[英]Restore a number from several its remainders (chinese remainder theorem)
I have a long integer number, but it is stored not in decimal form, but as set of remainders. 我有一个长整数,但它不是以十进制形式存储,而是存储为余数。
So, I have not the N
number, but set of such remainders: 所以,我没有
N
号码,而是有这样的余数:
r_1 = N % 2147483743
r_2 = N % 2147483713
r_3 = N % 2147483693
r_4 = N % 2147483659
r_5 = N % 2147483647
r_6 = N % 2147483629
I know, that N is less than multiplication of these primes, so chinese remainder theorem does work here ( http://en.wikipedia.org/wiki/Chinese_remainder_theorem ). 我知道,N小于这些素数的乘法,所以中文余数定理在这里起作用( http://en.wikipedia.org/wiki/Chinese_remainder_theorem )。
How can I restore N
in decimal, if I have this 6 remainders? 如果我有这6个剩余部分,如何以十进制恢复
N
? The wonderful will be any program to do this (C/C+GMP/C++/perl/java/bc). 精彩的将是任何程序(C / C + GMP / C ++ / perl / java / bc)。
For example, what minimal N can have this set of remainders: 例如,最小N可以有这组余数:
r_1 = 1246736738 (% 2147483743)
r_2 = 748761 (% 2147483713)
r_3 = 1829651881 (% 2147483693)
r_4 = 2008266397 (% 2147483659)
r_5 = 748030137 (% 2147483647)
r_6 = 1460049539 (% 2147483629)
The article you link already provides a constructive algorithm to find the solution . 您链接的文章已经提供了一个建设性的算法来找到解决方案 。
Basically, for each i
you solve integer equation ri*ni + si*(N/ni) = 1
where N = n1*n2*n3*...
. 基本上,对于每个
i
,求解整数方程ri*ni + si*(N/ni) = 1
,其中N = n1*n2*n3*...
The ri
and si
are unknowns here. ri
和si
在这里是未知的。 This can be solved by extended euclidean algorithm . 这可以通过扩展的欧几里德算法来解决。 It's very popular and you'll have no problem finding sample implementations in any language.
它非常受欢迎,您可以在任何语言中查找示例实现。
Then, assuming ei = si*(N/ni)
, the answer is sum(ei*ai)
for every i
. 然后,假设
ei = si*(N/ni)
,答案是每个i
sum(ei*ai)
。
All this is described in that article, with proof and example. 所有这些都在该文章中进行了描述,并附有证据和示例。
Here the code (C+GMP), based on this LGPL code by Ben Lynn blynn@github; 这里是代码(C + GMP),基于BenLen blynn @ github的 LGPL代码; stanford of Garner algorithm (found with RIP Google Code Search by query garner mpz_t): https://github.com/blynn/pbc/blob/master/guru/indexcalculus.c (Part of his The PBC (Pairing-Based Crypto) library)
加纳算法的stanford (通过查询garner mpz_t找到RIP谷歌代码搜索): https : //github.com/blynn/pbc/blob/master/guru/indexcalculus.c (他的部分PBC(基于配对的加密)图书馆)
Compile with gcc -std=c99 -lgmp
. 使用
gcc -std=c99 -lgmp
编译。 Also change size for your case. 还可以根据您的情况更改大小。
#include <gmp.h>
#include <stdlib.h>
#include <stdio.h>
#include <malloc.h>
// Garner's Algorithm.
// See Algorithm 14.71, Handbook of Cryptography.
// x - result v residuals m - primes t-size of vectors
static void CRT(mpz_t x, mpz_ptr *v, mpz_ptr *m, int t) {
mpz_t u;
mpz_t C[t];
int i, j;
mpz_init(u);
for (i=1; i<t; i++) {
mpz_init(C[i]);
mpz_set_ui(C[i], 1);
for (j=0; j<i; j++) {
mpz_invert(u, m[j], m[i]);
mpz_mul(C[i], C[i], u);
mpz_mod(C[i], C[i], m[i]);
}
}
mpz_set(u, v[0]);
mpz_set(x, u);
for (i=1; i<t; i++) {
mpz_sub(u, v[i], x);
mpz_mul(u, u, C[i]);
mpz_mod(u, u, m[i]);
for (j=0; j<i; j++) {
mpz_mul(u, u, m[j]);
}
mpz_add(x, x, u);
}
for (i=1; i<t; i++) mpz_clear(C[i]);
mpz_clear(u);
}
const int size=6; // Change this please
int main()
{
mpz_t res;
mpz_ptr t[size], p[size];
for(int i=0;i<size;i++) {
t[i]=(mpz_ptr)malloc(sizeof(mpz_t));
p[i]=(mpz_ptr)malloc(sizeof(mpz_t));
mpz_init(p[i]);
mpz_init(t[i]);
}
mpz_init(res);
for(int i=0;i<size;i++){
unsigned long rr,pp;
scanf("%*c%*c%*c = %lu (%% %lu)\n",&rr,&pp);
printf("Got %lu res on mod %% %lu \n",rr,pp);
mpz_set_ui(p[i],pp);
mpz_set_ui(t[i],rr);
}
CRT(res,t,p,size);
gmp_printf("N = %Zd\n", res);
}
Example is solved: 示例已解决:
$ ./a.out
r_1 = 1246736738 (% 2147483743)
r_2 = 748761 (% 2147483713)
r_3 = 1829651881 (% 2147483693)
r_4 = 2008266397 (% 2147483659)
r_5 = 748030137 (% 2147483647)
r_6 = 1460049539 (% 2147483629)
Got 1246736738 res on mod % 2147483743
Got 748761 res on mod % 2147483713
Got 1829651881 res on mod % 2147483693
Got 2008266397 res on mod % 2147483659
Got 748030137 res on mod % 2147483647
Got 1460049539 res on mod % 2147483629
N = 703066055325632897509116263399480311
N is 703066055325632897509116263399480311 N是703066055325632897509116263399480311
Here's a Python 3 implementation, based on this Rosetta Code task: https://rosettacode.org/wiki/Chinese_remainder_theorem 这是一个Python 3实现,基于此Rosetta Code任务: https : //rosettacode.org/wiki/Chinese_remainder_theorem
from functools import reduce
from operator import mul
def chinese_remainder(n, a):
"""
Chinese Remainder Theorem.
:param n: list of pairwise relatively prime integers
:param a: remainders when x is divided by n
"""
s = 0
prod = reduce(mul, n)
for n_i, a_i in zip(n, a):
p = prod // n_i
s += a_i * inverse(p, n_i) * p
return s % prod
def inverse(a, b):
"""
Modular multiplicative inverse.
"""
b0 = b
x0, x1 = 0, 1
if b == 1:
return 1
while a > 1:
q = a // b
a, b = b, a % b
x0, x1 = x1 - q * x0, x0
if x1 < 0:
x1 += b0
return x1
n = [2147483743, 2147483713, 2147483693, 2147483659, 2147483647, 2147483629]
a = [1246736738, 748761, 1829651881, 2008266397, 748030137, 1460049539]
print(chinese_remainder(n, a)) # 703066055325632897509116263399480311
A nice feature of Python is that it supports arbitrarily large integers naturally. Python的一个很好的特性是它自然地支持任意大的整数。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.