|
|
\documentclass{article}\usepackage[utf8]{inputenc}\usepackage{amsfonts}\usepackage{amsthm}\usepackage{amsmath}\usepackage{enumerate}\usepackage{hyperref}\hypersetup{ colorlinks, citecolor=black, filecolor=black, linkcolor=black, urlcolor=blue}\usepackage{xcolor}
% prevent warnings of underfull \hbox:
\usepackage{etoolbox}\apptocmd{\sloppy}{\hbadness 4000\relax}{}{}
\theoremstyle{definition}\newtheorem{definition}{Def}[section]\newtheorem{theorem}[definition]{Thm}
\title{Notes on Sonic}\author{arnaucube}\date{April 2022}
\begin{document}
\maketitle
\begin{abstract} Notes taken while reading Sonic paper \cite{cryptoeprint:2019/099}. Usually while reading papers I take handwritten notes, this document contains some of them re-written to $LaTeX$.
The notes are not complete, don't include all the steps neither all the proofs.\end{abstract}
\tableofcontents
\section{Sonic}
\subsection{Structured Reference String}$\{ \{g^{x^i}\}_{i=-d}^d, \{ g^{\alpha x^i} \}_{i=-d, i \neq 0}^d, \{ h^{x^i}, h^{\alpha x^i} \}_{i=-d}^d, e(g, h^\alpha) \}$
\subsection{System of constraints}Multiplication constraint: $a \cdot b = c$
$Q$ linear constraints:$$
a \cdot u_q + b \cdot v_q + c \cdot w_q = k_q$$
with $u_q, v_q, w_q \in \mathbb{F}^n$, and $k_q \in \mathbb{F}_p$.
\vspace{0.5cm}Example: $x^2 + y^2 = z$
$$a = (x, y), \qquad b = (x, y), \qquad c = (x^2, y^2)$$\begin{enumerate}[i.] \item $(x, y) \cdot (1, 0) + (x, y) \cdot (-1, 0) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow x - x = 0$ \item $(x, y) \cdot (0, 1) + (x, y) \cdot (0, -1) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow y - y = 0$ \item $(x, y) \cdot (0, 0) + (x, y) \cdot (0, 0) + (x^2, y^2) \cdot (1, 1) = z \longrightarrow x^2 + y^2 = z$\end{enumerate}
So,$$u_1 = (1, 0) \quad v_1=(-1, 0) \quad w_1=(0, 0) \quad k_1=0$$$$u_2 = (0, 1) \quad v_2=(0, -1) \quad w_2=(0, 0) \quad k_2=0$$$$u_3 = (0, 0) \quad v_3=(0, 0) \quad w_3=(1, 1) \quad k_2=z$$
\vspace{1cm}
Compress n multiplication constraints into an equation in formal indeterminate $Y$:$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^i = 0$$encode into negative exponents of $Y$:$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^-i = 0$$
Also, compress the $Q$ linear constraints, scaling by $Y^n$ to preserve linear independence:$$
\sum_{q=1}^Q (a \cdot u_q + b \cdot v_q + c \cdot w_q - k_q) \cdot Y^{q+n} = 0$$
Polys:
\begin{align}\nonumber & u_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot u_{q, i}\\\nonumber & v_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot v_{q, i}\\\nonumber & w_i(Y) = -Y^i - Y^{-1} + \sum_{q=1}^Q Y^{q+n} \cdot w_{q, i}\\\nonumber & k(Y) = \sum_{q=1}^Q Y^{q+n} \cdot k_q\end{align}
Combine the multiplicative and linear constraints to:
\begin{align}\nonumber & a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y)+ \sum_{i=1}^n a_i b_i (Y^i + Y^{-i}) - k(Y) = 0\end{align}
where $a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y)$ is embedded into the constant term of the polynomial $t(X, Y)$.
Define $r(X, Y)$ s.t. $r(X, Y) = r(XY, 1)$.
$$\Longrightarrow r(X, Y) = \sum_{i=1}^n (a_i X^i Y^i + b_i X^{-i} Y^{-i} + c_i X^{-i-n} Y^{-i-n})$$
$$s(X, Y) = \sum_{i=1}^n (u_i(Y) X^{-i} + v_i(Y) X^i + w_i(Y) X^{i+n})$$
$$r'(X, Y) = r(X, Y) + s(X, Y)$$$$t(X, Y) = r(X, Y) + r'(X, Y) - k(Y)$$
The coefficient of $X^0$ in $t(X, Y)$ is the left-hand side of the equation.
Sonic demonstrates that the constant term of $t(X, Y)$ is zero, thus demonstrating that our constraint system is satisfied.
\subsubsection{The basic Sonic protocol}
\begin{enumerate}[1.] \item Prover constructs $r(X, Y)$ using their hidden witness \item Prover commits to $r(X, 1)$, setting the maximum degree to n \item Verifier sends random challenge $y$ \item Prover commits to $t(X, y)$. The commitment scheme ensures that $t(X, y)$ has no constant term. \item Verifier sends random challenge $z$ \item Prover opens commitments to $r(z, 1), r(z, y), t(z, y)$ \item Verifier calculates $r'(z, y)$, and checks that $$r(z, y) \cdot r'(z, y) - k(y) == t(z, y)$$\end{enumerate}
Steps $3$ and $5$ can be made non-interactive by the Fiat-Shamir transformation.
\subsubsection{Polynomial Commitment Scheme}Sonic uses an adaptation of KZG \cite{kzg-tmp}, want:
\begin{enumerate}[i.] \item \emph{evaluation binding}, i.e. given a commitment $F$, an adversary cannot open F to two different evaluations $v_1$ and $v_2$ \item \emph{bounded polynomial extractable}, i.e. any algebraic adversary that opens a commitment $F$ knows an opening $f(X)$ with powers $-d \leq i \leq max, i \neq 0$.\end{enumerate}
\vspace{0.5cm}PC scheme (adaptation of KZG):
\begin{enumerate}[i.] \item Commit(info, $f(X)$) $\longrightarrow F$: $$F = g^{\alpha \cdot x^{d-max}} \cdot f(x)$$ \item Open(info, $F$, $z$, $f(x)$) $\longrightarrow (f(z), W)$: $$w(X) = \frac{f(X) - f(z)}{X-z}$$ $$W = g^{w(x)}$$ \item Verify(info, $F$, $z$, $(v, W)$) $\longrightarrow 0/1$:\\ Check: $$e(W, h^{\alpha \cdot x}) \cdot
e(g^v W^{-z}, h^{\alpha}) == e(F, h^{x^{-d+max}})$$
\end{enumerate}
\subsection{Succinct signatures of correct computation}Signature of correct computation to ensure that an element $s=s(z, y)$ for a known polynomial$$s(X, Y) = \sum_{i, j = -d}^d s_{i, j} \cdot X^i \cdot Y^i$$
Use the structure of $s(X, Y)$ to prove its correct calculation using a \emph{permutation argument} $\longrightarrow$ \emph{grand-product argument} inspired by Bayer and Groth, and Bootle et al.
Restrict to constraint systems where $s(X, Y)$ can be expressed as the sum of $M$ polynomials. Where $j-th$ poly is of the form:$$
\Psi_j(X, Y) = \sum_{i=1}^n \psi_{j, \sigma_{j, i}} \cdot X^i \cdot Y^{\sigma_{j, i}}$$
where $\sigma_j$ is the fixed polynomial permutation, and $\phi_{j, i} \in \mathbb{F}$ are the coefficients.
\vspace{1cm}\framebox{WIP}\vspace{1cm}
\bibliography{paper-notes.bib}\bibliographystyle{unsrt}
\end{document}
|
