# Message #2093

From: Charlie Mckiz <charliemckiz@rocketmail.com>

Subject: Re: [MC4D] About the number of permutations of MC4D calculating

Date: Mon, 30 Apr 2012 19:29:27 -0700

My solution is as follow

Rules that MC4D obey:

Position:

4-color pieces: The permutation of 4-color pieces is always

even

2/3-color pieces: The net permutation of 3-color and 2-color

is always even.

Orientation: Consider rotation of a piece is a permutation

of shading sides of that piece. So:

4-color pieces: The permutation of rotation of a 4-color

piece is always even. Besides, the net rotation of all 4-color pieces is 0.

3-color pieces: The net permutation of rotation of all

3-color pieces is even.

2-color pieces: The net permutation of rotation of all

2-color pieces is even.

Number of different kinds of pieces:

4-color pieces: 16

3-color pieces: 32

2-color pieces: 24

Based on the rules above:

Permutation of 4-color pieces: 16! ÷2

Orientation of 4-color pieces: (4! ÷2)15×3

Permutation of

2/3-color pieces: 32!×24! ÷2

Orientation of 3-color pieces: (3!)32 ÷2

Orientation of 2-color pieces: (2)24 ÷2

Total: 16! ÷2×(4! ÷2)15×3×32!×24! ÷2×(3!)32 ÷2×(2)24 ÷2

________________________________

From: Brandon Enright <bmenrigh@ucsd.edu>

To: 4D_Cubing@yahoogroups.com

Cc: charliemckiz@rocketmail.com; bmenrigh@ucsd.edu

Sent: Monday, April 30, 2012 3:38 PM

Subject: Re: [MC4D] About the number of permutations of MC4D calculating

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On Mon, 30 Apr 2012 15:31:05 -0000

"charliemckiz@rocketmail.com" <charliemckiz@rocketmail.com> wrote:

> Today I tried to calculate the number of MC4D permutations. My answer

> is a half of what posted in the website

> http://www.superliminal.com/cube/cube.htm . Does any one want to

> check the answer? I’m pretty sure about mine. And I see no mistake in

> the website solution.??

>

Well, I’ve aways taken the permutation calculation as gospel but I

decided to give it a go myself and I’ve arrived at the same result.

I find the sticker-based counting of permutations to be confusing so

I’ve done it piece-based. Here are my notes.

That is, there are:

Permutations:

* 8 centers with 192 "permutations" available. Fixing to one cancels

all re-orientation of the tesseract

* 24 face centers that can be in any permutation

* 32 3-color edges who’s parity is tied to the parity of the face

centers

* 16 4-color corners that can only be in an even permutation

Orientations:

* 8 centers with 24 orientations available (only 1 visually distinct)

* 24 face pieces with 8 orientations available (only 2 visually

distinct ones)

* 32 3-color edges with 6 orientations available

* 16 4-color edges with 12 orientations available

The calculation is further complicated in that:

* The total "flip" of the 24 face pieces is even (the 24th orientation

is determined by the previous 23)

* The orientation of the 32nd 2-color edge can only be in half (3) of

the orientations.

* The first 15 corners can be in any orientation but the last corner

can only be in 1/3 (4) orientations.

Taking all of that into account there should be:

(24! * 2^23) * ((32! / 2) * (6^31 * 3)) * ((16! / 2) * (12^15 * 4))

Which matches nicely with the website.

The only "tricky" portion of the calculation is counting the

orientations for the last 3-color and the last 4-color piece. I can

create a macro to demonstrate how to put a single 3-color piece or a

single 4-color piece in these reduced orientations if you’d like.

Regards,

Brandon

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