Odd layer cubes parity

Gafimp

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Hi, I am new to this forum so I don't know if this is in the right place.

My question is, why don't you get parity errors on big cubes with odd layers, like 5x5x5, 7x7x7?
Surely, because the centre pieces are indistinguishable from each other, two centre pieces could be swapped, and you could force a parity error when the cube is into 3x3x3 form, like you get on even layer cubes.
 
..two centre pieces could be swapped, and you could force a parity error when the cube is into 3x3x3 form, like you get on even layer cubes.

Not sure what you mean by that. The center pieces on a 5x5 are fixed to the core. This thread discusses parity. Apparently the way most people refer to parity is a misnomer, and it actually has to do with even vs. odd number of swaps? But, I still don't have a clue what I'm trying to say, so here is yet another thread.
 
Thanks for replies so far.

To my knowledge, a cube must have an even number of "pairs of pieces" swapped. So if a pair of pieces are swapped, another pair of pieces somewhere else on the cube must be swapped too to retain parity. So I guess a 3-cycle is just two swaps.

It is easier to explain with pictures.

nlsdie.png


On this picture, you can see a 4x4x4 and a 5x5x5.

You get a parity error on the 4x4x4 when A and C are swapped, and B and D are swapped. So the cube as a whole is parity free, but if you are solving it as if it's in 3x3x3 form, you cannot do it.

My question is, why can't pieces E and F be swapped, then the group of pieces G swap with the group of pieces H? Is it just that two centre pieces can't be swapped? 8 swaps in total (even number), so it should work.
 
Permutation isn't shared between centres and edges like it is between corners and midges.

EDIT: On a sidenote, this is what I meant when I was referring to parity on a 5x5x5;

IKcvu.png
 
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For your pictures, you can't switch H and G because the middle edges and corners have the same limits as a 3x3, yoiu cant swithc only 2 edges on a 3x3 so you can't on a 5x5 (the middles ones). Also 2 centres can be switched without affecting the rest of the cube.
 
Thanks for replies so far.

...

nlsdie.png


My question is, why can't pieces E and F be swapped, then the group of pieces G swap with the group of pieces H? Is it just that two centre pieces can't be swapped? 8 swaps in total (even number), so it should work.

To answer your question I would like you to consider a scenario. Assuming you did swap pieces E and F, why would that have any affect on the pieces from groups G or H? (Hint #1: it could affect some of those pieces, but not all).
Hint #2: Switching pieces E and F would also affect other pieces that you have not marked on the diagram

Also 2 centres can be switched without affecting the rest of the cube.

This is definitely false, no matter what size cube you consider, and no matter which center piece type you consider. I'm not trying to be harsh or bash you, and in the interest of good discussion, do you realize why your statement is false? To be very clear, I am considering your statement to have been made as if you consider switching two centers on a supercube.

Chris
 
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@cmhardw, is Gafimp wrong because he should have said that "two centers in the same orbit can be switched without affecting the rest of the cube"? Or is that also untrue? I am having similar problems..
 
To answer your question I would like you to consider a scenario. Assuming you did swap pieces E and F, why would that have any affect on the pieces from groups G or H? (Hint #1: it could affect some of those pieces, but not all).
Hint #2: Switching pieces E and F would also affect other pieces that you have not marked on the diagram
Chris

Swapping just E and F is impossible, it would result in a wrong parity on the whole cube wouldn't it? You can't swap just 2 pieces. An odd number of additional swaps would be in the cube somewhere. 3 is an odd number of additional swaps, swap the group of 3 pieces at G with the group of 3 pieces at H?

I assume this works because parity of edges on a 3x3x3 relies on parity of corners, and vice versa.
So then I assume parity of centre pieces on big cubes are dependant on the parity of the edge/corner pieces, is this wrong?
 
Also 2 centres can be switched without affecting the rest of the cube.

This is definitely false, no matter what size cube you consider, and no matter which center piece type you consider. I'm not trying to be harsh or bash you, and in the interest of good discussion, do you realize why your statement is false? To be very clear, I am considering your statement to have been made as if you consider switching two centers on a supercube.

Chris
Yes I realized that after Kirjava said that:
Technically, no.

Technically?

Yes its true I just tried it on a super cube in gabbasoft, and it did a 3 cycle.
Gafimp said:
...You can't swap just 2 pieces... is this wrong?

It would dpensd ion the peice, you can't switch only 2 centres, corners or middle edges. But you can have a 2x2 swap of corners and edges (J perm). You can however swap only 2 wing edges.
 
It would dpensd ion the peice, you can't switch only 2 centres, corners or middle edges. But you can have a 2x2 swap of corners and edges (J perm). You can however swap only 2 wing edges.

It gets more interesting on the larger sized cubes (larger than 4x4). Yes you can swap only two wings, but on a super5x5 or larger this is not the only effect you experience (i.e. more piece types, by necessity, have to change parity with the wings). Maybe you already knew this, but I wanted to include it for completeness for anyone following this thread.

Chris
 
It would dpensd ion the peice, you can't switch only 2 centres, corners or middle edges. But you can have a 2x2 swap of corners and edges (J perm). You can however swap only 2 wing edges.

It gets more interesting on the larger sized cubes (larger than 4x4). Yes you can swap only two wings, but on a super5x5 or larger this is not the only effect you experience (i.e. more piece types, by necessity, have to change parity with the wings). Maybe you already knew this, but I wanted to include it for completeness for anyone following this thread.

Chris
I know, it's very interesting. For a cube size n=4 or greater with r orbits with inner-layer odd parity, there are exactly \( 2r\left( n-2 \right)-4r^{2} \) center-edge pieces and/or obliques which cannot be solved back due to the 2r wing edges out of place caused by the odd permutation. What's interesting about this is it does not matter which orbits have parity, whether they are spaced out or together, etc. Only the number of orbits with odd parity matters. (I have a derivation to prove this strong statement too).

For example for a 1000X1000X1000 cube with 22 random orbits with odd parity, there are exactly 41, 976 center pieces which cannot be solved back (unless the wing edges are fixed back with an odd parity algorithm). But this is nothing compared to the maximum number of center pieces which cannot be corrected on the 1000X1000X1000 for some number of orbits with odd parity: 249, 000.
 
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Why is this still baffling/confusing people??

# of parities depends on the size of the cube, method and partially on parity definition...

Per
 
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