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149 lines
5.5 KiB
Python
149 lines
5.5 KiB
Python
from __future__ import annotations
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from math import gcd
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def pollard_rho(
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num: int,
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seed: int = 2,
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step: int = 1,
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attempts: int = 3,
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) -> int | None:
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"""
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Use Pollard's Rho algorithm to return a nontrivial factor of ``num``.
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The returned factor may be composite and require further factorization.
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If the algorithm will return None if it fails to find a factor within
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the specified number of attempts or within the specified number of steps.
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If ``num`` is prime, this algorithm is guaranteed to return None.
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https://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm
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>>> pollard_rho(18446744073709551617)
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274177
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>>> pollard_rho(97546105601219326301)
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9876543191
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>>> pollard_rho(100)
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2
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>>> pollard_rho(17)
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>>> pollard_rho(17**3)
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17
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>>> pollard_rho(17**3, attempts=1)
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>>> pollard_rho(3*5*7)
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21
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>>> pollard_rho(1)
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Traceback (most recent call last):
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...
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ValueError: The input value cannot be less than 2
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"""
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# A value less than 2 can cause an infinite loop in the algorithm.
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if num < 2:
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raise ValueError("The input value cannot be less than 2")
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# Because of the relationship between ``f(f(x))`` and ``f(x)``, this
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# algorithm struggles to find factors that are divisible by two.
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# As a workaround, we specifically check for two and even inputs.
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# See: https://math.stackexchange.com/a/2856214/165820
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if num > 2 and num % 2 == 0:
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return 2
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# Pollard's Rho algorithm requires a function that returns pseudorandom
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# values between 0 <= X < ``num``. It doesn't need to be random in the
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# sense that the output value is cryptographically secure or difficult
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# to calculate, it only needs to be random in the sense that all output
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# values should be equally likely to appear.
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# For this reason, Pollard suggested using ``f(x) = (x**2 - 1) % num``
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# However, the success of Pollard's algorithm isn't guaranteed and is
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# determined in part by the initial seed and the chosen random function.
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# To make retries easier, we will instead use ``f(x) = (x**2 + C) % num``
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# where ``C`` is a value that we can modify between each attempt.
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def rand_fn(value: int, step: int, modulus: int) -> int:
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"""
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Returns a pseudorandom value modulo ``modulus`` based on the
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input ``value`` and attempt-specific ``step`` size.
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>>> rand_fn(0, 0, 0)
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Traceback (most recent call last):
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...
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ZeroDivisionError: integer division or modulo by zero
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>>> rand_fn(1, 2, 3)
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0
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>>> rand_fn(0, 10, 7)
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3
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>>> rand_fn(1234, 1, 17)
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16
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"""
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return (pow(value, 2) + step) % modulus
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for _ in range(attempts):
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# These track the position within the cycle detection logic.
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tortoise = seed
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hare = seed
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while True:
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# At each iteration, the tortoise moves one step and the hare moves two.
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tortoise = rand_fn(tortoise, step, num)
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hare = rand_fn(hare, step, num)
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hare = rand_fn(hare, step, num)
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# At some point both the tortoise and the hare will enter a cycle whose
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# length ``p`` is a divisor of ``num``. Once in that cycle, at some point
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# the tortoise and hare will end up on the same value modulo ``p``.
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# We can detect when this happens because the position difference between
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# the tortoise and the hare will share a common divisor with ``num``.
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divisor = gcd(hare - tortoise, num)
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if divisor == 1:
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# No common divisor yet, just keep searching.
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continue
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# We found a common divisor!
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elif divisor == num:
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# Unfortunately, the divisor is ``num`` itself and is useless.
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break
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else:
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# The divisor is a nontrivial factor of ``num``!
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return divisor
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# If we made it here, then this attempt failed.
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# We need to pick a new starting seed for the tortoise and hare
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# in addition to a new step value for the random function.
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# To keep this example implementation deterministic, the
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# new values will be generated based on currently available
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# values instead of using something like ``random.randint``.
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# We can use the hare's position as the new seed.
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# This is actually what Richard Brent's the "optimized" variant does.
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seed = hare
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# The new step value for the random function can just be incremented.
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# At first the results will be similar to what the old function would
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# have produced, but the value will quickly diverge after a bit.
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step += 1
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# We haven't found a divisor within the requested number of attempts.
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# We were unlucky or ``num`` itself is actually prime.
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return None
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if __name__ == "__main__":
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import argparse
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parser = argparse.ArgumentParser()
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parser.add_argument(
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"num",
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type=int,
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help="The value to find a divisor of",
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)
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parser.add_argument(
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"--attempts",
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type=int,
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default=3,
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help="The number of attempts before giving up",
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)
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args = parser.parse_args()
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divisor = pollard_rho(args.num, attempts=args.attempts)
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if divisor is None:
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print(f"{args.num} is probably prime")
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else:
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quotient = args.num // divisor
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print(f"{args.num} = {divisor} * {quotient}")
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