/* ---------------------------------------------------------------- */
/* This program illustrates the impact of loss of significant       */
/* digits.  It solves a quadratic equation x^2 + bx + c = 0 using   */
/* two different ways.  The first is the one you learned in your    */
/* algebra course, which the other tries to minimize the effect.    */
/* ---------------------------------------------------------------- */

#include  <stdio.h>
#include  <stdlib.h>
#include  <math.h>

#define   TOL   (1.0e-07)

void  main(int argc, char *argv[])
{
     float  a, b, c;               /* input coefficients            */
     float  p, q;                  /* normalized coefficients       */
     float  d;                     /* discriminant                  */
     float  r1, r2;                /* roots                         */

     if (argc != 4) {
          printf("Use %s a b c\n", argv[0]);
          printf("Since you did not supply anything, I will use\n"
                 " a = 1, b = -20000 and c = 1.\n\n");
          a = 1.0;
          b = -20000.0;
          c = 1.0;
     }
     else {
          a = atof(argv[1]);
          b = atof(argv[2]);
          c = atof(argv[3]);
     }

     printf("Input equation: (%f)x^2 + (%f)x + (%f) = 0\n\n", a, b, c);
     if (a == 0.0) {
          printf("Coefficient a must not be zero\n");
          exit(1);
     }

     d = b*b - 4.0*a*c;
     if (d < 0.0) {
          printf("Equation has complex roots\n");
          exit(1);
     }
     d  = sqrt(d);
     r1 = (-b + d)/(a+a);
     r2 = (-b - d)/(a+a);
     printf("Roots by a naive method:\n\t%f and %f\n", r1, r2);

     /* ----------------------------------------------------------- */
     /* root = (-b +- sqrt(b^2-4ac))/(2a) is multiplied by          */
     /*    -b -+ sqrt(b^2-4ac)                                      */
     /*    -------------------                                      */
     /*    -b -+ sqrt(b^2-4ac)                                      */
     /* the roots are:                                              */
     /*           -2c                                               */
     /*    ------------------                                       */
     /*    b +- sqrt(b^2-4ac)                                       */
     /* thus, if b>0, + is chosen; otherwise, use -.  this makes    */
     /* sure that the denominator does not have subtraction!        */
     /* ----------------------------------------------------------- */

     d = sqrt(b*b - 4.0*a*c);
     if (b > 0.0)
          r1 = -2.0*c/(b + d);
     else
          r1 = -2.0*c/(b - d);
     r2 = (c/a)/r1;
     printf("Roots by a less well-known formula:\n");
     printf("\t%f and %f\n", r1, r2);

     /* ----------------------------------------------------------- */
     /* The given equation is transformed to x^2+2px+q=0, where p   */
     /* and q are defined as b/(2a) and c/a respectively.  Then, the*/
     /* roots are (-p +- sqrt(p*p-q)).  To avoid using subtraction, */
     /* do the following:                                           */
     /*   if p > 0: in this case, -p<0. The '-' is chosen and a root*/
     /*             is -p-sqrt(p*p-q)                               */
     /*   if p < 0: in this case, -p>0. The '+' is chosen and a root*/
     /*             is -p+sqrt(p*p-q)                               */
     /* ----------------------------------------------------------- */

     p = b/a/2.0;
     q = c/a;
     d = sqrt(p*p - q);
     if (p >= 0.0)
          r1 = (-p - d);
     else
          r1 = (-p + d);
     r2 = q/r1;
     printf("Roots by significant digits preserving method:\n");
     printf("\t%f and %f\n", r1, r2);

}