how manny parity cases does a 5x5 have?

[arckuss123]

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

thanks

 

[Stefan]

I estimate about 2.1 * 10^71.

 

[hr.mohr]

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

 

[Ton]

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.

 

[AvGalen]

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

 

[blah]

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

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.

So this was what I did: Defining the last two midges as solved, we have 4! = 24 cases. And after many many many hours of work, I've categorized them into 8 groups.

A

B

*

*

C

*

*

D

E

F

G

H

All cases within the same group can be transformed into the asterisked case either by doing a U2, or by flipping one of the tredges. You never have to do both, i.e. U2 and flip one of the tredges. And to flip one of the tredges, you need an average of exactly 2 moves, you can see how I arrived at this number later on.

So in other words, excluding the solved case, there are only 7 algorithms needed for the last two tredges, and there isn't even a need to mirror any of them. I suck at mirroring on the fly (be it L/R or F/B), so this is definitely good for me.

To flip one of the tredges:
0 moves: 1 case
1 move: 3 cases
2 moves: 13 cases
3 moves: 5 cases
That gives us an average of (0+3+26+15)/22 = 2 moves.

Now, AvG versus bigcubes.com. With the bigcubes.com tredge-pairing method, the probability of ending up with each case is obvious, you just have to count the number of cases in each group:
A: 1/3 (easiest, intuitive case )
B: 1/6
C: 1/6
D: 1/12
E: 1/12
F: 1/12
G: 1/24
H: 1/24

With AvG, it's a little different: Cases C, D, E, F and G can only occur if they were already like that right from the start, i.e. once the centers were done. So the combined probability of cases C, D, E, F and G occurring is 1/(24*23) + (1+4+2+2)/(24*23*22*21) = 0.00185, that's 0.2%. (Someone double check this for me please!)

So actually, the only cases that ever appear are A and B (and of course H). Solving A is intuitive. So you only need to learn one alg to solve B if you're doing AvG tredges.

If you get cases C, D, E, F or G, consider yourself unlucky and do the alg for either A or B to reduce the case to A or B. There really isn't any point in learning algs for these cases because they occur so rarely that you don't really get to practice them at all and you'll probably take longer to recall them than to do 2 algs.

Besides, for cases like D and G, you probably would've recognized them while pairing tredges and fixed them while it was still easy.

 

[mrCage]

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.

And what if i do NOT pair edges ?

Per

 

[AvGalen]

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.

And what if i do NOT pair edges ?

Per

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

 

[TMOY]

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.

 

[rachmaninovian]

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.

And what if i do NOT pair edges ?

Per

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

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..

 

[Paul Wagner]

I think there is only one.

 

[cmhardw]

[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]

 

[mrCage]

[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]

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

Per

 

[Novriil]

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.

 

[cmhardw]

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

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

 

[Robert-Y]

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

 

[Ethan Rosen]

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

 

[qqwref]

[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]

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.

 

[TMOY]

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.

 

[blah]

@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