Project Euler # 10 Java Solution Doesn't Work

Question

I am trying to find the sum of primes <2000000. This is my solution in Java, but I cannot get the correct answer. Please give some input on what might be wrong, and general code recommendations are appreciated.

Printing "sum" gives: 1308111344, which is incorrect.

Edit: Thanks for the help. Changed int to long and <to <= and it worked flawlessly, except it was an inefficient way to find primes :)

/*
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
*/
class Helper{
 public void run(){
  Integer sum = 0;
  for(int i = 2; i < 2000000; i++){
   if(isPrime(i))
    sum += i;   
  }
  System.out.println(sum);
 }

 private boolean isPrime(int nr){
  if(nr == 2)
   return true;
  else if(nr == 1)
   return false;
  if(nr % 2 == 0)
   return false;

  for(int i = 3; i < Math.sqrt(nr); i += 2){
   if(nr % i == 0)
    return false;
  }  
  return true;  
 }
}   
class Problem{
 public static void main(String[] args){
  Helper p = new Helper();
p.run();
}
}

Answer 1

, , overflow. BigInteger long . .

Answer 2

, (, 25). i < Math.sqrt(nr) i <= Math.sqrt(nr).

, .

Answer 3

isPrime . isPrime (9) true.

Answer 4

, :

• int, .. long
• < <=,

, , , (, Miller-Rabin), .. , , , .

cleaver: boolean , , . , , , , . couse , ( )

( " ", ):

    boolean[] numbers = new boolean[2000000];
    long sum = 0;

    for (int i = 0; i < numbers.length; ++i)
        numbers[i] = true;

    for (int i = 2; i < numbers.length; ++i)
        if (!numbers[i])
            continue;
        else {
            int j = i + i;
            while (j < 2000000) {                   
                numbers[j] = false;
                j += i;
            }           
        }

    for (int i = 2; i < 2000000; ++i)
        sum += numbers[i] ? i : 0;

    System.out.println(sum);

, - ( - ), .

Answer 5

Using the Sieve Effect of Eratosthenes , I solved the problem, here is my code

public class SumOfPrime {

    static void findSum()
    {
        long i=3;
        long sum=0;
        int count=0;
        boolean[] array = new boolean[2000000];
        for(long j=0;j<array.length;j++)
        {
         if((j&1)==0)
          array[(int)j]=false;   
         else    
         array[(int)j]=true;
        }
        array[1]=false;
        array[2]=true;
        for(;i<2000000;i+=2)
        { 
            if(array[(int)i] & isPrime(i))
            {   
                array[(int)i]=true;
                //Sieve of Eratosthenes
                for(long j=i+i;j<array.length;j+=i)
                    array[(int)j]=false;
            }
        }
        for(int j=0;j<array.length;j++)
        {
            if(array[j])
            {   
             //System.out.println(j);
             count++;   
             sum+=j;
            }
        }   
        System.out.println("Sum="+sum +" Count="+count);
    }
    public static boolean isPrime(long num)
    {
        boolean flag=false;
        long i=3;
        long limit=(long)Math.sqrt(num);
        for(;i<limit && !(flag);i+=2)
        {
            if(num%i==0)
            {
                flag=false;
                break;
            }   
        }
        if(i>=limit)
         flag=true;
        return flag;
    }

    public static void main(String args[])
    {
        long start=System.currentTimeMillis();
        findSum();
        long end=System.currentTimeMillis();
        System.out.println("Time for execution="+(end-start)+"ms");
    }

}

and the way out is

Sum=142913828922 Count=148933
Time for execution=2360ms

If in doubt please inform

Answer 6

Here is my solution

  public class ProjectEuler {

    public static boolean isPrime(int i) {
        if (i < 2) {
            return false;
        } else if (i % 2 == 0 && i != 2) {
            return false;
        } else {
            for (int j = 3; j <= Math.sqrt(i); j = j + 2) {
                if (i % j == 0) {
                    return false;
                }
            }

            return true;
        }
    }


    public static long sumOfAllPrime(int number){
        long sum = 2;

        for (int i = 3; i <= number; i += 2) {
            if (isPrime(i)) {
                sum += i;
            }
        }

        return sum;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        System.out.println(sumOfAllPrime(2000000));
    }
}