# 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?

>