I have a problem of using sqrt function which is used in a C program

Question

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

int main()
{
    int n, i, flag = 0;
    printf("Enter the number\n");
    scanf("%d", &n);
    for (i = 2; i <= sqrt(n); i++)
    {
        if (n % i == 0)
        {
            flag = 1;
            break;
        }
    }
    if (flag == 0)
    {
        printf("%d is a prime number", n);
    }
    else
    {
        printf("%d is not a prime number", n);
    }
    return 0;
}

In the for loop why is the sqrt function is used? Because for checking if the number n is a prime number, shouldn't we check till n ?

Answer 1

If n is evenly divisible by i , then both i and n/i are divisors. If i is greater than sqrt(n) , then n/i is smaller than sqrt(n) , ie: for any divisor i > sqrt(n) there is a divisor j < sqrt(n) . If sqrt(n) is an integer, then it is a divisor of n .

Hence it suffices to test integers up to and including sqrt(n) .

Note these issues:

• computing sqrt(n) at each iteration is wasteful and the compiler might not be smart enough to determine that the result is the same because n does not change. Computing the upper bound before the loop is advisable.

• on CPUs without fast floating point arithmetics, using a different test may be more efficient such as i * i <= n or n / i >= i .

• the program reports 0 and 1 as prime numbers.

• if the user enters a negative number, sqrt(n) returns NaN , conversion to int is undefined and may produce 0 , hence the test will fail and the number will be reported as prime.

• if the user types anything that is not a number, the behavior is undefined because scanf() will fail and n will remain uninitialized. You should test the return value of scan() , which will be 1 for a successful conversion and 0 for invalid input or EOF for end of file.

Here is a modified version:

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

int main() {
    int n, sq, i, flag = 0;

    printf("Enter the number: ");
    if (scanf("%d", &n) != 1) {
        printf("invalid input\n");
        return 1;
    }
    if (i < 2) {
        /* reject 0, 1 and negative numbers */
        flag = 1;
    } else {
        sq = round(sqrt(n));
        for (i = 2; i <= sq; i++) {
            if (n % i == 0) {
                flag = 1;
                break;
            }
        }
    }
    if (flag == 0) {
        printf("%d is a prime number\n", n);
    } else {
        printf("%d is not a prime number\n", n);
    }
    return 0;
}

Answer 2

If a number is not a prime a prime factor of it will always be less than or equal to its square root. This is why we only check factors up to a number's square root.

Consider checking the number 59 . Its square root is (approximately) 7.68 . If we're incrementing up from 2 , we will only have to iterate from 2 to 7 to determine if this is a prime number, rather than from 2 to 59 (or rather 58 ). This saves our loop many iterations.

Consider how that impacts a number like 7481 . We can iterate from 2 to 86 rather than from 2 to 7480 to find out that it's prime.

As the number you're checking gets bigger, this saves more and more computing time.