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I'm recently reading Berry Schoenmakers's Lecture Notes Cryptographic Protocols (page. 53) and confused about this following question:

Let $g$, $h$ denote generators of a group of large prime order $n$ such that $log_gh$ is unknown to anyone. Design Σ-protocols (and prove correctness) for the following relations:

{($A, B; x, y, z$) : $A = g^xh^y,B = g^{1/x}h^z, x \neq 0$};

My prove intuition is using Okamoto's protocol for $A = g^xh^y$ and $B = g^{1/x}h^z$, then how to prove the relation between $x$ in $A$ and $1/x$ in $B$? Or is there any other way to prove it?

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The two previous answers certainly solve the problem. Soundness of Geoffroy's protocol is fine indeed, but there is the appearance of the witness $x$ in the computation of the announcement $(A',B')$ as $B'=g^{r_x/x} h^r$. This can be avoided, however, and at the same time one can find the protocol maybe in a bit more natural way as follows.

Starting with $A=g^x h^y$ and $B=g^{1/x} h^z$, we see that $B$ is a Pedersen commitment to the (multiplicative) inverse $1/x$ of the committed value $x$ in the Pedersen commitment $A$. So, $x$ appears both in $A$ and $B$ and these two occurrences need to be connected somehow. EQ-composition is a very effective way to accomplish this, but we cannot directly apply it to exponents of the form $x$ and $1/x$. A simple way out is to move $x$ around a bit in the equation for $B$ by raising both sides to the power of $x$, such that we get: $$ A=g^x h^y,\qquad g=B^x h^{-zx}.$$ We can now apply EQ-composition to the factors $g^x$ and $B^x$, but one may wonder about the new factor $h^{-zx}$ which also depends on $x$. Fortunately, such a factor causes no problems because we can think of $h^{z'}=h^{-zx}$ as a factor that is independent of $x$; it's like replacing $z$ with $z'=-zx$, which is fine because this is a one-to-one transformation for nonzero $x$.

For the $\Sigma$-protocol we get:

  1. Prover sends announcement $(a,b)=(g^u h^v, B^u h^w)$ with $u,v,w\in_R\mathbb{Z}_n$.

  2. Verifier sends challenge $c\in_R\mathbb{Z}_n$.

  3. Prover sends response $(r,s,t)=(u+c\,x, v+c\,y, w-c\,z\,x) \bmod n$. Verifier accepts if $g^r h^s = a A^c$ and $B^r h^t = b g^c$.

It is instructive to see why special soundness holds. So, let's consider two accepting conversations $(a,b;c;r,s,t)$ and $(a,b;c';r',s',t')$ with $c\neq c'$. Then we find: $$\begin{array}{cl} & g^r h^s = a A^c,\ g^{r'} h^{s'} = a A^{c'},\quad B^r h^t = b g^c,\ B^{r'} h^{t'} = b g^{c'}\\ \Rightarrow& g^{r-r'} h^{s-s'} = A^{c-c'} ,\quad B^{r-r'} h^{t-t'} = g^{c-c'} \\ \Leftrightarrow& A = g^{\frac{r-r'}{c-c'}} h^{\frac{s-s'}{c-c'}},\quad B = g^{\frac{c-c'}{r-r'}} h^{-\frac{t-t'}{r-r'}}. \end{array}$$ Here, we are using that $r\neq r'$ holds as well: otherwise we see that $B^{r-r'} h^{t-t'} = g^{c-c'}$ is equivalent to $h^{t-t'} = g^{c-c'}$, and we would have $\log_g h = (t-t')/(c-c')$, contradicting the assumption that $\log_g h$ is unknown. Hence, a witness is obtained as $x=(r-r')/(c-c')$, $y=(s-s')/(c-c')$, and $z=-(t-t')/(r-r')$. Clearly, $x\neq0$ and $B=g^{1/x} h^z$ holds, as well as $A=g^x h^y$.

The same line of reasoning can be applied to show special soundness for Geoffroy's protocol.

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Yes, this is an exercise in a great textbook, so maybe giving just intuition could be appropriate. First, one would open both commitments (for $x$ and $1/x$) with standard responses that are linear in challenge. Next, one would multiply that responses to show that power-two (square) coefficient (challenge of verifier is the variable here) is exactly one. This kind of technique was suggested to prove a non-zero secret by existence of it's inverse at (shameless) the paper at MFCS 2012 on graph colorability.

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The natural way to prove this relation is to prove knowledge of committed values $(m,m')$ whose product is $1$. Intuitively, you want to do this by showing that the discrete logarithm of $g$ in base $B$ is also the discrete logarithm of $A$ in base $g$ (this is not correct as such but it gives an intuition).

I would proceed as follows: (I've not looked at the soundness proof so it could contain a mistake, but should suffice to clarify the intuition on how to proceed)

  • The prover sends $A' = g^{r_x}h^{r_y}$ and $B' = g^{r_x/x}h^r$ for random exponents $(r_x,r_y,r)$
  • The verifier sends a challenge $e$
  • The prover answers with $d_x = x\cdot e + r_x \bmod n, d_y = y\cdot e + r_y \bmod n$, and $d = r - zd_x \bmod n$
  • The verifier checks that $A^eA' = g^{d_x}h^{d_y}$ and $g^eB' = B^{d_x}h^{d}$

The two verification equations convince him that the prover can open $A$ to some value $x$ so that $B^x$ commits to $1$.

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  • $\begingroup$ One more question: In you construction, $x$ exists in $B'$, which is the first time i see a witness existing in prover's first round message. Besides, I can verify the construction, but I just cannot get the underlying intuition. intuitively, the same $r_x$ in $A'$ and $B'$ make some sense, besides that, I cannot figure out how it comes. maybe something important i missed. or this construction is the result of inspiration that has no fixed model to follow? $\endgroup$ – X.S. May 12 '17 at 11:21

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