# Is it possible I can open pedersen commitments without revealing r?

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?

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

• 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? – Jeff Lee Apr 11 at 17:58
• @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 $k$? – poncho Apr 11 at 20:04
• @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... – poncho Apr 11 at 20:10
• I am just thinking is it possible to use any other generator to replace K, seem it work too. – Jeff Lee Apr 12 at 4:19
• @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 $L$ – poncho Apr 12 at 12:18