在Mac OS 10.8上使用dev/urandom生成非常慢的随机数

Question

我正在编写一个使用Paillier Cryptosystem来加密矩阵的程序.要加密50 x 50矩阵,大约需要12秒!考虑到我打算加密大小为5000 x 5000及以上的矩阵,这是痛苦的.在Xcode上分析程序我发现这个paillier_get_rand_devurandom()是罪魁祸首.

这是调用跟踪快照:

这是此特定Paillier C库函数的源代码

    void paillier_get_rand_devurandom( void* buf, int len )
    {
          paillier_get_rand_file(buf, len, "/dev/urandom");
    }  


    void paillier_get_rand_file( void* buf, int len, char* file )
   {
         FILE* fp;
         void* p;

         fp = fopen(file, "r");

         p = buf;
         while( len )
         {
             size_t s;
             s = fread(p, 1, len, fp);
             p += s;
             len -= s;
         }

        fclose(fp);
    }

如果是的话,

    <http://www.en.wikipedia.org/wiki/Paillier_cryptosystem>
    Paillier Cryptosytem C library : <http://acsc.cs.utexas.edu/libpaillier/> 

我已经读过使用dev/random的随机数生成很慢而使用dev/urandom则更快.在我的情况下,两者都同样缓慢.这个随机数生成能否更快？

编辑:这是一个例子

    #include <stdio.h>
    #include <stdlib.h>
    #include <gmp.h>
    #include<paillier.h>


   int main(void)
   {

       paillier_pubkey_t* pub;//The public key
   paillier_prvkey_t* prv;//The private key 


       paillier_keygen(1024, &pub, &prv,paillier_get_rand_devurandom);


      paillier_ciphertext_t* ca;
      paillier_ciphertext_t* cb;
      paillier_ciphertext_t* res;


      paillier_plaintext_t* a;
      paillier_plaintext_t* b;
      paillier_plaintext_t* sum;



     a=paillier_plaintext_from_ui(45);
     b=paillier_plaintext_from_ui(100);


     //This is the encryption function 
     ca=paillier_enc(0, pub, a, paillier_get_rand_devurandom);
     cb=paillier_enc(0, pub, b, paillier_get_rand_devurandom);


     res=paillier_create_enc_zero();


     paillier_mul(pub, res,ca, cb);

     sum=paillier_dec(0, pub, prv, res);


     gmp_printf("The sum is : %Zd\n",sum);


     return 0;

   }

这是加密函数签名

    /*
Encrypt the given plaintext with the given public key using
randomness from get_rand for blinding. If res is not null, its
contents will be overwritten with the result. Otherwise, a new
paillier_ciphertext_t will be allocated and returned.
    */
    paillier_ciphertext_t* paillier_enc( paillier_ciphertext_t* res,
                                           paillier_pubkey_t* pub,
                        paillier_plaintext_t* pt,
                         paillier_get_rand_t get_rand );

对不起,这个问题只会越来越长

 The actual scale_encrypt_T()

 void scale_encrypt_T(char *scaledTfile)
{
    ...
    ...

//Printing scaled and then encrypted T[][] in a file 
for(i=0;i<n;i++)
{
    for(j=0;j<n;j++)
    {
        void *buf2;

        //T[][] is the 50 x 50 matrix 
        temp=(int)(T[i][j]*pow(10,6));  //Scale factor q = 10 to the power of 6

        p0=paillier_plaintext_from_ui(temp);

        //c0 is the cipher text
         /***So this is where its taking so long***/
        c0=paillier_enc(0, pub, p0, paillier_get_rand_devurandom);


        buf2=paillier_ciphertext_to_bytes(PAILLIER_BITS_TO_BYTES(pub->bits)*2, c0);

        fwrite(buf2, 1, PAILLIER_BITS_TO_BYTES(pub->bits)*2, fpt);
        free(buf2);

    }

}

Answer 1

假设n为 50，您的代码将执行 2500 次单独的加密。每次加密可能需要一些随机数据。

通过一次调用对整个数组进行加密来进行一次加密会更有效paillier_enc。我不熟悉 libpaillier API，但看起来您可以通过将 50×50 数组的所有数据安排在连续内存中来实现此目的，然后使用将paillier_plaintext_from_bytes整个数组打包到一个paillier_plaintext_t对象中，然后使用pallier_enc与那个物体。