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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