# Thread: Supercube centers and odd parity

1. ## Supercube centers and odd parity

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).

2. 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.

3. Originally Posted by cmowla
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

4. Originally Posted by Lucas
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).

5. Originally Posted by qqwref
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/v...hp?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).

6. 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

7. Originally Posted by cmowla
Do you know anyone who can?
Yes.

Originally Posted by cmowla
Has it ever been done before?
No.

Originally Posted by cmowla
This is like a serious math problem isn't it?
Not really.

Originally Posted by cmowla
It hurts my brain!
Yes.

Originally Posted by cmowla
Who ever figures that out should be given the "cube noble prize"!
No.

Originally Posted by cmowla
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.

8. 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).

9. 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).

10. Originally Posted by 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).
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.

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