Sieve of Eratosthenes

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

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)