Supercube centers and odd parity

[Christopher Mowla]

When solving a 5x5 supercube, I realize that if the cube has odd parity, the centers cannot be solved completely unless I first complete all of the edges with an odd parity break. For example, when executing an odd parity algorithm to a solved supercube, the center affected may or may not be able to be solved.

By experimentation, I find that if odd parity exists in the edges on the

4x4 supercube (all center pieces can be solved)
5x5 supercube (all but 2 center pieces can be solved),

etc.

Is there a general formula which states how many center pieces cannot be solved for any size cube if odd parity exists in the edges or can it only be found by experimentation?

The reason I ask this is because supercubes actually represent what happens to regular cubes when an odd parity algorithm is performed on them. In actuality, we are not really “solving” color cube centers. We just group the same colors together, but, when doing this same technique on a super cube, the cube will not be solved (the arrows will be pointing everywhere, even though they are of the same color).

 

[Lucas]

In 5x5 there is no parity. In 4x4, there is.

So in a (2n+1)^3 cube, you will only be able to have all but two centers solved if you have to swap two wings.
In a (2n)^3 cube, you can have all but two centers solved.

My first and second sentences are the most important.

 

[mrCage]

When solving a 5x5 supercube, I realize that if the cube has odd parity, the centers cannot be solved completely unless I first complete all of the edges with an odd parity break.

By experimentation, I find that if odd parity exists in the edges on the

4x4 supercube (all center pieces can be solved)
5x5 supercube (all but 2 center pieces can be solved),

etc.

Is there a general formula which states how many center pieces cannot be solved for any size cube if odd parity exists in the edges or can it only be found by experimentation?

Hi. I do not quite understand what you are describing. You may end up with a situation with a face whose centers cannot be solved independently of the other faces. Like a swap of centers on 2 faces. No need to rearrange the edges or whatever!! I guess it depends on your center solving strategy/method.

Per

 

[Christopher Mowla]

"In 5x5 there is no parity. In 4x4, there is."


Lucas,
I am saying if a cuber (for experimentation) does an odd parity algorithm to a solved 5x5 supercube, then the center distorted cannot be fully solved unless the edges are fully solved back. And yes, there is parity on a 5x5 supercube. Maybe the reason you never come across it is because you complete all of the centers prior to completing the edges (or vice versa), which is a good and smart way to solve a 5x5 S, but I am talking about the theory here, not just solving a cube in a certain way....

 

[qqwref]

In 5x5 there is no parity. In 4x4, there is.

Of course the 5x5 has parity.


cmowla: On large supercubes, each slice (and thus each set of wings) can have even or odd parity. The parity of a given set of wings is always equal to the parity of the corresponding set of +centers. Thus, on an odd NxNxN supercube, with (N-3)/2 possible parities, if you solve the centers the edges will not have any parity, and vice versa. Also, if you almost solve the edges, and they have a total of k parities, then you will be able to solve all of the centers with 3-cycles except for k sets of +centers (which you cannot get closer than a 2-cycle off on).

However, on an even NxNxN supercube, there are (N-2)/2 possible parities but the +centers are not there at all. Hence the position of the centers has nothing to do with the parity of the edges. That is, the centers are solvable with 3-cycles no matter what position the edges are in, and if you have solved the centers the edges may still have OLL parity on any or all of the possible layers.

So the answer to your question of "on an NxNxN supercube, if the edges are solved except for one 2-cycle on each set of wings, what is the minimum number of centers that we can have unsolved?" is: if N is even, 0; if N is odd, it's two times the number of slice layers, or (N-3).

 

[Christopher Mowla]

So the answer to your question of "on an NxNxN supercube, if the edges are solved except for one 2-cycle on each set of wings, what is the minimum number of centers that we can have unsolved?" is: if N is even, 0; if N is odd, it's two times the number of slice layers, or (N-3).

Thanks for your response. But, what if, for example, you do an odd parity algorithm to a solved 6x6x6 supercube, but only to one set symmetrical slice.

XXCCXX or XCXXCX instead of XCCCCX.

I know that there is more than 0 center pieces which cannot be solved with 3 cycles (commutators). I know for XCCCCX, all centers can be corrected, but not if it is just one set of inner layer slices. And this is different for all cube sizes and the number of symmetrical rows which an odd parity algorithm is executed on.

Also, look at your post at http://www.twistypuzzles.com/forum/viewtopic.php?f=8&t=12917


The proof, etc. has always got my attention on supercube centers (actually at the time I first saw that post, I did not even know how to solve a super cube).

 

[qqwref]

Ah, right, on the large ones parity happens to the obliques too. In that case since there are two sets of obliques you can get at best 4 pieces off.

In general, on a supercube larger than 5x5, the number of sets of obliques that are off (each set of obliques corresponds to some set of two different slice layers, and each set of obliques has two orbits of centers) is the number of slice layers with parity times the number of slice layers without parity. So on an NxNxN cube, if k slice layers have parity, out of m, then there are 2*(k)*(m-k) + k*(N mod 2) orbits of centers with odd parity, and you must have at least 4*k*(m-k) + 2*k*(N mod 2) pieces unsolved.

I guess this means that the most center pieces we can be left with is, for N going from 6 to 15: 4, 6, 8, 12, 16, 20, 24, 30, 36, 42. I'm not sure how to get a general formula out of this

 

[Christopher Mowla]

I appreciate your input, qqwref!

Do you know anyone who can? Has it ever been done before? This is like a serious math problem isn't it? It hurts my brain! Who ever figures that out should be given the "cube noble prize"! Don't you agree?

 

[Lucas Garron]

Do you know anyone who can?

Yes.

Has it ever been done before?

No.

This is like a serious math problem isn't it?

Not really.

It hurts my brain!

Yes.

Who ever figures that out should be given the "cube noble prize"!

No.

Don't you agree?

I don't agree, and I'm pretty sure no one who has an idea how to go about it would agree.

 

[qqwref]

Here, I'll do it. Look at the possibilities for 4*k*(m-k), this is at a maximum for one or two central values of k (depending on whether m is even or odd) and then drops off sharply (with at least a change of 4 for each increment/decrement of k) from there. Since the other term only goes up by two per increment of k, one of the two central values for k always yields the largest number of centers. If N is odd and there are two central values the later one is correct, otherwise any will work. Let's assume N>=6 then there are 4 cases (depending on N mod 4):
- If N = 4x, there are 2x-1 slice layers, and since N is even the expression is 4*k*((2x-1)-k). This is the largest for k=x or k=x-1 and it doesn't matter which is chosen - the number is 4*x*x-1.
- If N = 4x+1, there are 2x-1 slice layers. The expression 4*k*((2x-1)-k) is maximized for k=x or k=x-1, but since N is odd the k=x value is better, so the number is 4*x*(x-1) + 2*x.
- If N = 4x+2, there are 2x slice layers, and since N is even the expression is 4*k*(2x-k). This is the largest for k=x so the number is 4*x*x.
- If N = 4x+3, there are 2x slice layers. The expression 4*k*(2x-k) is maximized for k=x, and since N is odd our number is 4*x*x + 2*x.

This sequence of numbers goes up nicely by 2*x every time, so the number is (written in one way, there are probably other ways to write it):
4 * floor(N/4)^2 + 2 * floor(N/4) * ((N mod 4) - 2).

 

[Christopher Mowla]

This sequence of numbers goes up nicely by 2*x every time, so the number is (written in one way, there are probably other ways to write it):
4 * floor(N/4)^2 + 2 * floor(N/4) * ((N mod 4) - 2).

Thanks MG.

 

[Christopher Mowla]

This sequence of numbers goes up nicely by 2*x every time, so the number is (written in one way, there are probably other ways to write it):
4 * floor(N/4)^2 + 2 * floor(N/4) * ((N mod 4) - 2).

Thanks MG.

But...
your fomrula does not have the number of rows affected by odd parity in it. Doesn't that matter as well as the degree of the cube?:fp

(Click the attachment)

 

[qqwref]

I haven't looked at the math in quite a while (what's with you bumping all these old threads lately?) but I think my formula was supposed to represent the maximum over all possible values of r (for a given n).

 

[Christopher Mowla]

I haven't looked at the math in quite a while (what's with you bumping all these old threads lately?) but I think my formula was supposed to represent the maximum over all possible values of r (for a given n).

No. I did not mean the maximum, but for all values r and all degrees n.

For example, on the 9x9x9,

if there are 6 rows affected by odd parity, then 6 center pieces that cannot be solved back.
if there are 4 rows affected by odd parity, then there are 12 center pieces that cannot be solved back.

 

[qqwref]

No. I did not mean the maximum, but for all values r and all degrees n.

Actually, yes, that's what my formula is, unless I made an error. Anyway, when you ask "how many center pieces cannot be solved" the most reasonable interpretation is that you're looking for the extremal value, i.e. the largest number of centers you can be left with after solving as many pieces as possible with 3-cycles alone. If you want something more specific you have to ask for it (at the time, not a month later).

Since you've been an ass before I'm not going to help you out with finding the "general" formula, although it should be pretty obvious if you just take a few seconds to think about it. It's probably already posted in the topic.

 

[Christopher Mowla]

Since you've been an ass before I'm not going to help you out with finding the "general" formula, although it should be pretty obvious if you just take a few seconds to think about it. It's probably already posted in the topic.

Come on. I thought you were over that. Besides, this thread was after that and I didn't do anything (and it's not like I meant to "do anything" before: I was having an intellectual conversation with Chris until two members (not to mention names) intruded and started hell. Why? Because I told them that their reasoning behind their algorithms was not intuitive (to me at least). Then one guy made sense out of it to me. That's when I said that,hey, I have found a 24q. Then S kindled the flame. From there it went down hill. And, the mistake I made was trying to defend myself. Even Chris wasn't happy about their reactions. So be reasonable on my integrity). In this thread, I thought you and I were having an intellectual conversation. That's what I wanted in the other thread, but the others ruined that. What I initially implied about you I erased and apologized because I mistook you to be like the rest. I respect you. I didn't mean to be unclear in this thread. I really thought you knew what I meant. Sorry I took so long...I didn't mean to upset you.

And I didn't bump up the thread on "can you detect odd parity before cube is solved". Someone else did and then I replied.

As far as your formula goes, it is correct for the maximum amount (I have tested it by experimenting with various cube sizes).

No it isn't obvious to me, and no it is not in the thread anywhere.

 

[Christopher Mowla]

What puzzles me the most qqwref, is that, up to this point (after I “was an ass”) you freely helped me by answering all of my questions in this thread, as well as the 3X3X4 software without complaining, but now all of a sudden decide to cut off the communication? Why? Why now do you claim that you will not help me because of a misunderstanding in the past? There is no plausible transition to your change in alliance.

Is my question in this post not worth your intellectual energy? I think it is a good question, and, if you feel you have the capacity to find the formula, why hold that knowledge from the theory section? You contradict your own reaction to the very statement which you are most likely referring to me as a donkey because I was implying that I knew something that others didn’t…and I wasn’t going to share it freely. Well, you are doing the same thing here, only, you are name calling as well.

I don't know how or why I ever offended you...you started the dispute by stating something like, "...you [me] shouldn't be on this site if you are going to put others down." The problem with that assumption was, I wasn't putting anyone down: I was merely stating that computers and the brightest algorithm finders have not reached the optimal algorithm for the one edge flip odd parity algorithm (in quarter turn moves at least). THAT'S ALL I SAID! WHAT IS WRONG WITH THAT? HOW IS THAT DOWNING ANYONE? Why does that not make me fit to be on this site? The question I should be really asking you is do you only approve criticism and challenge from your over achiever peers? Don’t take me for a fool: I have seen some members say very harsh things (and no, I am not referring to you) (and not just saying harsh things to me, but other members too) and yet, they get away with it (you don’t tell them anything at all). However, I state a mere fact and am treated like I condemned speedsolving.com

I admit, you do appear to be a very intelligent being, based on your accomplishments, detailed and lush responses to other speedsolving.com members, and the mere fact that many other respected cubers regard you highly. And, if you are, then you should see my argument as logical and recant your refusal to help a poor ignorant member like me, who has not set an official world record on solving puzzles, who does not have many other seniorities on my side, etc.

What is speedsolving.com really about? It is just about your glory, Lucas Garron’s glory, […]’s glory, …, or is it about developing an entire cubing community world wide? So far I have seen that it is not an inviting community to new comers (by the ones who have seniority at least). I commend Stephan for actually being nicer to me than before, but others I cannot say the same decency for.

If I have forgotten some wrong that I did to you, please type it here, and I will apologize. Again, I apologize for any offenses to you if you feel that I owe you an apology. I want to get right with you (with Lucas, I can forget that...he appears to have made the decision to forever reject me and mock me...that's his problem, not mine: If he is reading this too, I would like to let him know that I wish to get right with him too, but I know it's no use).

And please, to all other members reading this, please do not comment on it (by saying negative things about me or qqwref). I hope this is the catalyst to resume his and my conservation about the formula.

 

[qqwref]

It's not really about seniority or world records or any of that stuff, but just the way people act. (A few of the senior people can get away with being mean, but they do walk a fine line - if they are mean and incorrect at the same time, they stand to lose a lot of respect because other people will think of them as just being jerks.) Most of the people who are widely respected and liked around here got there essentially by showing intelligence and diplomacy/politeness. If your goal is to become part of the community you should just focus on that; stuff like having records and accomplishments develops naturally as a result of experience, practice, and knowledge, and is in no way a prerequisite.

Anyway I'm pretty sure the post with the general formula is at http://www.speedsolving.com/forum/showthread.php?p=243603#post243603. I had to have that formula in order to figure out the maximum number of centers you can be off by. In the formula I used m for the number of separate slice depths on the NxNxN cube (or floor((N-2)/2)).

 

[Edward]

Please you guys, take it up in PM.

No I don't have anything useful to say in regards to the original question.

I would like to say though that you cant expect respect from "senior" members when you only have a few (although very good) posts. Im still trying to find that subtle, but drastic change in the STYLE of my posts. I feel that's what gets you respected(somewhat), a certain style of posting.
Once I do get it though, Im in the clear.

 

[ben1996123]

All I do is solve all the colours first, then orient the centres after, because it dosent mess anything else up.