this count? I think your baking in every parity problem during edge pairing and they could somehow be solved during pairing. Maybe fast maybe not fast and maybe not with one a classical edge pairing alg. It could be solved then, solved with far fewer moves because whatever alg/technique don't have to put the first 3 layers back together at this point. The OLL and PLL algs happen at a later point and have to put the first 3 layers back together in addition to solving their respective problems.

Edit: I think your right. Working backward I took a solved cube and couldn't induce a OLL error with just 3-cycle paring moves. PLL errors can be made with 3 3-cycle or 3 2-cycle pairing moves.

 

[aronpm]

You're wrong, SenileGenXer.

I actually do avoid OLL parity in my 4x4 solves, and it's completely unrelated to edge pairing. OLL parity is the only parity on the 4x4, and it's caused by odd/even number of quarter slice turns. If a scramble has 13 quarter slice turns, and you solve two adjacent centres with 8 quarter slice turns, you will have OLL parity. Anything after any two adjacent centres will be solved with *slice* *something* *unslice* which will always be an even number of quarter slice turns which won't change your parity. If you are avoiding OLL parity, you do it during centres.

As for "edge pairing parity", that's not a parity at all. It's only a parity on 5x5+, and then, it's actually OLL parity.

PLL parity is also not parity because it can be solved using 2 3-cycles, but yes it is determined during your final edge pairing.

 

[SenileGenXer]

Thanks. I realized on my own I was wrong about OLL parity - only. I really do appreciate the complete description. How do you avoid OLL parity if you aren't told or allowed to see the scramble?

I've heard cubers call edge pairing issues parity. You know exactly what I'm talking about. I think this is what the OP was talking about so the response was to him.

Parity as a word is jargon. It is highly specialized. The purpose of jargon is to exclude outsiders. Always seems sh***y to be a grammer Nazi especially on the use of jargon.

 

[aronpm]

How do you avoid OLL parity if you aren't told or allowed to see the scramble?

Inspect the parity of the edge cycles during inspection, a la BLD memo. It's tough ^^;

 

[Mike Hughey]

Inspect the parity of the edge cycles during inspection, a la BLD memo. It's tough ^^;

Wow, that sounds hard. Not only do you have to figure out the parity plus figure out your first few center solving moves - you also have to be able to figure out on the fly how to solve all centers without introducing parity. I did 4x4x4 fewest moves for a while back when we had it on the weekly competition, and sometimes it took me 5 to 10 minutes to figure out a decent way to solve centers without introducing parity. I guess it's true that I was going for fewest moves at the time, but is there a simple algorithmic approach for solving 4x4x4 centers with the proper amount of middle slice turns - something that's easy and doesn't require much thought?

(I'm figuring there probably is, and I'd love to learn it - it would be really fun to start doing this for 4x4x4 speedsolving, if I could do it practically!)

 

[Tim Reynolds]

Wow, that sounds hard. Not only do you have to figure out the parity plus figure out your first few center solving moves - you also have to be able to figure out on the fly how to solve all centers without introducing parity. I did 4x4x4 fewest moves for a while back when we had it on the weekly competition, and sometimes it took me 5 to 10 minutes to figure out a decent way to solve centers without introducing parity. I guess it's true that I was going for fewest moves at the time, but is there a simple algorithmic approach for solving 4x4x4 centers with the proper amount of middle slice turns - something that's easy and doesn't require much thought?

(I'm figuring there probably is, and I'd love to learn it - it would be really fun to start doing this for 4x4x4 speedsolving, if I could do it practically!)

I've never tried this, but it shouldn't be too hard I don't think, using normal center-solving methods. You just have to keep a running count of what the parity is, and switch it on every slice turn. You can stop after the third center, at which point you've probably only done like 20 moves, so it shouldn't be too hard to keep track for that long.

The second center should be easy since you don't have to count all of the times you do Rw U* Rw'. Those have two slice turns and don't affect the parity.

Once you're done with the third center, you have two adjacent centers, so your work is done. If the parity is bad, do Rw U2 Rw U2 Rw' (cancel the first few moves though. Seriously, you should be paying close enough attention to not have to do that whole sequence).

Then just don't do anything stupid on the last three centers or edge pairing

aronpm can probably explain in a more practical manner, since he actually does this. It seems to me that the hardest part is counting the edge parity; then again, I can't sub-1 memo a 4x4.

 

[Henrik]

I guess you have not seen this:

As Tim said: if you have parity you can avoid/fix it during your first centers.
Tim assume that you solve two opposite centers, then a 3rd.
So a rule of thumb: two adjacent centers fix/ or set the state of your parity.
This means that once you have two adj centers solved the core is set in a configuration that determines the rest of the centers thus maybe having parity.

So if you solve the first adj centers to avoid parity, it will also determine the rest of your solve, so to say.

Tim's approach is great.
Count quarter slice turns during your first centers, do the opposite.
If you know your cube has parity, then you have to solve the 3rd center so that to quarter slice turns become odd.
3 q-turns for 1st center makes it odd, then you have to solve the 3rd in an even number.
2 q-turns for 1st center makes it even, then the 3rd has to be odd!

I hope someone understood what I wrote!

 

[Mike Hughey]

Once you're done with the third center, you have two adjacent centers, so your work is done. If the parity is bad, do Rw U2 Rw U2 Rw' (cancel the first few moves though. Seriously, you should be paying close enough attention to not have to do that whole sequence).

I don't seem to be following this; maybe it's because I'm seeing it slightly differently. But I can see that if I hold the third center on top (not sure if that's what you meant), with the first and second centers on left and right, then I can do Rw U2 Rw' U2 Rw, and that would swap the parity. And again, it should be possible with practice to cancel most of that. Anyway, I see now that you'd only have to count through the first three centers, and then just a few moves that could be fairly braindead could fix parity, so yeah, I see how that works pretty well. I'm definitely going to try it!

aronpm can probably explain in a more practical manner, since he actually does this. It seems to me that the hardest part is counting the edge parity; then again, I can't sub-1 memo a 4x4.

I'm looking forward to trying this; I figure it might help me get a little faster at 4x4x4 BLD memo in the process.

Edit: I've tried it a few times now and it's worked, and yes, it's really quite easy to make the centers come out, now that I understand it. Now I just have to get my 50 second inspection time (for reasonably GOOD cases) down to sub-15.

Edit2: Even if it does take too long in inspection, it really is an amazing feeling to see that many solves in a row with no OLL parity! My best inspection time now is 32 seconds - still a long way to go.

 

[cubernya]

How do you guys trace the cycles so quickly? Do you just do a one pass and visually remember which piece you've already done?

 

[Mike Hughey]

How do you guys trace the cycles so quickly? Do you just do a one pass and visually remember which piece you've already done?

I'm obviously not fast enough (well, not yet), but that's basically what I do, yes. I trace the first cycle; if I'm lucky either the count is 23 (I don't count the buffer), meaning I'm done and I need to fix parity, or it's slightly less than that and I can see enough solved pieces to add up to 23. If that doesn't happen, then I have a standard order of wings that I use for 4x4x4 BLD as my "next buffer pieces"; I go through them in order and see if I remember them being in the cycle. If not, I start there for the next cycle; each added cycle increases the needed total amount by 1 (counting the new buffer twice). The 8 wings on the top of the cube are my first next alternate buffers; if I can manage to find all the cycles without running out of them, I can do fairly well at remembering whether those are done or not, but I'm not perfect. It's cases where I have to start extra cycles from pieces that are not on top where I really get in trouble; somehow I can remember the ones on top better than the others.

 

[Tim Reynolds]

I don't seem to be following this; maybe it's because I'm seeing it slightly differently. But I can see that if I hold the third center on top (not sure if that's what you meant), with the first and second centers on left and right, then I can do Rw U2 Rw' U2 Rw, and that would swap the parity. And again, it should be possible with practice to cancel most of that. Anyway, I see now that you'd only have to count through the first three centers, and then just a few moves that could be fairly braindead could fix parity, so yeah, I see how that works pretty well. I'm definitely going to try it!

Oh, yeah, I forgot to specify: third center on front when you do those moves I listed. But any variant that involves breaking the center, slicing, and putting it back works.

 

[Mike Hughey]

Oh, yeah, I forgot to specify: third center on front when you do those moves I listed. But any variant that involves breaking the center, slicing, and putting it back works.

Okay, now it makes sense. And after trying it for a while, I see how it usually involves more like one or two extra moves. It really is quite easy, now that I understand it.

But I do find that it's really hard for me to average much better than a minute inspection time! Aron is beast!

It sure is fun trying, though; this is the most fun I've ever had doing 4x4x4 speedsolving. It's just so cool when the cube never has OLL parity!