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Question:

Consider the following commitment scheme:
Public parameters: large primes $q$ and $p$ such that $p = 2\cdot q + 1$, and two generators $g, g'$ of a $q$-order subgroup of $\mathbb Z_p^*$.

  • Alice commits to $t$ in $\mathbb Z_q$ by uniformly picking $r$ from $\mathbb Z_q$, and sending $g^t \cdot g'^r$.
  • She opens the commitment by sending $s$ and $r$.
  • Suppose both parties are poly-time bounded.
  • If Bob gets to pick the public parameters, can he somehow cheat? (Alice verifies the parameters in poly time)

Attempted solution:

If he can cheat, I think Bob needs to choose $q$ and $p$ in a smart way, because even if he picks $g$ and $g'$ to be equal, he is stuck with computing discrete log, which isn't possible in poly-time.

I thought maybe Bob can pick $q$ to be a Carmichael number, then Alice will think it is actually prime (by checking with Miller–Rabin algorithm).

Two problem I ran into with this:

  1. I'm not sure how much it helps him to solve the problem — he can solve it modulo the prime factors of $q$ and use CRT, but if the factors aren't small enough its still exponential.
  2. Maybe there is no prime $p$ such that $p=2q+1$ for a Carmichael number $q$.

I think my solution fails because of the problems above... would appreciate a pointer in the right direction.

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  • $\begingroup$ Note that for Carmichael numbers, all numbers that are relatively prime are Fermat liars, but not necessarily strong liars, so the Miller-Rabin test is still likely to pick them out as composite numbers. $\endgroup$ – knbk Dec 31 '17 at 14:40
  • $\begingroup$ Yeah Alice can easily cheat if she picks the parameters, but about Bob I'm not sure. I think the point is to let Alice pick 1 generator (maybe q and p as well since it doesn't seem to help her) and have Bob pick the other, but I'm still not sure how Bob can cheat, even if he picks all the parameters. $\endgroup$ – Bar Dec 31 '17 at 15:24
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Assuming that Alice verifies that:

  • $p = 2q + 1$ with $p, q$ prime
  • $g, g'$ are generators of order $q$

then Bob cannot cheat.

First note that $g$ and $g'$ generate the same subgroup, such that $g' = g^\alpha$ for some integer $\alpha$. Given a signature value $s = g^t g'^r = g^{t + \alpha r}$, there are $q$ possible messages $t' \bmod q$, and for every $t'$ there exists a unique value $r' \bmod q$ such that

$$r' \equiv \frac{t - t' + \alpha r}{\alpha} \pmod q$$

and:

$$g^tg'^r \equiv g^{t'}g'^{r'} \pmod p$$

Without further knowledge of $t$ or $r$, every pair $(t', r')$ that satisfies the signature is equally likely. So if Bob only knows $(p, q, g, g', s)$, he cannot deduce any information from the signature.

Note that if Alice knows $\alpha$, she can cheat by calculating $r'$ for any message $t'$ and revealing $(t', r')$ instead of $(t, r)$.

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  • $\begingroup$ Clear, thanks. Is it always true that two generators of the same order generate the same subgroup? $\endgroup$ – Bar Dec 31 '17 at 16:23
  • $\begingroup$ @Gray Not for groups in general (e.g. $\langle 3 \rangle$ and $\langle 5 \rangle$ in $\mathbb{Z}_8^*$), but in the case that $p = 2q+1$, the only subgroup of order $q$ is the subgroup containing all quadratic residues $\bmod p$. $\endgroup$ – knbk Dec 31 '17 at 16:37
  • $\begingroup$ To complement this (nice) answer, note that this commitment scheme corresponds to the Pedersen commitment scheme, and stating that Bob cannot cheat is equivalent to the following statement: the parameters of the scheme are publicly verifiable, and the commitment scheme is perfectly hiding. Note that this implies that Bob could not cheat even if he was computationally unbounded. $\endgroup$ – Geoffroy Couteau Dec 31 '17 at 16:53
  • $\begingroup$ @Gray: "Is it always true that two generators of the same order generate the same subgroup?"; it is always true for $\mathbb{Z}_p^*$ for $p$ prime (but, as Gray mentioned, it is not necessarily true if we consider composite $p$) $\endgroup$ – poncho Dec 31 '17 at 17:14

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