Count prime number till N

Question

This code works fine to count prime number till n. The problem is when n value is large like 1000000 or more, then it takes long time to execute and print the output (more than 30 secs.). I want to fix this issue. Any help would be great. Below is the code:

public class PrimeNotillN {

        public static void main(String[] args) {

            int n = 1000;
            int count = 0;

            for (int i = 2; i < n; i++) {
                boolean res = checkprime(i);
                if (res == true)
                    count++;
            }
            System.out.println("total prime number till " + n + " is " + count);
        }

        private static boolean checkprime(int n) {
            if (n == 1)
                return false;
            else {
                for (int i = 2; i <= n / 2; i++) {
                    if (n % i == 0) {

                        return false;
                    }
                }

                return true;
            }
        }
    }

Answer 1

The easiest change to make is to end the for loop in checkprime after i has reached the square root of n . The reason is that if you have found a factor greater than the square root of n , that implies that there was a factor less than the square root of n that should have been found already.

for (int i = 2; i <= Math.sqrt(n); i++) {

Additionally, if you need more speed, the best algorithm for printing all prime numbers up to a limit is the Sieve of Eratosthenes . That involves replacing what you have. You'll need an array in which you'll mark which numbers are composite. Starting with 2, you will mark all multiples of 2 as composite. Moving through the array, if you find a number that isn't marked composite, it's prime and you can print it. With each prime you encounter, you will mark all of that prime's multiples as composite as well.

Answer 2

Primes, other than 2, are never even. Primes are only divisible by 1 and iteslef.

Also you can check up to it's sqrt. If you find nothing by then then it's a prime.

        public bool IsPrime(double num)
    {
        bool isprime = true;
        double limit = Math.Sqrt(num);

        if (num % 2 == 0)
            return num == 2;

        if (num == 3 || num == 5 || num == 7)
            return true;

        for (int i = 3; i < limit; i++)
            if (num % 1 == 0)
                return false;

        return isprime;
    }

Answer 3

Sieve of Eratosthenes is a very famous and efficient algorithm to generate all small prime numbers up to around 1-10 million. This is an ancient algorithm given by a Greek mathematician named Eratosthenes.

 public class CountPrimes {

  public static void main(String[] args) {
    System.out.println(countPrimes(1000000));
  }

  public static int countPrimes(int n) {
    boolean[] primes = new boolean[n +1];

    for(int i = 0 ; i <= n ; i++) {
        primes[i] = true;
    }

    primes[0] = false;
    primes[1] = false;

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

    int primeCounts = 0;
    for(int i = 2 ; i <= n ; i++) {
        if(primes[i]) {
            primeCounts++;
        }
    }

    return primeCounts;
  }

}

Answer 4

You can store found primes into list and make further checks only against them (Sieve of Eratosphenes):

import java.util.ArrayList;
import java.util.List;

public class PrimeNotillN {
    public static void main(String[] args) {
        int n = 1000000;
        int count = 0;
        List<Integer> primes = new ArrayList<>();

        for (int i = 2; i < n; i++) {
            boolean res = checkprime(primes, i);
            if (res) {
                count++;
                primes.add(i);
            }
        }
        System.out.println("total prime number till " + n + " is " + count);
    }

    private static boolean checkprime(List<Integer> primes, int n) {
        int fence = (int) Math.sqrt(n);
        for (int prime : primes) {
            if (n % prime == 0) {
                return false;
            }
            if (prime >= fence)
                break;
        }

        return true;
    }
}

Answer 5

Using the sieve of eratoshtenes approach and maintaining the cache of all prime numbers found

    public int countPrimes(int n) {
    if(n==0){
        return 0;
    }else{
        boolean[] isPrime=new boolean[n];
        for(int i=2;i<n;i++){
            isPrime[i]=true;
        }

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

        int counter=0;
        for(int i=2;i<n;i++){
            if(isPrime[i]){
                counter++;
            }
        }

        return counter;

    }
}