Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.
For example, if n is 10, the output should be “2, 3, 5, 7”. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”.

The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than n when n is smaller than 10 million

Following is the algorithm to find all the prime numbers less than or equal to a given integer n by Eratosthenes’ method:

  1. Create a list of consecutive integers from 2 to n: (2, 3, 4, …, n).
  2. Initially, let p equal 2, the first prime number.
  3. Starting from p, count up in increments of p and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked.
  4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this number (which is the next prime), and repeat from step 3.

When the algorithm terminates, all the numbers in the list that are not marked are prime.

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Explanation with Example:
Let us take an example when n = 50. So we need to print all print numbers smaller than or equal to 50.

We create a list of all numbers from 2 to 50

According to the algorithm we will mark all the numbers which are divisible by 2.

Now we move to our next unmarked number 3 and mark all the numbers which are multiples of 3.

We move to our next unmarked number 5 and mark all multiples of 5.

We continue this process and our final table will look like below:

So the prime numbers are the unmarked ones: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

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Implementation:
Following is C++ implementation of the above algorithm. In the following implementation, a boolean array arr[] of size n is used to mark multiples of prime numbers.

Java Program 
// Java program to print all primes smaller than or equal to
// n using Sieve of Eratosthenes
 
class SieveOfEratosthenes
{
    void sieveOfEratosthenes(int n)
    {
        // Create a boolean array "prime[0..n]" and initialize
        // all entries it as true. A value in prime[i] will
        // finally be false if i is Not a prime, else true.
        boolean prime[] = new boolean[n+1];
        for(int i=0;i<n;i++)
            prime[i] = true;
         
        for(int p = 2; p*p <=n; p++)
        {
            // If prime[p] is not changed, then it is a prime
            if(prime[p] == true)
            {
                // Update all multiples of p
                for(int i = p*2; i <= n; i += p)
                    prime[i] = false;
            }
        }
         
        // Print all prime numbers
        for(int i = 2; i <= n; i++)
        {
            if(prime[i] == true)
                System.out.print(i + " ");
        }
    }
     
    // Driver Program to test above function
    public static void main(String args[])
    {
        int n = 30;
        System.out.print("Following are the prime numbers ");
        System.out.println("smaller than or equal to " + n);
        SieveOfEratosthenes g = new SieveOfEratosthenes();
        g.sieveOfEratosthenes(n);
    }
}

Output:

Following are the prime numbers below 30
2 3 5 7 11 13 17 19 23 29

 

 

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AlgorithmCodingJAVAMathematical Algorithms

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