|
|
\documentclass{article}\usepackage[utf8]{inputenc}\usepackage{amsfonts}\usepackage{amsthm}\usepackage{amsmath}\usepackage{enumerate}\usepackage{hyperref}% \hypersetup{
% colorlinks,
% citecolor=blue,
% filecolor=blue,
% 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}
% custom lemma environment to set custom numbers
\newtheorem{innerlemma}{Lemma}\newenvironment{lemma}[1]{\renewcommand\theinnerlemma{#1}\innerlemma}{\endinnerlemma}
\title{Notes on Caulk and Caulk+}\author{arnaucube}\date{February 2023}
\begin{document}
\maketitle
\begin{abstract} Notes taken while reading about Caulk \cite{cryptoeprint:2022/621} and Caulk+ \cite{cryptoeprint:2022/957}.
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{Preliminaries}
\subsection{Lagrange Polynomials and Roots of Unity}
Let $\omega$ denote a root of unity, such that $\omega^N=1$. Set $\mathbb{H}=\{1, \omega, \omega^2, \ldots, \omega^{N^{-1}}\}$.
Let the $i^{th}$ Lagrange polynomial be $\lambda_i(X)=\prod_{s\neq i-1} \frac{X-\omega^s}{\omega^{i-1} -\omega}$.
Notice that $\lambda_i(\omega^{i-1})=1$ and $\lambda_i(w^j)=0,~\forall j\neq i-1$.
Let the vanishing polynomial of $\mathbb{H}$ be $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$.
\subsection{KZG Commitments}KZG as a Vector Commitment.
We have vector $\overrightarrow{c}=\{c_i\}_1^n$, which can be interpolated into $C(X)$ through Lagrange polynomials $\{ \lambda_i(X) \}$:$$C(X) = \sum^n_{i=1} c_i \cdot \lambda_i(X)$$so, $C(\omega^{i-1})=c_i$.
Commitment:$$C = [C(X)]_1 = \sum^n_{i=1} c_i \cdot [\lambda_i(X)]_1$$
Proof of \textbf{opening for single value} $v$ at position $i$:
$$Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$
$$\pi_{KZG} = Q =[Q(X)]_1$$
Verification:$$e(C - [v]_1, [1]_2) = e(\pi_{KZG},~[X-\omega^{i-1}]_2)$$unfold$$e([C(X)]_1 - [v]_1, [1]_2) = e([Q(X)]_1,~[X-\omega^{i-1}]_2)$$$$C(X) - v = Q(X) \cdot (X-\omega^{i-1}) \Longrightarrow Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$
Proof of \textbf{opening for a subset} of positions $I \subset [N]$:
$[H_I]_1$ such that for$$C_I(X) = \sum_{i \in I} c_i \cdot \tau_i(X)$$$$z_I(X) = \prod_{i \in I} (X - \omega^{i-1})$$
for $\{ \tau_i(X) \}_{i \in I}$ being the Lagrange interpolation polynomials over $\mathbb{H}_I = \{\omega^{i-1}\}_{i \in I}$.\emph{(recall, $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$))}
$H_I(X)$ can be computed by$$H_I(X) = \frac{C(X) - C_I(X)}{z_I(X)}$$
So, prover commits to $C_I(X)$ with $C_I = [C_I(X)]_1$, and computes $\pi_{KZG}$:$$\pi_{KZG} = H_I = [H_I(X)]_1$$
Then, verification checks:$$e(C - C_I, [1]_2) = e(\pi_{KZG}, [z_I(X)]_2)$$unfold$$e([C(X)]_1 - [C_I(X)]_1, [1]_2) = e([H_I(X)]_1, [z_I(X)]_2)$$$$C(X) - C_I(X) = H_I(X) \cdot z_I(X)$$$$C(X) - C_I(X) = \frac{C(X)-C_I(X)}{z_I(X)} \cdot z_I(X)$$
\subsection{Pedersen Commitments}\label{sec:pedersen}Commitment$$cm = v [1]_1 + r [h]_1 = [v + hr]_1$$
Prove knowledge of $v,~r$, Verifier sends challenge $\{s_1, s_2\}$. Prover computes:$$R = s_1 [1]_1 + s_2 [h]_1 = [s_1 + h s_2]_1$$$$c= H(cm, R)$$$$t_1 = s_1 + v c, ~~ t_2=s_2 + r c$$
Verification:$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1$$
unfold:$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1 = [t_1 + h t_2]$$$$[s_1 + h s_2]_1 + c \cdot [v + hr]_1 == [s_1 + vc + h(s_2 + rc)]_1$$$$[s_1 + h s_2 + cv + rch]_1 == [s_1 + vc + h s_2 + rch]_1$$
\section{Caulk}\subsection{Blinded Evaluation}Main idea: combine KZG commitments with Pedersen commitments to prove knowledge of a value $v$ which Pedersen commitment is committed in the KZG commitment.
Let $C(X) = \sum^N_{i=1} c_i \lambda_i(X)$, where $\overrightarrow{c} = \{c_i\}_{i \in I}$. In normal KZG, prover would compute $Q(X)=\frac{C(X)-v}{X-\omega^{i-1}}$, and send $[Q(X)]_1$ as proof. We will obfuscate the commitment:
rand $a \in \mathbb{F}$, blind commit to $z(X)=aX - b = a(X - \omega^{i-1})$, where $\omega^{i-1}=b/a$. Denote by $[z]_2$ the commitment to $[z(X)]_2$.
Prover computes:\begin{enumerate}[i.] \item $\pi_{ped}$, Pedersen proof that $cm$ is from $v, r$ (section \ref{sec:pedersen}) \item $\pi_{unity}$ (see \ref{sec:pi-unity}) \item For random $s$ computes: $$T(X)=\frac{Q(X)}{a} + hs \longrightarrow [T]_1=[T(X)]_1$$ $$S(X) = -r - s \cdot z(X) \longrightarrow [S]_2 = [S(X)]_2$$\end{enumerate}
i, ii, iii defines the \emph{zk proof of membership}, which proves that $(v, r)$ is a opening of $cm$, and $v$ opens $C$ at $\omega^{i-1}$.
Verifier checks proofs $\pi_{ped},~\pi_{unity}$ (i, ii), and checks$$e(C - cm, [1]_2) == e([T]_1, [z]_2) + e([h]_1, [S]_2)$$
unfold:\begin{align*} C(X) - cm &== T(X) \cdot z(X) + h \cdot S(X) \\ C(X) - v - hr &== (\frac{Q(X)}{a} + s h) \cdot z(X) + h (-r -s \cdot z(X)) \\ C(X) - v &== hr + (\frac{Q(X)}{a}) z(X) + sh \cdot z(X) - hr - sh \cdot z(X) \\ C(X) - v &== \frac{Q(X)}{a} \cdot z(X) \\ C(X) - v &== \frac{Q(X)}{a} \cdot a(X-\omega^{i-1}) \\ C(X) -v &== Q(X) \cdot (X - \omega^{i-1})\end{align*}Which matches with the definition of $Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$.
\subsubsection{\texorpdfstring{Correct computation of $z(x)$, $\pi_{unity}$}%
{Correct computation of proof unity}}\label{sec:pi-unity}
Want to prove that prover knows $a, b$ such that $[z]_2 = [a X - b]_2$, and $a^N = b^N$.
To prove $\frac{a}{b}$ is inside the evaluation domain (ie. $\frac{a}{b}$ is a N$^{th}$ root of unity), proves (in $log(N)$ time) that its N$^{th}$ is one ($\frac{a}{b} = 1$).
Conditions:\begin{enumerate}[i.] \item $f_0=\frac{a}{b}$ \item $f_i = f_{i-1}^2,~\forall~i=1, \ldots, log(N)$ \item $f_{log(N)} = 1$\end{enumerate}
Redefine i, and from there, redefine ii, iii:% \begin{minipage}[t]{0.45\textwidth}
% \begin{enumerate}[i.]
% \item \begin{align*}
% f_0 &= z(1) = a - b\\
% f_1 &= z(\sigma) a \sigma -b\\
% f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\
% f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\
% f_4 &= \frac{f_2}{f_3} = \frac{a}{b}
% \end{align*}
% \end{enumerate}
% \end{minipage}
% \begin{enumerate}[i.]
% \item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$
% \item $f_{4+log(N)} = 1$
% \end{enumerate}
% \begin{minipage}[t]{0.45\textwidth}
% \end{minipage}
\begin{enumerate}[i.] \item \begin{align*} f_0 &= z(1) = a - b\\ f_1 &= z(\sigma) a \sigma -b\\ f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\ f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\ f_4 &= \frac{f_2}{f_3} = \frac{a}{b} \end{align*} \item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$ \item $f_{4+log(N)} = 1$\end{enumerate}
\begin{lemma}{1} Let $z(X)$ $deg=1$, $n=log(N)+6$, $\sigma$ such that $\sigma^n =1$.
If $\exists$ $f(X) \in \mathbb{F}[X]$ such that \begin{enumerate}[1.] \item $f(X) = z(X)$, for $1, \sigma$ \item $f(\sigma^2)(1-\sigma)=f(1)-f(\sigma)$ \item $f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)$ \item $f(\sigma^4)f(\sigma^3)=f(\sigma^2)$ \item $f(\sigma^{4+i+1})=f(\sigma^{4+i})^2,~~\forall i= 0, \ldots, log(N)-1$ \item $f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1$ \end{enumerate} then, $z(X)=aX-b$, where $\frac{b}{a}$ is a N$^{th}$ root of unity.\end{lemma}
Let's see the relations between the conditions and the Lemma 1:\begin{scriptsize}\begin{align*} Conditions &\longrightarrow Lemma~1\\ \begin{array}{l} f_0 = z(1) = a - b\\ f_1 = z(\sigma) a \sigma -b \end{array} &\longrightarrow 1.~f(X) = z(X), for 1, \sigma\\ f_2 = \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a &\longrightarrow 2.~f(\sigma^2)(1-\sigma)=f(1)-f(\sigma)\\ f_3 = \sigma f_2 - f_1 = \sigma a - a \sigma + b = b &\longrightarrow 3.~f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)\\ f_4 = \frac{f_2}{f_3} = \frac{a}{b} &\longrightarrow 4.~f(\sigma^4)f(\sigma^3)=f(\sigma^2)\\ f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1 &\longrightarrow 5.~f(\sigma^{4+i+1})=f(\sigma^{4+i})^2,~\forall i= 0, \ldots, log(N)-1\\ f_{4+log(N)} = 1 &\longrightarrow 6.~f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1\end{align*}\end{scriptsize}
For succintness: aggregate $\{f_i\}$ in a polynomial $f(X)$, whose coefficients in Lagrange basis associated to $\mathbb{V}_n$ are the $f_i$ (ie. s.t. $f(\omega^i)=f_i$).
\begin{small}\begin{align*} f(X) &= (a-b) \rho_1(X) + (a \sigma - b) \rho_2(X) + a \rho_3(X) + b \rho_4(X) + \sum_{i=0}^{log(N)} (\frac{a}{b})^{2^i} \rho_{5+i}(X)\\ &= f_0 \rho_1(X) + f_1 \rho_2(X) + f_2 \rho_3(X) + f_3 \rho_4(X) + \sum_{i=0}^{log(N)} (f_4)^{2^i} \rho_{5+i}(X)\end{align*}\end{small}
Prover shows that $f(X)$ by comparing $f(\sigma^i)$ with the corresponding constraints from Lemma 1:
For rand $\alpha$ (set by Verifier), set $\alpha_1 = \sigma^{-1} \alpha,~\alpha_2= \sigma^{-2} \alpha$, and send $v_1=f(\alpha_1),~v_2=f(\alpha_2)$ with corresponding proofs of opening.
Given $v_1, v_2$, shows that $p_{\alpha}(X)$, which proves the constraints from Lemma 1, evaluates to $0$ at $\alpha$ (ie. $p_{\alpha}(\alpha)=0$).
\begin{align*} p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha) + [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\ &+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\ &+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\ &+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\ &+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ &+ [f(\alpha_1)-1] \rho_n(\alpha)\end{align*}
\subsubsection{\texorpdfstring{NIZK argument of knowledge for $R_{unity}$ and $deg(z)\leq 1$}%
{NIZK argument of knowledge for Runity and deg(z)<=1}}
Prover:\begin{small}\begin{align*} &r_0, r_1, r_2, r_3 \leftarrow^\$ \mathbb{F},~~~ r(X)=r_1+r_2 X + r_3 X^2\\ f(X) &= (a-b) \rho_1(X) + (a \sigma - b) \rho_2(X) + a \rho_3(X) + b \rho_4(X) + \sum_{i=0}^{log(N)} (\frac{a}{b})^{2^i} \rho_{5+i}(X)\\ &+ r_0 \rho_{5+log(N)}(X) + r(X) z_{V_n}(X)\\ \\p(X) &= [f(X) - (aX-b)](\rho_1(X) + \rho_2(X))\\ &+[(1-\sigma)f(X) - f(\sigma^{-1}X) + f(\sigma^{-1}X)] \rho_3(X)\\ &+ [f(X) + f(\sigma^{-2}X) - \sigma f(\sigma^{-1} X)] \rho_4(X)\\ &+ [f(X)f(\sigma^{-1}X)-f(\sigma^{-2}X)] \rho_5(X)\\ &+ [f(X)-f(\sigma^{-1}X)f(\sigma^{-1}X)] \prod_{i \notin [5, 4+log(N)]} (X-\sigma^i)\\ &+ [f(\sigma^{-1}X)-1] \rho_n(X)\end{align*}\end{small}
Set$$h'(X) = \frac{p(X)}{z_{V_n}(X)},~~h(X)=h'(X) + X^{d-1} z(X)$$output $([F]_1=[f(X)]_1, ~ [H]_1=[h(x)]_1)$.
\begin{footnotesize} Note that \begin{align*} h(x)&=h'(X)+X^{d-1}z(X)\\ &=\frac{p(X)}{z_{V_n}(X)} + X^{d-1} z(X) \longrightarrow p(X)+X^{d-1} z(X) = z_{V_n}(X) h(X) \end{align*}\end{footnotesize}
Verifier sets challenge $\alpha \in^{\$} \mathbb{F}$ (hash of transcript by Fiat-Shamir).
\begin{align*} p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha)\\ &+ [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\ &+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\ &+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\ &+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\ &+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ &+ [f(\alpha_1)-1] \rho_n(\alpha)\end{align*}
\begin{footnotesize} Note: for the check that $[z]_1$ has degree 1, we add $-h(X) z_{V_n}(\alpha)$, to include the term $X^{d-1} z(X)$ in $h(X)$. Later the Verifier will compute $[P]_1$ without the terms including $z(X)$ (ie. without $-X^{d-1} z(X)z_{V_n}(\alpha)-z(X)[\rho_1(\alpha)+\rho_2(\alpha)]$), which the Verifier will add via the pairing: \begin{align*} -&X^{d-1} z(X)z_{V_n}(\alpha)-z(X)(\rho_1(\alpha)+\rho_2(\alpha))\\ =~ &(-X^{d-1} z_{V_n}(\alpha) - (\rho_1(\alpha)+\rho_2(\alpha))) \cdot z(X)\\ \longrightarrow~ &e(- (\rho_1(\alpha)+\rho_2(\alpha)) - z_{V_n}(\alpha) [X^{d-1}]_1, [z]_2) \end{align*}\end{footnotesize}
Prover then generates KZG proofs\begin{align*} ((v_1, v_2), \pi_1) &\leftarrow KZG.Open(f(X), (\alpha_1, \alpha_2))\\ (0, \pi_2) &\leftarrow KZG.Open(p_{\alpha}(X), \alpha)\end{align*}
prover's output: $(v_1, v_2, \pi_1, \pi_2)$.
Verify:set $\alpha_1=\sigma^{-1}\alpha, ~\alpha_2=\sigma^{-2}\alpha$,
\begin{footnotesize}(notice that $f(X)\rightarrow [F]_1$, and $f(\alpha_1)=v_1,~f(\alpha_2)=v_2$)\end{footnotesize}
\begin{align*} [P]_1 = &-z_{V_n}(\alpha)[H]_1 + [F]_1 (\rho_1(\alpha) + \rho_2(\alpha))\\ &+ [(1-\sigma) [F]_1 - v_2 + v_1] \rho_3(\alpha)\\ &+ [[F]_1 + v_2 - \sigma v_1] \rho_4(\alpha)\\ &+ [[F]_1 v_1 - v_2] \rho_5(\alpha)\\ &+ [[F]_1 - v_1^2] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ &+ [v_1-1] \rho_n(\alpha)\end{align*}
$$KZG.Verify((\alpha_1, \alpha_2), (v_1, v_2), \pi_1)$$$$e([P]_1, [1]_2) + e(-(\rho_1(\alpha) + \rho_2(\alpha)) - z_{V_n}(\alpha) [x^{d-1}]_1, [z]_2) = e(\pi_2, [x-\alpha]_2)$$
\section{Caulk+}Main update from original Caulk: $R_{unity}$, $\pi_{unity}$ is replaced with a pairing check constraining the evaluation points to be roots of a polynomial dividing $X^n-1$.
KZG commitment $c$ to $C(X)$, with evaluation points in $\mathbb{H}$.\\KZG commitment $a$ to $A(X)$, with evaluation points in $\mathbb{V}$.
Witness:\\$I \subset [n], ~~ \{ c_i \}_{i \in I} ,~~ C(X), A(X) ,~~ u: [m] \rightarrow I$
Precomputed:\\$[W_1^i(x)]_2 ~~\forall i \in I$, where $W_1^i(X) = \frac{C(X) - c_i)}{X-\omega^i}$\\$[W_2^i(x)]_2 ~~\forall i \in I$, where $W_2^i(X) = \frac{Z_{\mathbb{H}}(X)}{X-\omega^i}$
\paragraph{Round 1}\begin{enumerate}[i.] \item rand blinding factors $r1, \ldots, r_6$ \item Lagrange basis polynomials $\{ \tau_i(X) \}_{i \in [m]}$ over $\omega^j_{j \in I}$ \item $Z_I'(X)= r_1 \prod_{i \in I} (X - \omega^i)$ \item $C_I(X)=\sum_{i \in I} c_i \tau_i(X)$ (unblinded) \item blinded $C_I'(X)=C_I(X) + (r_2 + r_3 X + r_4 X^2) Z_I'(X)$ \item set $U(x)$, being degree $m-1$ interpolation over $\mathbb{V}$ with $U(v_i)=\omega^{u(i)},~ \forall i\in [m]$ \item blinded $U'(X)= U(X) + (r_5 + r_6 X) Z_{\mathbb{V}}(X)$ \item return $z_I=[Z_I'(x)]_1,~ c_I=[C_I'(x)]_1,~ u=[U'(X)]_1$\end{enumerate}
Verifier sets random challenges $\chi_1, \chi_2$.
\paragraph{Round 2}\begin{enumerate}[i.] \item $[W_1(x)+ \chi_2 W_2(x)]_2 = \sum_{i \in I} \frac{[W_1^i(x)]_2 + \chi_2 [W_2^i(x)]_2}{\prod_{j \in I,~i \neq j} \omega^i - \omega^j}$ \item $H(X) = \frac{Z_I'(U'(X)) + \chi_1 (C_I'(U'(X)) - A(X))}{Z_{\mathbb{V}}(X)}$ \item return $w=r_1^{-1} [W_1(x) + \chi_2 W_2(x)]_2 - [r_2 + r_3 x + r_4 x^2]_2,~ h=[H(x)]_1$\end{enumerate}
Verifier sets random challenge $\alpha$.
\paragraph{Round 3}Output $v_1, v_2, \pi_1, \pi_2, \pi_3$, where\begin{align*} P_1(X) &\leftarrow Z_I'(X) + \chi_1 C_I'(X)\\ P_2(X) &\leftarrow Z_I'(U'(\alpha)) + \chi_1 (C_I'(U'(\alpha)) - A(X)) - Z_{\mathbb{V}}(\alpha) H(X)\\ (v_1, \pi_1) &\leftarrow KZG.Open(U'(X), \alpha)\\ (v_2, \pi_2) &\leftarrow KZG.Open(P_1(X), v_1)\\ (0, \pi_3) &\leftarrow KZG.Open(P_2(X), \alpha)\\\end{align*}
\paragraph{Verify}Compute $p_1 = z_I + \chi_1 c_I, ~~ p_2= [v_2]_1 - \chi_1 a - Z_{\mathbb{V}}(\alpha) h$, verify\begin{align*} 1 &\leftarrow KZG.Verify(u, \alpha, v_1, \pi_1)\\ 1 &\leftarrow KZG.Verify(p_1, v_1, v_2, \pi_2)\\ 1 &\leftarrow KZG.Verify(p_2, \alpha, 0, \pi_3)\\ e((C-c_I) &+ \chi_2[x^n -1]_1, [1]_2) = e(z_I, w)\end{align*}
\bibliography{paper-notes.bib}\bibliographystyle{unsrt}
\end{document}
|
