The Wikipedia article contains an explanation, but it’s not easy to find it immediately (also, procedure + proof don’t always answer the question “why it works”). Basically it comes down to the fact that for two integers a, b (assuming a >= b), it is always possible to write a = bq + r where … Read more
Without recursion: int result = numbers[0]; for(int i = 1; i < numbers.length; i++){ result = gcd(result, numbers[i]); } return result; For very large arrays, it might be faster to use the fork-join pattern, where you split your array and calculate gcds in parallel. Here is some pseudocode: int calculateGCD(int[] numbers){ if(numbers.length <= 2){ return … Read more
As far as I know, there isn’t any built-in method for primitives. But something as simple as this should do the trick: public int gcd(int a, int b) { if (b==0) return a; return gcd(b,a%b); } You can also one-line it if you’re into that sort of thing: public int gcd(int a, int b) { … Read more
Here is a recursive solution, using the Euclidean algorithm. var gcd = function(a, b) { if (!b) { return a; } return gcd(b, a % b); } Our base case is when b is equal to 0. In this case, we return a. When we’re recursing, we swap the input arguments but we pass the … Read more
Since GCD is associative, GCD(a,b,c,d) is the same as GCD(GCD(GCD(a,b),c),d). In this case, Python’s reduce function would be a good candidate for reducing the cases for which len(numbers) > 2 to a simple 2-number comparison. The code would look something like this: if len(numbers) > 2: return reduce(lambda x,y: GCD([x,y]), numbers) Reduce applies the given … Read more
I’ve used Euclid’s algorithm to find the greatest common divisor of two numbers; it can be iterated to obtain the GCD of a larger set of numbers. private static long gcd(long a, long b) { while (b > 0) { long temp = b; b = a % b; // % is remainder a = … Read more