Path: utzoo!attcan!uunet!lll-winken!lll-tis!helios.ee.lbl.gov!pasteur!ames!amdahl!ems!srcsip!shankar
From: shankar@srcsip.UUCP (Subash Shankar)
Newsgroups: comp.arch
Subject: Re: non-binary hardware
Message-ID: <8756@srcsip.UUCP>
Date: 16 Sep 88 16:13:21 GMT
References: <1285@mcgill-vision.UUCP> <3473@phri.UUCP> <5718@utah-cs.UUCP> <655@calvin.EE.CORNELL.EDU> <2997@pt.cs.cmu.edu> <17234@apple.Apple.COM>
Reply-To: shankar@ely.UUCP (Subash Shankar)
Organization: Honeywell Systems & Research Center, Camden, MN
Lines: 30

In article <17234@apple.Apple.COM> baum@apple.UUCP (Allen Baum) writes:
>
> [about multi-valued logic]
>There is some sort of proof, which I've seen, but don't remember details of,
>that the optimal base for computation is e (2.71828...). The proof wasn't
>very complicated, and I've forgotten what the measure is of 'best'. Anyway,
>bay in the 50's and early 60's, it was believed by some that 3 was closer to 
>'e' than 2, so the up and built some base 3 computers. 

The proof is as follows:

  Assume that hardware cost per digit is linearly proportional to the radix. 
    C = k1 * r   (C is cost per digit, r is radix)
  We know that the number of digits to represent a number, m, in radix r is
  approximately log(m) (base r) = ln(m)/ln(r)
    n = ln(m)/ln(r)  (n is number of digits, m is largest number (constant))
  So, the total cost is:
    Cn = k1 * r * ln(m) / ln(r)
  Differentiating (to maximize) with respect to r results in
                 ln(r)-1
    k1 * ln(m) * -------
                 ln(r)^2
  Setting equal to 0 and solving results in
    ln(r)=1   ==>  r=e

There is a serious flaw in this proof - it is not at all clear that the
hardware cost per digit is linearly proportional to the radix; hence the
controversy over the "optimal" radix remains an open question.  A second
flaw is that the above shows optimal radices for functional components such
as ALU's, but with interconnect being more important on chips, multi-valued
logic may be of great use.