A prime number, as we know, is a natural number that has exactly two distinct natural number divisors: the number 1 and itself.

We often get questions like the one given below:

Given a number n, print all primes smaller than or equal to n. — e.g. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”

Sieve of Eratosthenes is used to get all prime numbers in a given range and is a very efficient algorithm. You can read more about Sieve of Eratosthenes on Wikipedia.

In this algorithm, we follow the following steps to get prime numbers up to n (let’s suppose n = 12):

1. First, we create a boolean list of length n+1 filled with all True as:\ is_prime = [True, True, True, True, True, True, True, True, True, True, True, True, True]
2. Since, 0 and 1 are not primes, we make the first two elements of the list as False. So, now our list is:\ is_prime = [False, False, True, True, True, True, True, True, True, True, True, True, True]
3. Starting from 2, we make elements at index that are multiples of 2 as False, except itself. So, our list becomes:\ is_prime = [False, False, True, True, False, True, False, True, False, True, False, True, False]
4. Repeat the step 3 till square root of n (i.e. √12 in this example). At the end, our list becomes :\ is_prime = [False, False, True, True, False, True, False, True, False, False, False, True, False]
5. The remaining list contains True only at the index which is prime.

def seive_of_eratosthenes(n):
    is_prime = [True for _ in range(n+1)]
    is_prime[0], is_prime[1] = False, False

    for i in range(2, int(n**0.5)+1):
        for j in range(2*i, n+1, i):
            is_prime[j] = False

    return is_prime


if __name__ == "__main__":
    n = int(input("Enter the number upto which : "))
    li = seive_of_eratosthenes(n)
    for i in range(n+1):
        if li[i]:
            print(i, end=" ")

This code is a straightforward representation of the steps mentioned above.

We first make a boolean list with all values initialized with True. A value will be False if it is not prime else True. Then we start iterating from 2 to the square root of the number and updating the multiples as False as well. This is done using another ‘for’ loop from 2*i to n+1 with a step of j+1.

In the end, we get our final list and check the values. If the value at any index is True, it means the number is prime and we print the number.

Thank you for reading!