I cannot get my head around this.

In Paillier, the ciphertext is calculated using

$c = g^m.r^n\ mod\ n^2$

where $(n,g)$ forms the public key and $r$ is a random number $0<r<n$.

Assuming an attacker knows $c$, $r$, $n$. and $g$, what would it take - in average - to find $m$ ?


You don't need to guess, you can find $m$ for sure.

If you know $c,r,n,g$, then you can eliminate $r^n$ from the ciphertext and get $c'=g^m \bmod n^2$.

In $Z_{n^2}^*$, we have $(n+1)^x = 1+nx \bmod n^2$ ($x\in Z_n$). Therefore:

  • If $g=n+1$ is used, then $c'=g^m \bmod n^2 =1+mn$, then you can find $m=(c'-1)/n$.
  • If $g\ne n+1$, since the order of $g$ must be a multiple of $n$, we have $g= (1+n)^a=1+an \bmod n^2$ for some $a$, then we can find $a=(g-1)/n$. Then $c'=g^m \bmod n^2 =1+amn$, and we can compute $m=(c'-1)/an$.

Added based on Bruno's comment: the general forumula (for all $g$) is $m=(c'-1)/(g-1)$.

  • $\begingroup$ Doesn't $m=(c'-1)/n$ with $c'=c\,r^{-n}\bmod n^2$ silently assumes $g=n+1$ ? Otherwise, things are more complex, perhaps much more costly. The best I get so far is solving $g^m\bmod n$ for $m$ by baby-step/giant step, with $\approx 2\sqrt{m_\text{max}}$ modular multiplications modulo $n$. [update] Ah the new answer seems to address that! $\endgroup$
    – fgrieu
    Jul 11 '18 at 10:54
  • $\begingroup$ @fgrieu Thanks for pointing this out. The answer is edited to add the case where $g\ne n+1$. $\endgroup$ Jul 11 '18 at 10:59
  • $\begingroup$ Thanks. Generalizes to $m=(c'-1)/(g-1)$ for all values of $g$ then $\endgroup$
    – BGR
    Jul 11 '18 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.