Extended Binary Euclidean Algorithm

Core Idea

This code operates as follows:

1. It removes the common factor $2^k$ from $A = 2^k a$ and $B = 2^k b$.

2. Then, it applies the Binary GCD rules starting from $u = a$ and $v = b$, and reduces $u$ and $v$ until they become equal. The case where both $u$ and $v$ are even has already been handled in the previous step, so it's excluded here.

$$\gcd(u, v) = \begin{cases} \gcd(u, v) & \text{ if } u = v\\ \gcd \left(\frac{u}{2}, v \right) & \text{ if } u \equiv 0 \pmod 2 \land \ v \equiv 1 \pmod 2 \\ \gcd \left(u, \frac{v}{2} \right) & \text{ if } u \equiv 1 \pmod 2 \land \ v \equiv 0 \pmod 2 \\ \gcd \left(u - v, v \right) & \text{ if } u > v \land u \equiv 1 \pmod 2 \land \ v \equiv 1 \pmod 2 \\ \gcd \left(u, v - u \right) & \text{ if } v > u \land u \equiv 1 \pmod 2 \land \ v \equiv 1 \pmod 2 \\ \end{cases}$$

Meanwhile, we set the variables as $x_1 = 1, y_1 = 0, x_2 = 0, y_2 = 1$, and maintain the following two invariants:

$$ax_1 + by_1 = u, \quad ax_2 + by_2 = v$$

At each step, we update the expressions in a way that preserves these invariants. Let’s consider only case 2 and case 4 as examples.

Case 2: When $u$ is even and $v$ is odd

If $x_1$ and $y_1$ are both even, we update as follows:

$$ax_1 + by_1 = u \Leftrightarrow a\frac{x_1}{2} + b\frac{y_1}{2} = \frac{u}{2}$$

Otherwise (if either is odd), we update as:

$$ax_1 + by_1 = u \Leftrightarrow a\frac{(x_1 + b)}{2} + b\frac{(y_1 - a)}{2} = \frac{u}{2}$$

Why are $x_1 + b$ and $y_1 - a$ always even?

Since $u$ is even, the expression $a x_1 + b y_1$​ must also be even. A sum is even only if both terms are even or both are odd.

Now, note that $a$ and $b$ cannot both be even — because we already factored out the common power of 2 beforehand.

a	x₁	a * x₁	b	y₁	b * y₁
odd	odd	odd	odd	odd	odd
even	odd	even	odd	even	even
odd	even	even	even	odd	even

In each of these cases, the sum $a x_1 + b y_1$ is even, and both $x_1 + b$ and $y_1 - a$ are even as well.

Case 4: When $u > v$ and both $u$ and $v$ are odd

$$ax_1 + by_1 = u,\quad ax_2 + by_2 = v \Leftrightarrow a(x_1 - x_2) + b(y_1 - y_2) = u - v$$

Code

def binary_extended_gcd(a, b):
    # base case: if a or b is 0, return the other number as gcd, with appropriate coefficients
    if a == 0: return (b, 0, 1)
    if b == 0: return (a, 1, 0)

    # Step 1: Remove common factors of 2
    # gcd(2a, 2b) = 2 * gcd(a, b), so we factor out all common 2s
    shift = 0
    while a % 2 == 0 and b % 2 == 0:
        a //= 2
        b //= 2
        shift += 1  # keep track of how many times we factored out 2

    # Step 2: Initialize variables for the extended GCD
    u, v = a, b              # Working values to reduce down to GCD
    x1, y1 = 1, 0            # Bézout coefficients for u
    x2, y2 = 0, 1            # Bézout coefficients for v

    # Step 3: Main loop to compute GCD while updating Bézout coefficients
    while u != v:
        if u % 2 == 0:
            # If u is even, divide by 2
            u //= 2
            # Adjust (x1, y1) to keep Bézout identity valid
            if x1 % 2 == 0 and y1 % 2 == 0:
                x1 //= 2
                y1 //= 2
            else:
                # If odd, we adjust using the original a, b to keep x1, y1 as integers
                x1 = (x1 + b) // 2
                y1 = (y1 - a) // 2

        elif v % 2 == 0:
            # Same logic for v if it is even
            v //= 2
            if x2 % 2 == 0 and y2 % 2 == 0:
                x2 //= 2
                y2 //= 2
            else:
                x2 = (x2 + b) // 2
                y2 = (y2 - a) // 2

        elif u > v:
            # Reduce u by subtracting v
            u -= v
            x1 -= x2
            y1 -= y2
        else:
            # Reduce v by subtracting u
            v -= u
            x2 -= x1
            y2 -= y1

    # Step 4: Multiply back the common factor of 2 (if any)
    # At this point, u == v == gcd(a, b)
    return (u << shift, x1, y1)  # gcd * 2^shift, x, y

Written by ryan Kim from A41