Determining If a Number Is Prime

Check Number prime in JavaScript

Time complexity: O(sqrt(n))

Space complexity: O(1)

const isPrime = num => {
    for(let i = 2, s = Math.sqrt(num); i <= s; i++)
        if(num % i === 0) return false; 
    return num > 1;
}

To find a number is prime, Why checking till n/2 is better. What is the reason for avoiding numbres in second half of n

Because, the smallest multiple that will not make it a prime is 2. If you have checked all the numbers from 0 to n/2, what multiple is left that could possibly work? If multiple by 2 is bigger than n, then a multiple of 3 or 4 etc will also be bigger than n.

So the largest factor for any number N must be <= N/2

So yes take N/2, and check all integers smaller or equal to N/2. So for 11 you would check all integers smaller than 5.5, i.e. 1, 2, 3, 4 and 5.

The square root is explained here:
Why do we check up to the square root of a prime number to determine if it is prime?

And this question has been asked before.

Algorithm of checking if the number is prime

The algorithm can be improved further by observing that all primes are of the form 6k ± 1, with the exception of 2 and 3. This is because all integers can be expressed as (6k + i) for some integer k and for i = -1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k + 4); and 3 divides (6k + 3). So a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of form 6k ± 1

Implementation of the above:

#include <iostream>

bool isPrime(int n) {
    // Corner cases
    if(n <= 1) return false;
    if(n <= 3) return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if(n % 2 == 0 || n % 3 == 0) return false;

    for(int i = 5; i * i <= n; i = i + 6)
        if(n % i == 0 || n % (i + 2) == 0) return false;

    return true;
}

// Driver Program to test above function
int main() {
    std::cout << std::boolalpha
        << isPrime(11) << '\n'
        << isPrime(15) << '\n';
}

Why do we check up to the square root of a number to determine if the number is prime?

If a number n is not a prime, it can be factored into two factors a and b:

n = a * b

Now a and b can't be both greater than the square root of n, since then the product a * b would be greater than sqrt(n) * sqrt(n) = n. So in any factorization of n, at least one of the factors must be smaller than the square root of n, and if we can't find any factors less than or equal to the square root, n must be a prime.

How to determine if a number is prime

Your method for finding if your number is prime is the correct method.
To make it so that it does not consistently print out whether or not the number is prime, you could have an external variable, which represents whether or not the number is prime.

Such as

    boolean prime = true;
    for (int p = 2; p < sum; p++) {
        if (sum % p == 0) {
            prime = false;
            break;
        }
    }
    if (prime)
        System.out.println("The sum is a prime number.");
    else
        System.out.println("The sum is not a prime number.");

By doing this method the program will assume the number is prime until it proves that wrong. So when it finds it is not prime it sets the variable to false and breaks out of the loop.

Then after the loop finishes you just have to print whether or not the number was prime.

A way that you could make this loop faster is to go from when p = 2 to when p = the square root of sum. So using this method your for loop will look like this:

    double sq = Math.sqrt((double)sum);
    for (int p = 2; p < sq; p++) {
        //Rest of code goes here
    }

Hope this helps

C - determine if a number is prime

OK, so forget about C. Suppose I give you a number and ask you to determine if it's prime. How do you do it? Write down the steps clearly, then worry about translating them into code.

Once you have the algorithm determined, it will be much easier for you to figure out how to write a program, and for others to help you with it.

edit: Here's the C# code you posted:

static bool IsPrime(int number) {
    for (int i = 2; i < number; i++) {
        if (number % i == 0 && i != number) return false;
    }
    return true;
}

This is very nearly valid C as is; there's no bool type in C, and no true or false, so you need to modify it a little bit (edit: Kristopher Johnson correctly points out that C99 added the stdbool.h header). Since some people don't have access to a C99 environment (but you should use one!), let's make that very minor change:

int IsPrime(int number) {
    int i;
    for (i=2; i<number; i++) {
        if (number % i == 0 && i != number) return 0;
    }
    return 1;
}

This is a perfectly valid C program that does what you want. We can improve it a little bit without too much effort. First, note that i is always less than number, so the check that i != number always succeeds; we can get rid of it.

Also, you don't actually need to try divisors all the way up to number - 1; you can stop checking when you reach sqrt(number). Since sqrt is a floating-point operation and that brings a whole pile of subtleties, we won't actually compute sqrt(number). Instead, we can just check that i*i <= number:

int IsPrime(int number) {
    int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return 0;
    }
    return 1;
}

One last thing, though; there was a small bug in your original algorithm! If number is negative, or zero, or one, this function will claim that the number is prime. You likely want to handle that properly, and you may want to make number be unsigned, since you're more likely to care about positive values only:

int IsPrime(unsigned int number) {
    if (number <= 1) return 0; // zero and one are not prime
    unsigned int i;
    for (i=2; i*i<=number; i++) {
        if (number % i == 0) return 0;
    }
    return 1;
}

This definitely isn't the fastest way to check if a number is prime, but it works, and it's pretty straightforward. We barely had to modify your code at all!

How to check if a number is prime in a more efficient manner?

The lowest hanging fruit here is your stopping conditional

i <= x/2

which can be replaced with

i * i <= x

having taken care to ensure you don't overflow an int.This is because you only need to go up to the square root of x, rather than half way. Perhaps i <= x / i is better still as that avoids the overflow; albeit at the expense of a division which can be relatively costly on some platforms.

Your algorithm is also defective for x == 2 as you have the incorrect return value. It would be better if you dropped that extra test, as the ensuing loop covers it.

Is there a simple algorithm that can determine if X is prime?

The first algorithm is quite good and used a lot on Project Euler. If you know the maximum number that you want you can also research Eratosthenes's sieve.

If you maintain the list of primes you can also refine the first algo to divide only with primes until the square root of the number.

With these two algoritms (dividing and the sieve) you should be able to solve the problems.

Edit: fixed name as noted in comments

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