"Paul B. Webster VK2BZC" wrote:
>
> Scott Dattalo wrote:
>
> > One of the tricks that this routine does is to simultaneously multiply
> > and divide. This works well for two operands, but I don't see how to
> > make it work for three.  Also, the 'multiplication' is a bit-by-bit
> > either-or operation. In other words, at each stage of the
> > multiplication, the cumulative product is either being increased by
> > one coefficient or the other.  Short of going to trits, I don't see
> > how to handle three operands at the same time. Could you be more
> > specific?
>
>   Easey-peasey!  Let me find the original....

Cool, Paul CAN write PIC code. :)

>
> twist3:
>         movf    y,w
>         movwf   num
>
>         btfsc   A,0
>          movf   x,w
>         btfsc   B,0
>          movf   z,w
>         addwf   num,f
>         rrf     num,f
>
>         movf    y,w     ;A macro would work well here...
>         btfsc   A,1     ;All we're doing is an un-rolled
>          movf   x,w     ;multiplication. If a bit in 'A' is
>         btfsc   B,1     ;set, we add 'y' to the running sum
>          movf   z,w     ;or if a bit in 'B' is set, we add 'z'
>         addwf   num,f   ;otherwise we add 'x'. In either case
>         rrf     num,f   ;we divide by 16 overall.
>
>         movf    y,w
>         btfsc   A,2
>          movf   x,w
>         btfsc   B,2
>          movf   z,w
>         addwf   num,f
>         rrf     num,f
>
>         movf    y,w
>         btfsc   A,3
>          movf   x,w
>         btfsc   B,3
>          movf   z,w
>         addwf   num,f
>         rrf     num,f
>
>         return
>
>   Note that A & B = 0 (although the algorithm won't overflow)
> Coefficients for x, y and z are (16-A-B)/16, A/16 and B/16


But unfortunately, this won't work as expected.

The subtle trick is that the the coefficients for x and y are two's complements
of one another in the routine I posted. In your twist3, the coefficients for x
and for y are no longer necessarily two's complements. Consequently, there will
be stages in the multiplication where it is necessary to add x and add y (and
even maybe z).

As an example, suppose the coefficients for x,y, and z are 5, 7, and 4 (the sum
is 16 as required).

If you look at the bit patterns:

5  -  0101
7  -  0111
4  -  0100

It can be seen that for stage 0 (the lsb) that x & y need to be added, stage 1
only y, stage 2 x & y & z, and on stage 3 nothing.

Is this clear?

Scott