# how manny parity cases does a 5x5 have?



## arckuss123 (Jul 31, 2009)

i was just wondering that how many parity cases are there on a 5x5 rubiks cube.

thanks


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## Stefan (Jul 31, 2009)

I estimate about 2.1 * 10^71.


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## hr.mohr (Jul 31, 2009)

http://bigcubes.com/5x5x5/lastedges.html


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## Ton (Jul 31, 2009)

hr.mohr said:


> http://bigcubes.com/5x5x5/lastedges.html



A bit depending on your edge paring method how many you will encounter, all is in the link above as Mads wrote

For AvG you will encounter 2 , where you normally you can do with 1 (edge flip/swap), as the others (I only encounter a single edge flip) can be prevented in previous steps.


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## AvGalen (Jul 31, 2009)

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


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## blah (Jul 31, 2009)

Two: Odd and even 

No, obviously that didn't answer the question you had in mind (notice I didn't say your question).

I've found that people tend NOT to click on links when provided, so I've decided to make it a habit to just quote the entire post, if you want to see the entire thread, click on the little arrow next to "Originally Posted by blah". Anyway, hopefully this'll answer the question you had in mind 



blah said:


> So I finally got my V5 3 days ago  Turned out to be better than I expected
> 
> Anyway, I came across a couple of sources (including cutex.info, bigcubes.com, Erik's video, etc.) claiming different numbers of cases for the last two tredges: one said 2, another said 7, another said 12, another said 15... Point is, I was completely confused.
> 
> ...


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## mrCage (Jul 31, 2009)

Ton said:


> hr.mohr said:
> 
> 
> > http://bigcubes.com/5x5x5/lastedges.html
> ...


 
And what if i do NOT pair edges ?

Per


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## AvGalen (Jul 31, 2009)

mrCage said:


> Ton said:
> 
> 
> > hr.mohr said:
> ...


According to http://www.speedsolving.com/wiki/index.php/Parity :


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


 
I think Per should Pair


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## TMOY (Jul 31, 2009)

Well, the PLL parity alg (more exactly its faster, non-center-preserving version) is part of my system for the last 4 edges. But it's not a real parity case, just one of the permutation cases I can encounter.


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## rachmaninovian (Jul 31, 2009)

AvGalen said:


> mrCage said:
> 
> 
> > Ton said:
> ...


nah cage does not avoid parity. say for the LL, it is possible to get the "oll parity". is that not a parity? or, it is also possible to get the parity whereby the FUr and BUr edges need to be swapped. cage does not avoid parity, and the parity algorithms can be used for cage (in the pure form). even in the system that I use I get parities. and I do HATE parities :O though they are part of the 15 cases for opposite dedges I know..


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## Paul Wagner (Jul 31, 2009)

I think there is only one.


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## cmhardw (Aug 2, 2009)

[thread hijack]
There are 4 distinct scenarios considering the parities of all piece types on a supercube 5x5.

Can anyone correctly guess all 4? (Stefan, Lucas, Johannes, I know you guys would know this). Let's give others a chance for a time, then you guys can answer ;-)

--edit--
:fp
Oops! I made a major error in saying the supercube 5x5 had 6 distinct scenarios, it's actually only 4. Bonus points, though, to anyone who can list all 6 scenarios on the supercube 7x7 (the error I made to get that super 5x5 had 6 cases, I was thinking of the 7x7). By the way, there is a simple method to determine all 6 scenarios too, but I'm sure some of you already know that ;-)

Chris
[/thread hijack]


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## mrCage (Aug 2, 2009)

cmhardw said:


> [thread hijack]
> There are 4 distinct scenarios considering the parities of all piece types on a supercube 5x5.
> 
> Can anyone correctly guess all 4? (Stefan, Lucas, Johannes, I know you guys would know this). Let's give others a chance for a time, then you guys can answer ;-)
> ...


 
And exactly why wouldn't i know it also ?? 

Per


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## Novriil (Aug 2, 2009)

AvGalen said:


> 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.


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## cmhardw (Aug 3, 2009)

mrCage said:


> And exactly why wouldn't i know it also ??
> 
> Per



Per I know you would know this too, it's just that those others I mentioned have been more active in the past. I know you've been very active on the forum as of late, so you fit into that group too 
;-)

I say give others a day or two, then you guys can post 

Chris


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## Robert-Y (Aug 3, 2009)

Erm I'm hoping this isn't a stupid guess but here's my guess:

1. Two corners
2. Two wings
3. Two inner corners
4. Two inner edges


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## Ethan Rosen (Aug 3, 2009)

That's a tough one Chris
The only super 5x5 I've solved in a loooooong time is the sheperd's 5x5, so I'm probably way off, but my guess is:
2 x-centers
A single center group rotated by 90 degrees (there's absolutely no way this is right, but it happens on the sheperd's 5x5)
Two + centers
Two edges

These are all pretty stupid, but I wanted a guess


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## qqwref (Aug 3, 2009)

cmhardw said:


> [thread hijack]
> There are 4 distinct scenarios considering the parities of all piece types on a supercube 5x5.
> 
> Can anyone correctly guess all 4? (Stefan, Lucas, Johannes, I know you guys would know this). Let's give others a chance for a time, then you guys can answer ;-)
> ...



Here we go. I think it's OK for me to post because you didn't list my name  but still I will put it in spoiler tags to protect people who want to puzzle it out themselves.



Spoiler



The (supercube) 5x5x5 has "4" parities - that is, there are 4 types of pieces which can get a parity:
1. corners
2. midges
3. wings
4. +centers
but as it turns out #1 and #2 are equivalent, and #3 and #4 are equivalent. What I mean by that is that if you fix parity problem #1, parity problem #2 is automatically solved, and vice versa. (So, if you are solving a supercube 5x5, you actually never get the wing-transpose parity! Neat, huh? And a supercube 4x4 doesn't have +centers, so you can still get (both kinds of) parity after the centers are solved.) Normally the first set is only considered to be a parity in BLD, though.

Similarly, for supercube 7x7:
1. corners
2. midges
3. outer wings
4. outer +centers
5. inner wings
6. inner +centers
and similarly #1 and #2 are equivalent and so on.

And about that "simple method" - just consider the possible slice moves  For instance, for the supercube 5x5, you have R and 2R. (Assume the centers are fixed, so M is impossible. If you don't fix centers then you could consider an M move to have created parity on centers and midges, and it gets more complicated.) An R turn toggles parity of corners and midges, while a 2R turn toggles parity of wings and +centers. 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.


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## TMOY (Aug 3, 2009)

qqwref said:


> 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.


Spoiler



A quarter turn performs 4-cycles on all the pieces it involves. It will toggle parity of a piece if and only if the number of 4-cycles of that particular piece is odd. More precisely:
- on a 5^3, a face querter-turn will toggle parity of corners, midges, +centers and xcenters (there are 4 of each) but not wings (there are 8). An inner-slice quarter-turn will toggle parity of wings and +centers (4 of each) but not xcenters (8).
- on a 7^3, a face quarter-turn will toggle parity of everything but wings (about oblique centers, don't forget there are 2 distinct orbits), a slice quarter-turn will toggle parity of the wings and +centers it involves and of both orbits of oblique centers but not of xcenters.


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## blah (Aug 3, 2009)

@Chris, I'm not quite sure what you mean, or what kind of answer you're looking for, but I'm gonna try my luck anyway, and hopefully I understood you correctly and don't make myself look completely stupid 



Spoiler



Like I said, I don't exactly know what you mean by "4 distinct scenarios considering the parities of all piece types". But this is what I think it means: 4 distinct scenarios in which a 5x5x5 supercube can't be solved using only 3-cycle commutators of each piece type.

Each piece type on a solved cube has even permutation parity.

A quarter face turn causes the following to have odd parity: Corners, midges, x-centers and t-centers.

A quarter "non-middle slice" (don't know how else to name it) turn causes the following to have odd parity: Wings and t-centers.

So you can end up with one of the following four scenarios in a 5x5x5 supercube BLD solve if you use ONLY 3-cycle commutators of each piece type:
1. 2 corners, 2 midges, 2 x-centers and 2 t-centers unsolved (quarter face turn to fix parity)
2. 2 wings and 2 t-centers unsolved (quarter slice turn to fix parity)
3. 2 corners, 2 midges, 2 x-centers and 2 wings unsolved (quarter wide turn to fix parity)
4. All pieces solved

So was this the answer you were looking for? I haven't read the other spoilers so I don't know if my answer is completely off


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## jcuber (Aug 3, 2009)

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


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## Ton (Aug 3, 2009)

Novriil said:


> AvGalen said:
> 
> 
> > With AVG edge-pairing you only encounter 1 of them 50% of the time
> ...



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.


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## qqwref (Aug 3, 2009)

TMOY said:


> qqwref said:
> 
> 
> > Spoiler
> ...


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?)


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## cmhardw (Aug 4, 2009)

qqwref said:


> (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


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## qqwref (Aug 4, 2009)

cmhardw said:


> Spoiler
> 
> 
> 
> ...



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


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## cmhardw (Aug 4, 2009)

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


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## mrCage (Aug 4, 2009)

> *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


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## qqwref (Aug 4, 2009)

cmhardw said:


> Spoiler
> 
> 
> 
> ...



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


Spoiler



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


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