Sieve of Eratosthenes: Finding All Prime Numbers

Problem Statement

Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.

A prime number is a number that is divisible by only two numbers – themselves and 1

Example:

Input: n =10
Output: 2 3 5 7
Explanation: Only 4 prime numbers are there which are less than or equal to 10.

Input: n = 20 
Output: 2 3 5 7 11 13 17 19
Explanation: All above numbers are prime which are less than or equal to 20.

Simple Approach

Iterate from 2 to N, and check for prime. If it is a prime number, print the number.  Below is the implementation of the above approach

C++ Implementation

int checkPrime(int n) {
   if (n <= 1)
      return 0;

   for (int i = 2; i < n; i++)
      if (n % i == 0)
         return 0;

   return 1;
}
// Function to print all the prime numbers from 
// range 2 to n 
void printPrime(int n) {
   for (int i = 2; i <= n; i++) {
      if (checkPrime(i))
         cout << i << " ";
   }
}

Java Implementation

class Interviewbit {
   static int isPrime(int n) {
      for (int i = 2; i < n; i++)
         if (n % i == 0)
            return 0;

      return 1;
   }

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

Python Implementation

 def isPrime(n):
     
    # Corner case
    if n <= 1 :
        return False
 
    # check from 2 to n-1
    for i in range(2, n):
        if n % i == 0:
            return False
 
    return True
 
# Function to print primes
def printPrime(n):
    for i in range(2, n + 1):
        if isPrime(i):
            print(i, end = " ")

Time complexity: O(N*N), Where N is the number.
Space complexity: O(1)

Efficient Approach: Sieve of Eratosthenes

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. 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. 

• Firstly write all the numbers from  2,3,4…. n
• Now take the first prime number and mark all its multiples as visited.
• Now when you move forward take another number which is unvisited yet and then follow the same step-2 with that number.
• All numbers in the list left unmarked when the algorithm ends are referred to as prime numbers.

Dry run of the above approach

Step 1: The numbers between 1 and 100 are listed in the table below.

Step 2: The next step is to write in bold all the multiples of 2, except 2 itself.

Step 3: Now bold all multiples of 3, 5, and 7 and only leave these numbers.

Step 4: Since the multiples of 11, 13, 17, and 19 are not present on the list, 1 is finally shaded because it is not prime.

Step 5: The unshaded numbers are prime. They include:

2, 3, 5,7, 11, 13,17,19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

C++ Implementation Efficient Approach

void SieveOfEratosthenes(int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
            cout << p << " ";
}

Java Implementation of Efficient Approach

class SieveOfEratosthenes {
	void sieveOfEratosthenes(int n)
	{
		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] == true)
			{
				for (int i = p * p; i <= n; i += p)
					prime[i] = false;
			}
		}

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

Python Implementation of Efficient Approach

 def SieveOfEratosthenes(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        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,

Time complexity of Sieve of Eratosthenes:

Space complexity: O(1)

Analysis of Time Complexity

• PREREQUISITE: Time complexity of the Harmonic Series is O(logN), when N is tending to infinity.
• If we analyze our sieve for larger numbers then we can see .
  • The inner loop at each iteration can be expressed by:
  • i=2, the inner loop will be executed → upper bound is (n/2) times
  • i=3, the inner loop will be executed → upper bound is (n/3) times
  • i=5, the inner loop will be executed → upper bound is (n/5) times
  • …
  • i=n(if prime number), the inner loop will be executed → upper bound is 1 times.
• The above series time complexity is less than harmonic series: 
  • n/2 + n/4 + ..+ 1 → n (1/2 + 1/3 + 1/4 + … 1/n), 
• Hence, the run loop times should be 
  • n (1/2 + 1/3 + 1/5 + 1/7 + 1/ 11 + 1/13 + …).= logn That is why, in some article, 
• The time complexity is defined in O(nloglogn).

Conclusion

The outer loop is a general for-loop from 1 to N, the time complexity is O(N). The total time complexity is O(NloglogN).

Practice Questions

• All Factors
• Prime Sum
• Prime Numbers

Frequently Asked Questions

Q: How fast is Sieve of Eratosthenes?
A: It is very fast as it can store prime numbers up to the 1e7 range easily.

Q: How do you improve the sieve of Eratosthenes?
A: We can further improve the sieve to O(n) by writing it to an iterative sieve.

Q: How does the Sieve of Eratosthenes work?
A: It works on the principle of cancellation of prime factors.