Re: Calculating SQRT in ADA

[Hans Marqvardsen <hm@ddre.dk>]

John Herro wrote:
> 
> Here's a generic square root routine in Ada 83, taken from my AdaTutor program:
> 
> generic
>   type Dummy is digits <>;
> function Sqrt(X :in Dummy) return Dummy;
> function Sqrt(X :in Dummy) return Dummy is
>   Guess : Dummy := X;

    -- insert, just in case the compiler fail to recognize 
    -- a constant
    Tolerance : constant Dummy := 3.0*Dummy'epsilon*X;

> begin
>   if X < 0.0 then
>     raise Constraint_Error;
>   end if;
>   while X /= 0.0 and then
>        abs(Guess*Guess/X - 1.0) > 3.0*Dummy'Epsilon loop

    -- replace the above with simpler and faster:
    while (Guess*Guess - X) > tolerance loop


>     Guess := (X/Guess + Guess) * 0.5;
>   end loop;
>   return Guess;
> end Sqrt;
> 
> I tested this routine with Vax Ada 83, which has types Float, Long_Float, and
> Long_Long_Float.  That last type gives at least 33 decimal digits of precision.
>  I instantiated this function, as well as Text_IO.Float_IO, for all three
> types.  When I tested with the three types, all displayed digits of the answers
> were correct.  I hereby place this long and complex function :-) into the
> public domain.  Enjoy.
> - John Herro
> You can download a shareware AdaTutor program at
> http://members.aol.com/AdaTutor

Bravo, amazing that it is this simple to find a Square root
efficiently!

A few questions out of general interest:

Where doest the multiplier 3.0 come from ?

Cant you make the routine even simpler and faster, 
multiplying with X, after having established that X is 
not-negative? 
(At the same time avoiding a division 
and the need to test for X=0.0?)

Do you really need to use abs ? 
(Isn't always Guess >=  'true sqrt' anyway?)

Sincerely,
Hans Marqvardsen