# Message #2094

From: schuma <mananself@gmail.com>

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

Date: Tue, 01 May 2012 04:48:55 -0000

Charlie,

Your number is not half of the number on the website, but 3/4 of the number. And the difference is for the orientation of the last 4C piece.

When all the other pieces are solved, the last 4C piece has 4 possible orientations, but not 3. So the total number of orientations of 4C should be (4!/2)^15*4. Another way to get the same number is first counting all possible orientations for the last piece: (4!/2)^16, and then noting that one out of three is possible so /3: (4!/2)^16/3. I guess you may confused this two ways of counting so you wrote (4!/2)^15*3.

Nan

— In 4D_Cubing@yahoogroups.com, Charlie Mckiz <charliemckiz@…> wrote:

>

>

> 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

> Â

> Based on the rules above:

> Permutation of 4-color pieces: 16! Ã·2

> Orientation of 4-color pieces: (4! Ã·2)15Ã3

> Permutation of

> 2/3-color pieces: 32!Ã24! Ã·2

> Orientation of 3-color pieces: (3!)32 Ã·2

> Orientation of 2-color pieces: (2)24 Ã·2

> Total: 16! Ã·2Ã(4! Ã·2)15Ã3Ã32!Ã24! Ã·2Ã(3!)32 Ã·2Ã(2)24 Ã·2

>

>

> ________________________________

> From: Brandon Enright <bmenrigh@…>

> To: 4D_Cubing@yahoogroups.com

> Cc: charliemckiz@…; bmenrigh@…

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

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

>

>

> Â

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

> "charliemckiz@…" <charliemckiz@…> 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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