通过python脚本进行Php质因数分解

Question

有一个 python 脚本可以进行质因数分解。它非常快，不到一秒即可运行。但是 php 的一些函数运行起来很慢。它采用一个参数(一个长整型)，例如 1278426847636566097 并计算质因数分解。返回具有 2 个索引的数组。这个数字的结果是:

Array ( [0] => 1233387599 [1] => 1036516703 )

python 脚本: (getpq.py)

#!/usr/bin/env python
from __future__ import print_function
import prime
import sys
import json

def eprint(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)


pq = prime.primefactors(long(sys.argv[1]))

sys.stdout.write(json.dumps(pq))
sys.stdout.flush()

主要.py:

# NOTICE!!! This is copied from https://stackoverflow.com/questions/4643647/fast-prime-factorization-module

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 or n % 2 == 0:
        return False
    elif n < 1:
        raise ValueError("Out of bounds, first argument must be > 0")
    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(10000) # might seem low, but 1000*1000 = 1000000, so this will fully factor every composite < 1000000
def primefactors(n, sort=False):
    factors = []

    limit = int(n ** .5) + 1
    for checker in smallprimes:
        if checker > limit: break
        while n % checker == 0:
            factors.append(checker)
            n //= checker
            limit = int(n ** .5) + 1
            if checker > limit: 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 * b) // gcd(a, b)

因为我不懂 python，有什么办法可以做到这一点是 php 吗？

注意:我发现一个非常糟糕的算法，用 php 实现，大约需要 600 秒:

public function primefactor($num) {
    $sqrt = sqrt($num);
    for ($i = 2; $i <= $sqrt; $i++) {
        if ($num % $i == 0) {
            return array_merge($this->primefactor($num/$i), array($i));
        }
    }
    return array($num);
}

Answer 1

当然有。您需要研究“质因数分解”算法。如果您将 PHP 作为搜索的要求添加，您可以找到比上面的暴力方法更好的实现代码。开始here .

请注意，Python 方法不显示有关算法的任何内容:prime 包具有内置算法，您提供的代码只是调用它。我读过 Python 包使用概率方法(通过 300 位数字之类的东西测试为 100% 准确)。当你做你的研究时，请留意那些。将有几种通用语言的实现；我至少知道 Python、Java 和 C++。