With setup $p$ and $q$ where $p = 2q + 1$, and $g$ and $h$ is the generator with order $q$.

In Pedersen commitment, I commit the value m with $c=g^m h^r \bmod p$, then de-commit by revealing $(m, r)$. $c' = c$, then the commitment hold.

I am wondering if I don't need to reveal $r$ to verifier, instead I can reveal $m$ only and construct a proof of knowledge on $r$ to have the sample result. As $s = h^r$ if $s = {{c}\over{g^m}}$, I can use Schnorr Protocol to prove that I know $r$ in $s = h^r$ for $s = {{c}\over{g^m}}$.

If verification hold, mean that provers know $r$ with $m$ in a relationship $c=g^m h^r$. Am I right?

I would like to add follow-up question if the answer is yes.

To prove the knowledge of $r$ in $s = h^r$ for $s = {{c}\over{g^m}}$. I can use Schnorr Protocol to construct the proof. But if it is possible to make it only available for designated verifier to verify the proof?

I read a paper is about Designated Verifier Signature, but it is about creating proof on a signature $s=m^x$ where $m$ is the message and $x$ is the private key of signer.

So is it possible to make Schnorr Protocol only can be verified by a selected verifier with know public key?

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    $\begingroup$ Yes, you are correct. This feels too short to be an answer... $\endgroup$ – poncho Apr 11 at 12:48
  • $\begingroup$ @poncho If yes, what is the benefit of revealing r comparing with constructing a knowledge proof of r? $\endgroup$ – Jeff Lee Apr 11 at 14:08
  • $\begingroup$ @poncho Thanks for your answer!! I just edited by question. $\endgroup$ – Jeff Lee Apr 11 at 14:15

So is it possible to make Schnorr Protocol only can be verified by a selected verifier with know public key?

Here's the obvious way using a two dimensional Schnorr proof; this is a proof that, given $A^xB^y = C$, you know $x, y$. It's a straight-forward extension of the regular Schnorr proof:

  • The prover selects random $r, s$, and computes $T = A^rB^s$. He also computes $t = \text{Hash}(T)$ and publishes $T$, $u = x + rt$ and $v = y + st$.

  • The verifier checks whether $A^uB^v = C T^t$

We'll denote $K$ as the public key of the verifier, that is, she knows the value $k$ such that $G^k = K$.

Then, to do a Selected Verifier Proof that the commitment $C = G^m H^r$ is to the value $m$, the prover generates a two dimensional Schnorr proof that he knows the values $x, y$ such that $H^x K^y = C G^{-m}$. The valid prover can generate such a proof, because he knows such a pair $(x = r, y = 0)$. On the other hand, the verifier can not convince anyone else that this proves any specific value $m$, because for any $m$, she can construct a $y$ that allows her to generate such a proof.

Here's another idea that occurs to me; it appears to be a way to have a designated verifier Schnorr proof:

  • The prover wants to prove knowledge of a value $x$ s.t. $A^x = B$, for public $A, B$. We'll denote $K$ as the public key of the verifier.

  • The prover selects two random values $r_1, r_2$, and computes $T_1 = A^{r_1}, T_2 = K^{r_2}$ and $U = G^{r_2}$ and computes $t = T_1 + T_2 \bmod q$ (where $q$ is the size of the subgroup). Then, he publishes $T_1, U$ and $u = x + r_1t$

  • The designated verifier uses her private key $k$ to compute $T_2 = U^k$, and $t = T_1 + T_2 \bmod q$. Then, it proceeds like a standard Schnorr proof, checking whether $A^u = BT_1^t$

No one can verify this proof without the knowledge of $k$ (as they cannot compute $t$). The designated verifier knows no one else knows $k$, and hence the prover cannot select $t$ arbitrarily. And, if the verifier tried to forward this proof (possibly by forwarding the value $T$), this doesn't work (even if she exposed her private key $k$), because it is straight-forward to generate a validating $T_1, U, u$ set with the knowledge of $k$ (for arbitrary $A, B$)

Somebody should vet this 'designated Schnorr' proof before you use it; it looks like it meets the requirements. Here's the reasoning for the 'proof of knowledge' portion: a putative prover can set an arbitrary $T_1 = A^c B^d$ (for arbitrary $c, d$). However, in that case, the verification equation is $A^{ckt-u}B^{dkt+1} = 1$; this can be satisfied only if $dkt+1 \equiv 0$ (but to set the value $d$ appropriately, the prover would need to know $k$); otherwise, the prover would know that $x = (ckt-u)(dkt+1)^{-1}$, and so knowledge of $k$ (and $c, d$) would imply knowledge of the discrete log.

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  • $\begingroup$ Thanks for the reply. I would like to ask, in $𝐻_x 𝐾_y = 𝐶 𝐺^{-m} $, why need to use verifier's public key? In this case everyone can verify the proof because $K$ is the public key and everyone know it. Am I right? $\endgroup$ – Jeff Lee Apr 11 at 17:58
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    $\begingroup$ @JeffLee: the idea is that the verifier is convinced (because she knows the prover does not know the dlog of $K$, and hence the pair $(x=r, y=0)$ is the only possible option. However, if she forwards the proof to someone else, then (if she knew $r$), she could have constructed a proof with any $m$. Or, were you looking for a construction where the proof could not be validated without knowledge of $k$? $\endgroup$ – poncho Apr 11 at 20:04
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    $\begingroup$ @JeffLee: if the latter, just take the above proof, and encrypt it using $K$ as a public key. Only the verifier can read it, and if she tries to forward the decrypted version, well, the previous logic applies... $\endgroup$ – poncho Apr 11 at 20:10
  • $\begingroup$ I am just thinking is it possible to use any other generator to replace K, seem it work too. $\endgroup$ – Jeff Lee Apr 12 at 4:19
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    $\begingroup$ @JeffLee: $Hash( H^rK^s ) \ne Hash(H^rL^s)$ for a point $L \ne K$ and so, no, the verifier could not just change the $K$ used to another value $L$ $\endgroup$ – poncho Apr 12 at 12:18

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