Sieve of Eratosthenes in 0(n) time complexity

3.7

The classical Sieve of Eratosthenes algorithm takes O(N log (log N)) time to find all prime numbers less than N. In this article, a modified Sieve is discussed that works in O(N) time.

Example :

Given a number N, print all prime 
numbers smaller than N

Input :  int N = 15
Output : 2 3 5 7 11 13

Input : int N = 20
Output : 2 3 5 7 11 13 17 19

Manipulated Sieve of Eratosthenes algorithm works as following:

For every number i where i varies from 2 to N-1:
    Check if the number is prime. If the number
    is prime, store it in prime array.

For every prime numbers j less than the smallest  
prime factor p of i:
    Mark all numbers j*p as non_prime.
    Mark smallest prime factor of j*p as j

Below is implementation of above idea.

C++

// C++ program to generate all prime numbers
// less than N in O(N)
#include<bits/stdc++.h>
using namespace std;
const long long MAX_SIZE = 1000001;

// isPrime[] : isPrime[i] is true if number is prime 
// prime[] : stores all prime number less than N
// SPF[] that store smallest prime factor of number
// [for Exp : smallest prime factor of '8' and '16'
// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ]
vector<long long >isprime(MAX_SIZE , true);
vector<long long >prime;
vector<long long >SPF(MAX_SIZE);

// function generate all prime number less then N in O(n)
void manipulated_seive(int N)
{
    // 0 and 1 are not prime
    isprime[0] = isprime[1] = false ;

    // Fill rest of the entries
    for (long long int i=2; i<N ; i++)
    {
        // If isPrime[i] == True then i is
        // prime number
        if (isprime[i])
        {
            // put i into prime[] vector
            prime.push_back(i);

            // A prime number is its own smallest
            // prime factor
            SPF[i] = i;
        }

        // Remove all multiples of  i*prime[j] which are
        // not prime by making isPrime[i*prime[j]] = false
        // and put smallest prime factor of i*Prime[j] as prime[j]
        // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ]
        // so smallest prime factor of '10' is '2' that is prime[j] ]
        // this loop run only one time for number which are not prime
        for (long long int j=0;
             j < (int)prime.size() &&
             i*prime[j] < N && prime[j] <= SPF[i];
             j++)
        {
            isprime[i*prime[j]]=false;

            // put smallest prime factor of i*prime[j]
            SPF[i*prime[j]] = prime[j] ;
        }
    }
}

// driver  program to test above function
int main()
{
    int N = 13 ; // Must be less than MAX_SIZE

    manipulated_seive(N);

    // pint all prime number less then N
    for (int i=0; i<prime.size() && prime[i] <= N ; i++)
        cout << prime[i] << " ";

    return 0;
}

Java

// Java program to generate all prime numbers
// less than N in O(N)


import java.util.Vector;

class Test
{
	static final int MAX_SIZE = 1000001;
	// isPrime[] : isPrime[i] is true if number is prime 
	// prime[] : stores all prime number less than N
	// SPF[] that store smallest prime factor of number
	// [for Exp : smallest prime factor of '8' and '16'
	// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ]
	static Vector<Boolean>isprime = new Vector<>(MAX_SIZE);
	static Vector<Integer>prime = new Vector<>();
	static Vector<Integer>SPF = new Vector<>(MAX_SIZE);
	 
	// method generate all prime number less then N in O(n)
	static void manipulated_seive(int N)
	{
	    // 0 and 1 are not prime
	    isprime.set(0, false);
	    isprime.set(1, false);
	    
	    // Fill rest of the entries
	    for (int i=2; i<N ; i++)
	    {
	        // If isPrime[i] == True then i is
	        // prime number
	        if (isprime.get(i))
	        {
	            // put i into prime[] vector
	            prime.add(i);
	 
	            // A prime number is its own smallest
	            // prime factor
	            SPF.set(i,i);
	        }
	 
	        // Remove all multiples of  i*prime[j] which are
	        // not prime by making isPrime[i*prime[j]] = false
	        // and put smallest prime factor of i*Prime[j] as prime[j]
	        // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ]
	        // so smallest prime factor of '10' is '2' that is prime[j] ]
	        // this loop run only one time for number which are not prime
	        for (int j=0;
	             j < prime.size() &&
	             i*prime.get(j) < N && prime.get(j) <= SPF.get(i);
	             j++)
	        {
	            isprime.set(i*prime.get(j),false);
	 
	            // put smallest prime factor of i*prime[j]
	            SPF.set(i*prime.get(j),prime.get(j)) ;
	        }
	    }
	}
    
	// Driver method
	public static void main(String args[]) 
	{
		int N = 13 ; // Must be less than MAX_SIZE
		
		// initializing isprime and spf
		for (int i = 0; i < MAX_SIZE; i++){
			isprime.add(true);
			SPF.add(2);
		}

		
	    manipulated_seive(N);
	 
	    // pint all prime number less then N
	    for (int i=0; i<prime.size() && prime.get(i) <= N ; i++)
	        System.out.print(prime.get(i) + " ");
	}
}


Output :
2 3 5 7 11

Illustration:

isPrime[0] = isPrime[1] = 0

After i = 2 iteration :
isPrime[]   [F, F, T, T, F, T, T, T]
SPF[]       [0, 0, 2, 0, 2, 0, 2, 0]
     index   0  1  2  3  4  5  6  7

After i = 3 iteration :
isPrime[]  [F, F, T, T, F, T, F, T, T, F ]
SPF[]      [0, 0, 2, 3, 2, 0, 2, 0, 0, 3 ]
  index     0  1  2  3  4  5  6  7  8  9

After i = 4 iteration :
isPrime[]  [F, F, T, T, F, T, F, T, F, F]
SPF[]      [0, 0, 2, 3, 2, 0, 2, 0, 2, 3]
  index     0  1  2  3  4  5  6  7  8  9

This article is contributed by Divyanshu Srivastava and Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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