Date   : Wed, 15 Jun 2005 22:54:56 +0200
From   : Carlo <radiorama@...>
Subject: Re: Faster 6502 multiplication

Richard_Talbot-Watkins@... wrote:

 > Your first method would be more efficient than the second if the CPU was
 > having to divide the term(s) by 4 itself.
 >
 > But what I am trying to do is get rid of the division by 4 altogether by
 > looking up "pre-divided-by-4" values from the table.  Remember, that
 > (a+b)^2 - (a-b)^2 is an 18-bit result, and so would require more byte
 > lookups and/or some additional bit-shifting to get the result.  I am trying
 > to create code which is lean on CPU cycles here!

Sorry for reading your message superficially. I didn't enter deeply into the 
description of the algorithm...

> *bangs forehead against desk*
> 
> D'oh, I should've considered that more closely before my last reply.
> 
> Of course, since it's a multiple of 4, the bottom two bits will always be
> zero.  Which means the bottom two bits of (a+b)^2 and (a-b)^2 have to be
> equal.  So we can discard them straight away as we are not expecting
> anything other than a result of 00 from them.

That's exactly what I meant to say of course :)

Ciao,
Carlo.