The greatest common divisor of two positive integers python

Question

def mike(a,b):
    d=min(a,b)
    e=d
    if max(a,b)%d==0 and e%d==0:
        return d
    else:
        return mike(max(a,b),d-1)
print(mike(12,18))

I cannot find a way to have the value of 'e' as a fixed value of the minimum among the two integers, which in case is 12. As the code moves on so does the value of 'e'. But I want it to have a fixed value of the least among the two. Sorry for the noob question. Help please.

Answer 1

So you don't need to funcions, you can just add e as an optional argument

def mike(a,b,e=None):
    d=min(a,b)
    if not e:
        e=d
    if max(a,b)%d==0 and e%d==0:
    return d,e
    else:
        return mike(max(a,b),d-1,e)
print(mike(12,18))

That way you can call it recursively, and still track and output the minimum of the original function

Answer 2

I don't understand why you set e=d and why you need d-1 in recursion

def mike(a,b):
    rest = max(a, b) % min(a, b)
    if rest == 0:
        return min(a, b)
    else:
        return mike(min(a, b), rest)

Calc of gcd.

(48,30)

Divide the biggest by the lowest and get the rest. 48 / 30 = 1 (rest 18)

Divide the previous divisor by the rest 30 / 18 = 1 (rest 12)

Divide the previous divisor by the rest again 18 / 12 = 1 (rest 6)

Divide the previous divisor by the rest again 12 / 6 = 2 (rest 0)

gcd is the divisor when rest == 0

Answer 3

I eventually did this at the end. Thanks for all the answers though. I know I made it a way more complicated.

def mike(x,y):
    e=min(x,y)
    def mi(a,b):
        d=min(a,b)
        if max(a,b)%d==0 and e%d==0:
            return d
        else:
            return mi(max(a,b),d-1)
    zz=mi(x,y)
    return zz

print(mike(12,18))