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

Originally Posted by cmhardw

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,

Spoiler:

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.

Thread: The interrelation of piece parities on the n x n x n supercube

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