Date: Sat, 27 Jan 1996 18:23:18 -0500
X-Sender: Keith_Adams@postoffice.brown.edu Mime-Version: 1.0
To: Thomas_Banchoff@postoffice.brown.edu From: Keith_Adams@brown.edu (Keith Adams) Subject: Re: How Many Edges Hath a Hypercube? 

>>edges of a cube of n dimensions:

>>_
>>/ n>1: 2 * E(n-1) + V(n-1)
>>|
>>E(n) = <
>>|
>>\ n<=1: n
>>-

I have the lousiest idea of how to actually go about snapping recursive functions into closed form. I am usually only able to do it when it's an intuitively obvious leap, such as V(n)=2^n. However, I think it's possible to use data from this recursive function to make an intuitive clutch at an answer.

Dimension	Edges	Vertices
0	0	1
1	1	2
2	4	4
3	8	12
4	16	32
5	32	80

All these vertices, with the exception of the one for dimension zero, are divisible by their dimension. It looks, in fact, like they're all half the edge number times the number of their dimension. I'm at a loss to argue why that should be, though. Is there some reasonable, procedural way of looking at the recursive function and smooshing it down into E(n)=n*2^n/2 ? 

Keith Adams