how manny parity cases does a 5x5 have?

[jcuber]

I use bigcubes then freestyle or AVG, and it is possible by slective pairing, to only encounter two unsolved possibilities.

 

[Ton]

With AVG edge-pairing you only encounter 1 of them 50% of the time

How the hell??
only 1??
you mean the two opposite wing change? I don't know but I've gotten the Dw R U R' F R' F R Dw' too. Well I can avoid it but avoiding it takes more time. And I've gotten those one-center-flipped things too.

Yep 1, the single edge flip you will only encounter if it was there from the beginning. Still you can solve it with 1, a little optimization is to use 2 for AvG.

 

[qqwref]

Spoiler

A quarter turn will only toggle parity of a piece that is either in the middle of a face (along that slice) or on an edge, which is why it's the +centers and midges but not the xcenters or obliques that get parity.

Sorry but this is wrong. I put the correct statement in spoiler tags too.

I was trying to only state what happened to parity with regards to cubes, but you're right, I got it a bit wrong. Obliques do indeed get parity on a supercube. (But Chris, wouldn't that mean there are then 8 orbits of pieces which have a parity on the 7x7x7 supercube? Or did I read your question wrong?)

 

[cmhardw]

(But Chris, wouldn't that mean there are then 8 orbits of pieces which have a parity on the 7x7x7 supercube? Or did I read your question wrong?)

Ha, yeah it does. I totally put my foot in my mouth on this one. I apparently don't know how to exponentiate.

Here is the formula I use to determine the number of distinct parity scenarios on an n x n x n supercube. This comes from studying the parities of the larger cubes when coming up with my formulas on this page.

Spoiler

For every natural number n where n > 1
There exist 2^floor(n/2) distinct parity scenarios of the pieces on the n x n x n supercube. By floor I mean the function that outputs the greatest integer less than or equal to the argument.

So for example if n = 2 then we are looking at the 2x2x2 supercube = 2x2x2 cube. 2^floor(2/2) = 2^1 = 2 parity scenarios. These scenarios are:
1) Corner parity is even
2) Corner parity is odd

By parity scenario I mean considering all possible combinations of even and odd parity when considering all orbits of pieces together.

Using that formula you should get 8 scenarios for the 7x7x7 supercube. I was totally off, and apparently can't multiply. I also do have a really easy way to determine all 8 scenarios. All 4 scenarios have been listed in this thread for the 5x5x5 supercube. I should have phrased the question better, but blah more or less said the answer I had in mind. Here is the answer I was more or less looking for, but did not word my question well enough for:

Spoiler

The 4 parity scenarios for the 5x5x5 supercube are:
1) All piece types have even parity
2) corners, midges, x-centers, t-centers have odd parity ; wings have even parity
3) t-centers, wings have odd parity ; corners, midges, x-centers have even parity
4) corners, midges, x-centers, wings have odd parity; t-centers have even parity

Also sorry if I left any names out in my list of those who could answer easily. This forum is too big to make such a list quickly now. I was trying to do a fun puzzle, and in doing so did a miserable fail. :-S

Chris

 

[qqwref]

Spoiler

The 4 parity scenarios for the 5x5x5 supercube are:
1) All piece types have even parity
2) corners, midges, x-centers, t-centers have odd parity ; wings have even parity
3) t-centers, wings have odd parity ; corners, midges, x-centers have even parity
4) corners, midges, x-centers, wings have odd parity; t-centers have even parity

Ah, I see what you meant now. The number of parity scenarios for a given odd cube is indeed what you said.

 

[cmhardw]

Using reduction method solving, you can experience the following number of parity scenarios on the n x n x n cube (non-supercube)

Spoiler

For every even natural number n
There exist 2^(n/2) possible parity scenarios when solving the even n x n x n cube (non-supercube) using the reduction method.

For every odd natural number n, where n > 1
There exist 2^[(n-3)/2] possible parity scenarios when solving the odd n x n x n cube (non-supercube) using the reduction method.

------

To clarify the 8 cases on 6x6x6 would be
1) outer wing parity, and PLL parity
2) inner wing parity, and PLL parity
3) both wing types have parity, and PLL parity
4) only PLL parity
5) outer wing parity
6) inner wing parity
7) both wing types have parity
8) no parity

And the 4 cases for 7x7x7 would be:
1) inner wing parity
2) outer wing parity
3) both wing types have parity
4) no parity

Problem solved ;-) The 5x5x5 has 2 parity situations. Either wings have parity, or they don't. Also the 4x4x4 has 4 parity situations, no parity, OLL parity, PLL parity, both OLL and PLL parity.

Chris

P.S. Does anyone have a way using only the floor and ceiling functions to condense these into 1 formula? I can see how to do it using modular arithmetic, but that's boring and easy

 

[mrCage]

Parity Cases

'USING THE CAGE METHOD AVOIDS THESE PARITIES. THESE PARITY ALGORITHMS ARE ONLY FOR THE REDUCTION METHOD FOR BIG CUBES'

Well it avoids THOSE cases (normally). But other ones DO occur as pointed out already

Per

 

[qqwref]

Spoiler

For every even natural number n
There exist 2^(n/2) possible parity scenarios when solving the even n x n x n cube (non-supercube) using the reduction method.

For every odd natural number n, where n > 1
There exist 2^[(n-3)/2] possible parity scenarios when solving the odd n x n x n cube (non-supercube) using the reduction method.

P.S. Does anyone have a way using only the floor and ceiling functions to condense these into 1 formula? I can see how to do it using modular arithmetic, but that's boring and easy

It's ugly, but you can do it like this:

Spoiler

2^(3*floor[n/2] - n)