Question

完成本教程后

http://tommd.github.io/

使用HElib库： https://github.com/shaih/HElib

我得到以下输出：

输出已损坏。鉴于该示例具有级别16，应该有足够的空间来执行这些操作。

参数有问题吗？

代码：

#include "FHE.h"
#include "EncryptedArray.h"
#include <NTL/lzz_pXFactoring.h>
#include <fstream>
#include <sstream>
#include <sys/time.h>

using namespace std;
/**
 *
 */
int main(int argc, char** argv) {

     /* On our trusted system we generate a new key
     * (or read one in) and encrypt the secret data set.
     */

    long m=0, p=2, r=1; // Native plaintext space
                        // Computations will be 'modulo p'
    long L=16;          // Levels
    long c=3;           // Columns in key switching matrix
    long w=64;          // Hamming weight of secret key
    long d=0;
    long security = 128;
    ZZX G;
    m = FindM(security,L,c,p, d, 0, 0);

    FHEcontext context(m, p, r);
    // initialize context
    buildModChain(context, L, c);
    // modify the context, adding primes to the modulus chain
    FHESecKey secretKey(context);
    // construct a secret key structure
    const FHEPubKey& publicKey = secretKey;
    // an "upcast": FHESecKey is a subclass of FHEPubKey

    //if(0 == d)
    G = context.alMod.getFactorsOverZZ()[0];

   secretKey.GenSecKey(w);
   // actually generate a secret key with Hamming weight w

   addSome1DMatrices(secretKey);
   cout << "Generated key" << endl;


   EncryptedArray ea(context, G);
   // constuct an Encrypted array object ea that is
   // associated with the given context and the polynomial G

   long nslots = ea.size();

   vector<long> v1;
   for(int i = 0 ; i < nslots; i++) {
       v1.push_back(i*2);
   }
   Ctxt ct1(publicKey);
   ea.encrypt(ct1, publicKey, v1);

   vector<long> v2;
   Ctxt ct2(publicKey);
   for(int i = 0 ; i < nslots; i++) {
       v2.push_back(i*3);
   }
   ea.encrypt(ct2, publicKey, v2);

     // On the public (untrusted) system we
   // can now perform our computation

   Ctxt ctSum = ct1;
   Ctxt ctProd = ct1;

   ctSum += ct2;
   ctProd *= ct2;


     vector<long> res;
    ea.decrypt(ctSum, secretKey, res);

    cout << "All computations are modulo " << p << "." << endl;
    for(int i = 0; i < res.size(); i ++) {
        cout << v1[i] << " + " << v2[i] << " = " << res[i] << endl;
    }

    ea.decrypt(ctProd, secretKey, res);
    for(int i = 0; i < res.size(); i ++) {
        cout << v1[i] << " * " << v2[i] << " = " << res[i] << endl;
    }

    return 0;
}

生成的密钥
所有计算都是模2。

0 + 0 = 0
2 + 3 = 1
4 + 6 = 0
6 + 9 = 1
8 + 12 = 0
10 + 15 = 1
12 + 18 = 0
14 + 21 = 1
16 + 24 = 0
18 + 27 = 1
20 + 30 = 0
22 + 33 = 1
24 + 36 = 0
26 + 39 = 1
28 + 42 = 0
30 + 45 = 1
32 + 48 = 0
34 + 51 = 1
36 + 54 = 0
38 + 57 = 1
40 + 60 = 0
42 + 63 = 1
44 + 66 = 0
46 + 69 = 1
48 + 72 = 0
50 + 75 = 1
52 + 78 = 0
54 + 81 = 1
56 + 84 = 0
58 + 87 = 1
60 + 90 = 0

...省略了一些总和输出

0 * 0 = 0 2 * 3 = 0 4 * 6 = 0 6 * 9 = 0 8 * 12 = 0 10 * 15 = 0 12 * 18 = 0 14 * 21 = 0 16 * 24 = 0 18 * 27 = 0 20 * 30 = 0 22 * 33 = 0 24 * 36 = 0 26 * 39 = 0 28 * 42 = 0 30 * 45 = 0 32 * 48 = 0 34 * 51 = 0 36 * 54 = 0 38 * 57 = 0 40 * 60 = 0 42 * 63 = 0 44 * 66 = 0 46 * 69 = 0 48 * 72 = 0 50 * 75 = 0 52 * 78 = 0 54 * 81 = 0 56 * 84 = 0 58 * 87 = 0 60 * 90 = 0 62 * 93 = 0 64 * 96 = 0 66 * 99 = 0 68 * 102 = 0 70 * 105 = 0 72 * 108 = 0 74 * 111 = 0 76 * 114 = 0 78 * 117 = 0 80 * 120 = 0 82 * 123 = 0 84 * 126 = 0 86 * 129 = 0

...

Answer 1

啊，这是对正在执行的操作的误解。注意常量p=2。我有文字All computations are modulo 2.。也许还说明All inputs are modulo 2将有助于敲定这一点。让我们看看我们的一些计算：

0 + 0 mod 2 = 0
2 + 3 mod 2 = 1
4 + 6 mod 2 = 0
6 + 9 mod 2 = 1

一切看起来都不错 - 加成环2只是异或。乘法怎么样？在环2（二进制）中，它只是AND：

0 * 0 = 0
2 * 3 = 6 mod 2 = 0
4 * 6 = 24 mod 2 = 0
6 * 9 = 54 mod 2 = 0

所以所有人都检查出来。最后，回顾一下博客，看看我再次打电话给你，让你有办法对你认为更令人愉悦的事情进行操作：

在这种情况下，我正在构建GF（2） - 所以我的同态添加是XOR，乘法是AND。改变这一点就像改变一样简单 p的值。想要看到2 + 2 = 4的人应该将p设置为某事匹配他们想要的域，例如257以获得8位Ints。

然而，HELIB在这方面已经退步了 - 设置p等于大于2的任何内容在我上次尝试时都不起作用。 Shai证实这是一个已知的回归。

运行一个hello world HElib程序

1 个答案: