Question

I am given the function gcd, which is defined as follows:

def gcd(a, b):
    if (0 == a % b):
        return b
    return gcd(b, a%b)

Now I am asked to write a recursive function gcd2(a,b) that returns a list of three numbers (g, s, t) where g = gcd(a, b) and g = s*a + t*b.

This means that you would enter two values (a and b) into the gcd(a, b) function. The value it returns equals g in the next function.

These same a and b values are then called into gcd2(a, b). The recursive part is then used to find the values for s and t so that g = s*a + t*b.

I am not sure how to approach this because I can’t really envision what the “stopping-condition” would be, or what exactly I’d be looping through recursively to actually find s and t. Can anyone help me out?

Answer 1

The key insight is that we can work backwards, finding s and t for each a and b in the recursion. So say we have a = 21 and b = 15. We need to work through each iteration, using several values — a, b, b % a, and c where a = c * b + a % b. First, let’s consider each step of the basic GCD algorithm:

21 = 1 * 15 + 6
15 = 2 * 6  + 3
6  = 2 * 3  + 0 -> end recursion

So our gcd (g) is 3. Once we have that, we determine s and t for 6 and 3. To do so, we begin with g, expressing it in terms of (a, b, s, t = 3, 0, 1, -1):

3  = 1 * 3 + -1 * 0

Now we want to eliminate the 0 term. From the last line of the basic algorithm, we know that 0 = 6 – 2 * 3:

3 = 1 * 3 + -1 * (6 - 2 * 3)

Simplifying, we get

3 = 1 * 3 + -1 * 6 + 2 * 3
3 = 3 * 3 + -1 * 6

Now we swap the terms:

3 = -1 * 6 + 3 * 3

So we have s == -1 and t == 3 for a = 6 and b = 3. So given those values of a and b, gcd2 should return (3, -1, 3).

Now we step back up through the recursion, and we want to eliminate the 3 term. From the next-to-last line of the basic algorithm, we know that 3 = 15 – 2 * 6. Simplifying and swapping again (slowly, so that you can see the steps clearly…):

3 = -1 * 6 + 3 * (15 - 2 * 6)
3 = -1 * 6 + 3 * 15 - 6 * 6
3 = -7 * 6 + 3 * 15
3 = 3 * 15 + -7 * 6

So for this level of recursion, we return (3, 3, -7). Now we want to eliminate the 6 term.

3 = 3 * 15 + -7 * (21 - 1 * 15)
3 = 3 * 15 + 7 * 15 - 7 * 21
3 = 10 * 15 - 7 * 21
3 = -7 * 21 + 10 * 15

And voila, we have calculated s and t for 21 and 15.

So schematically, the recursive function will look like this:

def gcd2(a, b):
    if (0 == a % b):
        # calculate s and t
        return b, s, t
    else:
        g, s, t = gcd2(b, a % b)
        # calculate new_s and new_t
        return g, new_s, new_t

Note that for our purposes here, using a slightly different base case simplifies things:

def gcd2(a, b):
    if (0 == b):
        return a, 1, -1
    else:
        g, s, t = gcd2(b, a % b)
        # calculate new_s and new_t
        return g, new_s, new_t