Sieve of Eratosthenes

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 or so (Ref Wiki).

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 p2, count up in increments of p and mark each of these numbers greater than or equal to p2 itself in the list. These numbers will be p(p+1), p(p+2), p(p+3), etc..
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.

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 and are greater than or equal to the square of it.

Now we move to our next unmarked number 3 and mark all the numbers which are multiples of 3 and are greater than or equal to the square of it.

We move to our next unmarked number 5 and mark all multiples of 5 and are greater than or equal to the square of it.

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.

Thanks to Krishan Kumar for providing above explanation.

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 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 Program to test above function int main() {     int n = 30;     cout << "Following are the prime numbers smaller "          << " than or equal to " << n << endl;     SieveOfEratosthenes(n);     return 0; }

 // 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

 # 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):         if prime[p]:             print p,    # driver program if __name__=='__main__':     n = 30     print "Following are the prime numbers smaller",     print "than or equal to", n     SieveOfEratosthenes(n)

 // C# program to print all primes // smaller than or equal to n // using Sieve of Eratosthenes using System;    namespace prime {     public class GFG     {                                 public static 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 = new bool[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)                 Console.Write(i + " ");         }                        }                    // Driver Code         public static void Main()         {             int n = 30;             Console.WriteLine("Following are the prime numbers");             Console.WriteLine("smaller than or equal to " + n);             SieveOfEratosthenes(n);                        }     } }    // This code is contributed by Sam007.



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

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

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Segmented Sieve.
Sieve of Eratosthenes in 0(n) time complexity