# Square-1 parity problem / parity states description



## bcube (Nov 28, 2015)

Hi,

according to Chris´ calculations and Christopher´s calculations, I perceive the 3x3x3 cube as having 43 252 003 274 489 856 000 / 2 parity problems and the same number of parity problem free positions. There are two parity states for 3x3x3, namely (in Christopher´s notation) C(3,0,0) and C(3,0,1). Similarly, for 4x4x4 there are 4 parity states: C(4,0,0), C(4,0,1), C(4,1,0) and C(4,1,1). 

Now, I am wondering if it is also possible to describe parity problems and parity states this way in case of square-1? Even in cube shape alone it seems to be kind of complicated because of a combination of both "slash" move (which behaves as R2 on 3x3x3 cube, i.e. always an even permutation of corners and even permutation of edges) and "numbered" moves (which behave as quarter moves on 3x3x3, i.e. sometimes an even permutation for both corners and edges and sometimes an odd permutation for both corners and edges)...


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## bobthegiraffemonkey (Nov 28, 2015)

For the cube shape, there are two clear, well-defined parity states, it's just less simple to change the parity than it is for a 3x3 or a 4x4. For most other shapes, there are two clear parity states, but the labeling of these as even/odd is arbitrary (but such an arbitrary labelling is useful for a system which solves cubeshape into a state of even parity, which I developed and use). For the shapes with 6 corners on one layer, I would argue that it is not sensible to have two distinct parity states, but perhaps someone will disagree with me.


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## bcube (Nov 29, 2015)

I am not sure about it but shouldn't each shape have 4 parity states? Namely

- both corners and edges are in even permutation
- both corners and edges are in odd permutation
- corners are in odd permutation, edges are in even permutation
- edges are in odd permutation, corners are in even permutation

The first two options are not called "parity problem" by speedcubers, although following Christopher's and Chris' idea, only the positions which meet the first option criterion shouldn't be called parity problems.

There is also a catch how to define parity problem in a different shape than cubeshape. Would it be possible to simply compare two states in one particular shape (first state would be a reference and second state in the same shape (after aligning U and D layers, if needed) would be scrambled) to get either first or second option mentioned above (i.e. a situation which is not called a parity problem by speedcubers)?

Numbered moves and slash move would have different meaning in terms of parity problem (as introduced by Christopher and Chris) in different shapes then, so they cannot be generalized (like saying that R2 is always an even permutation for both corners and edges in case of 3x3x3. I cannot simply say that slash move is always an even permutation for both corners and edges in case of square-1).

In case of NxNxN cubes, half of the positions are in odd permutation and the second half is in even permutation. Can it be also said for a square-1?


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## bobthegiraffemonkey (Nov 29, 2015)

Parity on sq-1 is rather tricky, and how you choose to define it in each situation (and also which situations you choose to define it) will likely depend on why you are considering parity. I have worked out the details for what I need, but you're probably looking at parity for a different reason so you might prefer a different approach.



bcube said:


> I am not sure about it but shouldn't each shape have 4 parity states? Namely
> 
> - both corners and edges are in even permutation
> - both corners and edges are in odd permutation
> ...


I suppose so. at least for the shapes where this makes sense. It really depends on the context: for doing sq-1 BLD for example it would be helpful, and I assume it's something that you're interested in too. Personally, I'm only concerned about the total parity, rather than the parity of edges and corners individually, so the first two options are equivalent (even parity) and so are the last two (odd parity).



bcube said:


> There is also a catch how to define parity problem in a different shape than cubeshape. Would it be possible to simply compare two states in one particular shape (first state would be a reference and second state in the same shape (after aligning U and D layers, if needed) would be scrambled) to get either first or second option mentioned above (i.e. a situation which is not called a parity problem by speedcubers)?


This is exactly how it should work in my opinion, but obviously the choice of reference is arbitrary for each shape (that isn't cubeshape).



bcube said:


> Numbered moves and slash move would have different meaning in terms of parity problem (as introduced by Christopher and Chris) in different shapes then, so they cannot be generalized (although is there a shape in which slash move doesn't toggle a permutation of either corners or edges?).


For most shapes, I think it makes more sense to define parity only with regards to the shape of each layer, ignoring how they are rotated. For the cases with 6 corners on one layer this is different, and you could meaningfully define parity depending on how this layer is rotated (and in the one case where the other layer has 180 degree rotational symmetry, it can also be done for that layer to some extent), which is exactly why in the way I define parity there is a problem for these shapes.



bcube said:


> In case of NxNxN cubes, half of the positions are in odd permutation and the second half is in even permutation. Can it be also said for a square-1?


Yes, for any sensible definition there will be the same number of even parity states and odd parity states.


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## bcube (Nov 29, 2015)

bobthegiraffemonkey said:


> Parity on sq-1 is rather tricky, and how you choose to define it in each situation (and also which situations you choose to define it) will likely depend on why you are considering parity. I have worked out the details for what I need, but you're probably looking at parity for a different reason so you might prefer a different approach.



Long story short, I call a situation in which at least one piece orbit is in odd permutation a parity problem. In case of NxNxN cubes, these parity problems can be grouped into several parity states. All parity states but C(n,0,0) represent at least one piece orbit in odd permutation. I would like to have similar system for other puzzles too, for example square-1. Currently I have a trouble to distinguish between "both corners and edges are in odd permutation" (odd permutation by the definition on which I am operating) and "corners are in odd permutation but edges are in even permutation" (also an odd permutation by the definition on which I am operating) when I try to describe some position.

For NxNxN cubes, C(n,w,c) function and parity states work well for every position which can be reached. I would like to have such a general system for square-1 too (based on the mathematical meaning of the parity of permutation). 



bobthegiraffemonkey said:


> I suppose so. at least for the shapes where this makes sense.



I fail to see how this (i.e. some of the shapes) is optional. I would like to have a general system, describing every position available.



bobthegiraffemonkey said:


> It really depends on the context: for doing sq-1 BLD for example it would be helpful, and I assume it's something that you're interested in too.



Not as much as about how to describe which position can be called a parity problem ;-) 



bobthegiraffemonkey said:


> Personally, I'm only concerned about the total parity, rather than the parity of edges and corners individually, so the first two options are equivalent (even parity) and so are the last two (odd parity).



In that case it leads to a conclusion that on 3x3x3, one quarter move is an even permutation (which I think is in conflict with a statement that half of its positions is in even permutation and half of its positions is in odd permutation). In my first post can be seen that I am not operating on this definition - but that is another story. 



bobthegiraffemonkey said:


> This is exactly how it should work in my opinion, but obviously the choice of reference is arbitrary for each shape (that isn't cubeshape).



Well, I would say it is not completely arbitrary - it should lead to a cubeshape in which a permutation of corners and a permutation of edges are equal, right? From that point, square-1 can be described in a way as 3x3x3 cube I think, slash move being R2, but still looking for a good expression of an equivalent of quarter move on 3x3x3 (especially when something like (1,-1) is made). 



bobthegiraffemonkey said:


> For most shapes, I think it makes more sense to define parity only with regards to the shape of each layer, ignoring how they are rotated.



Hmmm, it makes sense to me. On the other hand, a rotating of U and D layers cannot be ignored completely beacuse they need to be aligned in a way that slash move is possible to make. Several sensible "cases" by this aligning can be made and it would be nice to tell about all of them if it is a parity problem or not. 



bobthegiraffemonkey said:


> For the cases with 6 corners on one layer this is different, and you could meaningfully define parity depending on how this layer is rotated (and in the one case where the other layer has 180 degree rotational symmetry, it can also be done for that layer to some extent), which is exactly why in the way I define parity there is a problem for these shapes.



As said above, I would like to have a tool for describing all positions available, similarly to NxNxN cubes.



bobthegiraffemonkey said:


> Yes, for any sensible definition there will be the same number of even parity states and odd parity states.



A "parity state" is probably understood differently by us (I hope I use this term in agreement with Christopher and Chris), but you probably meant this "I'm only concerned about the total parity, rather than the parity of edges and corners individually, so the first two options are equivalent (even parity) and so are the last two (odd parity)". I understand it as you saying "number of positions which are in odd permutation is equal to number of positions which are in even permutation". Not that I don´t believe you, but could you please defend/prove this statement? I think it would require to tell about each available position if it is even or odd permutation (which is my goal).


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