On Tuesday, 1994-11-08  Rob Basler wrote to All about "Need sqrt()
algorithm" as follows:

RB> I need to do a distance calculation between two points 
RB> (x0,z0) and (x1,z1).  The way I know to do this is to use 
RB> the distance formula, but it requires a square root 
RB> calculation.

Hi Rob,

I presume you want this in fairly standard C. The following function
returns the square root of an integer, but the root is multiplied by
100 to provide 2 decimal places. For example, the square root of 10 is
3.16 (to 2 places) and this routine returns 316. You place the decimal
point yourself when your calculations are complete. The places are
gained by multiplying the argument by 10000; to get more places
increase the number of trailing zeroes by 2 for each additional place
you need (e.g. for 3 places multiply by 1000000).

Your distance would then be:

  dist = (isqrt((x1-x0)**2+(z1-z0)**2) + 50)/100;

The addition of 50 provides rounding instead of truncation. In 3
dimensions it would be:

  dist = (isqrt((x1-x0)**2+(y1-y0)**2+(z1-z0)**2) +50)/100;

Note that stdlib.h is needed for abs(). You can replace this if you
want.

I hope I understood your problem correctly. If not, provide me with
more details and I will try again.

Regards

Dave

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#include 

/* The function isqrt() returns the square root
   of an integer, multiplied by 100. This provides
   2 decimal places; the decimal point is assumed
   to be between the hundreds digit and the tens
   digit.

   It uses the Newton-Raphson algorithm, but in an
   integer domain instead of real numbers. */

unsigned long isqrt(unsigned long N)

{
   unsigned long t0,t1,t2;

   t0 = N*10000; /* dilation for 2 decimal places */
   t1 = t0/2;

   if (t1 > 0)
   do
   {
      t2 = t1;
      t1 = (t0/t1 + t1)/2;
   } while(abs(t2-t1) > 1);

   return t1;

}

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