From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=0.2 required=5.0 tests=BAYES_00,INVALID_MSGID, REPLYTO_WITHOUT_TO_CC autolearn=no autolearn_force=no version=3.4.4 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 103376,fedc2d05e82c9174 X-Google-Attributes: gid103376,public From: Hans Marqvardsen Subject: Re: Calculating SQRT in ADA Date: 1999/03/24 Message-ID: <36F8C60D.26AA@ddre.dk>#1/1 X-Deja-AN: 458871277 Content-Transfer-Encoding: 7bit References: <36F913E0.75F51763@lmco.com> <19990324125518.01416.00000432@ngol08.aol.com> Content-Type: text/plain; charset=us-ascii Organization: DDRE Mime-Version: 1.0 Reply-To: hm@ddre.dk_nospam Newsgroups: comp.lang.ada Date: 1999-03-24T00:00:00+00:00 List-Id: 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