"Domain error" from sqrt() when drawing an ellipse?

Question

The program to draw an ellipse using polynominal algoritm.

#include< graphics.h>        

#include< stdio.h>    

#include< math.h>    

#include< conio.h>    

void main() {   

int gd=DETECT,gm;    

float a,b,h,k;    

float x,xend,y;     

clrscr();    

initgraph(&gd,&gm,"C:\\TC\\BGI");    

printf("Enter the semi major axis and semi minor axis:\n");    

scanf("%f%f",&a,&b);    

printf("\nEnter the center coordinates, (h,k):");    

scanf("%f%f",&h,&k);    

xend=a;    

for(x=0;x<=xend;x = x+0.1){    

    y=b*sqrt(1-(x*x)/(a));    

    putpixel(x+h,y+k,RED);    

    putpixel(-x+h,y+k,RED);    

    putpixel(-x+h,-y+k,RED);    

    putpixel(x+h,-y+k,RED);    

}    

getch();    

//closegraph();    

}    

/*

output

Enter the semi major axis and semi minor axis:

200 130

Enter the center coordinates, (h,k):

200 230

*/

The output of the program shows

sqrt: Domain error

How to solve this error.

Answer 1

There are several issues here:

• The domain error results from passing an illegal argument to sqrt. The sqrt function takes a non-negative argument and returns the non-negative square root. It is a good idea to check the argument of the square root, the discriminant, before calling sqrt unless you can be absolutely sure that the argument is valid.

• In your loop, you accumulate floating-point numbers. Such additions are inaccurate, especially if you do many accumulations. It is better to control your loop with integers and then calculate a floating-point number from these integers.

• Your formula isn't correct. An ellipsis is described by

  (x/a)² + (y/b)² = 1
  

  You are missing one a in the denominator here. (It may be better to use a normalised rnning variable.)

Putting all this together:

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

int main()
{
    float a = 200.0f;
    float b = 130.0f;

    int i;
    int n = 50;

    for (i = 0; i <= n; i++) {
        float c = 1.0f * i / n;          // normalised x
        float x = a * c;                 // actual x
        float discr = 1.0f - c*c;        // calc. discriminat separately ...
        float y = (discr <= 0) ? 0       // ... so we can check it
                               : b * sqrtf(discr);

        printf("%20.12f%20.12f\n", x, y);
    }

    return 0;
}

(The third point is the most important, of course. When you deal with discrete pixels, it might also be questionable to increment your running variable by fractions of 1.0.)

Answer 2

Your problem is here:

for(x=0;x<=xend;x = x+0.1){    

    y=b*sqrt(1-(x*x)/(a));

X is 0 at the start and 1 - (x * x) / (a) can be negative if the division results in a value less than 1. 1 - (0 * 0) / (a) will be 1 in most cases (a == 0 is a special case). So, you pass a negative value to sqrt, which expects positive values. In reality one can calculate the square root of a negative number as well if the domain of the result is the set of Complex numbers, but, unfortunately sqrt is not applicable for that purpose.