\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsfonts} \usepackage{amsthm} % theorems structure \usepackage{enumerate} \usepackage{hyperref} % links \usepackage{tikz} % diagrams \usepackage{pgf-umlsd} % diagrams \usepackage{amsmath} % arrows \theoremstyle{definition} \newtheorem{example}{Example}[section] \title{Sigma protocol and OR proofs - notes} \author{arnaucube} \date{March 2022} \begin{document} \maketitle \begin{abstract} This document contains the notes taken during the \emph{Cryptography Seminars} given by \href{https://github.com/rbkhmrcr}{Rebekah Mercer}. \end{abstract} \tableofcontents \section{Sigma protocol} \subsection{The protocol} Let $q$ be a prime, $q$ a prime divisor in $p-1$, and $g$ and element of order $q$ in $\mathbb{Z}_p^a$. Then we have $G = \langle g \rangle$. \\ We assume that computationally for a given $A$ it's hard to find $a \in \mathbb{F}$ such that $A = g^a$. \\ Alice wants to prove that knows the \emph{witness} $w \in \mathbb{F}$, such that the \emph{statement} $X = g^w$, without revealing $w$. \begin{enumerate}[1.] \item Alice generates a random $a \xleftarrow{r} \mathbb{F}$, and computes $A=g^a$. And sends $A$ to Bob. \item Bob generates a challenge $c \xleftarrow{r} \mathbb{F}$, and sends it to Alice. \item Alice computes $z=a + c \cdot w$, and sends it to Bob. \item Bob verifies it by checking that $g^z == X^c \cdot A$. \end{enumerate} We can unfold Bob's verification and see that: $$g^z == X^c \cdot A$$ $$g^{a+cw} == g^{wc} g^a$$ $$g^{a+cw} == g^{wc+a}$$ \begin{center} \begin{sequencediagram} \newinst[1]{a}{Alice} \newinst[3]{b}{Bob} \mess[0]{a}{$A$}{b} \mess[2]{b}{$c$}{a} \mess[2]{a}{$z$}{b} \mess[0]{b}{$ok$}{a} \end{sequencediagram} \end{center} Properties: \begin{enumerate}[i.] \item \emph{correctness/completness}: if Alice know the witness for the statement, then they can create a valid proof. \item \emph{soundness}: if someone does not have knowledge of the witness, can not form a valid proof (verifier will always reject). \item \emph{zero knowledge}: nobody gains knowledge of anything new with the proof. prior knowledge + proof = prior knowledge \end{enumerate} \subsection{Non interactive protocol} With the \emph{Fiat-Shamir Heuristic}, we model a hash function as a random oracle, thus we can replace Bob's role by a hash function in order to obtain the challenge $c \in \mathbb{F}$. \\ So, we replace the step 2 from the described protocol by $c = H(X || A)$ (where $H$ is a known hash function). \subsection{What could go wrong (Simulator)} If the verifier (Bob) sends $c \in \mathbb{F}$, prior to the prover committed to $A$, the prover could create a proof about a public key which they don't know $w$. \begin{enumerate}[1.] \item Bob sends $c \xleftarrow{r} \mathbb{F}$ to Alice \item Alice generates $z \xleftarrow{r} \mathbb{F}$ \item Alice then computes $A = g^z X^{-c}$, and sends $z, A$ to Bob \item Bob would check that $g^z == X^c A$ and it would pass the verification, as $g^z== X^c \cdot A \Rightarrow g^z==X^c \cdot g^z X^{-c} \Rightarrow g^z == g^z$. \end{enumerate} As we've seen, it's really important the order of the steps, so Alice must commit to $A$ before knowing $c$.\\ This 'fake' proof generation is often called the \emph{simulator} and used for further constructions. \section{OR proof} \emph{OR proofs} allows the prover to prove that they know the witness $w$ of one of the two known \emph{public keys} $X_0, X_1 \in \mathbb{F}$, without revealing which one. It uses the construction seen in the \emph{sigma protocols} together with the idea of the \emph{simulator}. A similar construction is used for $n$ statements in the \emph{ring signatures} scheme (used for example in \emph{Monero}). In our case, we will work with $n=2$. \subsection{The protocol} \subsubsection{Simulator} We can assume that the simulator is a box that for given the inputs $(g, X)$, it will output $(A_s, c_s, z_s)$, such that verification succeeds ($g^{z_s}==X^{c_s} \cdot A_s$). \begin{center} \begin{tikzpicture} \node [draw, minimum width=2cm, minimum height=1.2cm, right=1cm ] (simulator) {simulator}; \draw[-stealth] ++(-1,0) -- (simulator.west) node[midway,above]{$g, X$}; \draw[-stealth] (simulator.east) -- ++(+2,0) node[midway,above]{$A_s, c_s, z_s$}; \end{tikzpicture} \end{center} Internally, the simulator computes $$z_s \xleftarrow{r} \mathbb{F},~c_s \xleftarrow{r} \mathbb{F},~A_s = g^{z_s} \cdot X^{c_s}$$ \subsection{Flow} For two known \emph{public keys} $X_0, X_1 \in G$, Alice knows $w_b \in \mathbb{F}$, for $b \in \{0, 1\}$, such that $g^{w_b} = X_0$ or $g^{w_b} = X_1$. As we don't know if Alice controls $0$ or $1$, from now on, we will use $b$ and $1-b$. \\ So, Alice knows $w_b \in \mathbb{F}$ such that $X_b = g^{w_b}$, and does not know $w_{1-b}$ for $X_{1-b}=g^{w_{1-b}}$. \begin{enumerate} \item First of all, as in the \emph{Sigma protocol}, Alice generates a random \emph{commitment} $a_b \xleftarrow{r} \mathbb{F}$, and computes $A_b = g^{a_b}$. \item Then, Alice will run the \emph{simulator} for $1-b$. \begin{list}{} \item Sets a random $c_{1-b} \xleftarrow{r} \mathbb{F}$, and runs the simulator with inputs\\$(c_{1-b}, X_{1-b})$, and outputs $(A_{1-b}, c_{1-b}, z_{1-b})$. \item Remember that internally the \emph{simulator} will set random\\ $z_{1-b}, c_{1-b} \xleftarrow{r} \mathbb{F}$, and compute an $A_{1-b}$ such that\\ $A_{1-b} = g^{z_{1-b}} \cdot X_{1-b}^{c_{1-b}}$. \end{list} \item Now, Alice sends $A_b, A_{1-b}$ to Bob \item And Bob sends back the \emph{challenge} $s \xleftarrow{r} \mathbb{F}$. \item Alice then splits the challenge $s$ into $c_b, c_{1-b}$, by $s = c_{1-b} \oplus c_b$. So Alice can compute $c_b = s \oplus c_{1-b}$. \item Then Alice computes $z_b = a_b \cdot w_b + c_b$. And sends to Bob $(c_b, c_{1-b}, z_b, z_{1-b})$. \item Bob can perform the verification by checking that: \begin{enumerate}[i.] \item $s == c_b \oplus c_{1-b}$ \item $g_{z_{1-b}} == A_{1-b} \cdot X_{1-b}^{-c_{1-b}}$ \item $g_{z_b} == A_b \cdot X_b^{-c_b}$ \end{enumerate} \end{enumerate} \begin{center} \begin{sequencediagram} \newinst[1]{a}{Alice} \newinst[3]{b}{Bob} \mess[1]{a}{$A_b, A_{1-b}$}{b} \mess[1]{b}{$s$}{a} \mess[1]{a}{$c_b, c_{1-b}, z_b, z_{1-b}$}{b} \end{sequencediagram} \end{center} \section{Resources} \begin{enumerate} \item \href{https://cs.au.dk/~ivan/Sigma.pdf}{https://cs.au.dk/~ivan/Sigma.pdf} \item \emph{Cryptography Made Simple}, Nigel Smart. Section 21.3. \end{enumerate} \end{document}