Calculating Permutations on nxnxn Rubik's cube

[cmhardw]

Hi cmowla,

Very cool post! I like your approach for how you listed this formula, particularly your mention of using trig functions to avoid the step functions.

I did find a couple of points from your post that I don't fully agree with, though:

c) \( \frac{1}{2^{\left\lfloor \frac{\left( n-2 \right)^{2}}{4} \right\rfloor }} \)
If corners are in an odd permutation, then all non-fixed (NF) center orbits are in an odd permutation. (For those who can't see this, just do R to a supercube and see that it does a circular 4-cycle of every NF center piece orbit. Furthermore, corners in an odd permutation does not make the permutation of wing edges odd because R does 2 4-cycles of each and every orbit of wing edges, which is an even permutation).

The bolded line is only true if every wing orbit has even parity.

Assume the corners are in an even permutation now. Then assume that either the NF center pieces or the wings (it doesn't really matter) are in an odd permutation. Then the other is in an odd permutation too, that is, for cube sizes \( \ge \)the 5x5x5 supercube.

If the corners are in an even permutation, then it is impossible for all non-fixed centers to be in an odd permutation. This is because the parity of every x-center orbit will always match the parity of the corners. So if the corners have even parity, so too will every x-center orbit.

Also, what about cases where only some of the center orbits have even parity, and others have odd? Or cases where some of the wings orbits have even party, and others have odd?

More generally, and this result comes from my center parity matrix method, there are 2 situations where we can infer information about the parities of various orbits on the supercube:

1) If the parities of the corners and every wing orbit are known, then you can deduce the parity of every non-fixed center orbit.

2) If the parity of every non-fixed center orbit is known, then you can deduce the parity of the corners AND every wing orbit.

I know it's very possible I have misinterprented the given values in my formula, but if not (meaning that my explanation of each factor of the formula is correct), shouldn't there be another odd permutation reduction factor of \( \frac{1}{2^{\left\lfloor \frac{n-2}{2} \right\rfloor }} \) for the 5x5x5 and greater cubes since the parity state of \( f\left( n,r \right)=n-2-2r\text{ }\left( n\ge 4,\text{ }r\ge 1,\text{ }n,r\in \mathbb{Z} \right) \) 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

Your formula:

\( f\left( n,r \right)=n-2-2r\text{ }\left( n\ge 4,\text{ }r\ge 1,\text{ }n,r\in \mathbb{Z} \right) \)

doesn't work for n=8 and r=2. On the 8x8x8 cube, if you change the parity of 2 wings orbits, then you change the parity of 4 center orbits. This can easily be seen using the 3x3 center parity matrix that represents the 8x8x8 cube like I detail in this thread. However, your formula would imply that if n=8 and r=2 we would change the parity of:
\( f\left( 8,2 \right)=8-2-2*2=2 \) center orbits.

As to your question, I don't think any special reduction is necessary for cubes greater than or equal to the 5x5x5 cube. The general result I listed earlier about deducing the parities of various pieces given certain information would prove this if I'm not mistaken.

As a side note, since not all parity orbit situations are possible (for example, the parities of each x-center orbit must match that of every other x-center orbit), then it's more effective to first permute the corners and the wings into any of their possible permutations (which as you mention these orbits are all independent of each other). However, due to the center parity matrix method, this then determines the parities of every non-fixed center orbit (and also the parity of the center twists of the fixed centers for odd cubes).

 

[Christopher Mowla]

Very cool post! I like your approach for how you listed this formula, particularly your mention of using trig functions to avoid the step functions.

Thanks! I knew you would like it.

I did find a couple of points from your post that I don't fully agree with, though:

c) \( \frac{1}{2^{\left\lfloor \frac{\left( n-2 \right)^{2}}{4} \right\rfloor }} \)
If corners are in an odd permutation, then all non-fixed (NF) center orbits are in an odd permutation. (For those who can't see this, just do R to a supercube and see that it does a circular 4-cycle of every NF center piece orbit. Furthermore, corners in an odd permutation does not make the permutation of wing edges odd because R does 2 4-cycles of each and every orbit of wing edges, which is an even permutation).

The bolded line is only true if every wing orbit has even parity.

You're absolutely right. I can't believe I overlooked the fact that there can be no more than \( \frac{1}{2^{\left\lfloor \frac{\left( n-2 \right)^{2}}{4} \right\rfloor }} \) NF center orbits in an odd permutation because that IS the number of center orbits, period.

However, I did read your center parity matrix method, and I definitely agree with everything about it.

Your formula:

\( f\left( n,r \right)=n-2-2r\text{ }\left( n\ge 4,\text{ }r\ge 1,\text{ }n,r\in \mathbb{Z} \right) \)

doesn't work for n=8 and r=2. On the 8x8x8 cube, if you change the parity of 2 wings orbits, then you change the parity of 4 center orbits. This can easily be seen using the 3x3 center parity matrix that represents the 8x8x8 cube like I detail in this thread. However, your formula would imply that if n=8 and r=2 we would change the parity of:
\( f\left( 8,2 \right)=8-2-2*2=2 \) center orbits.

Yeah, I can't believe I forgot to multiply that by r! So the correct formula is:
\( r\left( n-2-2r \right) \).

This formula actually is directly related to (but I found this formula by trying to find the number of center pieces in an odd permutation along with the wings) the concept you mention how the "click" of an X-center piece affects the parity of pieces in its intersecting row and column of the matrix. In fact, here is a good visual of how I came up with that formula. I know it's probably very basic for you, but I thought others might appreciate the simplicity of this explanation of why my formula for the number of NF center orbits affected by an odd permutation of wings is correct.

Derivation

Spoiler

If we take a solved 13x13x13, focusing on the NF center pieces only (we shade in the wings, corners, and fixed center piece), we have (n-2) = 13 - 2 = 11 center pieces in slice 2R.

Let’s start out by doing a quarter turn to slice 2R. Clearly the orbit of wing edges slice 2R has an odd permutation.
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[/FONT]Now we highlight the rest of slice 2R.
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If we imagine doing the move U, (not that we actually make another move), we can clearly see that the center orbits affected will transfer to the bottom as well.

NOTE: The row shown below (and for the remaining iterations) is a visual only. I understand perfectly well that a 2 4-cycle of X-center pieces is done in each (M+1)R slice we do and that there is a 4-cycle of the remaining center pieces in that slice. The representation does however accurately show how the parity state of inner layer slices are related with minimal thought.
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[/FONT] At the intersection of the yellow row and yellow column is an even permutation, namely, the X-center piece. You can think of the column and the row representing an odd permutation: where they intersect becomes an even permutation (analogous to your \( 0_{L} \) idea). Indicated by the color orange.
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[/FONT] If we reflect this now to the top, clearly (n-2)-2 center orbits in an odd permutation now in the yellow column (just focus on the yellow column, not its corresponding yellow row).
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Suppose we now choose another wing edge orbit.
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[/FONT] Now we identify the row and column affected by it (the intersection of the new colored row and the previous yellow row is indicated by the color light green).
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[/FONT] Again, if we imagine doing U, the slice affected gets transferred to the bottom, and their intersection is an X-center piece.
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But this transferring row also intersects a piece in the first slice (slice 2R) we did a quarter turn to.
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[/FONT] Reflecting this to the top, clearly we have (n-2)-2-2 center orbits left with an odd permutation in both slices. That is, 2(n-2-2(2)).
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[/FONT] If we choose yet another orbit of wings,
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And identify its column (its intersection of the previous two rows is indicated with the color light blue).
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[/FONT] And we also find its corresponding row by visualizing doing U (its intersection with the previously affected columns is indicated by light blue),
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[/FONT] As always, we take note of the intersection of the current row and column to be the X-center piece (pink),
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Finally, reflecting this to the top,
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[/FONT]Note that every color besides the yellow, red and dark blue represent NF center pieces in an even permutation. They are analogous to your \( 0_{L} \). In fact, the amount of these pieces (not the orbit number) for r orbits of wings (no matter which combination we choose) can be represented by \( 8\sum\limits_{i=1}^{r}{2i-1}=8r^{2} \).


Just focusing on the indicated portion below,
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We now have exactly (13-2)-2-2-2=5 NF center orbits in an odd permutation in each wing edge orbit we chose. There are 5 dark blue center pieces (each indicating a NF center orbit) in slice 5R, 5 red center pieces (each indicating a NF center orbit) in slice 3R, and 5 yellow center pieces (each indicating a NF center orbit) in slice 2R.

Clearly there is a pattern and if we were to choose arbitrary wing edge orbits, the number of NF center orbits in an odd permutation with the wings would be the same for any combination of the same number of wing edge orbits.

For example,
\( \left( \begin{matrix}
5 \\
3 \\
\end{matrix} \right)=10 \) different ways we could have selected 3 orbits of wings for this particular cube size because we have a total of 5 orbits of wings taking 3 at a time.

Hence the formula is \( r\left( n-2-2r \right) \).

.Therefore the minimum number of NF center orbits in an odd permutation along with wings is \( \left\lfloor \frac{n-2}{2} \right\rfloor \left( n-2-2\left\lfloor \frac{n-2}{2} \right\rfloor \right) \), and, more importantly, the maximum number of center orbits in an odd permutation with wings is \( \left\lfloor \frac{n}{4} \right\rfloor \left( n-2-2\left\lfloor \frac{n}{4} \right\rfloor \right)=\left\lfloor \frac{\left\lceil \left( \frac{n-2}{2} \right)^{2} \right\rceil }{2} \right\rfloor \).

(I'll eventually prove these in great detail.)

Now having given it some careful thought, I have come up with a way to verify that the reducing factor \( \frac{1}{2^{\left\lfloor \left( \frac{n-2}{2} \right)^{2} \right\rfloor }} \) is indeed correct (I was wrong to think we needed more).

Our goal is to determine the maximum number of parity situations on an nxnxn supercube. Specifically, how the combination of the wings and corners affect the parity of the NF center pieces (since I apparently covered the other parity cases well enough).

First of all, we consider the independent cases.

1) We need to know the maximum number of center orbits affected by wings (assuming the corners are in an even permutation).

I have already said that that amount is equal to \( \left\lfloor \frac{n}{4} \right\rfloor \left( n-2-2\left\lfloor \frac{n}{4} \right\rfloor \right)=\left\lfloor \frac{\left\lceil \left( \frac{n-2}{2} \right)^{2} \right\rceil }{2} \right\rfloor \).

2) The corners affect all \( \left\lfloor \frac{n-2}{2} \right\rfloor \) X-center piece orbits.

This leaves \( \left\lfloor \frac{n-2}{4} \right\rfloor \left( \left( n-2 \right)-2-2\left\lfloor \frac{n-2}{4} \right\rfloor \right)=\left\lfloor \frac{\left\lceil \left( \frac{n-4}{2} \right)^{2} \right\rceil }{2} \right\rfloor \) non X-center NF center orbits to be unable to be affected by even the worst odd permutation combination of wings (this amount is zero until you get to the 7x7x7 and higher supercubes).

By the way, the number of orbits of wings which cause the worst (maximal center orbits in an odd permutation) combination for the NF center orbits is:

\( f\left( n \right)=\left\{ \begin{matrix}
\left\lfloor \frac{n}{4} \right\rfloor ,\text{ if }n\text{ is not divisible by 4,} \\
\left\{ \left\lfloor \frac{n}{4} \right\rfloor -1,\left\lfloor \frac{n}{4} \right\rfloor \right\}\text{ if }n\text{ is divisible by 4}\text{.} \\
\end{matrix} \right. \)

(Thus two values for r yields the maximum for cube sizes evenly divisible by 4.)

Regardless of where these extra NF center orbits are (the number is only important, and their number is very well defined since we know the maximum number of center orbits which wings can affect and the number of X-center orbits and we know the total amount of center orbits there are), if they have an odd permutation, it will be from the corners (and consequently, along with the X-center pieces, fixed center pieces on odd cubes, and middle edges on odd cubes). This is unquestionably true since the parity of the corners has the potential to affect the parity of all center orbits, if all orbits of wings are in an even permutation, that is.

 

[Christopher Mowla]

Rate of Change for the Number of Permutations Formula

As I have mentioned in my post before the last one is that we can represent the formula for the nxnxn without using step functions. We can use continuous functions, .i.e., trig functions.

Formula for the Regular nxnxn. [Link]

\( \left( 8!\times 3^{7} \right)\left( \frac{24!}{4!^{6}} \right)^{\frac{1}{4}\left[ \left( n-1 \right)\left( n-3 \right)+\cos ^{2}\left( \frac{n\pi }{2} \right) \right]}24!^{\frac{n-2-\sin ^{2}\left( \frac{n\pi }{2} \right)}{2}}\left( \frac{12!\times 2^{11}}{2} \right)^{\sin ^{2}\left( \frac{n\pi }{2} \right)}\frac{1}{24^{\cos ^{2}\left( \frac{n\pi }{2} \right)}} \)
​

If you clicked that link and scrolled down, you would have eventually come across the (first) derivative of the formula. For those who haven't had calculus, all we need to do to find the rate of change to a function once we have the derivative is to substitute the x-value (here, it's n, the degree of the cube) and obtain the rate of change/slope of the function (in our case, the formula for the number of permutations) at that specific point (in our case, the specific cube size).

Obviously that derivative is quite lengthy, and Mathematica (not Wolfram, the program Mathematica itself) gets overloaded on my computer for cube size n = 999.

For convenience to everyone who has Mathematica, here is the derivative ready to go. Just substitute values for x (I'd recommend using Ctrl+H in Wordpad)and evaluate the cell, and you'll get the same results I show below (and other values for other cube sizes). (Note: I have the //N at the end, but you can remove it to get the exact values.)

Spoiler

2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) (1 + 2 x) Log[3] + 2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) Log[5] (-2 + 2 x - Pi Cos[(x Pi)/2] Sin[(x Pi)/2]) + 2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) Log[2] (7 + 2 x + 11 Pi Cos[(x Pi)/2] Sin[(x Pi)/2]) + 2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) Log[96577] ((1 - Pi Cos[(x Pi)/2] Sin[(x Pi)/2])/2 + (-4 + 2 x - Pi Cos[(x Pi)/2] Sin[(x Pi)/2])/4) + 2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) Log[11] (1 + (-4 + 2 x - Pi Cos[(x Pi)/2] Sin[(x Pi)/2])/2) + 2^(-15 + (-3 + x) (-1 + x) + 11 x - 2 Cos[(x Pi)/2]^2 + 9 Sin[(x Pi)/2]^2) 3^(-1 + (-3 + x) (-1 + x) + 5 x) 5^(-3 + (-3 + x) (-1 + x) + 2 x + Cos[(x Pi)/2]^2) 7^(-1 + x + (3 ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2))/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) 11^(-2 + x + ((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/2) 96577^(((-3 + x) (-1 + x) + Cos[(x Pi)/2]^2)/4 + (-2 + x - Sin[(x Pi)/2]^2)/2) Log[7] (1 + (1 - Pi Cos[(x Pi)/2] Sin[(x Pi)/2])/2 + (3 (-4 + 2 x - Pi Cos[(x Pi)/2] Sin[(x Pi)/2]))/4)//N

With these very large numbers, we mainly pay attention to the power of ten they are.

Here are some of the rates of change (decimal form of the exact numbers).

Spoiler

n=2: 1.00644*10^8
n=3: 1.95718*10^21
n=4: 4.6708*10^47
n=5: 2.29032*10^76
n=6: 1.55307*10^118
n=7: 2.27539*10^162
n=8: 4.73234*10^219
n=9: 2.15956*10^279
n=10: 1.412862071588920*10^352
n=11: 2.041829565969861*10^427
n=12: 4.250648102022849*10^515
n=13: 1.961573176908497*10^606
n=14: 1.307459144395711*10^710
n=15: 1.935819473534397*10^816
n=16: 4.146563061346539*10^935
n=17: 1.975599053252146*10^1057
n=18: 1.363218034588532*10^1192
n=19: 2.094148907347992*10^1329
n=20: 4.662632788088312*10^1479
n=21: 2.312628415760626*10^1632

I deduced the following pattern from the numbers above by:
1) Taking the list of powers of 10:
{8,21,47,76,118,162,219,279,352,...}
2) Finding delta list (a list showing the distance between each number)
{13, 26, 29, 42, 44, 57, 60, 73,...}
3) And then taking the difference of that list:
{13, 3, 13, 2, 13, 3, 13, ...}
which obviously has a pattern involving the numbers 13, 3 and 2.

This is what I came up with:

Spoiler

\( \begin{matrix}
2 & 8+13\left( 0 \right)=8 \\
3 & 8+13\left( 1 \right)=21 \\
4 & 8+13\left( 3 \right)+3\left( 0 \right)=47 \\
5 & 8+13\left( 5 \right)+3\left( 1 \right)=76 \\
6 & 8+13\left( 8 \right)+3\left( 2 \right)+2\left( 0 \right)=118 \\
\end{matrix} \)
\( \begin{matrix}
7 & 8+13\left( 11 \right)+3\left( 3 \right)+2\left( 1 \right)=162 \\
8 & 8+13\left( 15 \right)+3\left( 4 \right)+2\left( 2 \right)=219 \\
9 & 8+13\left( 19 \right)+3\left( 6 \right)+2\left( 3 \right)=279 \\
10 & 8+13\left( 24 \right)+3\left( 8 \right)+2\left( 4 \right)=352 \\
11 & 8+13\left( 29 \right)+3\left( 10 \right)+2\left( 6 \right)=427 \\
\end{matrix} \)
\( \begin{matrix}
12 & 8+13\left( 35 \right)+3\left( 12 \right)+2\left( 8 \right)=515 \\
13 & 8+13\left( 41 \right)+3\left( 15 \right)+2\left( 10 \right)=606 \\
14 & 8+13\left( 48 \right)+3\left( 18 \right)+2\left( 12 \right)=710 \\
15 & 8+13\left( 55 \right)+3\left( 21 \right)+2\left( 15 \right)=816 \\
\end{matrix} \)
\( \begin{matrix}
16 & 8+13\left( 63 \right)+3\left( 24 \right)+2\left( 18 \right)=935 \\
17 & 8+13\left( 71 \right)+3\left( 28 \right)+2\left( 21 \right)=1057 \\
18 & 8+13\left( 80 \right)+3\left( 32 \right)+2\left( 24 \right)=1192 \\
19 & 8+13\left( 89 \right)+3\left( 36 \right)+2\left( 28 \right)=1329 \\
\end{matrix} \)
\( \begin{matrix}
20 & 8+13\left( 99 \right)+3\left( 40 \right)+2\left( 32 \right)=1479 \\
21 & 8+13\left( 109 \right)+3\left( 45 \right)+2\left( 36 \right)=1632 \\
\end{matrix} \)
...
=
8+
13*{0,1,3,5,8,11,15,19,24,29,35,41,48,55,63,71,80,89,99,109,...}+
3*{0,0,0,1,2,3,4,6,8,10,12,15,18,21,24,28,32,36,40,45,...}+
2*{0,0,0,0,0,1,2,3,4,6,8,10,12,15,18,21,24,28,32,36,...}

We can find formulas for each of the three sequences just copying and pasting the first 20 values we have into Wolfram Alpha (in list form) and scroll down to the "possibly closed form" formula. For example, a formula for the sequence multiplied by 13.
(The formula that link gives is a little different from the one I used, but they both work.)

We can use the same formula for the sequences multiplied by 2 and 3, just shifting n by an additional 2 for the sequence multiplied by the 2. When all of this is done,

\( 8+ \)
\( 13\left( \frac{1}{8}\left( 2n^{2}+\cos \left( n\pi \right)-9 \right) \right)+ \)
\( 3\left( -\frac{1}{8}\left( \sin \left( \frac{\left( n-3 \right)\pi }{2} \right)-\left( n-3 \right) \right)\left( \sin \left( \frac{\left( n-3 \right)\pi }{2} \right)+\left( n-3 \right)+2 \right) \right)+ \)
\( 2\left( -\frac{1}{8}\left( \sin \left( \frac{\left( n-5 \right)\pi }{2} \right)-\left( n-5 \right) \right)\left( \sin \left( \frac{\left( n-5 \right)\pi }{2} \right)+\left( n-5 \right)+2 \right) \right) \)

We can simplify and get the approximation formula.

My Approximation Formula
for the Power of Ten of the Rate of Change Values
of the Regular nxnxn Cube
\( n\ge 2 \)[Link]

\( \frac{1}{16}\left( 62n^{2}-56n-4\cos \left( \frac{n\pi }{2} \right)+21\cos \left( n\pi \right)-33 \right) \)
​

My approximation formula yields the correct values up to n=31, I believe. However, the real values become larger than the approximation formula for larger n.

But for reasonable size n, the approximation formula is 99.9% accurate.

For example, for n=900,
\( \frac{1}{16}\left( 62\left( 900 \right)^{2}-56\left( 900 \right)-4\cos \left( \frac{\left( 900 \right)\pi }{2} \right)+21\cos \left( \left( 900 \right)\pi \right)-33 \right)=\text{3135599} \)
and Mathematica spat out 1.420456649789211*10^3137815, (thus 3137815)., but

\( \frac{\text{3135599}}{\text{3137815}}\times 100=99.92937761... \)%.
​

I'm not sure how useful the rate of change is, but we have a formula for it now, and those who don't have sophisticated computers can at least have a good approximation for even very large values of n.

Although what I am about to say may not be true, maybe somehow the rate of change and the God's number sequences are related somehow. Speaking of which, although I have been told that this formula most likely gives values higher than what God's number is, here is a formula I made! (For entertainment purposes only.)

\( 3n^{2}-2n\sin ^{2}\left( \frac{n\pi }{2} \right)-1 \)
[Link]
{0,11,20,47,64,107,132,191,224,299,340,431,480,587,644,...}
​

 

[Christopher Mowla]

I took the derivative on my Ti-89 Titanium Calculator and did more simplification myself. This is a much shorter formula for the actual rate of change of the formula for the number of permutations of the regular nxnxn cube. I made an effort to keep the original factorials from f (n) as well. Perhaps if someone studies this formula enough, we can come up with an explanation how the rate of change works on the nxnxn cube. [Link]

\( {f}'\left( n \right)=\frac{\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)24!^{\frac{1}{4}\left( \left( n+1 \right)\left( n-3 \right)+3\cos ^{2}\left( \frac{n\pi }{2} \right) \right)}\ln \left( 24!^{4\left( n-1 \right)}4!^{-24\left( n-2 \right)} \right)}{{}} \)
\( \text{ }8\left( 24\times 12!\times 2^{10} \right)^{\cos ^{2}\left( \frac{n\pi }{2} \right)}4!^{\frac{3}{2}\left( \left( n-1 \right)\left( n-3 \right)+\cos ^{2}\left( \frac{n\pi }{2} \right) \right)} \)
​

and
\( \cos ^{2}\left( \frac{n\pi }{2} \right)=\left( n+1 \right)\bmod 2=(n+1)-2\left\lfloor \frac{n+1}{2} \right\rfloor \).
​

​

(Note that this version of the first derivative of f (n) has been simplified AND has been additionally simplified for n being a positive integer).
​

With this version of the derivative, I was able to successfully compute n = 5 000.

\( {f}'\left( 5\text{ 000} \right)=4.809141984700355\times 10^{96\text{ }928\text{ }416} \).


My approximation formula, yielded 96 857 499. So the percent error even at this size n is
​

\( \left( 1-\frac{96\text{ 857 499}}{\text{96 928 416}} \right)\times 100=0.073164303 \)%.

Not that it's important, but if someone has a better computer than I have, could you compute very large values for n and post the results. Any results from cube sizes in the millions (or larger) would be interesting to see.



EDIT:
We can rewrite the derivative as [Link]

\( {f}'\left( n \right)=\frac{\left( 8!\times 3^{7} \right)\left( 12!\times 2^{10} \right)\left( 24! \right)^{\frac{1}{4}\left( n+1 \right)\left( n-3 \right)}\ln \left( 24!^{4\left( n-1 \right)}4!^{-24\left( n-2 \right)} \right)}{8\left( 4! \right)^{\frac{3}{2}\left( n-1 \right)\left( n-3 \right)}} \)\( \times \left( \frac{\left( 24! \right)^{\frac{3}{4}}}{24\times 4!^{\frac{3}{2}}\times 12!\times 2^{10}} \right)^{\cos ^{2}\left( \frac{n\pi }{2} \right)} \)

, where the portion raised to \( \cos ^{2}\left( \frac{n\pi }{2} \right)=\left( n+1 \right)\bmod 2 \) is an extra factor (approximately 505) for even cubes.
​

 

[mrCage]

Hi,

Bump warning!!!!!!

I made a simple script for calculating permutations on nxnxn cubes of various sizes.

Here is an example with size=6.

http://fmc.mustcube.net/Cube_Permutations.php?size=6

Obviously you just change the size parameter in the querystring for other sizes. If the size is missing it will default to 3

Per

PS! I modified the code to no longer be using querystring, but session variables. As a downside direct linking to a calculation of other sizes than 3 (default) is no longer possible. The original version was simply too easy to "hack" and was prone to errrors for that reason ...

Actually you can link directly like so: http://fmc.mustcube.net/Cube_Permutations_middle.php?size=6

 

[WBCube]

Here is an example with size=6.

http://fmc.mustcube.net/Cube_Permutations.php?size=6

I'm getting a 404, for other sizes too

 

[ben1996123]

I'm getting a 404, for other sizes too

I'm not. Nice script, works for really large sizes too.

100x100:

Spoiler

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0,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

 

[WBCube]

This is what I get:
Shame

 

[Godmil]

That's pretty cool, it's nice seeing just how quickly the number gets huge.

 

[tim]

This is what I get: View attachment 2497
Shame

You miss a ".php" at the end.

 

[WBCube]

You miss a ".php" at the end.

Ah, thank you. Not sure why that was happening, I was just clicking the hyperlink.

 

[mrCage]

Using the bcmath library in PHP is a lot of fun. It does maths on numbers represented as (long) strings. I made my own functions for exponeentiation and factorial to make it work

 

[Stefan]

I made a simple script for calculating permutations on nxnxn cubes of various sizes.

Bet it's not simpler than in Python
http://codepad.org/RCtDpfk0

 

[Ranzha]

Bet it's not simpler than in Python
http://codepad.org/RCtDpfk0

And Tweetable, too.

 

[mrCage]

Bet it's not simpler than in Python
http://codepad.org/RCtDpfk0

That was really compact!! But it has no user input and no output formatting. My code can actually group the digits in blocks of desired size and with chosen delimiter. For better readability
I did not know that Python works "out of the box" with really huge numbers. Will it also do decimal number calculations like needed for pi or the golden ratio ??

Per

PS! My current hosting has a rather short script timeout limit that i cannot overcome. It is not allowed, because the hosting is not on a dedicated server ...