Largest prime factor program takes aaaages - Java

Question

So this is problem 3 from project Euler. For those who don\'t know, I have to find out the largest prime factor of 600851475143. I have the below code:

impor         


        

Answer 1

change

for(long i = 2; i <= limit; i++)

to

// add the one for rounding errors in the sqrt function
new_limit = sqrt(limit) + 1;
// all even numbers are not prime 
for(long i = 3; i <= new_limit; i+=2)
{
...
}

Factoring 1,000,000 for example instead of iterating 1,000,000 times the thing only needs to do around 500 iterations.

Answer 2

I have done several dozen of the challenges on Project Euler. Some of the questions can be solved with brute force (they recommend not to do this) but others require "out of the box" thinking. You cannot solve that by problem with brute force.

There is lots of help on the web to lead you in the right direction, for example: http://thetaoishere.blogspot.com.au/2008/05/largest-prime-factor-of-number.html

Answer 3

try this ..

public class LargestPrimeFactor{
public static int largestPrimeFactor(long number) {
    int i;
    for (i = 2; i <= number; i++) {
        if (number % i == 0) {
            number /= i;
            i--;
        }
    }
    return i;
}

/*  change according to ur requirement. 
public static long getLong(String prompt) {
    System.out.print(prompt + " ");
    long mrlong = input.nextLong();
    input.nextLine();
    return mrlong;
}
 */

public static void main(String[] args) {
    //long num = getLong("What number do you want to analyse? ");
    System.out.println(largestPrimeFactor(600851475143l));
}
}

Answer 4

public static void main(String[] args) {

    long number = 600851475143L;

    long highestPrime = -1;
    for (long i = 2; i <= number; ++i) {
        if (number % i == 0) {
            highestPrime = i;
            number /= i;
            --i;
        }
    }

    System.out.println(highestPrime);
}

Answer 5

The number of prime factors a number can have is always less than sqrt of that number so that there is no need to iterate through the number n to find its largest prime factor.

See this code.

    public class LargestPrimeFactor {
        public static void main(String[] args) {

            Scanner sc=new Scanner(System.in);
            long num=sc.nextLong();

            if(num>0 && num<=2)
            {
                System.out.println("largest prime is:-" + num);
                System.exit(0);
            }

            int i=((Double)Math.sqrt(num)).intValue();
            int j=3;
            int x=0;

            //used for looping through the j value which can also be a prime. for e.g in case of 100 we might get 9 as a divisor. we need to make sure divisor is also a prime number.
            int z=0;
//same function as j but for divisor
            int y=3;
            int max=2;
//divisor is divisible
            boolean flag=false;
//we found prime factors
            boolean found=false;

            while(x<=i)
            {
                y=3;
                flag=false;

                if(num % j ==0)
                {
                    if(j>max)
                    {
                        for(z=0;z<Math.sqrt(j);z++)
                        {
                            if(j!=y && j % y==0)
                            {
                                flag=true;
                            }
                            y+=2;
                        }
                        if(!flag)
                        {
                            found=true;
                            max=j;
                        }
                    }
                }
                j+=2;
                x++;
            }
            if(found){
                System.out.println("The maximum prime is :- " + max);
            }
            else
            {
                System.out.println("The maximum prime is :- " + num);   
            }
        }
    }

Answer 6

public class LargestPrimeFactor {

public static boolean isPrime(long num){
    int count = 0;
    for(long i = 1; i<=num/2 ; i++){
        if(num % i==0){
            count++;
        }
    }
    if(count==1){
        return true;
    }
    return false;
}

public static String largestPrimeFactor(long num){
    String factor = "none";
    for(long i = 2; i<= num/2 ; i++){
        if(num % i==0 && isPrime(i)){
           factor = Long.toString(i); 
        }
    }
    return factor;     
}
public static void main(String[] args) {
    System.out.println(largestPrimeFactor(13195));
}

}