How many times do we execute the loop in the previous algorithm?

*best case:*1 time, when*x*divides*y*or vice-versa

E.g., gcd(500, 1000)*worst case:*min(*x*,*y*) times, when gcd(*x*,*y*) = 1

Ex . gcd(500, 1001)

Hence, the previous algorithm behaves bad when *x* and *y* are big and
gcd(*x*, *y*) is small.