Solving modular linear congruences for large numbers

Question

I\'m looking for a better algorithm than one I found on stackoverflow to handle 4096 byte numbers, i\'m hitting a maximum recursion depth.

Code from stackoverlow post, i

Answer 1

If you have Python 3.8 or later, you can do everything you need to with a very small number of lines of code.

First some mathematics: I'm assuming that you want to solve ax = b (mod m) for an integer x, given integers a, b and m. I'm also assuming that m is positive.

The first thing you need to compute is the greatest common divisor g of a and m. There are two cases:

• if b is not a multiple of g, then the congruence has no solutions (if ax + my = b for some integers x and y, then any common divisor of a and m must also be a divisor of b)

• if b is a multiple of g, then the congruence is exactly equivalent to (a/g)x = (b/g) (mod (m/g)). Now a/g and m/g are relatively prime, so we can compute an inverse to a/g modulo m/g. Multiplying that inverse by b/g gives a solution, and the general solution can be obtained by adding an arbitrary multiple of m/g to that solution.

Python's math module has had a gcd function since Python 3.5, and the built-in pow function can be used to compute modular inverses since Python 3.8.

Putting it all together, here's some code. First a function that finds the general solution, or raises an exception if no solution exists. If it succeeds, it returns two integers. The first gives a particular solution; the second gives the modulus that provides the general solution.

def solve_linear_congruence(a, b, m):
    """ Describe all solutions to ax = b  (mod m), or raise ValueError. """
    g = math.gcd(a, m)
    if b % g:
        raise ValueError("No solutions")
    a, b, m = a//g, b//g, m//g
    return pow(a, -1, m) * b % m, m

And then some driver code, to demonstrate how to use the above.

def print_solutions(a, b, m):
    print(f"Solving the congruence: {a}x = {b}  (mod {m})")
    try:
        x, mx = solve_linear_congruence(a, b, m)
    except ValueError:
        print("No solutions")
    else:
        print(f"Particular solution: x = {x}")
        print(f"General solution: x = {x}  (mod {mx})")

Example use:

>>> print_solutions(272, 256, 1009)
Solving the congruence: 272x = 256  (mod 1009)
Particular solution: x = 179
General solution: x = 179  (mod 1009)
>>> print_solutions(98, 105, 1001)
Solving the congruence: 98x = 105  (mod 1001)
Particular solution: x = 93
General solution: x = 93  (mod 143)
>>> print_solutions(98, 107, 1001)
Solving the congruence: 98x = 107  (mod 1001)
No solutions