[algogeeks] Re: Implement division without using divide operator

[Atamurad Hezretkuliyev]

Here's working code, I used binary search.

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

#define EPS 0.00000001

double div(double a, double b)
{
    if( fabs(a-b) < EPS)
        return 1;
    if( fabs(b) < EPS )
        return NAN;
    if( fabs(a) < EPS )
        return 0;

    double sign = (a<0 ? 1 : -1) * (b<0 ? 1 : -1);

    a = fabs(a);
    b = fabs(b);
   
    double l = 0;
    double r = 1;
    while( b*r <= a ) { l=r; r=r*2; }

    double x;
    do {
        x = (l+r)/2;

        if( fabs(x*b-a) < EPS)     return sign * x;

        if( x*b > a )
            r = x;
        else
            l = x;

    } while(1);
}

int main()
{
    srand(time(NULL));
    for(int i=0; i<1000; i++) {
        double a = 0.001 * (rand() % 1000000) - rand() % 1000;
        double b = 0.001 * (rand() % 1000000) - rand() % 1000;

        double x = div(a,b);

        printf("%.3lf \\ %.3lf = %.3lf (%.3lf) %.3lf\n", a, b, x, a/b, 
fabs(x-(a/b)));

        if( fabs(x-(a/b)) > EPS) printf("Error!!!\n");
    }
}

megha wrote:
> Hi all,
>
> I am looking for the program/algorithm to implement division without
> using divide operator?
>
> Any idea?
>
> Thanks
>
>
> >
>
>   


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

#define EPS 0.00000001

double div(double a, double b)
{
	if( fabs(a-b) < EPS) 
		return 1;
	if( fabs(b) < EPS )
		return NAN;
	if( fabs(a) < EPS )
		return 0;

	double sign = (a<0 ? 1 : -1) * (b<0 ? 1 : -1);

	a = fabs(a);
	b = fabs(b);
	
	double l = 0;
	double r = 1;
	while( b*r <= a ) { l=r; r=r*2; }

	double x;
	do {
		x = (l+r)/2;

		if( fabs(x*b-a) < EPS) 	return sign * x;

		if( x*b > a ) 
			r = x;
		else
			l = x;

	} while(1);
}

int main()
{
	srand(time(NULL));
	for(int i=0; i<1000; i++) {
		double a = 0.001 * (rand() % 1000000) - rand() % 1000;
		double b = 0.001 * (rand() % 1000000) - rand() % 1000;

		double x = div(a,b);

		printf("%.3lf \\ %.3lf = %.3lf (%.3lf) %.3lf\n", a, b, x, a/b, fabs(x-(a/b)));

		if( fabs(x-(a/b)) > EPS) printf("Error!!!\n");
	}
}