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