C 或 C++:用于分解整数的库?

发布于 2024-07-21 07:05:11 字数 1817 浏览 9 评论 0原文

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无力看清 2024-07-28 07:05:11

查看 Jason Papadopoulos 编写的用于分解大整数的 MSIEVE 库。

Msieve 是我努力理解和优化的结果
使用最强大的现代算法对整数进行因式分解。

本文档对应于 Msieve 库的 1.46 版本。
不要指望仅仅通过阅读它就能成为保理专家。 我有
包括一个相对完整的参考列表,您可以和
如果您不想将代码视为黑匣子,则应该查找
解决您的保理问题。

Check out MSIEVE library for factoring large integers by Jason Papadopoulos.

Msieve is the result of my efforts to understand and optimize how
integers are factored using the most powerful modern algorithms.

This documentation corresponds to version 1.46 of the Msieve library.
Do not expect to become a factoring expert just by reading it. I've
included a relatively complete list of references that you can and
should look up if you want to treat the code as more than a black box
to solve your factoring problems.

寻找一个思念的角度 2024-07-28 07:05:11

要在 C 中因式分解,您可以尝试使用概率方法:

我的命题的标题:

#include <stdlib.h>
#include <sys/time.h>

typedef unsigned long long int positive_number; // __uint128_t
static inline positive_number multiplication_modulo(positive_number lhs, positive_number rhs, positive_number mod);
static int is_prime(positive_number n, int k); // prime checker
positive_number factor_worker(positive_number n);
positive_number factor(positive_number n, int timeout);

因式分解过程经理,因为有一个timeout:

double microtime() {
    struct timeval time; gettimeofday(&time, 0);
    return (double) time.tv_sec + (double) time.tv_usec / 1e6;
}

// This is the master function you can call, expecting a number and a timeout(s)
positive_number factor(positive_number n, int timeout) {
    if (n < 4) return n;
    if (n != (n | 1)) return 2;
    double begin = microtime();
    int steps = 8; // primality check iterations
    positive_number a, b;
    for (a = n >> 1, b = (a + n / a) >> 1; b < a; a = b, b = (a + n / a) >> 1, ++steps);
    if (b * b == n) return b ; // we just computed b = sqrt(n)
    if (is_prime(n, steps)) return n;
    do { positive_number res = factor_worker(n);
        if (res != n) return res;
        if (++steps % 96 == 0 && is_prime(n, 32)) return n ; // adjust it
    } while (microtime() - begin < timeout);
    return n;
}

乘数帮助器,因为计算是以 N 为模完成的:

static inline positive_number multiplication_modulo(positive_number lhs, positive_number rhs, positive_number mod) {
    positive_number res = 0; // we avoid overflow in modular multiplication
    for (lhs %= mod, rhs%= mod; rhs; (rhs & 1) ? (res = (res + lhs) % mod) : 0, lhs = (lhs << 1) % mod, rhs >>= 1);
    return res; // <= (lhs * rhs) % mod
}

质数检查器帮助器,当然:

static int is_prime(positive_number n, int k) {
    positive_number a = 0, b, c, d, e, f, g; int h, i;
    if ((n == 1) == (n & 1)) return n == 2;
    if (n < 51529) // fast constexpr check for small primes (removable)
        return (n & 1) & ((n < 6) * 42 + 0x208A2882) >> n % 30 && (n < 49 || (n % 7 && n % 11 && n % 13 && n % 17 && n % 19 && n % 23 && n % 29 && n % 31 && n % 37 && (n < 1369 || (n % 41 && n % 43 && n % 47 && n % 53 && n % 59 && n % 61 && n % 67 && n % 71 && n % 73 && ( n < 6241 || (n % 79 && n % 83 && n % 89 && n % 97 && n % 101 && n % 103 && n % 107 && n % 109 && n % 113 && ( n < 16129 || (n % 127 && n % 131 && n % 137 && n % 139 && n % 149 && n % 151 && n % 157 && n % 163 && n % 167 && ( n < 29929 || (n % 173 && n % 179 && n % 181 && n % 191 && n % 193 && n % 197 && n % 199 && n % 211 && n % 223))))))))));
    for (b = c = n - 1, h = 0; !(b & 1); b >>= 1, ++h);
    for (; k--;) {
        for (g = 0; g < sizeof(positive_number); ((char*)&a)[g++] = rand()); // random number
        do for (d = e = 1 + a % c, f = n; (d %= f) && (f %= d););
        while (d > 1 && f > 1);
        for (d = f = 1; f <= b; f <<= 1);
        for (; f >>= 1; d = multiplication_modulo(d, d, n), f & b && (d = multiplication_modulo(e, d, n)));
        if (d == 1) continue;
        for (i = h; i-- && d != c; d = multiplication_modulo(d, d, n));
        if (d != c) return 0;
    }
    return 1;
}

分解worker,单个调用不会保证成功,这是一次概率尝试:

positive_number factor_worker(positive_number n) {
    size_t a; positive_number b = 0, c, d, e, f;
    for (a = 0; a < sizeof(positive_number); ((char*)&b)[a++] = rand()); // pick random polynomial
    c = b %= n;
    do {
        b = multiplication_modulo(b, b, n); if(++b == n) b = 0;
        c = multiplication_modulo(c, c, n); if(++c == n) c = 0;
        c = multiplication_modulo(c, c, n); if(++c == n) c = 0;
        for (d = n, e = b > c ? b - c : c - b; e; f = e, e = multiplication_modulo(d / f, f, n), e = (d - e) % n, d = f);
        // handle your precise timeout in the for loop body
    } while (d == 1);
    return d;
}

使用示例:

#include <stdio.h>

positive_number exec(positive_number n) {
    positive_number res = factor(n, 2); // 2 seconds
    if (res == n) return res + printf("%llu * ", n) * fflush(stdout) ;
    return exec(res) * exec(n / res);
}

int main() {
    positive_number n = 0, mask = -1, res;
    for (int i = 0; i < 1000;) {
        int bits = 4 + rand() % 60; // adjust the modulo for tests
        for (size_t k = 0; k < sizeof(positive_number); ((char*)&n)[k++] = rand());
        // slice a random number with the "bits" variable
        n &= mask >> (8 * sizeof (positive_number) - bits); n += 4;
        printf("%5d. (%2d bits) %llu = ", ++i, bits, n);
        res = exec(n);
        if (res != n) return 1;
        printf("1\n");
    }
}

您可以将其放入 primes.c 文件中,然后编译 + 执行:

gcc -O3 -std=c99 -Wall -pedantic primes.c ; ./a.out ;

另外, 128 位整数 GCC 扩展 扩展可能可用。

示例输出:

  358205873110913227 =    380003149 * 942639223    took 0.01s
  195482582293315223 =    242470939 * 806210357    took 0.0021s
  107179818338278057 =    139812461 * 766597037    took 0.0023s
   44636597321924407 =    182540669 * 244529603    took 0s
  747503348409771887 =    865588487 * 863578201    took 0.016s

// 128-bit extension available output :
170141183460469231731687303715506697937 =
13602473 * 230287853 * 54315095311400476747373 took 0.646652s

信息:此 C99 100 行概率因式分解软件命题基于 Miller–Rabin 素性测试,后跟或不跟随 >Pollard 的 rho 算法。 和你一样,我最初的目标是分解一点long long int。 根据我的测试,它的运行速度对我来说足够快,甚至对于一些更大的人来说也是如此。 谢谢。

该软件按“原样”提供,不提供任何形式的保证。

To factor integers in C you can try to use a probabilistic approach :

The headers of my proposition :

#include <stdlib.h>
#include <sys/time.h>

typedef unsigned long long int positive_number; // __uint128_t
static inline positive_number multiplication_modulo(positive_number lhs, positive_number rhs, positive_number mod);
static int is_prime(positive_number n, int k); // prime checker
positive_number factor_worker(positive_number n);
positive_number factor(positive_number n, int timeout);

The factorization process manager, because there is a timeout:

double microtime() {
    struct timeval time; gettimeofday(&time, 0);
    return (double) time.tv_sec + (double) time.tv_usec / 1e6;
}

// This is the master function you can call, expecting a number and a timeout(s)
positive_number factor(positive_number n, int timeout) {
    if (n < 4) return n;
    if (n != (n | 1)) return 2;
    double begin = microtime();
    int steps = 8; // primality check iterations
    positive_number a, b;
    for (a = n >> 1, b = (a + n / a) >> 1; b < a; a = b, b = (a + n / a) >> 1, ++steps);
    if (b * b == n) return b ; // we just computed b = sqrt(n)
    if (is_prime(n, steps)) return n;
    do { positive_number res = factor_worker(n);
        if (res != n) return res;
        if (++steps % 96 == 0 && is_prime(n, 32)) return n ; // adjust it
    } while (microtime() - begin < timeout);
    return n;
}

The multiplier helper because computations are done modulo N :

static inline positive_number multiplication_modulo(positive_number lhs, positive_number rhs, positive_number mod) {
    positive_number res = 0; // we avoid overflow in modular multiplication
    for (lhs %= mod, rhs%= mod; rhs; (rhs & 1) ? (res = (res + lhs) % mod) : 0, lhs = (lhs << 1) % mod, rhs >>= 1);
    return res; // <= (lhs * rhs) % mod
}

The prime checker helper, of course :

static int is_prime(positive_number n, int k) {
    positive_number a = 0, b, c, d, e, f, g; int h, i;
    if ((n == 1) == (n & 1)) return n == 2;
    if (n < 51529) // fast constexpr check for small primes (removable)
        return (n & 1) & ((n < 6) * 42 + 0x208A2882) >> n % 30 && (n < 49 || (n % 7 && n % 11 && n % 13 && n % 17 && n % 19 && n % 23 && n % 29 && n % 31 && n % 37 && (n < 1369 || (n % 41 && n % 43 && n % 47 && n % 53 && n % 59 && n % 61 && n % 67 && n % 71 && n % 73 && ( n < 6241 || (n % 79 && n % 83 && n % 89 && n % 97 && n % 101 && n % 103 && n % 107 && n % 109 && n % 113 && ( n < 16129 || (n % 127 && n % 131 && n % 137 && n % 139 && n % 149 && n % 151 && n % 157 && n % 163 && n % 167 && ( n < 29929 || (n % 173 && n % 179 && n % 181 && n % 191 && n % 193 && n % 197 && n % 199 && n % 211 && n % 223))))))))));
    for (b = c = n - 1, h = 0; !(b & 1); b >>= 1, ++h);
    for (; k--;) {
        for (g = 0; g < sizeof(positive_number); ((char*)&a)[g++] = rand()); // random number
        do for (d = e = 1 + a % c, f = n; (d %= f) && (f %= d););
        while (d > 1 && f > 1);
        for (d = f = 1; f <= b; f <<= 1);
        for (; f >>= 1; d = multiplication_modulo(d, d, n), f & b && (d = multiplication_modulo(e, d, n)));
        if (d == 1) continue;
        for (i = h; i-- && d != c; d = multiplication_modulo(d, d, n));
        if (d != c) return 0;
    }
    return 1;
}

The factorization worker, a single call does not guarantee a success, it's a probabilistic try :

positive_number factor_worker(positive_number n) {
    size_t a; positive_number b = 0, c, d, e, f;
    for (a = 0; a < sizeof(positive_number); ((char*)&b)[a++] = rand()); // pick random polynomial
    c = b %= n;
    do {
        b = multiplication_modulo(b, b, n); if(++b == n) b = 0;
        c = multiplication_modulo(c, c, n); if(++c == n) c = 0;
        c = multiplication_modulo(c, c, n); if(++c == n) c = 0;
        for (d = n, e = b > c ? b - c : c - b; e; f = e, e = multiplication_modulo(d / f, f, n), e = (d - e) % n, d = f);
        // handle your precise timeout in the for loop body
    } while (d == 1);
    return d;
}

Example of usage :

#include <stdio.h>

positive_number exec(positive_number n) {
    positive_number res = factor(n, 2); // 2 seconds
    if (res == n) return res + printf("%llu * ", n) * fflush(stdout) ;
    return exec(res) * exec(n / res);
}

int main() {
    positive_number n = 0, mask = -1, res;
    for (int i = 0; i < 1000;) {
        int bits = 4 + rand() % 60; // adjust the modulo for tests
        for (size_t k = 0; k < sizeof(positive_number); ((char*)&n)[k++] = rand());
        // slice a random number with the "bits" variable
        n &= mask >> (8 * sizeof (positive_number) - bits); n += 4;
        printf("%5d. (%2d bits) %llu = ", ++i, bits, n);
        res = exec(n);
        if (res != n) return 1;
        printf("1\n");
    }
}

You can put it into a primes.c file then compile + execute :

gcc -O3 -std=c99 -Wall -pedantic primes.c ; ./a.out ;

Also, 128-bit integers GCC extension extension may be available.

Example output :

  358205873110913227 =    380003149 * 942639223    took 0.01s
  195482582293315223 =    242470939 * 806210357    took 0.0021s
  107179818338278057 =    139812461 * 766597037    took 0.0023s
   44636597321924407 =    182540669 * 244529603    took 0s
  747503348409771887 =    865588487 * 863578201    took 0.016s

// 128-bit extension available output :
170141183460469231731687303715506697937 =
13602473 * 230287853 * 54315095311400476747373 took 0.646652s

Info : This C99 100 lines probabilistic factorization software proposition is based on a Miller–Rabin primality test followed or not by a Pollard's rho algo. Like you, i initially aimed to factorize just a little long long int. By my tests it's working fast enough to me, even for some larger. Thank you.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND.

汹涌人海 2024-07-28 07:05:11

GMP-ECM(整数分解的椭圆曲线法)怎么样?

如果 Inria 官方项目页面的链接不可用,您可以检查 网络存档上的最新版本

How about GMP-ECM (Elliptic Curve Method for Integer Factorization)?

If the link to the official project page at Inria is unavailable, you can check a recent version on the web archive.

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