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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.

-------------------

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.

Last edited: Jan 2, 2011