A typical fast solution to the gcd algorithm would be some iterative version like this one: def gcd(x, y): while y != 0: (x, y) = (y, x % y) return x In fact, in python you’d just use directly the gcd function provided by fractions module. And if you wanted such function to deal … Read more
I think you might be looking for something like this: combine :: Int ->Int -> (Int,Int,Int) combine n m = (x1, y1, gcd n m) where (x1, y1) = gcdext n m gcdext :: Int -> Int -> (Int, Int) gcdext n m = gcdexthelper n m 1 0 0 1 where gcdexthelper n m … Read more
(Note that when computing gcd(a,b), we can assume that a <= b; if that were not true, we could just swap them.) The Euclidean algorithm is certainly the most efficient way to compute the gcd. But if you need two other (inefficient) ways to compute gcd(a,b), there are plenty: Starting from a and going down, … Read more