I'm usually a lurker, but here's my two cents for someone with more
     time than me:

     Consider (A - B)^2 = A^2 + B^2 - 2AB

     so AB = (A^2 + B^2 - [A-B]^2)/2

     Since 101 > A > B, a small lookup table will suffice for all the
     squares. (Note that A - B > 101 too!)

     I'm not sure that this would end up being much faster, but at least
     it's different...

     --- phd


______________________________ Reply Separator _________________________________
Subject: Re: This year's programming challenge
Author:  fastfwd@ix.netcom.com at WDI-INTERNET
Date:    9/17/97 4:50 AM


Scott Dattalo <sdattalo@unix.SRI.COM> wrote:

> Just so you don't get the whole beer, here's one that's 34 cycles
> too (and probably the same as yours).

Guys:

Are you just gonna let Scott run away with the prize this year?  So
far, his is the only response I've received.

For what it's worth, Scott's routine couldn't BE more different from
the (as-yet-undisclosed) one that I wrote.  There are LOTS of other
ways to solve the problem I posed... Anyone else care to give it a
try?

-Andy
=== Meet other PICLIST members at the Embedded Systems Conference:
=== 6:30 pm on Wednesday, 1 October, at Bytecraft Limited's booth.
===
=== For more information on the Embedded Systems Conference,
=== see: http://www.embedsyscon.com/

=== Andrew Warren - fastfwd@ix.netcom.com
=== Fast Forward Engineering - Vista, California
=== http://www.geocities.com/SiliconValley/2499