# Message #2097

From: Charlie Mckiz <charliemckiz@rocketmail.com>

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

Date: Tue, 01 May 2012 08:51:18 -0400

I see. It was my mistake. The only problem is the orientation of 4C pieces.

Sent from my iPad

On May 1, 2012, at 12:48 AM, "schuma" <mananself@gmail.com> wrote:

> 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)15Ã—3

> > Permutation of

> > 2/3-color pieces: 32!Ã—24! 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)15Ã—3Ã—32!Ã—24! 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

> > —–BEGIN PGP SIGNED MESSAGE—–

> > Hash: SHA1

> >

> > 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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> >

>

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