What is the use of prime(n, n/2) here. If I dont use it it doesn't print the output

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#include<stdio.h>
main()
{
    int n, p;
    printf("Enter a number ");
    scanf("%d", &n);
    p= prime(n,n/2);
    if(p == 1)
    {
        printf("Yes");
    }
    else
    {
        printf("No");
	}
}
int prime(int n, int i)
{
    if(i == 1)
        printf("Yes");
    else
    {
        if(n%i == 0)
            printf("No");
        else
            prime(n, i-1); 
    }
}

What I have tried:

So this program is about finding prime numbers using recursion method
I have tried to run the program without the n/2 but it won't print any clarification why please
it would help me if there is any other method to get prime numbers using recursion method

Posted 28-Oct-22 3:34am

Member 15808958

Updated 28-Oct-22 4:53am

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Comments

jeron1 28-Oct-22 10:18am

prime() is supposed to return an int, nothing is being return right now.

Member 15808958 28-Oct-22 10:57am

yea but in the question they asked if it is a prime number yes should be printed if it is not a prime no should be printed using recursion

merano99 28-Oct-22 11:39am

yes should be printed if it is NOT a prime ? With you it is currently exactly the other way around (which would also be usual). A recursion without return is also rather unusual.

jeron1 28-Oct-22 13:54pm

True, but at some point you are counting on a return value;

p= prime(n,n/2); // you are expecting a return value here
if(p == 1) // in order to test it here
{
printf("Yes");
}

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OriginalGriff · Answer 1 · 2022-10-28T04:14:00

n/2 provides the starting point for the prime check.
If you think about it, n cannot be divisible by any number larger than n/2 as 2 is the smallest divisor that allows a zero remainder.
If you want to see is 11 is prime, then 11 / 2 is 5 (remember, integer division discards fractional parts) if you try to divide 11 by 6 or above, you are guaranteed not to get a zero remainder.

But ... a more efficient way to do it is to start by checking low numbers: they eliminate more non-prime values faster: 2 eliminates all the even numbers (half the possible ones)for example, then 3 eliminates one third, and so on. So check for two, then only ever check the odd values up to n/2 If it's not prime, you will get the result a lot quicker than going from the high values down.

But the big question is "Why?" - prime number calculations aren't naturally recursive - they are iterative. So an iterative solution is going to work better than a recursive one, because it will cope with larger "might be prime" values. Stacks are pretty small, and each time you recurse you use a significant amount of stack memory. It will run out, and pretty quickly!

merano99 · Answer 2 · 2022-10-28T04:53:00

The subroutine should actually have a return value and should not output anything itself, so that the main program makes sense.
Currently your code looks like this:

C

int prime(int n, int i)
{
    if(i == 1)
        printf("Yes");
    else
    {
        if(n%i == 0)
            printf("No");
        else
            prime(n, i-1); 
    }
}

Instead of the output of printf("Yes") there should be return 1. The following else could then be omitted. Accordingly, printf("No") should be replaced.

Unfortunately, your design crashes at both 0 and -1 and for other negative numbers the result is wrong.

I agree with OriginalGriff's suggestion not to start testing at n/2 and then count down, but rather to test the other way around.

If you want you can write the function isprime() with one parameter like this:

C

int isprime(int num)
{
	return prime(num, num / 2);
}

What is the use of prime(n, n/2) here. If I dont use it it doesn't print the output

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What is the use of prime(n, n/2) here. If I dont use it it doesn't print the output

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