Given a number n, print all primes smaller than or equal to n.

Examples:

Input: n = 10 Output: 2, 3, 5, 7 Input: n = 20 Output: 2, 3, 5, 7, 11, 13, 17, 19

We have discussed Sieve of Eratosthenes algorithm for the above task.

Below is Sieve of Sundaram algorithm.

printPrimes(n)[Prints all prime numbers smaller than n] 1) In general Sieve of Sundaram, produces primes smaller than (2*x + 2) for a number given number x. Since we want primes smaller than n, we reduce n-2 to half. We call it nNew. nNew = (n-2)/2; For example, if n = 102, then nNew = 50. 2) Create an arraymarked[n]that is going to be used to separate numbers of the form i+j+2ij from others where 1 <= i <= j 3) Initialize all entries of marked[] as false. 4) // Mark all numbers of the form i + j + 2ij as true // where 1 <= i <= j Loop for i=1 to nNew a) j = i; b) Loop While (i + j + 2*i*j) <= nNew (i) primes[i + j + 2*i*j] = true; (ii) j++ 5) If n > 2, then print 2 as first prime. 6) Remaining primes are of the form 2i + 1 where i is index of NOT marked numbers. So print 2i + 1 for all i such that marked[i] isfalse.

Below is C++ implementation of above algorithm:

## CPP

// C++ program to print primes smaller than n using // Sieve of Sundaram. #include <bits/stdc++.h> using namespace std; // Prints all prime numbers smaller int SieveOfSundaram(int n) { // In general Sieve of Sundaram, produces primes smaller // than (2*x + 2) for a number given number x. // Since we want primes smaller than n, we reduce n to half int nNew = (n-2)/2; // This array is used to separate numbers of the form i+j+2ij // from others where 1 <= i <= j bool marked[nNew + 1]; // Initalize all elements as not marked memset(marked, false, sizeof(marked)); // Main logic of Sundaram. Mark all numbers of the // form i + j + 2ij as true where 1 <= i <= j for (int i=1; i<=nNew; i++) for (int j=i; (i + j + 2*i*j) <= nNew; j++) marked[i + j + 2*i*j] = true; // Since 2 is a prime number if (n > 2) cout << 2 << " "; // Print other primes. Remaining primes are of the form // 2*i + 1 such that marked[i] is false. for (int i=1; i<=nNew; i++) if (marked[i] == false) cout << 2*i + 1 << " "; } // Driver program to test above int main(void) { int n = 20; SieveOfSundaram(n); return 0; }

## Java

// Java program to print primes smaller // than n using Sieve of Sundaram. import java.util.Arrays; class GFG { // Prints all prime numbers smaller static int SieveOfSundaram(int n) { // In general Sieve of Sundaram, produces // primes smaller than (2*x + 2) for a number // given number x. Since we want primes // smaller than n, we reduce n to half int nNew = (n - 2) / 2; // This array is used to separate numbers of the // form i+j+2ij from others where 1 <= i <= j boolean marked[] = new boolean[nNew + 1]; // Initalize all elements as not marked Arrays.fill(marked, false); // Main logic of Sundaram. Mark all numbers of the // form i + j + 2ij as true where 1 <= i <= j for (int i = 1; i <= nNew; i++) for (int j = i; (i + j + 2 * i * j) <= nNew; j++) marked[i + j + 2 * i * j] = true; // Since 2 is a prime number if (n > 2) System.out.print(2 + " "); // Print other primes. Remaining primes are of // the form 2*i + 1 such that marked[i] is false. for (int i = 1; i <= nNew; i++) if (marked[i] == false) System.out.print(2 * i + 1 + " "); return -1; } // Driver code public static void main(String[] args) { int n = 20; SieveOfSundaram(n); } } // This code is contributed by Anant Agarwal.

2 3 5 7 11 13 17 19

**Illustration:**

All red entries in below illustration are marked entries. For every remaining (or black) entry x, the number 2x+1 is prime.

Lets see how it works for n=102, we will have the sieve for (n-2)/2 as follows:

Mark all the numbers which can be represented as i + j + 2ij

Now for all the unmarked numbers in the list, find 2x+1 and that will be the prime:

Like 2*1+1=3

2*3+1=7

2*5+1=11

2*6+1=13

2*8+1=17 and so on..

**How does this work?**

When we produce our final output, we produce all integers of the form 2x+1 (i.e., they are odd) except 2 which is handled separately.

Let q be an integer of the form 2x + 1. q is excluded if and only if x is of the form i + j + 2ij. That means, q = 2(i + j + 2ij) + 1 = (2i + 1)(2j + 1) So, an odd integer is excluded from the final list if and only if it has a factorization of the form (2i + 1)(2j + 1) which is to say, if it has a non-trivial odd factor. Source: Wiki

**Reference: **

https://en.wikipedia.org/wiki/Sieve_of_Sundaram

This article is contributed by **Anuj Rathore**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above