Sieve of Eratosthenes - print all primes smaller than or equal to number n.

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Sun Feb 06 2022 05:18:47 GMT+0000 (UTC)

Saved by @Uttam #java #mathematics #lecture #gfg #geeksforgeeks #efficientmethod #naivemethod #sieveoferatosthenes #prime #sieve #eratosthenes #isprime

// Shorter Implementation of the optimized sieve : 

import java.io.*;
import java.util.*;

public class Main {
	
	static void sieve(int n)
	{
		if(n <= 1)
			return;

		boolean isPrime[] = new boolean[n+1];

		Arrays.fill(isPrime, true);

		for(int i=2; i <= n; i++)
		{
			if(isPrime[i])
			{
        System.out.print(i+" ");
				for(int j = i*i; j <= n; j = j+i)
				{
					isPrime[j] = false;
				}
			}
		}
	}

	public static void main (String[] args) {
		
		int n = 23;

		sieve(n);

	}
}

//Optimized Implementation : Time Complexity : O(nloglogn), Auxiliary Space : O(n)
	
	static void sieve(int n)
	{
		if(n <= 1)
			return;

		boolean isPrime[] = new boolean[n+1];

		Arrays.fill(isPrime, true);

		for(int i=2; i*i <= n; i++)
		{
			if(isPrime[i])
			{
				for(int j = i*i; j <= n; j = j+i) // Replaced 2*i by i*i
				{
					isPrime[j] = false;
				}
			}
		}

		for(int i = 2; i<=n; i++)
		{
			if(isPrime[i])
				System.out.print(i+" ");
		}
	}


//Simple Implementation of Sieve : 
	
	static void sieve(int n)
	{
		if(n <= 1)
			return;

		boolean isPrime[] = new boolean[n+1];

		Arrays.fill(isPrime, true);

		for(int i=2; i*i <= n; i++)
		{
			if(isPrime[i])
			{
				for(int j = 2*i; j <= n; j = j+i) 
				{
					isPrime[j] = false;
				}
			}
		}

		for(int i = 2; i<=n; i++)
		{
			if(isPrime[i])
				System.out.print(i+" ");
		}
	}


//Naive Solution : Time Complexity : O(n(sqrt(n))
  
	static boolean isPrime(int n)
	{
		if(n==1)
			return false;

		if(n==2 || n==3)
			return true;

		if(n%2==0 || n%3==0)
			return false;

		for(int i=5; i*i<=n; i=i+6)
		{
			if(n % i == 0 || n % (i + 2) == 0)
				return false; 
		}

		return true;
	}
	
	static void printPrimes(int n)
	{
		for(int i = 2; i<=n; i++)
		{
			if(isPrime[i])
				System.out.print(i+" ");
		}
	}
content_copyCOPY

Given a number n, print all primes smaller than or equal to n. Following is the algorithm to find all the prime numbers less than or equal to a given integer n by the Eratosthenes method: 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 prime numbers smaller than or equal to 50. We create a list of all numbers from 2 to 50. STEP-1: 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. STEP-2: 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. STEP-3: 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. STEP-4: We continue this process and our final table will look like below: STEP-5: So, the prime numbers are the unmarked ones: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.