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! ÷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@…>
> > 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—–
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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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>
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