Solving modular linear congruences for large numbers

问题

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 copy/pasted it but lost the original link:

def linear_congruence(a, b, m):
    if b == 0:
        return 0

    if a < 0:
        a = -a
        b = -b

    b %= m
    while a > m:
        a -= m

    return (m * linear_congruence(m, -b, a) + b) // a

This works fine for smaller numbers, for example:

In [167]: pow_mod(8261, 63, 4033)                                                                                                                             
63 1 8261 4033
31 195 1728 4033
15 2221 1564 4033
7 1231 2098 4033
3 1518 1601 4033
1 2452 2246 4033
0 2147 3266 4033
Out[167]: 2147

And the linear congruence works:

linear_congruence(8261, 3266, 4033):
2147

But i hit maximum recursion depth with larger numbers. Is there a better algorithm or non recursive algorithm of the linear_congruence algorithm i provided?

Based on Eric Postpischil's remark, i wrote the pseudocode from the wikipedia entry and created a very fast linear congruence algorithm utilizing the method from here: http://gauss.math.luc.edu/greicius/Math201/Fall2012/Lectures/linear-congruences.article.pdf .

This works nicely on pows with a powers of 2-1, to get the answer. I'm looking into how offsetting from this changes the answer and hope to incorporate it to work for those answers as well, but for now, i have what i need since i'm working with powers of 2 -1 for y in pow(x, y, z):

 def fastlinearcongruencex(powx, divmodx, N, withstats=False):
   x, y, z = egcditerx(powx, N, withstats)
   if x > 1:
      powx//=x
      divmodx//=x
      N//=x
      if withstats == True:
        print(f"powx = {powx}, divmodx = {divmodx}, N = {N}")
      x, y, z = egcditerx(powx, N)
      if withstats == True:
        print(f"x = {x}, y = {y}, z = {z}")
   answer = (y*divmodx)%N
   if withstats == True:
      print(f"answer = {answer}")
   return answer

def egcditerx(a, b, withstats=False):
  s = 0
  r = b
  old_s = 1
  old_r = a
  while r!= 0:
    quotient = old_r // r
    old_r, r = r, old_r - quotient * r
    old_s, s = s, old_s - quotient * s
    if withstats == True:
      print(f"quotient = {quotient}, old_r = {old_r}, r = {r}, old_s = {old_s}, s = {s}")
  if b != 0:
    bezout_t = quotient = (old_r - old_s * a) // b
    if withstats == True:
      print(f"bezout_t = {bezout_t}")
  else:
    bezout_t = 0
  if withstats == True:
    print("Bézout coefficients:", (old_s, bezout_t))
    print("greatest common divisor:", old_r)
  return old_r, old_s, bezout_t

It even works instantaneously on 4096 byte numbers which is great:

In [19036]: rpowxxxwithbitlength(1009,offset=0, withstats=True, withx=True, withbl=True)                                                                  
63 1 272 1009
31 272 327 1009
15 152 984 1009
7 236 625 1009
3 186 142 1009
1 178 993 1009
0 179 256 1009
Out[19036]: (179, 256, True, 272)

In [19037]: fastlinearcongruencex(272,256,1009)                                                                                                           
Out[19037]: 179

Thank you Eric for pointing out what this was, i wrote an extremely fast linear congruence algorithm utilizing egcd and the procedure from the pdf above. If any stackoverflowers need a fast algorithm, please point them to this one. I also learned that congruence is always maintained when the pow(x,y,z) has a y off of the powers of 2-1. I will look into this further to see if an offset change exists to keep the answers intact and will follow up in the future if found.

回答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

来源：https://stackoverflow.com/questions/63021828/solving-modular-linear-congruences-for-large-numbers