why parity?

[Kirjava]

I'll rephrase it.

Why is the 4x4 OLL parity? Anybody?

I think you accidently a word or something.

It's caused by building the centres in the wrong places.

 

[Kirjava]

What algorithm do you use to solve parity in human kociemba, Kirjava?

There are a few. N perm works.

I think the optimal alg for it would be very short, but I have no idea how to work it out.

 

[freshcuber]

Thanks, fixed the post.

So the placement of individual center cubies causes OLL edge parity? How does the permutation of the centers affect misorientation of the edges?

 

[Kirjava]

Actually, the real thing that creates OLL parity is an odd number of inner slice quarter turns. It's just that after centres are done the 'parity' of number of quarter turns on inner slices is fixed.

Protip; the dedge isn't misoriented, the two adjacent edges are swapped.

 

[DeCubeRob]

lol, i have a feeling for how the 3x3 works, but by reading all this i do understand more and more on how the 4x4 works!
i always found a 4x4 hard to understand from a 3x3 perspective, but my perspective is changing now, i see that things are linked,
besides i never though of linking a center to an edge before. hmm interesting lol

i hope i wrote it so you understand lol.
atleast i know what i meen.

 

[Christopher Mowla]

So the placement of individual center cubies causes OLL edge parity? How does the permutation of the centers affect misorientation of the edges?

If you are asking about the 4x4x4, the permutation of the center pieces does not affect the permutation of the wing edges.

It has been said time and time again on these forums that the reason we can be certain of this fact is because a single inner layer quarter turn does 2 4-cycles of the 4x4x4's center pieces in that slice. For example, if we did r' to a solved 4x4x4 cube, the center pieces (commonly referred to as x-center pieces) are permuted in the following manner:

​

Note that each of the two 4-cycles are an odd permutation of the x-center pieces. Two of them makes an even permutation of the x-center pieces. (For example, 2 x odd number = even number.)

My point is, the only way center pieces on a big cube cause an odd permutation in the wing edges is if they too have an odd permutation in them. Of course, I have to be careful to mention that x-center pieces never cause an odd permutation in the wing edges of a big cube, but other center pieces can.

For example, you can have only two x-center pieces swapped on a 4x4x4 (an odd permutation of the x-center pieces), but it does not cause OLL parity. Rather, it simultaneously is in an odd permutation with a minimum of two corners.

Here is an algorithm I made which swaps just two x-center pieces and two corners on a 4x4x4:
r2 y r U l' u2 r U' 2R L2 x' 3d' 2R2 3d L' F' L 3d' 2R2 3d L' x U m U r2 y' (29q/23h)
I made it for this thread.
​

​

Now for the 5x5x5 and larger supercubes, if you solve back all non-x-center pieces to their original positions (which can easily be seen on a supercube) before the edge pairing process, then you are guaranteed to not have an odd permutation in the wing edges. Just as the corners are in an odd permutation simultaneously with an odd permutation of x-center pieces (the above example), wing edges and non-x-center pieces are in an odd permutation simultaneously. (You can see why by analyzing the cycles a single inner layer quarter turn does to the non-x-center pieces on a big supercube larger than the 4x4x4).

Of course, this does not apply to the 6x6x6 and higher order even cubes which are bandaged 4x4x4s: they behave just like the 4x4x4.

And, maybe you have realized by now that, for the 3x3x3, edges and corners are in an odd permutation simultaneously (e.g. T-Perm), along with at least one (but an odd number of) center facelet twisted 90 degrees, which can be seen on a 3x3x3 supercube.

 

[freshcuber]

If you are asking about the 4x4x4, the permutation of the center pieces does not affect the permutation of the wing edges.
...

...okay?

I appreciate your response, it was clearly well thought out and structured I just didn't understand most of it.

 

[Zbox95]

Since the 3x3x3 puzzle is considered the "normal" puzzle and the original, ALL cases on a 3x3x3 is considered normal for a cube. With a puzzle the 4x4x4 you can get parity on the 3x3x3 step since it wont look like it would on 3x3x3. Plus, you can't get LastLayer parity on 5x5x5 and 7x7x7 cubes since they have fixed corners and edges, like 3x3x3 cube, however, you can get edgepairing parity whan pairing up edges.

That's all! Bye

 

[300SpartanX]

For any cube, only multiples of 2 edges are flipped. In 4x4 OLL parity, 2 edges are flipped.

 

[vcuber13]

thats not true at all, you cant flip edges on a 4x4
and parity is usually thought of as a 2 cycle (or 4 or whatever)

 

[5BLD]

Basically, when you solve a 4x4x4 with reduction, there are some cases, called pariy, that you cannot reach on a regular 3x3x3. This is due to pairing centres up wrongly in the first step, I think. The parity algorithm swaps two edges, and two centres, or something like that. You cannot flip a wing in place- and the edges on a 4x4x4 are all wings. The algorithm swaps the two edges that make up the misoriented dedge, but there also has to be another swap to make it an even number, which are the centers because there is mo visible change.
(I might be very very wrong but this is just what I think...)

 

[Stefan]

This is due to pairing centres up wrongly in the first step

there also has to be another swap to make it an even number

Nope. Two wings can be swapped without anything else changing.

 

[MaeLSTRoM]

Nope. Two wings can be swapped without anything else changing.

But the "normal" parity algs also swaps 2 sets of 2 centres?
Do (F l' F' [Parity] F l F') to see what I mean.
So you can't use standard parity on a super-4x4x4?

 

[Stefan]

So you can't use standard parity on a super-4x4x4?

U2 "fixes" it, no?

 

[MaeLSTRoM]

Didn't think of that...

On a 6x6x6 can the OLL parity be on both sets of wing edges?
Same for PLL?

 

[Stefan]

On a 6x6x6 can the OLL parity be on both sets of wing edges?
Same for PLL?

Like when you simulate a 4x4x4 on a 6x6x6 (1-2-2-1 layers) and get 4x4x4 parities?

 

[MaeLSTRoM]

Like when you simulate a 4x4x4 on a 6x6x6 (1-2-2-1 layers) and get 4x4x4 parities?

I meant can it occur on either set of edges individually of each other

 

[Stefan]

Ah. Yes, for "OLL parity" just do 2R or 3R to induce the parity, then fix everything but two wings with 3-cycle commutators. And "PLL parity" isn't really parity in the first place (just two swaps).

 

[Christopher Mowla]

...

Based on what you wrote, and not about a concept that you could be thinking, did you even attempt to read my former post in this thread?

Nope. Two wings can be swapped without anything else changing.

You are definitely correct, but maybe you should tell him that this is true only because you solve back everything on the 4x4x4 super except for those two wings in order for this to be possible.

Although what he stated is incorrect and conflicted with my previous post, I believe he might be thinking about another concept he is not yet aware of (but will be after this post, if I can manage to be clear enough). He might be thinking about what optimal pure algorithms for non-supercubes do to a cube rather than thinking about what a series of algorithms do to a cube. Even with supercube pure two wing edge swaps (for the 4x4x4 only, of course), some x-center pieces must be messed up during the odd permutation part of the algorithm (.i.e., with the extra quarter turn).

@5BLD,
As an example besides trivial U2 fixes (to correct the x permutation of x-center pieces by certain single edge "flip" algorithms

Spoiler

Not all algorithms which make an x-permutation of the x-center pieces in the same face need to be single slice turn based. For example, Bw' R' u Rw' l' Uw Rw' Uw' l Uw y Lw Dw r' Dw' Lw' y' Rw Uw' Rw u' R Bw is wide turn-based single edge "flip" algorithm I created which does an x-permutation of the x-center pieces in the top face of a 4x4x4. For more details about it, see this post in my Methods Thread (note that the algorithm is in SiGN notation near the end of that long post)

), here is one TMOY showed Chris Hardwick in another thread:
U l U' l' U' l' U l U l' U' l' U' l D x' d2 l' d2 l x U D'

Of course to make this the briefest 4x4x4 supercube pure single edge "flip" algorithm it can be, since that is the specific 2-cycle of wings being discussed right now, as I stated in another thread:

if we shift the base algorithm and then conjugate, we have two half turn moves less:

x' r2 U2 l' U' l D x' d2 l' d2 l x U2 D' l U' l' U' l' U l U l' U r2 x (23 btm)

But analyzing the base/2 opposite wing edge swapper, as I later stated in the thread that TMOY showed it in,

U l U' l' U' l' U l U l' U' l' U' l D x' d2 l' d2 l x U D' = [U l U' l' U', l'] (l') [l', U'] [U' D: [f2, l']]

It swaps two wings and permutes 6 X-center pieces (with a 2 3-cycle),
[U l U' l' U', l'] (l') [l', U']
and then restores the 6 X-center pieces with a very brief 2 3-cycle [f2, l']
, whose set-up moves, U' D, (to move all X-center pieces affected into slice 2L) partially cancel with the first piece!

This algorithm's odd permutation part, i.e. [U l U' l' U', l'] (l') [l', U'], is quite genius, as it promotes a very brief (I would think optimal, but I cannot claim that) pure algorithm for the 4x4x4 supercube, but even it cannot get away with messing up X-center pieces first.

 

[uberCuber]

2x2 doesnt really have a parity, the parity you come across is not actually a parity in terms of edges... as there are no edges.

Do a U move, and try to solve the corners using 3-cycles.

a 3x3x3 cant have a parity as it comes with the rules of the rubiks cube

lolwut