# The interrelation of piece parities on the n x n x n supercube

#### cmhardw

I've been wanting to post this for some time, and after being prompted to by a PM message, I finally wrote it out. Here is a constructive method for showing the interrelation of the permutation parities of the centers, corners, and edges of the n x n x n supercube. Using this method I was able to come up with the supercube and super-supercube combinations formulas, and prove that they were accurate, on my website.

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The Supercube Parity States Matrix:
To aid in our discussion I define the Supercube parity states matrix. The goal of this matrix is to completely describe the parity of every piece type on the n x n x n supercube, and to map all possible parity states.

For every even n x n x n cube we will define a square matrix C, called the center parity matrix. This matrix will have dimensions $$\frac{n-2}{2}$$ x $$\frac{n-2}{2}$$. Throughout this example I will use the 6x6x6 cube as an example for the even case. The 6x6x6 center parity matrix has dimensions 2x2.

I will refer to pieces using SiGN notation. The entry $$C_{mn}$$ in the matrix ($$m^{th}$$ row and $$n^{th}$$ column) represents the parity of all piece types that may occupy the position is at the intersection of the layers (m+1)B and (n+1)L (SiGN notation). For example, the entry $$C_{11}$$ will be the piece on the U face intersected by the 2B and 2L layers. Please use the applet at alg.garron.us to test other entries. For clarity, in the 6x6x6 center parity matrix I will define each entry by it's intersecting faces.

$$C_{11}$$ : U, 2B, 2L
$$C_{12}$$ : U, 2B, 3L
$$C_{21}$$ : U, 3B, 2L
$$C_{22}$$ : U, 3B, 3L

For each entry in the center parity matrix a value of 1 implies that the center parity in question has odd permutation parity, and a value of 0 implies that it has even permutation parity.

Now we need to see how doing turns on our supercube will affect the center parity matrix. The solved, even, n x n x n cube has all entries 0 for even parity. Turning an outer layer will toggle the parity of every center orbit by performing a 4-cycle on all piece types. Every value in the matrix changes either from 0 to 1 or vice versa. This one is fairly easy to see.

Turning the inner layer (m+1)B will toggle all the values in row m and column m of the matrix. This is a bit harder to see, but notice that when an inner layer is turned, all the x-centers on that face are cycled via two disjoint 4-cycles. This is an overall even permutation of that particular x-center orbit. All the oblique and t-center orbits on that slice are 4-cycled, changing the parity state. If you look at where the representative pieces are for each of these orbits in the center parity matrix, you will find that they are the row and column that intersect at the particular x-center orbit contained on that inner layer slice.

Notice that this operation in the matrix toggles the value in location $$C_{mm}$$ twice, thus leaving it unchanged. For example, turning the layer 2B on the solved 6x6x6 supercube will toggle all values in row 1 and all values in column 1. This toggles the cell $$C_{11}$$ twice, leaving it unchanged. This will result in the following Matrix:
$$\left( \begin{array}{ccc} 0_{L} & 1 \\ 1 & 0 \\ \end{array} \right)$$

This means that there are now two center parity orbits that are odd, and two that are even. The subscript L in the entry $$C_{11}$$ will be defined momentarily.

Name the operation of toggling all entries in row m, and column m, as "clicking" cell $$C_{mm}$$ on the matrix. The click operation on entry $$C_{mm}$$ will toggle all entires in row m, and column m. Notice again that this toggles the entry $$C_{mm}$$ twice, leaving it unchanged. If it helps you to visualize, you can imagine that this will light up the entry $$C_{mm}$$ as well. This is, in fact, what the subscript L stands for, that this entry is lit and the click operation has been performed on that entry. "Clicking" a second time will restore all affected entries to their original values, and remove the L subscript from entry $$C_{mm}$$.

I will define the operation of toggling every entry in the matrix (the effect of turning one outer layer by one quarter turn) as the "toggle" operation. The toggling operation does not effect the "lit" or "unlit" state of any of the $$C_{mm}$$ entries. The "lit" state corresponds to the parity of the wing orbit contained on the slice (m+1)B or (m+1)L. If the $$C_{mm}$$ entry is lit with the subscript L, then the wing orbit at (m+1)L has odd permutation parity. If the entry at $$C_{mm}$$ has no subscript, then the wing orbit contained on the slice (m+1)L has even permutation parity. The reason the toggle operation does not affect the lit states is because a quarter turn on an outer layer (the toggle operation) will perform two disjoint 4-cycles on each wing orbit. This overall effect is to perform an even permutation on all wing orbits, leaving their parity states unchanged.

Lemma 1: All click entries from $$C_{11}$$ to $$C_{\frac{n-2}{2} \frac{n-2}{2}}$$ are commutative.

Proof: You can verify this by showing that any commutator made up of the click operation of any two entries $$C_{jj}$$ and $$C_{kk}$$ results in the identity operation.

Lemma 2: The toggle operation commutes with any and all click operations.

Proof: You can verify this by showing that a commutator composed of any click operation of entry $$C_{kk}$$ with the toggle operation results in the identity operation.

Now, by lemma 1 and lemma 2, since all operations commute, then there are $$2^{\frac{n}{2}}$$ possible states for the even n x n x n cube center parity matrix.

Now, if you know the lit or unlit status of all $$C_{mm}$$ entries, as well as whether or not the toggle operation was used, then you can deduce the value of every other entry in the matrix by starting with the solved matrix and applying the toggle operation (if necessary) and performing the click operation on all "lit" $$C_{mm}$$ cells. A good question to ask here is how to know if the toggle operation has been applied, i.e. whether the corner permutation is even or odd? The corner permutation will always match the parity of every single x-center orbit on the n x n x n supercube. This can be verified by noticing that the only turn that will change the parity of the corner permutation is an outer layer quarter turn - likewise for all x-center orbits. If the entries in the diagonal from $$C_{11}$$ to $$C_{\frac{n-2}{2} \frac{n-2}{2}}$$ are all 1's, then the corner parity is odd, and the toggle operation has been applied to the matrix. If the same entires are all 0's, then the corner parity is even and the toggle operation has not been applied.

This concludes the proof that knowing the parity of every wing orbit (which of the $$C_{mm}$$ cells are "lit" or not), as well as the parity of the corner permutation (whether or not the toggle operation was applied), you can deduce the parity of every other center orbit.

For the odd n x n x n cube you end up with a matrix of dimensions $$\frac{n-3}{2}$$ x $$\frac{n-1}{2}$$. As an example, for the 7x7x7 cube this would have dimensions 2x3. All of the operations are defined the same way as for the even case. You can verify that all the lemmas are true for the odd case as well.

Examples:
Let's determine the parity of all center orbits on an 11x11x11 supercube with odd corner parity, and odd wing parity in the slices 3L, 4L only.

We know that the toggle operation has been applied, and the click operation has been applied to entries $$C_{22}$$ and $$C_{33}$$. Start with the matrix C for the solved 11x11x11 supercube:

$$\left( \begin{array}{ccccccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$

All of the following operations are commutative, but let's first apply the toggle operation to represent the odd corner parity:

$$\left( \begin{array}{ccccccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{array} \right)$$

Now apply the click operation on cells $$C_{22}$$ and $$C_{33}$$ in any order.
$$\left( \begin{array}{ccccccccc} 1 & 0 & 0 & 1 & 1 \\ 0 & 1_L & 1 & 0 & 0 \\ 0 & 1 & 1_L & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 \\ \end{array} \right)$$

As an example, the entry in $$C_{32}$$ is a 1. This tells us that the parity of the center pieces that may pass through the location crossed by the U, 4B, and 3L faces has odd permutation parity.

Final notes:
In order for the matrix to completely describe the parity of every piece type on the n x n x n supercube, then we must include how to read the parity of the central most center orientations. We will define even parity in the central most center orientations to mean that the number of clockwise twists necessary to return all orientations to the solved state is congruent to 0 (mod 2). Odd parity means that the number of clockwise twists necessary is congruent to 1 (mod 2).

Lemma 3:
The central most center parity is always the same as the parity of every x-center orbit.

Proof:
Let's examine the solved supercube. The only way to change the parity of the central most center orientations is to turn an outer layer by an odd number of quarter turns. Doing this will change the parity of the center most center orientations, but also that of the x-center orbits. Recall that the only way to change the parity of the x-center orbit is also to turn an outer layer by an odd number of quarter turns. Since the only way to change the parity state of either the center most center orientations or the x-centers is the outer layer quarter turn, then the parity of both the center most center twists and the x-enter permutation states are always the same.

The Supercube Parity State Matrix can be used as a constructive proof for the center parity theorem, listed below.

N x N x N supercube center parity Theorem:
The combined parities of the corner orbit and all wing edge orbits work together to uniquely determine the parity of each and every center piece orbit. There are $$2^{floor(\frac{n}{2})}$$ possible parity situations on the N x N x N supercube, as is shown by using the center parity state matrix method.

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#### hic0057

##### Member
You put a great deal of effort into your post. Being 14 makes it hard to comprehend all of this because I haven't learnt about mathematical stuff you shown. Speedsolving you never be the same without the effort you put into cubing theory and other stuff.

#### TMOY

##### Member
Lemma 1: All click entries from $$C_{11}$$ to $$C_{\frac{n-2}{2} \frac{n-2}{2}}$$ are commutative.

Proof: You can verify this by showing that any commutator made up of the click operation of any two entries $$C_{jj}$$ and $$C_{kk}$$ results in the identity operation.

Lemma 2: The toggle operation commutes with any and all click operations.

Proof: You can verify this by showing that a commutator composed of any click operation of entry $$C_{kk}$$ with the toggle operation results in the identity operation.

Now, by lemma 1 and lemma 2, since all operations commute, then there are $$2^{\frac{n}{2}}$$ possible states for the even n x n x n cube center parity matrix.

Sorry but this is wrong. As an obvious counterexample, look at the case n=4: the matrix is 1x1, hence with only 2 possible states, and your result would imply $$2^{\frac{n}{2}}=4$$ different parities, which is clearly impossible.

For the general case when n is even, let's call $$f_0$$ the operation "turning the outer layer" on the matrix, and for $$i=1,\dots,\frac{n-2}2$$, let's call $$f_i$$ the operation "turning the i+1-th inner layer"., and look at the effect of the product $$f_1f_2\dots f_{\frac{n-2}2}$$ on the matrix. Every entry on the main diagonal is unaffected, and the entry $$C_{ij}$$, $$i\neq j$$, is toggled by both $$f_i$$ and [/math]f_jj[/math], hence is also unaffected overall. Hence $$f_1f_2\dots f_{\frac{n-2}2}$$ doesn't change the parity state of the centers, which proves that the number of such parity states is at most $$2^{\frac{n-2}2}$$.

To prove that it's exactly $$2^{\frac{n-2}2}$$, we still have to check that the elements $$f_0,f_1,\dots,f_{\frac{n-4}2}$$ are unlinked. Consider the elements on the last row of the matrix: the one on column $$\frac{n-2}2$$ is toggled by $$f_0$$ and only by $$f_0$$, and for every $$i<\frac{n-2}2$$, the one on column i is toggled by $$f_0$$ and $$f_i$$, and only by those two. Hence there is a bijection between the group generated by the $$f_i$$, $$0\leq i\leq\frac{n-4}2$$, and the possible states of the last row of the matrix, which proves that they are not linked.

Your proof works when n is odd, though. The number of possible parity situations of the centers of the n*n*n supercube, $$n\geq 4$$, in the general case is then $$2^{\mathrm{floor}(\frac{n-1}2)}$$.

#### cmhardw

Sorry but this is wrong. As an obvious counterexample, look at the case n=4: the matrix is 1x1, hence with only 2 possible states, and your result would imply $$2^{\frac{n}{2}}=4$$ different parities, which is clearly impossible.
I apologize that my post is so long, but saying that the 4x4x4 supercube has four parity states was not a mistake, and it is correct.

The four matrices for the 4x4x4 supercube, as described by my original post, are:

$$\left( \begin{array}{ccc} 0 \\ \end{array} \right)$$
Even parity in the corner orbit, even parity in the wing orbit, even parity in the x-center orbit.

$$\left( \begin{array}{ccc} 0_L \\ \end{array} \right)$$
Even parity in the corner orbit, even parity in the center orbit, odd parity in the wing orbit.

$$\left( \begin{array}{ccc} 1 \\ \end{array} \right)$$
odd parity in the corner orbit, odd parity in the center orbit, even parity in the wing orbit.

$$\left( \begin{array}{ccc} 1_L \\ \end{array} \right)$$
odd parity in the corner orbit, odd parity in the center orbit, odd parity in the wing orbit.

For the general case when n is even, let's call $$f_0$$ the operation "turning the outer layer" on the matrix, and for $$i=1,\dots,\frac{n-2}2$$, let's call $$f_i$$ the operation "turning the i+1-th inner layer"., and look at the effect of the product $$f_1f_2\dots f_{\frac{n-2}2}$$ on the matrix. Every entry on the main diagonal is unaffected, and the entry $$C_{ij}$$, $$i\neq j$$, is toggled by both $$f_i$$ and [/math]f_jj[/math], hence is also unaffected overall. Hence $$f_1f_2\dots f_{\frac{n-2}2}$$ doesn't change the parity state of the centers, which proves that the number of such parity states is at most $$2^{\frac{n-2}2}$$.
This product would light each of the x-center orbits, changing the parity states of the wings, and producing a new matrix state. On the 11x11x11 example, the resulting matrix would be:

$$\left( \begin{array}{ccccc} 0_L & 0 & 0 & 0 & 1 \\ 0 & 0_L & 0 & 0 & 1 \\ 0 & 0 & 0_L & 0 & 1 \\ 0 & 0 & 0 & 0_L & 1 \\ \end{array} \right)$$

This matrix provides the following information about the parity state of the 11x11x11 supercube. The quickest way to simulate this would be to take a solved cube and turn every different inner layer by one quarter turn.

All 4 wing orbits have odd parity because every x-center orbit is lit with the subscript L. Every oblique center orbit has even parity, every x-center orbit has even parity, every t-center orbit has odd parity, corners and central edges have even parity (they always match the x-center parity state). The centralmost center clockwise twists are congruent to 0 (mod 2) since the corner parity is even. I did not mention this in my first post, but it can be added to the lemmas. Central most center twists are congruent to 0 (mod 2) if the corner parity (and central edges, and x-centers) is even, and congruent to 1 (mod 2) if corner parity is odd.

On the 6x6x6 supercube the result of your described operation would be:
$$\left( \begin{array}{ccc} 0_L & 0 \\ 0 & 0_L \\ \end{array} \right)$$

This tells us that every center orbit has even parity, corners have even parity, and both wing orbits have odd parity.

To prove that it's exactly $$2^{\frac{n-2}2}$$, we still have to check that the elements $$f_0,f_1,\dots,f_{\frac{n-4}2}$$ are unlinked. Consider the elements on the last row of the matrix: the one on column $$\frac{n-2}2$$ is toggled by $$f_0$$ and only by $$f_0$$, and for every $$i<\frac{n-2}2$$, the one on column i is toggled by $$f_0$$ and $$f_i$$, and only by those two. Hence there is a bijection between the group generated by the $$f_i$$, $$0\leq i\leq\frac{n-4}2$$, and the possible states of the last row of the matrix, which proves that they are not linked.
Yes this is true, but the 0 or 1 value of every element on the main diagonal of an even matrix (for even cubes) can only be toggled by turning an outer layer. The L subscript can be added by performing the click operation. This is true for every element on an odd matrix where the row and column are the same. This conforms with the fact that an x-center orbit cannot be made to have odd parity other than by turning an outer layer by an odd number of quarter turns. This entry, however, can be lit if you perform the click operation (which is the same as turning the inner layer that intersects that particular x-center orbit). Any further matrix theory needs to be looked into by someone more knowledgeable than me, but let me remind you that every element in the main diagonal of an even matrix can take on 4 possible values, $$0$$, $$0_L$$, $$1$$, $$1_L$$. See the example about for the 4x4x4 matrix above, or my original post, for a description of each of these values.

Your proof works when n is odd, though. The number of possible parity situations of the centers of the n*n*n supercube, $$n\geq 4$$, in the general case is then $$2^{\mathrm{floor}(\frac{n-1}2)}$$.
The number of parity situations on the centers of the cube is $$2^{\mathrm{floor}(\frac{n-1}2)}$$, but my claim was not this. My claim was that the number of parity situations on the entire n x n x n cube, including all piece types, is $$2^{\mathrm{floor}(\frac{n}2)}$$.

Perhaps this matrix would be more aptly named the supercube parity state matrix, and not the center party state matrix? The 1's and 0's represent the parity of the center orbits, but the L subscripts and the value of the main diagonal entries let you infer the parity states of every piece orbit on the cube.

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#### TMOY

##### Member
I
The number of parity situations on the centers of the cube is $$2^{\mathrm{floor}(\frac{n-1}2)}$$, but my claim was not this. My claim was that the number of parity situations on the entire n x n x n cube, including all piece types, is $$2^{\mathrm{floor}(\frac{n}2)}$$.
In this case, your proof is really unnecessarily complicated. Your claim is completely straightforward if you only consider parities on corners and wings and completely forget about the centers.

You only need to consider the center parity matrix if you actually want to compute the number of parities on centers. That's why I assumed it was your goal.

#### cmhardw

In this case, your proof is really unnecessarily complicated. Your claim is completely straightforward if you only consider parities on corners and wings and completely forget about the centers.

You only need to consider the center parity matrix if you actually want to compute the number of parities on centers. That's why I assumed it was your goal.
Don't consider this as solely a proof of the center parity theorem. I intended this as a simple way to compute the entire parity state of the n x n x n supercube in detail, and to cover all possibilities. I would argue that it is relatively straightforward to use this method to catalogue all possible parity states on any sized supercube larger than the 3x3x3. This could be useful for solving a larger cube like a 5x5x5 or 7x7x7 supercube blindfolded. In fact, I used this method to prepare how to solve each parity state for the 5x5x5 supercube blindfolded. One consequence of the method is that it does prove the theorem, even if not elegantly.

I also found this method very useful to try to discover the parity types of the n x n x n super-supercube. The purpose of my post was to present this material to the community. I've been referring to it literally since a couple years ago. Now I have finally written it up in a post. If no one likes the method, then don't use it. I simply found it incredibly helpful for my own uses.

Chris

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#### cmhardw

Thanks for the input everyone. I edited my original post to call the matrix the "Supercube parity state matrix" rather than the "center parity matrix." I also made it more clear that the purpose of the matrix is to provide a simple to read, concise way to fully describe the parity of all piece types on the n x n x n supercube, and to map all possible parity states. I also included the part about how the center most center twists can be read off the matrix. Lastly, I made it more clear that the matrix happens to be a constructive proof for the center parity theorem, but that it is not intended to be only a proof of this theorem. This is not an elegant proof of the theorem, the matrix method just happens to prove it constructively.

#### mrCage

##### Member
Only the 4x4x4 supercube has any parities. Any other cube would have visible side effect if some "dedge" flip occurs. And i dont really like to call swapped corners for parity. Oh well, not everyone likes same kind of tea. And some prefer coffee or chocolate. Hehe ...

Per

Sorry for the bump!!

PS! Any even sized cube (n>4) could also have similar parity ("OLL parity") since the centers could be fully restored as well after the normal parity alg. These would be "block" OLL parities ...

Wait, i have to investigate this a bit further ...

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#### cmhardw

Only the 4x4x4 supercube has any parities. Any other cube would have visible side effect if some "dedge" flip occurs. And i dont really like to call swapped corners for parity. Oh well, not everyone likes same kind of tea. And some prefer coffee or chocolate. Hehe ...
I'm pretty sure I know what you're referring to, and I agree. The 4x4x4 supercube is the only n x n x n cube (n > 3) where the permutation parity of the wing orbit is completely independent of all other piece orbits on the cube.

As for the term "parity", I would feel better if we had a precise definition of what exactly that means. Presently I read it as "odd permutation parity [in a specified, or implied, piece orbit]"

PS! Any even sized cube could also have similar parity ("OLL parity") since the centers could be fully restored as well after the normal parity alg. These would be "block" OLL parities ...
Changing the parity to any wing orbit on the n x n x n cube (n > 3) would change the parity of n-4 center orbits. So, for example, on the 17x17x17 cube, changing the parity of any wing orbit will change the parity of 13 center piece orbits. On the 6x6x6 cube, changing the parity of one of the wing orbits will change the parity of 2 of the center orbits (both oblique center piece orbits). Notice that on the 4x4x4 cube that changing the parity of one of the wing orbits (the only orbit) will change the parity of 0 center orbits (or will not change the parity of the only center orbit).

So I do agree with your statement about the 4x4x4 supercube, however I would like us to come up with a more precise definition of what you mean when you say "parity".

#### Cielo

##### Member
When turning some layer by 90°, some kinds of pieces will be permutated in odd times of 4-cycles simultaneously.

But in most situations, one piece will show up in more than one layer, so the "number" of different parities needs a more clear definition.

#### Christopher Mowla

Changing the parity to any wing orbit on the n x n x n cube (n > 3) would change the parity of n-4 center orbits.
Sorry I didn't post this earlier, but I thought I should add that (not that you didn't know this), in general, the parity state of $$f\left( n,r \right)=r\left( n-2-2r \right)\text{ }\left( n\ge 4,\text{ }r\ge 1,\text{ }n,r\in \mathbb{Z}^{+} \right)$$ center orbits is changed to odd if the parity state of r orbits of wings (in any of the $$2^{\left\lfloor \frac{n-2}{2} \right\rfloor }-1$$ possible combinations) is changed to an odd permutation.

Also, $$f\left( n \right)=\left\lfloor \frac{n}{4} \right\rfloor \left( n-2-2\left\lfloor \frac{n}{4} \right\rfloor \right)$$ is the maximum number of center orbits that can be in an odd permutation (well, ones in an odd permutation simultaneously with the wings) in a cube size $$n\ge 4$$ $$\left( n\in \mathbb{Z} \right)$$, and $$f\left( n \right)=\left\lfloor \frac{n-2}{2} \right\rfloor \left( n-2-2\left\lfloor \frac{n-2}{2} \right\rfloor \right)$$ is the minimum number.

EDIT:
Perhaps the most compact maximum formula that can exist is $$f\left( n \right)=\left\lfloor \frac{1}{2}\left( \frac{n-2}{2} \right)^{2} \right\rfloor$$

There are several other maximum formulas. Here's another one: $$f\left( n \right)=\frac{\left( n-1 \right)\left( n-3 \right)-\cos \left( \frac{n\pi }{2} \right)\left( \cos \left( \frac{n\pi }{2} \right)+2 \right)}{8}$$

This tells us that: $$f\left( n \right)=\left\{ \begin{matrix} \frac{\left( n-1 \right)\left( n-3 \right)}{8},\text{ if }n\text{ is odd,} \\ \frac{\left( n-1 \right)\left( n-3 \right)}{8}-\frac{3}{8},\text{ if }n\text{ is divisible by 4,} \\ \frac{\left( n-1 \right)\left( n-3 \right)}{8}+\frac{1}{8},\text{ if }n\text{ is even, but not divisible by 4}\text{.} \\ \end{matrix} \right.$$

So there are 3 different categorizations as far as the number of center orbits affected by wings is.

Since the total number of orbits of non fixed center pieces is: $$\left\{ \begin{matrix} \frac{\left( n-1 \right)\left( n-3 \right)}{4},\text{ if }n\text{ is odd,} \\ \left( \frac{n-2}{2} \right)^{2},\text{ if }n\text{ is even}\text{.} \\ \end{matrix} \right.$$, the maximum formula above tells us that exactly half of the center orbits is the maximum number which can possibly be in an odd permutation with the wings, and slightly less than half for even cube sizes.

Just to verify this formally,
1) For odd cube sizes,
The total number of non fixed center orbits is $$\frac{\left( n-1 \right)\left( n-3 \right)}{4}$$ and the maximum number of non fixed center orbits that possibly can be in an odd permutation with wings is from $$r=\left\lfloor \frac{n}{4} \right\rfloor$$ orbits of wings in an odd permutation. The percentage is therefore (maximum)/(total) * 100.

$$\frac{\frac{\left( n-1 \right)\left( n-3 \right)}{8}}{\frac{\left( n-1 \right)\left( n-3 \right)}{4}}=\frac{1}{2}$$.

2) For even cubes divisible by 4, the maximum number of non fixed center orbits which are affected by $$r=\frac{n}{4}-1\text{ or }r=\frac{n}{4}$$ wing orbits is $$\frac{\left( n-1 \right)\left( n-3 \right)}{8}-\frac{3}{8}$$, and the total number of non-fixed center orbits in even cubes is $$\left( \frac{n-2}{2} \right)^{2}$$.

Therefore the percentage is: $$\frac{\frac{\left( n-1 \right)\left( n-3 \right)}{8}-\frac{3}{8}}{\left( \frac{n-2}{2} \right)^{2}}=\frac{1}{2}-\frac{2}{\left( n-2 \right)^{2}}$$ and $$\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{2}-\frac{2}{\left( n-2 \right)^{2}} \right]=\frac{1}{2}$$, which tells us (those who know calculus) that the percentage gets arbitrarily close to, but never reaches 1/2. The percentage starts at 0% with the 4x4x4, but approaches 50% as n gets very large.

3) For even cubes not divisible by 4, the maximum number of non fixed center orbits which are affected by $$r=\frac{n-2}{4}$$ wing orbits is $$\frac{\left( n-1 \right)\left( n-3 \right)}{8}-\frac{1}{8}$$, and the total number of non-fixed center orbits in even cubes is $$\left( \frac{n-2}{2} \right)^{2}$$. Therefore, the percentage for this category of the nxnxn cube is: $$\frac{\frac{\left( n-1 \right)\left( n-3 \right)}{8}-\frac{1}{8}}{\left( \frac{n-2}{2} \right)^{2}}=\frac{1}{2}-\frac{1}{\left( n-2 \right)^{2}}$$ and $$\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{2}-\frac{1}{\left( n-2 \right)^{2}} \right]=\frac{1}{2}$$, which tells us (those who know calculus) that the percentage gets arbitrarily close to, but never reaches 1/2. The percentage starts at 43.75% with the 6x6x6, but approaches 50% as n gets very large.

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