# Message #638

From: Roice Nelson <roice3@gmail.com>

Subject: a short diversion into sticker and cubie counts

Date: Mon, 02 Feb 2009 23:12:13 -0600

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/477>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?