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Can I make it faster to find perfect numbers? I tried to make it faster with array and another algorithm, but none made it faster.

public class Perfect{
    static long perfectNumber;
    static long startTime = System.nanoTime();
    static long endTime;
    static long mersenne;

    public static void main(String[] args) {

        long p = 2;
        while (p < 32) {
            if( p % 2 == 0&&p!=2){
                    p++;
            }
            else{
            if (isPrime(p) == true) {

                mersenne = (long) (Math.pow(2, p) - 1);
                if (isPrime(mersenne) == true) {
                    perfectNumber = (long) Math.pow(2, (p - 1)) * mersenne;
                    System.out.println(perfectNumber);
                }
            }
            p++;
        }

    }
        endTime = System.nanoTime();
        System.out.println("Time:   " + (endTime - startTime) + "ns"
                );
    }
    private static boolean isPrime(long testPrime) {


            for (long i = 3; i < Math.sqrt(testPrime); i += 2) {

                if (testPrime % i == 0) {

                    return false;
                }
        }

        return true;
    }
}
Jongware
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user3056796
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1 Answers1

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There are a number of minor improvements you could make - probably none of which will make any difference to the run time:

  1. Using p % 2 may result in a division - p & 1 will not and therefore should be a fraction faster.
  2. You could easily pre-calculate all primes up to a reasonable limit.
  3. Math.pow(2,x) is equaivalent to 2 << x in all cases and may well be much quicker.
  4. Printing inside your loop is wasteful - you should collect all results in a list and print them at the end.

Apart from those trivial and probably premature optimisations - you should run the calculation a few hundred times and take the average time before attempting any optimisations.

BTW - Just because you are using nanotime does not mean you are getting nanosecond resolution - far from it.

Also -

As the Wikipedia article points out - you could avoid much of your calculation by enumerating the common binary pattern mentioned there:

for ( int i = 0; i < 10; i++ ) {
  long q = ((1 << (i+2)) - 1) << (i+1);
  // Printing BigIntegers in binary is easy.
  BigInteger bq = BigInteger.valueOf(q);
  System.out.println(q+" = "+bq.toString(2));
}

6 = 110
28 = 11100
120 = 1111000
496 = 111110000
2016 = 11111100000
8128 = 1111111000000
32640 = 111111110000000
130816 = 11111111100000000
523776 = 1111111111000000000
2096128 = 111111111110000000000

Obviously you still have to test them but you don't have to test nearly as many.

Also - if you want to take it one step further into BigInteger you could start with something like:

for ( int i = 0; i < 10; i++ ) {
  BigInteger p = BigInteger.ONE.shiftLeft(i+2).subtract(BigInteger.ONE).shiftLeft(i+1);
  System.out.println(p.toString(10)+" = "+p.toString(2));
  //long q = ((1 << (i+2)) - 1) << (i+1);
  //BigInteger bq = BigInteger.valueOf(q);
  //System.out.println("    "+bq.toString(10)+" = "+bq.toString(2));
}
OldCurmudgeon
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