Sieve of Eratosthenes

Software Development
#1

Hello Everyone,

Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.

Example:

Input : n =10
Output : 2 3 5 7

Input : n = 20
Output: 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 the Eratosthene’s method:
When the algorithm terminates, all the numbers in the list that are not marked are prime.

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

// C++ program to print all primes

// smaller than or equal to

// n using Sieve of Eratosthenes

#include <bits/stdc++.h>

using namespace std;

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.

bool prime[n + 1];

memset (prime, true , sizeof (prime));

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 greater than or

// equal to the square of it

// numbers which are multiple

// of p and are less than p^2

// are already been marked.

for ( int i = p * p; i <= n; i += p)

prime[i] = false ;

}

}

// Print all prime numbers

for ( int p = 2; p <= n; p++)

if (prime[p])

cout << p << " " ;

}

// Driver Code

int main()

{

int n = 30;

cout << "Following are the prime numbers smaller "

<< " than or equal to " << n << endl;

SieveOfEratosthenes(n);

return 0;

}

Output

Following are the prime numbers smaller than or equal to 30 2 3 5 7 11 13 17 19 23 29

Time complexity : O(n*log(log(n)))

*JAVA

// 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 * p; 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 Code
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);
}

}

*Python

Python program to print all

primes smaller than or equal to

n using Sieve of Eratosthenes

def SieveOfEratosthenes(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.
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):

	# If prime[p] is not
	# changed, then it is a prime
	if (prime[p] == True):

		# Update all multiples of p
		for i in range(p * p, n+1, p):
			prime[i] = False
	p += 1

# Print all prime numbers
for p in range(2, n+1):
	if prime[p]:
		print p,

Driver code

if name == ‘main’:
n = 30
print “Following are the prime numbers smaller”,
print “than or equal to”, n
SieveOfEratosthenes(n)