Hi Rob,

On  Rob Landley wrote to Rob Basler:

 RL> To be serious, the answer you are looking for involves the application
 RL> of the Newton-Raphsen Iteration. 

The iterative approximation, as posted by Ronald Van Iwaarden:
x(new) = {x(old) +a/x(old)}/2 is Newton Raphson in its pure form.

x(new) =  x(old) - f(x(old))/f'(x(old))   (NR)

with f'(x) = df/dx.

f(x) = x*x - a, so f'(x) = 2x. If you fill in these results in (NR), you get:

x(new) = x(old) + x(old)*x(old)/{2x(old)} - a / {2x(old)} =
         x(old) - (1/2) x(old) + a/{2x(old)} = {x(old) + a/x(old)} /2

QED.


 RL>         E=.001 'DESIRED ACCURACY

Robs shoot2kill problem concearns pixels on a screen, so the result is
likely to be in unsigned integers. Since the difference between n*n and
(n+1)*(n+1) is 2n+1, it will be sufficient to ask whether abs(f(n)) <
2*n as stopping criterion. 


73 es cuagn,
 Henk

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