Message #640
From: David Smith <djs314djs314@yahoo.com>
Subject: Re: [MC4D] a short diversion into sticker and cubie counts
Date: Tue, 03 Feb 2009 11:52:43 -0800
You’re welcome, Roice! I’m glad my observations helped, and great
to hear you are off to solve another problem.
As for your observation, check another unique insight off the list for
Roice! ;) I thought of a cool way to visualize why your fact must be
true, and it could even be used to intuitively prove the result without
solving the equation you gave.
When the number of cubies per edge equals the number faces, this
of course also means that the number of faces equals the number of
layers ((n-1)-dimensional slices of an n-dimensional cube). Now just
mentally reposition each sticker on a particular face to the inside of
a corresponding cubie on a particular layer of the cube. Do this for
each face, and we see that each sticker rests within each cubie,
establishing a 1-to-1 correspondence of cubies to stickers. This is
a lot easier to visualize than to describe in words. This "proof
without words" can also be used to establish the inequalities you
gave.
As for the terminology of higher-dimensional Rubik’s cubes, I prefer
n^d, which is what I use in my formulas. (At least I will use d
eventually! ;)) Strangely however, I also use the term n-dimensional,
as you can see above. I believe the deference to n as the dimension
(perhaps unconsciously) is because this is used almost everywhere
for dimensions in mathematics. As an example, consider the
standard term, "Euclidian n-space" for the set of all n-tuples of real
numbers, denoted by R^n.
Anyway, I hope this helps establish the a priori justification you were
looking for, and good luck with your checkerboard problem!
All the best,
David
— On Tue, 2/3/09, Roice Nelson <roice3@gmail.com> wrote:
From: Roice Nelson <roice3@gmail.com>
Subject: [MC4D] a short diversion into sticker and cubie counts
To: 4D_Cubing@yahoogroups.com
Date: Tuesday, February 3, 2009, 12:12 AM
David, thank you for the excellent email response on the parity discussion with Levi (and for the fix of my incorrect 4-cycle claim about corners on MC5D). It did indeed provide more insight for me by laying things out for general dimensions, and I think I have a good picture of things, with confidence to look at configurations and determine their possibility. (I’d like to soon use this new found knowledge gained from Levi and you to prove to myself that my experimental conclusions on the possible 4D checkerboards were right.)
Anyway, I ran across the following small surprise thinking about the parity discussion (though it isn’t related to that), and figured I’d share with everybody…
A d-dimensional Rubik’s cube with n cubies per side has the same number of cubies as stickers when n is equal to the number of faces (that is, when n = f = 2d). This occurs when n = 6 on a 3D cube, when n = 8 on a 4D cube, and so on for the 10^5, 12^6, 14^7, etc. The math showing this is simple. You just solve for n in the following equation, where the left side represents the number of cubies and the right the number of stickers.
n^d = 2d * n^(d-1)
when n < 2d, the number of stickers is greater than the number of pieces.
when n > 2d, the number of stickers is less than the number of pieces.
This reversal comes into play since the number of hidden, 0-colored cubies grows more quickly with increasing n than other piece types, adding to the piece count but not to the sticker count. 1C cubie numbers are a wash for the difference of the two counts. Higher colored cubies, which contribute more stickers than pieces, don’t grow in number fast enough to keep up with the 0Cs, even with all 2C … dC cubie types combined.
Why do the number of 0-colored pieces grow so much faster than the others, even taken together? Consider a puzzle with very large n. In the limit, 0-coloreds are the only piece type that are filling up the full dimension of the cube. 1-coloreds fill up the faces of dimension d-1, 2-coloreds fill up the d-2 spaces, etc. And higher dimensional spaces are more voluminous, so it makes sense 0C will win out in the end.
I found n = f neat because a priori, why should the number per side have any relationship whatsoever to the the number of faces? (maybe this surprise is just the fact that the number of faces = 2d in disguise.) I also wonder why the existence of puzzles where the number of stickers and cubies coincide should even be guaranteed, another fact not a priori obvious to me. A non-existence conclusion that can be drawn is that no puzzle (of any dimension) with odd n can have the the same number of stickers and cubies, since the n = 2d constraint would imply fractional dimension.
Take Care All,
Roice
P.S. I want to defer to the group on the use of m^n verses n^d. In this email, I wanted to say "n-dimensional" at one point, but that would have conflicted with my usual labels. It got me second-guessing myself. Any opinions? Maybe we should make a poll?