python - 快速素数分解模块

Question

我正在寻找一种实现或清晰的算法，以便在 python、伪代码或其他任何东西中获得 N 的主要因素 -可读。有一些要求/限制:

• N 介于 1 到 ~20 位之间
• 没有预先计算的查找表，但记忆化很好
• 不需要经过数学证明(例如，如果需要，可以依赖哥德巴赫猜想)
• 不需要精确，如果需要可以是概率/确定性

我需要一个快速的素数分解算法，不仅是为了它自己，而且是为了在许多其他算法中使用，比如计算欧拉 phi(n)。

我尝试过来自 Wikipedia 等的其他算法，但要么我无法理解它们 (ECM)，要么我无法从该算法 (Pollard-Brent) 创建一个有效的实现。

我对 Pollard-Brent 算法真的很感兴趣，所以如果有更多关于它的信息/实现会非常好。

谢谢!

编辑

经过一番折腾后，我创建了一个非常快速的素数/因式分解模块。它结合了优化的试除法算法、Pollard-Brent 算法、米勒-拉宾素数测试和我在互联网上找到的最快的素数筛。 gcd 是常规 Euclid 的 GCD 实现(二进制 Euclid 的 GCD 比常规的慢得多)。

赏金

哦，快乐，可以获得赏金!但是我怎样才能赢呢？

• 在我的模块中查找优化或错误。
• 提供替代/更好的算法/实现。

最完整/最有建设性的答案将获得赏金。

最后是模块本身:

import random

def primesbelow(N):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    #""" Input N>=6, Returns a list of primes, 2 <= p < N """
    correction = N % 6 > 1
    N = {0:N, 1:N-1, 2:N+4, 3:N+3, 4:N+2, 5:N+1}[N%6]
    sieve = [True] * (N // 3)
    sieve[0] = False
    for i in range(int(N ** .5) // 3 + 1):
        if sieve[i]:
            k = (3 * i + 1) | 1
            sieve[k*k // 3::2*k] = [False] * ((N//6 - (k*k)//6 - 1)//k + 1)
            sieve[(k*k + 4*k - 2*k*(i%2)) // 3::2*k] = [False] * ((N // 6 - (k*k + 4*k - 2*k*(i%2))//6 - 1) // k + 1)
    return [2, 3] + [(3 * i + 1) | 1 for i in range(1, N//3 - correction) if sieve[i]]

smallprimeset = set(primesbelow(100000))
_smallprimeset = 100000
def isprime(n, precision=7):
    # http://en.wikipedia.org/wiki/Miller-Rabin_primality_test#Algorithm_and_running_time
    if n < 1:
        raise ValueError("Out of bounds, first argument must be > 0")
    elif n <= 3:
        return n >= 2
    elif n % 2 == 0:
        return False
    elif n < _smallprimeset:
        return n in smallprimeset


    d = n - 1
    s = 0
    while d % 2 == 0:
        d //= 2
        s += 1

    for repeat in range(precision):
        a = random.randrange(2, n - 2)
        x = pow(a, d, n)
    
        if x == 1 or x == n - 1: continue
    
        for r in range(s - 1):
            x = pow(x, 2, n)
            if x == 1: return False
            if x == n - 1: break
        else: return False

    return True

# https://comeoncodeon.wordpress.com/2010/09/18/pollard-rho-brent-integer-factorization/
def pollard_brent(n):
    if n % 2 == 0: return 2
    if n % 3 == 0: return 3

    y, c, m = random.randint(1, n-1), random.randint(1, n-1), random.randint(1, n-1)
    g, r, q = 1, 1, 1
    while g == 1:
        x = y
        for i in range(r):
            y = (pow(y, 2, n) + c) % n

        k = 0
        while k < r and g==1:
            ys = y
            for i in range(min(m, r-k)):
                y = (pow(y, 2, n) + c) % n
                q = q * abs(x-y) % n
            g = gcd(q, n)
            k += m
        r *= 2
    if g == n:
        while True:
            ys = (pow(ys, 2, n) + c) % n
            g = gcd(abs(x - ys), n)
            if g > 1:
                break

    return g

smallprimes = primesbelow(1000) # might seem low, but 1000*1000 = 1000000, so this will fully factor every composite < 1000000
def primefactors(n, sort=False):
    factors = []

    for checker in smallprimes:
        while n % checker == 0:
            factors.append(checker)
            n //= checker
        if checker > n: break

    if n < 2: return factors

    while n > 1:
        if isprime(n):
            factors.append(n)
            break
        factor = pollard_brent(n) # trial division did not fully factor, switch to pollard-brent
        factors.extend(primefactors(factor)) # recurse to factor the not necessarily prime factor returned by pollard-brent
        n //= factor

    if sort: factors.sort()

    return factors

def factorization(n):
    factors = {}
    for p1 in primefactors(n):
        try:
            factors[p1] += 1
        except KeyError:
            factors[p1] = 1
    return factors

totients = {}
def totient(n):
    if n == 0: return 1

    try: return totients[n]
    except KeyError: pass

    tot = 1
    for p, exp in factorization(n).items():
        tot *= (p - 1)  *  p ** (exp - 1)

    totients[n] = tot
    return tot

def gcd(a, b):
    if a == b: return a
    while b > 0: a, b = b, a % b
    return a

def lcm(a, b):
    return abs((a // gcd(a, b)) * b)

Answer 1

如果您不想重新发明轮子，请使用库 sympy

pip install sympy

使用函数sympy.ntheory.factorint

Given a positive integer n, factorint(n) returns a dict containing the prime factors of n as keys and their respective multiplicities as values. For example:

例子:

>>> from sympy.ntheory import factorint
>>> factorint(10**20+1)
{73: 1, 5964848081: 1, 1676321: 1, 137: 1}

您可以分解一些非常大的数字:

>>> factorint(10**100+1)
{401: 1, 5964848081: 1, 1676321: 1, 1601: 1, 1201: 1, 137: 1, 73: 1, 129694419029057750551385771184564274499075700947656757821537291527196801: 1}