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+21 -0README.md
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BINnotes_caulk.pdf
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+347 -0notes_caulk.tex
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+18 -0paper-notes.bib
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BINweil-pairing.pdf
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+65 -10weil-pairing.tex
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# math |
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Notes, code and documents done while reading books and papers. |
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### Sage code |
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- [blind signatures over elliptic curve](blind-sign-over-ec.sage) |
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- [BLS signatures](bls-sigs.sage) |
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- [FFT](fft.sage) |
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- [IPA](ipa.sage) |
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- [KZG commitments](kzg.sage) |
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- [Powers of Tau](powersoftau.sage) |
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- [Ring signatures](ring-signatures.sage) |
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- [Sigma protocol](sigma.sage) |
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|
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### Notes |
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- [Notes on "Abstract Algebra" book, by Charles C. Pinter](abstract-algebra-charles-pinter-notes.pdf) |
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- [Notes on Caulk paper](notes_caulk.pdf) |
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- [Notes on the DFT & FFT](fft-notes.pdf) |
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- [Notes on the BLS signatures](notes_bls-sig.pdf) |
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- [Notes on IPA from Halo paper](notes_halo.pdf) |
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- [Notes on Sonic paper](notes_sonic.pdf) |
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- [Notes on Weil pairing](weil-pairing.pdf) |
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- [Notes on Sigma protocol and OR proofs](sigma-or-notes.pdf) |
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- [Notes on Reed-Solomon codes](notes_reed-solomon.pdf) |
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BIN
notes_caulk.pdf
BIN
notes_caulk.pdf
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\documentclass{article} |
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\usepackage[utf8]{inputenc} |
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\usepackage{amsfonts} |
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\usepackage{amsthm} |
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\usepackage{amsmath} |
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\usepackage{enumerate} |
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\usepackage{hyperref} |
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% \hypersetup{ |
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% colorlinks, |
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% citecolor=blue, |
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% filecolor=blue, |
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% linkcolor=black, |
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% urlcolor=blue |
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% } |
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\usepackage{xcolor} |
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% prevent warnings of underfull \hbox: |
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\usepackage{etoolbox} |
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\apptocmd{\sloppy}{\hbadness 4000\relax}{}{} |
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\theoremstyle{definition} |
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\newtheorem{definition}{Def}[section] |
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\newtheorem{theorem}[definition]{Thm} |
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% custom lemma environment to set custom numbers |
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\newtheorem{innerlemma}{Lemma} |
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\newenvironment{lemma}[1] |
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{\renewcommand\theinnerlemma{#1}\innerlemma} |
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{\endinnerlemma} |
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\title{Notes on Caulk and Caulk+} |
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\author{arnaucube} |
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\date{February 2023} |
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\begin{document} |
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\maketitle |
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\begin{abstract} |
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Notes taken while reading about Caulk \cite{cryptoeprint:2022/621} and Caulk+ \cite{cryptoeprint:2022/957}. |
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Usually while reading papers I take handwritten notes, this document contains some of them re-written to $LaTeX$. |
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The notes are not complete, don't include all the steps neither all the proofs. |
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\end{abstract} |
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\tableofcontents |
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\section{Preliminaries} |
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\subsection{Lagrange Polynomials and Roots of Unity} |
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Let $\omega$ denote a root of unity, such that $\omega^N=1$. Set $\mathbb{H}=\{1, \omega, \omega^2, \ldots, \omega^{N^{-1}}\}$. |
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Let the $i^{th}$ Lagrange polynomial be $\lambda_i(X)=\prod_{s\neq i-1} \frac{X-\omega^s}{\omega^{i-1} -\omega}$. |
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Notice that $\lambda_i(\omega^{i-1})=1$ and $\lambda_i(w^j)=0,~\forall j\neq i-1$. |
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Let the vanishing polynomial of $\mathbb{H}$ be $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$. |
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\subsection{KZG Commitments} |
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KZG as a Vector Commitment. |
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We have vector $\overrightarrow{c}=\{c_i\}_1^n$, which can be interpolated into $C(X)$ through Lagrange polynomials $\{ \lambda_i(X) \}$: |
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$$C(X) = \sum^n_{i=1} c_i \cdot \lambda_i(X)$$ |
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so, $C(\omega^{i-1})=c_i$. |
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Commitment: |
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$$C = [C(X)]_1 = \sum^n_{i=1} c_i \cdot [\lambda_i(X)]_1$$ |
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Proof of \textbf{opening for single value} $v$ at position $i$: |
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$$Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$ |
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$$\pi_{KZG} = Q =[Q(X)]_1$$ |
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Verification: |
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$$e(C - [v]_1, [1]_2) = e(\pi_{KZG},~[X-\omega^{i-1}]_2)$$ |
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unfold |
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$$e([C(X)]_1 - [v]_1, [1]_2) = e([Q(X)]_1,~[X-\omega^{i-1}]_2)$$ |
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$$C(X) - v = Q(X) \cdot (X-\omega^{i-1}) \Longrightarrow Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$$ |
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Proof of \textbf{opening for a subset} of positions $I \subset [N]$: |
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$[H_I]_1$ such that for |
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$$C_I(X) = \sum_{i \in I} c_i \cdot \tau_i(X)$$ |
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$$z_I(X) = \prod_{i \in I} (X - \omega^{i-1})$$ |
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for $\{ \tau_i(X) \}_{i \in I}$ being the Lagrange interpolation polynomials over $\mathbb{H}_I = \{\omega^{i-1}\}_{i \in I}$. |
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\emph{(recall, $z_H(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1$))} |
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$H_I(X)$ can be computed by |
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$$H_I(X) = \frac{C(X) - C_I(X)}{z_I(X)}$$ |
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So, prover commits to $C_I(X)$ with $C_I = [C_I(X)]_1$, and computes $\pi_{KZG}$: |
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$$\pi_{KZG} = H_I = [H_I(X)]_1$$ |
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Then, verification checks: |
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$$e(C - C_I, [1]_2) = e(\pi_{KZG}, [z_I(X)]_2)$$ |
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unfold |
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$$e([C(X)]_1 - [C_I(X)]_1, [1]_2) = e([H_I(X)]_1, [z_I(X)]_2)$$ |
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$$C(X) - C_I(X) = H_I(X) \cdot z_I(X)$$ |
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$$C(X) - C_I(X) = \frac{C(X)-C_I(X)}{z_I(X)} \cdot z_I(X)$$ |
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\subsection{Pedersen Commitments}\label{sec:pedersen} |
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Commitment |
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$$cm = v [1]_1 + r [h]_1 = [v + hr]_1$$ |
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Prove knowledge of $v,~r$, Verifier sends challenge $\{s_1, s_2\}$. Prover computes: |
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$$R = s_1 [1]_1 + s_2 [h]_1 = [s_1 + h s_2]_1$$ |
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$$c= H(cm, R)$$ |
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$$t_1 = s_1 + v c, ~~ t_2=s_2 + r c$$ |
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Verification: |
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$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1$$ |
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unfold: |
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$$R + c \cdot cm == t_1 [1]_1 + t_2 [h]_1 = [t_1 + h t_2]$$ |
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$$[s_1 + h s_2]_1 + c \cdot [v + hr]_1 == [s_1 + vc + h(s_2 + rc)]_1$$ |
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$$[s_1 + h s_2 + cv + rch]_1 == [s_1 + vc + h s_2 + rch]_1$$ |
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\section{Caulk} |
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\subsection{Blinded Evaluation} |
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Main idea: combine KZG commitments with Pedersen commitments to prove knowledge of a value $v$ which Pedersen commitment is committed in the KZG commitment. |
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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: |
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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 $[z(X)]_2$. |
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Prover computes: |
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\begin{enumerate}[i.] |
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\item $\pi_{ped}$, Pedersen proof that $cm$ is from $v, r$ (section \ref{sec:pedersen}) |
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\item $\pi_{unity}$ (see \ref{sec:pi-unity}) |
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\item For random $s$ computes: |
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$$T(X)=\frac{Q(X)}{a} + hs \longrightarrow [T]_1=[T(X)]_1$$ |
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$$S(X) = -r - s \cdot z(X) \longrightarrow [S]_2 = [S(X)]_2$$ |
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\end{enumerate} |
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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 $\omega^{i-1}$. |
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Verifier checks proofs $\pi_{ped},~\pi_{unity}$ (i, ii), and checks |
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$$e(C - cm, [1]_2) == e([T]_1, [z]_2) + e([h]_1, [S]_2)$$ |
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unfold: |
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\begin{align*} |
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C(X) - cm &== T(X) \cdot z(X) + h \cdot S(X) \\ |
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C(X) - v - hr &== (\frac{Q(X)}{a} + s h) \cdot z(X) + h (-r -s \cdot z(X)) \\ |
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C(X) - v &== hr + (\frac{Q(X)}{a}) z(X) + sh \cdot z(X) - hr - sh \cdot z(X) \\ |
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C(X) - v &== \frac{Q(X)}{a} \cdot z(X) \\ |
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C(X) - v &== \frac{Q(X)}{a} \cdot a(X-\omega^{i-1}) \\ |
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C(X) -v &== Q(X) \cdot (X - \omega^{i-1}) |
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\end{align*} |
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Which matches with the definition of $Q(X) = \frac{C(X) - v}{X-\omega^{i-1}}$. |
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\subsubsection{\texorpdfstring{Correct computation of $z(x)$, $\pi_{unity}$}% |
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{Correct computation of proof unity}}\label{sec:pi-unity} |
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Want to prove that prover knows $a, b$ such that $[z]_2 = [a X - b]_2$, and $a^N = b^N$. |
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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 ($\frac{a}{b} = 1$). |
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Conditions: |
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\begin{enumerate}[i.] |
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\item $f_0=\frac{a}{b}$ |
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\item $f_i = f_{i-1}^2,~\forall~i=1, \ldots, log(N)$ |
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\item $f_{log(N)} = 1$ |
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\end{enumerate} |
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Redefine i, and from there, redefine ii, iii: |
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% \begin{minipage}[t]{0.45\textwidth} |
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% \begin{enumerate}[i.] |
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% \item \begin{align*} |
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% f_0 &= z(1) = a - b\\ |
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% f_1 &= z(\sigma) a \sigma -b\\ |
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% f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\ |
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% f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\ |
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% f_4 &= \frac{f_2}{f_3} = \frac{a}{b} |
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% \end{align*} |
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% \end{enumerate} |
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% \end{minipage} |
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% \begin{enumerate}[i.] |
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% \item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$ |
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% \item $f_{4+log(N)} = 1$ |
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% \end{enumerate} |
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% \begin{minipage}[t]{0.45\textwidth} |
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% \end{minipage} |
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\begin{enumerate}[i.] |
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\item \begin{align*} |
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f_0 &= z(1) = a - b\\ |
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f_1 &= z(\sigma) a \sigma -b\\ |
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f_2 &= \frac{f_0 - f_1}{1 - \sigma} = \frac{a(1-\sigma)}{1-\sigma} = a\\ |
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f_3 &= \sigma f_2 - f_1 = \sigma a - a \sigma + b = b\\ |
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f_4 &= \frac{f_2}{f_3} = \frac{a}{b} |
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\end{align*} |
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\item $f_{5+i} = f_{4+i}^2,~~\forall i=0, \ldots, log(N)-1$ |
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\item $f_{4+log(N)} = 1$ |
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\end{enumerate} |
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\begin{lemma}{1} |
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Let $z(X)$ $deg=1$, $n=log(N)+6$, $\sigma$ such that $\sigma^n =1$. |
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If $\exists$ $f(X) \in \mathbb{F}[X]$ such that |
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\begin{enumerate}[1.] |
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\item $f(X) = z(X)$, for $1, \sigma$ |
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\item $f(\sigma^2)(1-\sigma)=f(1)-f(\sigma)$ |
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\item $f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)$ |
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\item $f(\sigma^4)f(\sigma^3)=f(\sigma^2)$ |
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\item $f(\sigma^{4+i+1})=f(\sigma^{4+i})^2,~~\forall i= 0, \ldots, log(N)-1$ |
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\item $f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1$ |
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\end{enumerate} |
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then, $z(X)=aX-b$, where $\frac{b}{a}$ is a N$^{th}$ root of unity. |
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\end{lemma} |
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Let's see the relations between the conditions and the Lemma 1: |
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\begin{scriptsize} |
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\begin{align*} |
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Conditions &\longrightarrow Lemma~1\\ |
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\begin{array}{l} |
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f_0 = z(1) = a - b\\ f_1 = z(\sigma) a \sigma -b |
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\end{array} &\longrightarrow 1.~f(X) = z(X), for 1, \sigma\\ |
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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)\\ |
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f_3 = \sigma f_2 - f_1 = \sigma a - a \sigma + b = b &\longrightarrow 3.~f(\sigma^3)=\sigma f(\sigma^2)-f(\sigma)\\ |
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f_4 = \frac{f_2}{f_3} = \frac{a}{b} &\longrightarrow 4.~f(\sigma^4)f(\sigma^3)=f(\sigma^2)\\ |
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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\\ |
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f_{4+log(N)} = 1 &\longrightarrow 6.~f(\sigma^{5+log(N)} \cdot \sigma^{-1})=1 |
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\end{align*} |
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\end{scriptsize} |
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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. $f(\omega^i)=f_i$). |
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\begin{small} |
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\begin{align*} |
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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)\\ |
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&= 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) |
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\end{align*} |
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\end{small} |
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Prover shows that $f(X)$ by comparing $f(\sigma^i)$ with the corresponding constraints from Lemma 1: |
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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. |
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Given $v_1, v_2$, shows that $p_{\alpha}(X)$, which proves the constraints from Lemma 1, evaluates to $0$ at $\alpha$ (ie. $p_{\alpha}(\alpha)=0$). |
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\begin{align*} |
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p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha) + [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\ |
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&+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\ |
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&+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\ |
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&+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\ |
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&+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ |
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&+ [f(\alpha_1)-1] \rho_n(\alpha) |
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\end{align*} |
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\subsubsection{\texorpdfstring{NIZK argument of knowledge for $R_{unity}$ and $deg(z)\leq 1$}% |
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{NIZK argument of knowledge for Runity and deg(z)<=1}} |
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Prover: |
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\begin{small} |
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\begin{align*} |
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&r_0, r_1, r_2, r_3 \leftarrow^\$ \mathbb{F},~~~ r(X)=r_1+r_2 X + r_3 X^2\\ |
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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)\\ |
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&+ r_0 \rho_{5+log(N)}(X) + r(X) z_{V_n}(X)\\ |
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\\ |
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p(X) &= [f(X) - (aX-b)](\rho_1(X) + \rho_2(X))\\ |
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&+[(1-\sigma)f(X) - f(\sigma^{-1}X) + f(\sigma^{-1}X)] \rho_3(X)\\ |
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&+ [f(X) + f(\sigma^{-2}X) - \sigma f(\sigma^{-1} X)] \rho_4(X)\\ |
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&+ [f(X)f(\sigma^{-1}X)-f(\sigma^{-2}X)] \rho_5(X)\\ |
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&+ [f(X)-f(\sigma^{-1}X)f(\sigma^{-1}X)] \prod_{i \notin [5, 4+log(N)]} (X-\sigma^i)\\ |
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&+ [f(\sigma^{-1}X)-1] \rho_n(X) |
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\end{align*} |
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\end{small} |
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Set |
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$$h'(X) = \frac{p(X)}{z_{V_n}(X)},~~h(X)=h'(X) + X^{d-1} z(X)$$ |
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output $([F]_1=[f(X)]_1, ~ [H]_1=[h(x)]_1)$. |
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\begin{footnotesize} |
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Note that |
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\begin{align*} |
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h(x)&=h'(X)+X^{d-1}z(X)\\ |
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&=\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) |
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\end{align*} |
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\end{footnotesize} |
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Verifier sets challenge $\alpha \in^{\$} \mathbb{F}$ (hash of transcript by Fiat-Shamir). |
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\begin{align*} |
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p_{\alpha}(X) = &-h(X) z_{V_n}(\alpha)\\ |
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&+ [f(X)-z(X)]\cdot (\rho_1(\alpha) + \rho_2(\alpha))\\ |
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&+ [(1-\sigma) f(X) - f(\alpha_2) + f(\alpha_1)] \rho_3(\alpha)\\ |
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&+ [f(X) + f(\alpha_2) - \sigma f(\alpha_1)] \rho_4(\alpha)\\ |
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&+ [f(X) f(\alpha_1) - f(\alpha_2)] \rho_5(\alpha)\\ |
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&+ [f(X) - f(\alpha_1) f(\alpha_1)] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ |
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&+ [f(\alpha_1)-1] \rho_n(\alpha) |
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\end{align*} |
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\begin{footnotesize} |
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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: |
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\begin{align*} |
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-&X^{d-1} z(X)z_{V_n}(\alpha)-z(X)(\rho_1(\alpha)+\rho_2(\alpha))\\ |
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=~ &(-X^{d-1} z_{V_n}(\alpha) - (\rho_1(\alpha)+\rho_2(\alpha))) \cdot z(X)\\ |
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\longrightarrow~ &e(- (\rho_1(\alpha)+\rho_2(\alpha)) - z_{V_n}(\alpha) [X^{d-1}]_1, [z]_2) |
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\end{align*} |
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\end{footnotesize} |
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Prover then generates KZG proofs |
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\begin{align*} |
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((v_1, v_2), \pi_1) &\leftarrow KZG.Open(f(X), (\alpha_1, \alpha_2))\\ |
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(0, \pi_2) &\leftarrow KZG.Open(p_{\alpha}(X), \alpha) |
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\end{align*} |
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prover's output: $(v_1, v_2, \pi_1, \pi_2)$. |
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Verify: |
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set $\alpha_1=\sigma^{-1}\alpha, ~\alpha_2=\sigma^{-2}\alpha$, |
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|
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\begin{footnotesize}(notice that $f(X)\rightarrow [F]_1$, and $f(\alpha_1)=v_1,~f(\alpha_2)=v_2$)\end{footnotesize} |
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|
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\begin{align*} |
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[P]_1 = &-z_{V_n}(\alpha)[H]_1 + [F]_1 (\rho_1(\alpha) + \rho_2(\alpha))\\ |
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&+ [(1-\sigma) [F]_1 - v_2 + v_1] \rho_3(\alpha)\\ |
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&+ [[F]_1 + v_2 - \sigma v_1] \rho_4(\alpha)\\ |
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&+ [[F]_1 v_1 - v_2] \rho_5(\alpha)\\ |
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&+ [[F]_1 - v_1^2] \prod_{i \notin [5, \ldots, 4+log(N)]} (\alpha - \sigma^i)\\ |
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&+ [v_1-1] \rho_n(\alpha) |
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\end{align*} |
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|
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$$KZG.Verify((\alpha_1, \alpha_2), (v_1, v_2), \pi_1)$$ |
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$$e([P]_1, [1]_2) + e(-(\rho_1(\alpha) + \rho_2(\alpha)) - z_{V_n}(\alpha) [x^{d-1}]_1, [z]_2) = e(\pi_2, [x-\alpha]_2)$$ |
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|
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|
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\section{Caulk+} |
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\emph{WIP} |
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|
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\bibliography{paper-notes.bib} |
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\bibliographystyle{unsrt} |
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|
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\end{document} |
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weil-pairing.pdf
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weil-pairing.pdf