Generating Number Pairs and GCD in Java

Question

Problem: Given A,B print the number of pairs (a,b) such that GCD(a,b)=1 and 1<=a<=A and 1<=b<=B.

Solution (Brute Force Approach) In the below code, i have used brute force approach and it works fine. However the execution time is more 10 sec if A & B > 10^5

Alternative Solution From my research i found out that finding prime factors of A & B will reduce the execution time considerably (< 3 sec), but i'm not sure how to apply it.

Need Help: Can anyone help me to arrive at the result with < 3 sec execution time?

class GCD {
public static void main(String[] args) {
    int A = 0, B = 0, GCD = 0, count = 0;
    BigInteger B1, B2 = null;

    A = Integer.parseInt(args[0]);
    B = Integer.parseInt(args[1]);

    for (int a = 1; a <= A; a++) {
        for (int b = 1; b <= B; b++) {
            B1 = BigInteger.valueOf(a);
            B2 = BigInteger.valueOf(b);
            GCD = calculateGCD(B1, B2);
            if (GCD == 1) {
                count++;
            }
        }
    }
    System.out.println(count);
}

public static int calculateGCD(BigInteger number1, BigInteger number2) {
    return (number1.gcd(number2)).intValue();
}
}

here is an idea how to calculate the gcd efficiently: http://stackoverflow.com/questions/4885537/what-is-the-fastest-way-to-find-the-gcd-of-n-numbers — cruxi, May 30 '14 at 10:00
Do you know how to find the prime factorization of a number? Start by writing a method `factor` that returns a `List` of the appropriate numeric type. Additionally, is using `BigInteger` required? It's a good bit slower than `long`. — chrylis -cautiouslyoptimistic-, May 30 '14 at 10:02
Now i have 2 Lists which contains the prime factors of A and B. But im not sure how to proceed further with those prime factors. — user3690689, May 30 '14 at 13:58

score 0 · Answer 1 · answered Jun 01 '14 at 10:30

I do not want to write a complete programm or something, but I want to give you some tips for speeding up your program:

gcd(a,b) = gcd(b,a) so only compute pairs (a, b) with a < b. gcd(a,a) = 1 holds only for a = 1. Also gcd(1,b) = 1 for all b, so you can start with a = 2 and a count = 1 + 2*(B-1).
compute all primefactors for all 1 < a <= A at once by using something like the Sieve of Eratosthenes. E.g. every secound number contains primefactor 2, every third the primfactor 3.
You do not need to compute the gcd. Let a contain the distinct primfactors p and q. Then you know:
- There are B-a numbers to test.
- Every p-th number contains also the primfactor p, every q-th primenumber also contains the primfactor q. floor( (B-a)/p ) numbers have a gcd >= p and floor( (B-a)/q ) numbers have a gcd >= q and floor( (B-a)/(p*q) ) numbers have you counted twice. So you can get the number of pairs (a,b) with a < b as
  (B-a) - floor( (B-a)/p ) - floor( (B-a)/q ) + floor( (B-a)/(p*q) )
- if you want also the pairs it self, you can use a for-loop and jump every step where i (for-loop counter) is divisable by any primefactor of a

I think this should speed up your program as much as you need to reach less than a second.

Generating Number Pairs and GCD in Java

1 Answers1