Message #639

From: rev_16_4 <rev_16_4@yahoo.com>
Subject: Re: a short diversion into sticker and cubie counts
Date: Tue, 03 Feb 2009 06:31:56 -0000

That’s a delightful finding. If n=2d, then stickers=cubies. I’ve
always thought the 6^3 had a sort of hidden beauty to it.

As for the m^n or n^d. My preference for m^n stems from my use of
single letters to label my algoriths, from a to g. But n^d uses more
descriptive variable names. I think developing a standard is
essential for simplifying discussions.

-Levi

— In 4D_Cubing@yahoogroups.com, Roice Nelson <roice3@…> wrote:
>
> 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
> proveto myself that my experimental
> conclusions on the possible 4D
>
checkerboards<http://games.groups.yahoo.com/group/4D_Cubing/message/47
7>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?
>