A better program to evaluate square root of a number

Question

I am a beginner in C programming language and I was given this task to write a program that calculates the square root of a natural number without using math.h

Answer 1

You appear to be using the bisection method. This will work, and is a stable approximation method, but is very slow.

A better approach would be the Newton-Raphson method. There are instances where it can become unstable (ie: nth order polynomial root solving where multiple roots are very close to each other, eg: y=(x-1)(x-1.1)(x-1.1), but it gets the job done. By faster, I mean that it will require fewer iterations to arrive at a desired number of significant figures compared to your approach.

Edit: Since you must use the divide-and-conquer approach, I have put both algorithms in the code.

Code Listing

---

/* Preprocessor inclusions */
#include <stdio.h>

/* Preprocessor Macros */
#define TOL (0.001)     /* Error tolerance bound */
#define FABS(a)   ((a)>0.0?(a):-(a))

/* Function Prototypes */
double sq_root(double x);
int divideAndConquerAndTruncateForSquareRoot(int n);

/* Function Definitions */
double sq_root(double x) {
    double rt = 1, ort = 0;

    while(FABS(ort-rt) > TOL) {
        ort = rt;
        rt = ((x/rt) + rt) / 2;
    }
    return rt;
}

int divideAndConquerAndTruncateForSquareRoot(int n) {
   int low = 0;
   int high = n+1;
   int mid;
   while ((high-low) > 1) {
      mid = (low+high) / 2;
      if ((mid*mid) <= n) {
         low = mid;
      } else {
         high = mid;
      }
   }
   return low;
}

int main(void) {
    int i;
    printf("Calculating floating point square roots with Newton-Raphson method with a tolerance of %lf\n", TOL);
    for(i = 2; i<10; i++) {
       printf("square root of %d is %f\n",i, sq_root(i));
    }
    printf("Calculating truncated integer square roots with divide and conquer approach\n");
    for(i = 2; i<10; i++) {
       printf("square root of %d is %d\n",i, divideAndConquerAndTruncateForSquareRoot(i));
    }

    return 0;
}

Sample Output

---

Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.001000
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating truncated integer square roots with divide and conquer approach
square root of 2 is 1
square root of 3 is 1
square root of 4 is 2
square root of 5 is 2
square root of 6 is 2
square root of 7 is 2
square root of 8 is 2
square root of 9 is 3

Also, here's a demonstration of the important of defining error bounds and tolerances for a root-solving algorithm. Notice how much the error bounds change:

Code Listing

/* Preprocessor inclusions */
#include <stdio.h>

/* Preprocessor Macros */
#define FABS(a)   ((a)>0.0?(a):-(a))
double TOL = 1.0;

/* Function Prototypes */
double sq_root(double x);

/* Function Definitions */
double sq_root(double x) {
    double rt = 1, ort = 0;

    while(FABS(ort-rt) > TOL) {
        ort = rt;
        rt = ((x/rt) + rt) / 2;
        diff = ort-rt;
    }
    return rt;
}

int main(void) {
    int i;
    int j;
    TOL = 1.0;
    for (i=0; i<5; i++, TOL /= 10) {
       printf("Calculating floating point square roots with Newton-Raphson method with a tolerance of %lf\n", TOL);
       for(j = 2; j<10; j++) {
          printf("square root of %d is %f\n", j, sq_root(j));
       }
    }
    return 0;
}

Sample Output

---

Calculating floating point square roots with Newton-Raphson method with a tolerance of 1.000000
square root of 2 is 1.000000
square root of 3 is 1.000000
square root of 4 is 1.000000
square root of 5 is 1.000000
square root of 6 is 1.000000
square root of 7 is 1.000000
square root of 8 is 1.000000
square root of 9 is 1.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.100000
square root of 2 is 1.416667
square root of 3 is 1.732143
square root of 4 is 2.000610
square root of 5 is 2.238095
square root of 6 is 2.449494
square root of 7 is 2.645767
square root of 8 is 2.828469
square root of 9 is 3.000092
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.010000
square root of 2 is 1.414216
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236069
square root of 6 is 2.449494
square root of 7 is 2.645767
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.001000
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.000100
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.000010
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.000001
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000
Calculating floating point square roots with Newton-Raphson method with a tolerance of 0.000000
square root of 2 is 1.414214
square root of 3 is 1.732051
square root of 4 is 2.000000
square root of 5 is 2.236068
square root of 6 is 2.449490
square root of 7 is 2.645751
square root of 8 is 2.828427
square root of 9 is 3.000000

Finally, here's one last example for comparing the number of iterations each approach uses to generate the same results.

Code Listing

---

/*******************************************************************************
 * Preprocessor Inclusions and Definitions
 ******************************************************************************/
#include <stdio.h>

#define FABS(a)   ((a)>0.0?(a):-(a))
typedef double (*fcnPtr)(double, double, int*);

/*******************************************************************************
 * Function Prototypes
 ******************************************************************************/
double sqrt_bisection(double x, double tolerance, int* iterations);
double sqrt_newton_raphson(double x, double tolerance, int* iterations);

/*******************************************************************************
 * Function Definitions
 ******************************************************************************/
/******************************************************************************/
double sqrt_newton_raphson(double x, double tolerance, int* iterations) {
   if (!iterations || tolerance < 0.0) {
      printf("Invalid input\n");
      return 0.0;
   }
   *iterations = 0;
   double rt = 1, ort = 0;

   while(FABS(ort-rt) > tolerance) {
      (*iterations)++;
      ort = rt;
      rt = ((x/rt) + rt) / 2;
   }
   return rt;
}

/******************************************************************************/
double sqrt_bisection(double x, double tolerance, int* iterations) {
   if (!iterations || tolerance < 0.0) {
      printf("Invalid input\n");
      return 0.0;
   }
   *iterations = 0;
   double low = 0;
   double high = x+1;
   double mid;
   while (FABS(high-low) > tolerance) {
      (*iterations)++;
      mid = (low+high) / 2;
      if ((mid*mid) <= x) {
         low = mid;
      } else {
         high = mid;
      }
   }
   return low;
}

/******************************************************************************/
int main(void) {
   int i, j, k;
   fcnPtr fn = NULL;
   double tolerance;
   int iterations;
   fcnPtr functions[2] = { sqrt_bisection, sqrt_newton_raphson };
   char* functionNames[2] = { "Bisection Method", "Newton Raphson Method" };

   for (i=0; i<2; i++) {
      fn = functions[i];
      for(j = 2; j<10; j++) {
         tolerance = 0.1;
         for (k=0; k<10; k++) {
            printf("square root of %d is %lf (%d iterations for tolerance of %e) using %s\n", j,
                  fn((double)j, tolerance, &iterations), iterations, tolerance, functionNames[i]);
            tolerance /= 10;
         }
      }
   }

   return 0;
}

Sample Output

---

square root of 2 is 1.406250 (5 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 2 is 1.412109 (9 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 2 is 1.413574 (12 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 2 is 1.414124 (15 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 2 is 1.414209 (19 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 2 is 1.414213 (22 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 2 is 1.414213 (25 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 2 is 1.414214 (29 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 2 is 1.414214 (32 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 2 is 1.414214 (35 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 3 is 1.687500 (6 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 3 is 1.726562 (9 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 3 is 1.731445 (12 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 3 is 1.731995 (16 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 3 is 1.732048 (19 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 3 is 1.732050 (22 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 3 is 1.732051 (26 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 3 is 1.732051 (29 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 3 is 1.732051 (32 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 3 is 1.732051 (36 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 4 is 1.953125 (6 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 4 is 1.992188 (9 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 4 is 1.999512 (13 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 4 is 1.999969 (16 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 4 is 1.999998 (19 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 4 is 2.000000 (23 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 4 is 2.000000 (26 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 4 is 2.000000 (29 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 4 is 2.000000 (33 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 4 is 2.000000 (36 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 5 is 2.156250 (6 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 5 is 2.232422 (10 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 5 is 2.235352 (13 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 5 is 2.235992 (16 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 5 is 2.236067 (20 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 5 is 2.236068 (23 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 5 is 2.236068 (26 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 5 is 2.236068 (30 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 5 is 2.236068 (33 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 5 is 2.236068 (36 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 6 is 2.406250 (7 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 6 is 2.447266 (10 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 6 is 2.448975 (13 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 6 is 2.449455 (17 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 6 is 2.449489 (20 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 6 is 2.449489 (23 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 6 is 2.449490 (27 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 6 is 2.449490 (30 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 6 is 2.449490 (33 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 6 is 2.449490 (37 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 7 is 2.625000 (7 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 7 is 2.640625 (10 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 7 is 2.645508 (13 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 7 is 2.645691 (17 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 7 is 2.645744 (20 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 7 is 2.645751 (23 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 7 is 2.645751 (27 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 7 is 2.645751 (30 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 7 is 2.645751 (33 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 7 is 2.645751 (37 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 8 is 2.812500 (7 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 8 is 2.821289 (10 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 8 is 2.827881 (14 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 8 is 2.828362 (17 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 8 is 2.828422 (20 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 8 is 2.828427 (24 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 8 is 2.828427 (27 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 8 is 2.828427 (30 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 8 is 2.828427 (34 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 8 is 2.828427 (37 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 9 is 2.968750 (7 iterations for tolerance of 1.000000e-01) using Bisection Method
square root of 9 is 2.998047 (10 iterations for tolerance of 1.000000e-02) using Bisection Method
square root of 9 is 2.999878 (14 iterations for tolerance of 1.000000e-03) using Bisection Method
square root of 9 is 2.999954 (17 iterations for tolerance of 1.000000e-04) using Bisection Method
square root of 9 is 2.999992 (20 iterations for tolerance of 1.000000e-05) using Bisection Method
square root of 9 is 3.000000 (24 iterations for tolerance of 1.000000e-06) using Bisection Method
square root of 9 is 3.000000 (27 iterations for tolerance of 1.000000e-07) using Bisection Method
square root of 9 is 3.000000 (30 iterations for tolerance of 1.000000e-08) using Bisection Method
square root of 9 is 3.000000 (34 iterations for tolerance of 1.000000e-09) using Bisection Method
square root of 9 is 3.000000 (37 iterations for tolerance of 1.000000e-10) using Bisection Method
square root of 2 is 1.416667 (2 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 2 is 1.414216 (3 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 2 is 1.414214 (4 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 2 is 1.414214 (4 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 2 is 1.414214 (4 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 2 is 1.414214 (5 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 2 is 1.414214 (5 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 2 is 1.414214 (5 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 2 is 1.414214 (5 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 2 is 1.414214 (5 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 3 is 1.732143 (3 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 3 is 1.732051 (4 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 3 is 1.732051 (4 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 3 is 1.732051 (4 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 3 is 1.732051 (5 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 3 is 1.732051 (5 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 3 is 1.732051 (5 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 3 is 1.732051 (5 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 3 is 1.732051 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 3 is 1.732051 (6 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 4 is 2.000610 (3 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 4 is 2.000000 (4 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 4 is 2.000000 (4 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 4 is 2.000000 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 4 is 2.000000 (5 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 4 is 2.000000 (5 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 4 is 2.000000 (5 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 4 is 2.000000 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 4 is 2.000000 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 4 is 2.000000 (6 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 5 is 2.238095 (3 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 5 is 2.236069 (4 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 5 is 2.236068 (5 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 5 is 2.236068 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 5 is 2.236068 (5 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 5 is 2.236068 (5 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 5 is 2.236068 (6 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 5 is 2.236068 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 5 is 2.236068 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 5 is 2.236068 (6 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 6 is 2.449494 (4 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 6 is 2.449494 (4 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 6 is 2.449490 (5 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 6 is 2.449490 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 6 is 2.449490 (5 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 6 is 2.449490 (6 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 6 is 2.449490 (6 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 6 is 2.449490 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 6 is 2.449490 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 6 is 2.449490 (6 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 7 is 2.645767 (4 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 7 is 2.645767 (4 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 7 is 2.645751 (5 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 7 is 2.645751 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 7 is 2.645751 (6 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 8 is 2.828469 (4 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 8 is 2.828427 (5 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 8 is 2.828427 (5 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 8 is 2.828427 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 8 is 2.828427 (6 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 8 is 2.828427 (6 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 8 is 2.828427 (6 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 8 is 2.828427 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 8 is 2.828427 (6 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 8 is 2.828427 (7 iterations for tolerance of 1.000000e-10) using Newton Raphson Method
square root of 9 is 3.000092 (4 iterations for tolerance of 1.000000e-01) using Newton Raphson Method
square root of 9 is 3.000000 (5 iterations for tolerance of 1.000000e-02) using Newton Raphson Method
square root of 9 is 3.000000 (5 iterations for tolerance of 1.000000e-03) using Newton Raphson Method
square root of 9 is 3.000000 (5 iterations for tolerance of 1.000000e-04) using Newton Raphson Method
square root of 9 is 3.000000 (6 iterations for tolerance of 1.000000e-05) using Newton Raphson Method
square root of 9 is 3.000000 (6 iterations for tolerance of 1.000000e-06) using Newton Raphson Method
square root of 9 is 3.000000 (6 iterations for tolerance of 1.000000e-07) using Newton Raphson Method
square root of 9 is 3.000000 (6 iterations for tolerance of 1.000000e-08) using Newton Raphson Method
square root of 9 is 3.000000 (7 iterations for tolerance of 1.000000e-09) using Newton Raphson Method
square root of 9 is 3.000000 (7 iterations for tolerance of 1.000000e-10) using Newton Raphson Method

References

---

1. 1.1.7 Example: Square Roots by Newton's Method, Accessed 2014-08-19, <http://mitpress.mit.edu/sicp/chapter1/node9.html>
2. Square root in C using Newton-Raphson method, Accessed 2014-08-19, <https://stackoverflow.com/questions/14038456/square-root-in-c-using-newton-raphson-method>
3. Newton's Method, Accessed 2014-08-19, <http://en.wikipedia.org/wiki/Newton%27s_method>

Answer 2

OP's method can be fixed with the following. The method only works for int in the range 0 <= x < sqrt(INT_MAX)*2.

scanf("%d", &a);
int s = 1, l = a, mid = 0;
while (s <= l) {
  mid = (s + l) / 2;
  if ((mid * mid) == a) {
    break;
  }
  if ((mid * mid) < a) {
    s = mid + 1;
  } else {
    l = mid - 1;
  }
}
mid = (s + l) / 2;
printf("%d\n", mid);

Wiki has a nice solution. Following is an unsigned version that works for all unsigned. It is fast and simple. On embedded processors without a 1-instruction clock multiply, likely the fastest.

[Edit 2017] Added portable greatest power-of-4 code

#include <stdio.h>
#include <limits.h>

// greatest power of 4 <= a power-of-2 minus 1
#define POW4_LE_POW2M1(pow2m1) ((((pow2m1)/2) + 1) >> ((pow2m1)%3==0))

unsigned isqrt(unsigned num) {
  unsigned res = 0;

  // The greatest power-of-4
  unsigned bit = POW4_LE_POW2M1(UINT_MAX);

  // "bit" starts at the highest power of four <= the argument.
  while (bit > num)
    bit >>= 2;

  while (bit != 0) {
    if (num >= res + bit) {
      num -= res + bit;
      res = (res >> 1) + bit;
    } else
      res >>= 1;
    bit >>= 2;
  }
  return res;
}

---

[Edit] In a comment in another answer, @Nils Pipenbrinck requested a integer sqrt function based on Newton-Raphson. The following performs a bit faster than requested as it makes a better first guess than assuming 1. For 32-bit unsigned, it does 5 max iterations. The tricky bit was dealing with a proper terminating condition.

unsigned isqr_Newton(unsigned x) {
  unsigned rt[3] = {1,0,0};  // current and previous 2 root candidates

  unsigned x2 = x; // Form initial guess
  while (x2 > rt[0]) {
    x2 >>= 1;
    rt[0] <<= 1;
  }

  do {
    rt[2] = rt[1];
    rt[1] = rt[0];
    rt[0] = (x/rt[0] + rt[0])/2;
    } while (rt[1] != rt[0] && rt[2] != rt[0]);
  return (rt[0] + rt[1])/2;
}