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  1. Look at the code below for Find how many Cardano Triplets exist such that a+b+c<=n

  2. The function is to find out the total cardano triplet, help me optimizing it for larger values Integer

         private static boolean isTrue(long a, long b, long c) {
         long res = ((8 * a * a * a) + (15 * a * a) + (6 * a) - (27 * b * b * c));
         //double res=((Math.cbrt(a+(b*Math.sqrt(c))))+(Math.cbrt(a-(b*Math.sqrt(c)))));
         return res == 1;
     }

3.The below function returning triplet to function isTrue(...) and updating global counter variable private static int counter=0;

private static long countCardano(long n) {
    long c;
    boolean b;
    for (long i = 2; i <= n; i++) {
        for (long j = 1; j <= n; j++) {
            for (long k = 5; k <= n; k++) {
                if ((i + j + k) <= n) {
                    if (isTrue(i, j, k)) {
                        //System.out.println("("+i+","+j+","+k+")");
                        counter++;
                    }
                }
            }
        }

    }
    return counter;
}
Mikheil Zhghenti
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Arijeet Mukherjee
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  • It works, but it's O(n^3) so the time it's executing will be massive – Morph21 Sep 14 '20 at 06:54
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    If it works for some numbers, then it pretty much can't be a "compile time issue". It's an issue with the run time of your application. – Joachim Sauer Sep 14 '20 at 07:10
  • Based on my experience: When I done sorting on 1M records the same issue happens. After then I had run my code on an online platform and run fine but it consume time as required. So might be your hardware is not have enough tendency to compute up to that extent. – Vishal Sheth Sep 14 '20 at 07:22
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    There is a limit to what you can do with O(n^3) algorithm. To improve beyond that limit you need to improve the algorithm e.g. instead of brute force, something like: https://math.stackexchange.com/questions/1885095/parametrization-of-cardano-triplet – David Soroko Sep 14 '20 at 08:16

1 Answers1

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I would try to reduce the number of iterations based on the if((i+j+k)<=n) condition. So, if this condition is false, clearly increasing k won't respect the condition neither, so i would add a break on the else branch. Similar to that, i would add a condition for (i+j)<n, since adding a positive k to this won't respect the condition neither, so on the else branch of that, i would break again.

So the code would look like that :

private static long countCardano(long n) {
    long c;
    boolean b;
    for(long i=2;i<= n-6;i++) //n-6 because j starts from 1 and k from 5 {
        for(long j=1;j<=n-2;j++) {
            if ((i+j) > (n-5)) break;
            for(long k=5;k<=n-3;k++) {
                if((i+j+k)<=n) {
                  if(isTrue(i,j,k)) {
                    counter++;
                  }
                } else {
                  break;
                }
            }
        }
        
    }
    return counter;
}
Vladimir Stanciu
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