Square root of bignum using GMP
Asked Answered
G

3

7

I need to get the square root of a 210 digit number accurately, I thought GMP was the right tool for the job, what am I doing wrong?

#include <stdlib.h>
#include <stdio.h>
#include "gmp.h"

int
main (int argc, char *argv[])
{
  mpz_t sq_me, sq_out, test;
  mpz_init(sq_me);
  mpz_init(sq_out);
  mpz_init(test);
  mpz_set_str (sq_me, argv[1], 10);

  mpz_sqrt(sq_out, sq_me);
  mpz_mul(test,sq_out,sq_out);

  gmp_printf ("%Zd\n\n", sq_out);
  gmp_printf ("%Zd\n\n", test);

  return 0;
}

Input:

24524664490027821197651766357308801846702678767833275974341445171506160083003858 72169522083993320715491036268271916798640797767232430056005920356312465612184658 17904100131859299619933817012149335034875870551067

Output:

49522383313031109809242226159886283348695660460381271324714928680654813093947239 9634016783775955618921028

24524664490027821197651766357308801846702678767833275974341445171506160083003858 72169522083993320715491034366358025027526868495267716284867043049443779615862887 47102011391915422793532619329760963626718900576784

Gizzard answered 5/5, 2009 at 0:10 Comment(0)
P
12

Here's the code you need for floating point square root, you can see that the initial input and final output are identical.

#include <stdlib.h>
#include <stdio.h>
#include "gmp.h"

int main (int argc, char *argv[]) {
    mpf_t sq_me, sq_out, test;
    mpf_set_default_prec (10000);
    mpf_init(sq_me);
    mpf_init(sq_out);
    mpf_init(test);
    mpf_set_str (sq_me, argv[1], 10);

    mpf_sqrt(sq_out, sq_me);
    mpf_mul(test,sq_out,sq_out);

    gmp_printf ("Input:       %Ff\n\n", sq_me);
    gmp_printf ("Square root: %.200Ff\n\n", sq_out);
    gmp_printf ("Re-squared:  %Ff\n\n", test);

    return 0;
}

Here's the output with your parameter:

Input:       2452466449002782119765176635730880184670267876783327597434144
51715061600830038587216952208399332071549103626827191679864079776723243005
60059203563124656121846581790410013185929961993381701214933503487587055106
7.000000

Square root: 4952238331303110980924222615988628334869566046038127132471492
86806548130939472399634016783775955618921028.19202568258368255653837168412
92356432661548614332014106174638951390596672950394981098992388116308833260
04535647648563996144250924277757344248059826024201642748515325655438898558
17807282091590722890002

Re-squared:  2452466449002782119765176635730880184670267876783327597434144
51715061600830038587216952208399332071549103626827191679864079776723243005
60059203563124656121846581790410013185929961993381701214933503487587055106
7.000000
Pedestrianism answered 5/5, 2009 at 2:26 Comment(0)
R
5

You're getting the result as an integer and then squaring that. The input number must not be a perfect square, so it is truncating the decimals and decreasing the precision of the number. Look into the 'mpf' category of functions for floats rather than 'mpz' for integers.

Ramonaramonda answered 5/5, 2009 at 0:14 Comment(1)
"The input number must not be a perfect square" - what with the last digit being "7", and all...Dragster
S
0

I am using a libgmp-3.dll

In vb.net I wrote this function for a high precision square root. I haven't tested it thoroughly but it gets me the Square Root of 3 to any precision I need

 Public Shared Function SquareRoot(ByVal Value As BigInt, ByVal Precision As Integer) As String
    Dim Ten As New BigInt(10)
    Dim DecLen As Integer = Value.Sqrt.ToString.Length
    Dim RootDigits As String = (Value * Ten.Power(Precision * 2)).Sqrt

    'Add trailing zeros
    RootDigits = RootDigits.PadRight((DecLen + Precision), "0")
    Return RootDigits.Substring(0, DecLen) & "." & RootDigits.Substring(DecLen)
End Function
Sorensen answered 27/1, 2010 at 23:29 Comment(0)

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