8 changed files with 413 additions and 340 deletions
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BINnotes_bls-sig.pdf
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+87 -0notes_bls-sig.tex
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BINnotes_halo.pdf
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+145 -0notes_halo.tex
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BINnotes_sonic.pdf
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+181 -0notes_sonic.tex
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BINpaper-notes.pdf
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+0 -340paper-notes.tex
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notes_bls-sig.pdf
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notes_bls-sig.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=black, |
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filecolor=black, |
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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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|
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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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|
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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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|
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|
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\title{Notes on BLS Signatures} |
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\author{arnaucube} |
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\date{} |
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|
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\begin{document} |
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\maketitle |
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|
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\begin{abstract} |
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Notes taken while reading about BLS signatures \cite{bls-sig-eth2}. Usually while reading papers I take handwritten notes, this document contains some of them re-written to $LaTeX$. |
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|
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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{BLS signatures} |
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|
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\paragraph{Key generation} |
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$sk \in \mathbb{Z}_q$, $pk = [sk] \cdot g_1$, where $g_1 \in G_1$, and is the generator. |
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|
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\paragraph{Signature} |
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$$\sigma = [sk] \cdot H(m)$$ |
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where $H$ is a function that maps to a point in $G_2$. So $H(m), \sigma \in G_2$. |
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|
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\paragraph{Verification} |
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$$e(g_1, \sigma) == e(pk, H(m))$$ |
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|
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Unfold: |
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$$e(pk, H(m)) = e([sk] \cdot g_1, H(m) = e(g_1, H(m))^{sk} = e(g_1, [sk] \cdot H(m)) = e(g_1, \sigma))$$ |
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|
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\paragraph{Aggregation} |
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Signatures aggregation: |
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$$\sigma_{aggr} = \sigma_1 + \sigma_2 + \ldots + \sigma_n$$ |
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where $\sigma_{aggr} \in G_2$, and an aggregated signatures is indistinguishible from a non-aggregated signature. |
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|
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\vspace{0.5cm} |
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Public keys aggregation: |
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$$pk_{aggr} = pk_1 + pk_2 + \ldots + pk_n$$ |
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where $pk_{aggr} \in G_1$, and an aggregated public keys is indistinguishible from a non-aggregated public key. |
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|
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\paragraph{Verification of aggregated signatures} |
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Identical to verification of a normal signature as long as we use the same corresponding aggregated public key: |
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$$e(g_1, \sigma_{aggr})==e(pk_{aggr}, H(m))$$ |
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|
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Unfold: |
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$$\fbox{e(pk_{aggr}, H(m))}= e(pk_1 + pk_2 + \ldots + pk_n, H(m)) =$$ |
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$$=e([sk_1] \cdot g_1 + [sk_2] \cdot g_1 + \ldots + [sk_n] \cdot g_1, H(m))=$$ |
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$$=e([sk_1 + sk_2 + \ldots + sk_n] \cdot g_1, H(m))=$$ |
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$$=e(g_1, H(m))^{(sk_1 + sk_2 + \ldots + sk_n)}=$$ |
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$$=e(g_1, [sk_1 + sk_2 + \ldots + sk_n] \cdot H(m))=$$ |
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$$=e(g_1, [sk_1] \cdot H(m) + [sk_2] \cdot H(m) + \ldots + [sk_n] \cdot H(m))=$$ |
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$$=e(g_1, \sigma_1 + \sigma_2 + \ldots + \sigma_n)= \fbox{e(g_1, \sigma_{aggr})}$$ |
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Note: in the current notes $pk \in G_1$ and $\sigma, H(m) \in G_2$, but we could use $\sigma, H(m) \in G_1$ and $pk \in G_2$. |
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\bibliography{paper-notes.bib} |
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\bibliographystyle{unsrt} |
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\end{document} |
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BIN
notes_halo.pdf
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notes_halo.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=black, |
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filecolor=black, |
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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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|
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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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|
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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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\title{Notes on Halo} |
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\author{arnaucube} |
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\date{} |
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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 Halo paper \cite{cryptoeprint:2019/1021}. 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{modified IPA (from Halo paper)} |
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Notes taken while reading about the modified Inner Product Argument (IPA) from the Halo paper \cite{cryptoeprint:2019/1021}. |
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\subsection{Notation} |
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\begin{description} |
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\item[Scalar mul] $[a]G$, where $a$ is a scalar and $G \in \mathbb{G}$ |
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\item[Inner product] $<\overrightarrow{a}, \overrightarrow{b}> = a_0 b_0 + a_1 b_1 + \ldots + a_{n-1} b_{n-1}$ |
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\item[Multiscalar mul] $<\overrightarrow{a}, \overrightarrow{b}> = [a_0] G_0 + [a_1] G_1 + \ldots [a_{n-1}] G_{n-1}$ |
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\end{description} |
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\subsection{Transparent setup} |
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$\overrightarrow{G} \in^r \mathbb{G}^d$, $H \in^r \mathbb{G}$ |
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Prover wants to commit to $p(x)=a_0$ |
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\subsection{Protocol} |
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Prover: |
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$$P=<\overrightarrow{a}, \overrightarrow{G}> + [r]H$$ |
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$$v=<\overrightarrow{a}, \{1, x, x^2, \ldots, x^{d-1} \} >$$ |
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where $\{1, x, x^2, \ldots, x^{d-1} \} = \overrightarrow{b}$. |
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We can see that computing $v$ is the equivalent to evaluating $p(x)$ at $x$ ($p(x)=v$). |
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We will prove: |
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\begin{enumerate}[i.] |
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\item polynomial $p(X) = \sum a_i X^i$\\ |
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$p(x) = v$ (that $p(X)$ evaluates $x$ to $v$). |
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\item $deg(p(X)) \leq d-1$ |
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\end{enumerate} |
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Both parties know $P$, point $x$ and claimed evaluation $v$. For $U \in^r \mathbb{G}$, |
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$$P' = P + [v] U = <\overrightarrow{a}, G> + [r]H + [v] U$$ |
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Now, for $k$ rounds ($d=2^k$, from $j=k$ to $j=1$): |
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\begin{itemize} |
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\item random blinding factors: $l_j, r_j \in \mathbb{F}_p$ |
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\item |
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$$L_j = < \overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [< \overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U$$ |
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$$L_j = < \overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [< \overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U$$ |
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\item Verifier sends random challenge $u_j \in \mathbb{I}$ |
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\item Prover computes the halved vectors for next round: |
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$$\overrightarrow{a} \leftarrow \overrightarrow{a}_{hi} \cdot u_j^{-1} + \overrightarrow{a}_{lo} \cdot u_j$$ |
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$$\overrightarrow{b} \leftarrow \overrightarrow{b}_{lo} \cdot u_j^{-1} + \overrightarrow{b}_{hi} \cdot u_j$$ |
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$$\overrightarrow{G} \leftarrow \overrightarrow{G}_{lo} \cdot u_j^{-1} + \overrightarrow{G}_{hi} \cdot u_j$$ |
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\end{itemize} |
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After final round, $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{G}$ are each of length 1. |
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Verifier can compute |
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$$G = \overrightarrow{G}_0 = < \overrightarrow{s}, \overrightarrow{G} >$$ |
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and $$b = \overrightarrow{b}_0 = < \overrightarrow{s}, \overrightarrow{b} >$$ |
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where $\overrightarrow{s}$ is the binary counting structure: |
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\begin{align*} |
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&s = (u_1^{-1} ~ u_2^{-1} \cdots ~u_k^{-1},\\ |
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&~~~~~~u_1 ~~~ u_2^{-1} ~\cdots ~u_k^{-1},\\ |
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&~~~~~~u_1^{-1} ~~ u_2 ~~\cdots ~u_k^{-1},\\ |
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&~~~~~~~~~~~~~~\vdots\\ |
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&~~~~~~u_1 ~~~~ u_2 ~~\cdots ~u_k) |
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\end{align*} |
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|
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And verifier checks: |
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$$[a]G + [r'] H + [ab] U == P' + \sum_{j=1}^k ( [u_j^2] L_j + [u_j^{-2}] R_j)$$ |
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where the synthetic blinding factor $r'$ is $r' = r + \sum_{j=1}^k (l_j u_j^2 + r_j u_j^{-2})$. |
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\vspace{1cm} |
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Unfold: |
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$$ |
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\textcolor{brown}{[a]G} + \textcolor{cyan}{[r'] H} + \textcolor{magenta}{[ab] U} |
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== |
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\textcolor{blue}{P'} + \sum_{j=1}^k ( \textcolor{violet}{[u_j^2] L_j} + \textcolor{orange}{[u_j^{-2}] R_j}) |
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$$ |
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\begin{align*} |
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&Right~side = \textcolor{blue}{P'} + \sum_{j=1}^k ( \textcolor{violet}{[u_j^2] L_j} + \textcolor{orange}{[u_j^{-2}] R_j})\\ |
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&= \textcolor{blue}{< \overrightarrow{a}, \overrightarrow{G}> + [r] H + [v] U}\\ |
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&+ \sum_{j=1}^k (\\ |
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&\textcolor{violet}{[u_j^2] \cdot <\overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [<\overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U}\\ |
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&\textcolor{orange}{+ [u_j^{-2}] \cdot <\overrightarrow{a}_{hi}, \overrightarrow{G}_{lo}> + [r_j] H + [<\overrightarrow{a}_{hi}, \overrightarrow{b}_{lo}>] U} |
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) |
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\end{align*} |
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\begin{align*} |
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&Left~side = \textcolor{brown}{[a]G} + \textcolor{cyan}{[r'] H} + \textcolor{magenta}{[ab] U}\\ |
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& = \textcolor{brown}{< \overrightarrow{a}, \overrightarrow{G} >}\\ |
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&+ \textcolor{cyan}{[r + \sum_{j=1}^k (l_j \cdot u_j^2 + r_j u_j^{-2})] \cdot H}\\ |
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&+ \textcolor{magenta}{< \overrightarrow{a}, \overrightarrow{b} > U} |
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\end{align*} |
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\section{Amortization Strategy} |
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TODO |
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\bibliography{paper-notes.bib} |
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\bibliographystyle{unsrt} |
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\end{document} |
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notes_sonic.pdf
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notes_sonic.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=black, |
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filecolor=black, |
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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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|
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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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|
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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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|
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|
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\title{Notes on Sonic} |
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\author{arnaucube} |
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\date{} |
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|
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\begin{document} |
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\maketitle |
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|
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\begin{abstract} |
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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 $LaTeX$. |
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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{Sonic} |
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\subsection{Structured Reference String} |
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$\{ \{g^{x^i}\}_{i=-d}^d, \{ g^{\alpha x^i} \}_{i=-d, i \neq 0}^d, \{ h^{x^i}, h^{\alpha x^i} \}_{i=-d}^d, e(g, h^\alpha) \}$ |
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\subsection{System of constraints} |
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Multiplication constraint: $a \cdot b = c$ |
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$Q$ linear constraints: |
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$$ |
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a \cdot u_q + b \cdot v_q + c \cdot w_q = k_q |
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$$ |
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with $u_q, v_q, w_q \in \mathbb{F}^n$, and $k_q \in \mathbb{F}_p$. |
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\vspace{0.5cm} |
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Example: $x^2 + y^2 = z$ |
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$$a = (x, y), \qquad b = (x, y), \qquad c = (x^2, y^2)$$ |
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\begin{enumerate}[i.] |
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\item $(x, y) \cdot (1, 0) + (x, y) \cdot (-1, 0) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow x - x = 0$ |
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\item $(x, y) \cdot (0, 1) + (x, y) \cdot (0, -1) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow y - y = 0$ |
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\item $(x, y) \cdot (0, 0) + (x, y) \cdot (0, 0) + (x^2, y^2) \cdot (1, 1) = z \longrightarrow x^2 + y^2 = z$ |
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\end{enumerate} |
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So, |
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$$u_1 = (1, 0) \quad v_1=(-1, 0) \quad w_1=(0, 0) \quad k_1=0$$ |
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$$u_2 = (0, 1) \quad v_2=(0, -1) \quad w_2=(0, 0) \quad k_2=0$$ |
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$$u_3 = (0, 0) \quad v_3=(0, 0) \quad w_3=(1, 1) \quad k_2=z$$ |
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\vspace{1cm} |
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Compress n multiplication constraints into an equation in formal indeterminate $Y$: |
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$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^i = 0$$ |
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encode into negative exponents of $Y$: |
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$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^-i = 0$$ |
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Also, compress the $Q$ linear constraints, scaling by $Y^n$ to preserve linear independence: |
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$$ |
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\sum_{q=1}^Q (a \cdot u_q + b \cdot v_q + c \cdot w_q - k_q) \cdot Y^{q+n} = 0 |
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$$ |
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Polys: |
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\begin{align} |
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\nonumber & u_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot u_{q, i}\\ |
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\nonumber & v_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot v_{q, i}\\ |
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\nonumber & w_i(Y) = -Y^i - Y^{-1} + \sum_{q=1}^Q Y^{q+n} \cdot w_{q, i}\\ |
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\nonumber & k(Y) = \sum_{q=1}^Q Y^{q+n} \cdot k_q |
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\end{align} |
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Combine the multiplicative and linear constraints to: |
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\begin{align} |
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\nonumber & a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y) |
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+ \sum_{i=1}^n a_i b_i (Y^i + Y^{-i}) - k(Y) = 0 |
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\end{align} |
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where $a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y)$ is embeded into the constant term of the polynomial $t(X, Y)$. |
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Define $r(X, Y)$ s.t. $r(X, Y) = r(XY, 1)$. |
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$$\Longrightarrow r(X, Y) = \sum_{i=1}^n (a_i X^i Y^i + b_i X^{-i} Y^{-i} + c_i X^{-i-n} Y^{-i-n})$$ |
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$$s(X, Y) = \sum_{i=1}^n (u_i(Y) X^{-i} + v_i(Y) X^i + w_i(Y) X^{i+n})$$ |
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$$r'(X, Y) = r(X, Y) + s(X, Y)$$ |
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$$t(X, Y) = r(X, Y) + r'(X, Y) - k(Y)$$ |
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The coefficient of $X^0$ in $t(X, Y)$ is the left-hand side of the equation. |
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Sonic demonstrates that the constant term of $t(X, Y)$ is zero, thus demonstrating that our constraint system is satisfied. |
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\subsubsection{The basic Sonic protocol} |
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\begin{enumerate}[1.] |
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\item Prover constructs $r(X, Y)$ using their hidden witness |
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\item Prover commits to $r(X, 1)$, setting the maximum degree to n |
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\item Verifier sends random challenge $y$ |
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\item Prover commits to $t(X, y)$. The commitment scheme ensures that $t(X, y)$ has no constant term. |
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\item Verifier sends random challenge $z$ |
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\item Prover opens commitments to $r(z, 1), r(z, y), t(z, y)$ |
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\item Verifier calculates $r'(z, y)$, and checks that |
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$$r(z, y) \cdot r'(z, y) - k(y) == t(z, y)$$ |
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\end{enumerate} |
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Steps $3$ and $5$ can be made non-interactive by the Fiat-Shamir transformation. |
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\subsubsection{Polynomial Commitment Scheme} |
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Sonic uses an adaptation of KZG \cite{kzg-tmp}, want: |
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\begin{enumerate}[i.] |
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\item \emph{evaluation binding}, i.e. given a commitment $F$, an adversary cannot open F to two different evaluations $v_1$ and $v_2$ |
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\item \emph{bounded polynomial extractable}, i.e. any algebraic adversary that opens a commitment $F$ knows an opening $f(X)$ with powers $-d \leq i \leq max, i \neq 0$. |
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\end{enumerate} |
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\vspace{0.5cm} |
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PC scheme (adaptation of KZG): |
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\begin{enumerate}[i.] |
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\item Commit(info, $f(X)$) $\longrightarrow F$: |
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$$F = g^{\alpha \cdot x^{d-max}} \cdot f(x)$$ |
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\item Open(info, $F$, $z$, $f(x)$) $\longrightarrow (f(z), W)$: |
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$$w(X) = \frac{f(X) - f(z)}{X-z}$$ |
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$$W = g^{w(x)}$$ |
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\item Verify(info, $F$, $z$, $(v, W)$) $\longrightarrow 0/1$:\\ |
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Check: |
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$$e(W, h^{\alpha \cdot x}) \cdot |
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e(g^v W^{-z}, h^{\alpha}) |
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== e(F, h^{x^{-d+max}})$$ |
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\end{enumerate} |
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|
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\subsection{Succint signatures of correct computation} |
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Signature of correct computation to ensure that an element $s=s(z, y)$ for a known polynomial |
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$$s(X, Y) = \sum_{i, j = -d}^d s_{i, j} \cdot X^i \cdot Y^i$$ |
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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. |
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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: |
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$$ |
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\Psi_j(X, Y) = |
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\sum_{i=1}^n \psi_{j, \sigma_{j, i}} |
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\cdot X^i \cdot Y^{\sigma_{j, i}} |
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$$ |
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where $\sigma_j$ is the fixed polynomial permutation, and $\phi_{j, i} \in \mathbb{F}$ are the coefficients. |
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\vspace{1cm} |
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\framebox{WIP} |
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\vspace{1cm} |
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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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BIN
paper-notes.pdf
BIN
paper-notes.pdf
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\documentclass{article} |
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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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\title{Paper notes} |
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\author{arnaucube} |
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\date{} |
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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 papers. 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{SnarkPack} |
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Notes taken while reading SnarkPack paper \cite{cryptoeprint:2021/529}. |
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Groth16 proof aggregation. |
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\begin{enumerate}[i.] |
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\item Simple verification:\\ |
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Proof: $\pi_i = (A_i, B_i, C_i)$\\ |
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Verifier checks: $e(A_i, B_i) == e(C_i, D)$\\ |
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Where $D$ is the $CRS$. |
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\item Batch verification: |
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$r \in^\$ F_q$\\ |
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$r^i \cdot e(A_i, B_i) == e(C_i, D)$\\ |
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$\Longrightarrow \prod e(A_i, B_i)^{r^i} == \prod e(C_i, D)^{r^i}$\\ |
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$\Longrightarrow \prod e(A_i, B_i^{r^i}) == \prod e(C_i^{r^i}, D)$ |
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\item Snark Aggregation verification:\\ |
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$z_{AB} = \prod e(A_i, B_i^{r^i})$\\ |
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$z_C = \prod C_i^{r^i}$\\ |
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Verification: $z_{AB} == e(z_C, D)$ |
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\end{enumerate} |
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\section{Sonic} |
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Notes taken while reading Sonic paper \cite{cryptoeprint:2019/099}. Does not include all the steps, neither the proofs. |
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\subsection{Structured Reference String} |
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$\{ \{g^{x^i}\}_{i=-d}^d, \{ g^{\alpha x^i} \}_{i=-d, i \neq 0}^d, \{ h^{x^i}, h^{\alpha x^i} \}_{i=-d}^d, e(g, h^\alpha) \}$ |
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\subsection{System of constraints} |
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Multiplication constraint: $a \cdot b = c$ |
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$Q$ linear constraints: |
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$$ |
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a \cdot u_q + b \cdot v_q + c \cdot w_q = k_q |
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$$ |
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with $u_q, v_q, w_q \in \mathbb{F}^n$, and $k_q \in \mathbb{F}_p$. |
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\vspace{0.5cm} |
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Example: $x^2 + y^2 = z$ |
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$$a = (x, y), \qquad b = (x, y), \qquad c = (x^2, y^2)$$ |
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\begin{enumerate}[i.] |
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\item $(x, y) \cdot (1, 0) + (x, y) \cdot (-1, 0) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow x - x = 0$ |
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\item $(x, y) \cdot (0, 1) + (x, y) \cdot (0, -1) + (x^2, y^2) \cdot (0, 0) = 0 \longrightarrow y - y = 0$ |
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\item $(x, y) \cdot (0, 0) + (x, y) \cdot (0, 0) + (x^2, y^2) \cdot (1, 1) = z \longrightarrow x^2 + y^2 = z$ |
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\end{enumerate} |
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So, |
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$$u_1 = (1, 0) \quad v_1=(-1, 0) \quad w_1=(0, 0) \quad k_1=0$$ |
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$$u_2 = (0, 1) \quad v_2=(0, -1) \quad w_2=(0, 0) \quad k_2=0$$ |
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$$u_3 = (0, 0) \quad v_3=(0, 0) \quad w_3=(1, 1) \quad k_2=z$$ |
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\vspace{1cm} |
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Compress n multiplication constraints into an equation in formal indeterminate $Y$: |
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$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^i = 0$$ |
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encode into negative exponents of $Y$: |
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$$\sum_{i=1}^n (a_i b_i - c_i) \cdot Y^-i = 0$$ |
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Also, compress the $Q$ linear constraints, scaling by $Y^n$ to preserve linear independence: |
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$$ |
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\sum_{q=1}^Q (a \cdot u_q + b \cdot v_q + c \cdot w_q - k_q) \cdot Y^{q+n} = 0 |
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$$ |
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Polys: |
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\begin{align} |
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\nonumber & u_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot u_{q, i}\\ |
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\nonumber & v_i(Y) = \sum_{q=1}^Q Y^{q+n} \cdot v_{q, i}\\ |
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\nonumber & w_i(Y) = -Y^i - Y^{-1} + \sum_{q=1}^Q Y^{q+n} \cdot w_{q, i}\\ |
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\nonumber & k(Y) = \sum_{q=1}^Q Y^{q+n} \cdot k_q |
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\end{align} |
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Combine the multiplicative and linear constraints to: |
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\begin{align} |
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\nonumber & a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y) |
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+ \sum_{i=1}^n a_i b_i (Y^i + Y^{-i}) - k(Y) = 0 |
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\end{align} |
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where $a \cdot u(Y) + b \cdot v(Y) + c \cdot w(Y)$ is embeded into the constant term of the polynomial $t(X, Y)$. |
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Define $r(X, Y)$ s.t. $r(X, Y) = r(XY, 1)$. |
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$$\Longrightarrow r(X, Y) = \sum_{i=1}^n (a_i X^i Y^i + b_i X^{-i} Y^{-i} + c_i X^{-i-n} Y^{-i-n})$$ |
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$$s(X, Y) = \sum_{i=1}^n (u_i(Y) X^{-i} + v_i(Y) X^i + w_i(Y) X^{i+n})$$ |
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$$r'(X, Y) = r(X, Y) + s(X, Y)$$ |
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$$t(X, Y) = r(X, Y) + r'(X, Y) - k(Y)$$ |
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The coefficient of $X^0$ in $t(X, Y)$ is the left-hand side of the equation. |
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Sonic demonstrates that the constant term of $t(X, Y)$ is zero, thus demonstrating that our constraint system is satisfied. |
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\subsubsection{The basic Sonic protocol} |
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\begin{enumerate}[1.] |
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\item Prover constructs $r(X, Y)$ using their hidden witness |
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\item Prover commits to $r(X, 1)$, setting the maximum degree to n |
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\item Verifier sends random challenge $y$ |
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\item Prover commits to $t(X, y)$. The commitment scheme ensures that $t(X, y)$ has no constant term. |
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\item Verifier sends random challenge $z$ |
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\item Prover opens commitments to $r(z, 1), r(z, y), t(z, y)$ |
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\item Verifier calculates $r'(z, y)$, and checks that |
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$$r(z, y) \cdot r'(z, y) - k(y) == t(z, y)$$ |
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\end{enumerate} |
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Steps $3$ and $5$ can be made non-interactive by the Fiat-Shamir transformation. |
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\subsubsection{Polynomial Commitment Scheme} |
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Sonic uses an adaptation of KZG \cite{kzg-tmp}, want: |
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\begin{enumerate}[i.] |
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\item \emph{evaluation binding}, i.e. given a commitment $F$, an adversary cannot open F to two different evaluations $v_1$ and $v_2$ |
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\item \emph{bounded polynomial extractable}, i.e. any algebraic adversary that opens a commitment $F$ knows an opening $f(X)$ with powers $-d \leq i \leq max, i \neq 0$. |
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\end{enumerate} |
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\vspace{0.5cm} |
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PC scheme (adaptation of KZG): |
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\begin{enumerate}[i.] |
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\item Commit(info, $f(X)$) $\longrightarrow F$: |
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$$F = g^{\alpha \cdot x^{d-max}} \cdot f(x)$$ |
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\item Open(info, $F$, $z$, $f(x)$) $\longrightarrow (f(z), W)$: |
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$$w(X) = \frac{f(X) - f(z)}{X-z}$$ |
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$$W = g^{w(x)}$$ |
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\item Verify(info, $F$, $z$, $(v, W)$) $\longrightarrow 0/1$:\\ |
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Check: |
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$$e(W, h^{\alpha \cdot x}) \cdot |
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e(g^v W^{-z}, h^{\alpha}) |
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== e(F, h^{x^{-d+max}})$$ |
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\end{enumerate} |
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\subsection{Succint signatures of correct computation} |
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Signature of correct computation to ensure that an element $s=s(z, y)$ for a known polynomial |
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$$s(X, Y) = \sum_{i, j = -d}^d s_{i, j} \cdot X^i \cdot Y^i$$ |
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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. |
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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: |
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$$ |
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\Psi_j(X, Y) = |
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\sum_{i=1}^n \psi_{j, \sigma_{j, i}} |
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\cdot X^i \cdot Y^{\sigma_{j, i}} |
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$$ |
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where $\sigma_j$ is the fixed polynomial permutation, and $\phi_{j, i} \in \mathbb{F}$ are the coefficients. |
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\vspace{1cm} |
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\framebox{WIP} |
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\vspace{1cm} |
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\section{BLS signatures} |
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Notes taken while reading about BLS signatures \cite{bls-sig-eth2}. |
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\paragraph{Key generation} |
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$sk \in \mathbb{Z}_q$, $pk = [sk] \cdot g_1$, where $g_1 \in G_1$, and is the generator. |
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\paragraph{Signature} |
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$$\sigma = [sk] \cdot H(m)$$ |
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where $H$ is a function that maps to a point in $G_2$. So $H(m), \sigma \in G_2$. |
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\paragraph{Verification} |
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$$e(g_1, \sigma) == e(pk, H(m))$$ |
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Unfold: |
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$$e(pk, H(m)) = e([sk] \cdot g_1, H(m) = e(g_1, H(m))^{sk} = e(g_1, [sk] \cdot H(m)) = e(g_1, \sigma))$$ |
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\paragraph{Aggregation} |
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Signatures aggregation: |
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$$\sigma_{aggr} = \sigma_1 + \sigma_2 + \ldots + \sigma_n$$ |
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where $\sigma_{aggr} \in G_2$, and an aggregated signatures is indistinguishible from a non-aggregated signature. |
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\paragraph{Public keys aggregation} |
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$$pk_{aggr} = pk_1 + pk_2 + \ldots + pk_n$$ |
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where $pk_{aggr} \in G_1$, and an aggregated public keys is indistinguishible from a non-aggregated public key. |
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\paragraph{Verification of aggregated signatures} |
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Identical to verification of a normal signature as long as we use the same corresponding aggregated public key: |
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$$e(g_1, \sigma_{aggr})==e(pk_{aggr}, H(m))$$ |
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Unfold: |
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$$\fbox{e(pk_{aggr}, H(m))}= e(pk_1 + pk_2 + \ldots + pk_n, H(m)) =$$ |
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$$=e([sk_1] \cdot g_1 + [sk_2] \cdot g_1 + \ldots + [sk_n] \cdot g_1, H(m))=$$ |
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$$=e([sk_1 + sk_2 + \ldots + sk_n] \cdot g_1, H(m))=$$ |
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$$=e(g_1, H(m))^{(sk_1 + sk_2 + \ldots + sk_n)}=$$ |
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$$=e(g_1, [sk_1 + sk_2 + \ldots + sk_n] \cdot H(m))=$$ |
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$$=e(g_1, [sk_1] \cdot H(m) + [sk_2] \cdot H(m) + \ldots + [sk_n] \cdot H(m))=$$ |
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$$=e(g_1, \sigma_1 + \sigma_2 + \ldots + \sigma_n)= \fbox{e(g_1, \sigma_{aggr})}$$ |
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\section{modified IPA (from Halo)} |
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Notes taken while reading about the modified Inner Product Argument (IPA) from the Halo paper \cite{cryptoeprint:2019/1021}. |
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\subsection{Notation} |
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\begin{description} |
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\item[Scalar mul] $[a]G$, where $a$ is a scalar and $G \in \mathbb{G}$ |
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\item[Inner product] $<\overrightarrow{a}, \overrightarrow{b}> = a_0 b_0 + a_1 b_1 + \ldots + a_{n-1} b_{n-1}$ |
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\item[Multiscalar mul] $<\overrightarrow{a}, \overrightarrow{b}> = [a_0] G_0 + [a_1] G_1 + \ldots [a_{n-1}] G_{n-1}$ |
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\end{description} |
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\subsection{Transparent setup} |
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$\overrightarrow{G} \in^r \mathbb{G}^d$, $H \in^r \mathbb{G}$ |
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Prover wants to commit to $p(x)=a_0$ |
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\subsection{Protocol} |
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Prover: |
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$$P=<\overrightarrow{a}, \overrightarrow{G}> + [r]H$$ |
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$$v=<\overrightarrow{a}, \{1, x, x^2, \ldots, x^{d-1} \} >$$ |
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where $\{1, x, x^2, \ldots, x^{d-1} \} = \overrightarrow{b}$. |
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We can see that computing $v$ is the equivalent to evaluating $p(x)$ at $x$ ($p(x)=v$). |
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We will prove: |
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\begin{enumerate}[i.] |
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\item polynomial $p(X) = \sum a_i X^i$\\ |
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$p(x) = v$ (that $p(X)$ evaluates $x$ to $v$). |
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\item $deg(p(X)) \leq d-1$ |
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\end{enumerate} |
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Both parties know $P$, point $x$ and claimed evaluation $v$. For $U \in^r \mathbb{G}$, |
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$$P' = P + [v] U = <\overrightarrow{a}, G> + [r]H + [v] U$$ |
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Now, for $k$ rounds ($d=2^k$, from $j=k$ to $j=1$): |
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\begin{itemize} |
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\item random blinding factors: $l_j, r_j \in \mathbb{F}_p$ |
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\item |
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$$L_j = < \overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [< \overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U$$ |
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$$L_j = < \overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [< \overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U$$ |
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\item Verifier sends random challenge $u_j \in \mathbb{I}$ |
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\item Prover computes the halved vectors for next round: |
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$$\overrightarrow{a} \leftarrow \overrightarrow{a}_{hi} \cdot u_j^{-1} + \overrightarrow{a}_{lo} \cdot u_j$$ |
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$$\overrightarrow{b} \leftarrow \overrightarrow{b}_{lo} \cdot u_j^{-1} + \overrightarrow{b}_{hi} \cdot u_j$$ |
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$$\overrightarrow{G} \leftarrow \overrightarrow{G}_{lo} \cdot u_j^{-1} + \overrightarrow{G}_{hi} \cdot u_j$$ |
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\end{itemize} |
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After final round, $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{G}$ are each of length 1. |
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Verifier can compute |
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$$G = \overrightarrow{G}_0 = < \overrightarrow{s}, \overrightarrow{G} >$$ |
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and $$b = \overrightarrow{b}_0 = < \overrightarrow{s}, \overrightarrow{b} >$$ |
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where $\overrightarrow{s}$ is the binary counting structure: |
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\begin{align*} |
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&s = (u_1^{-1} ~ u_2^{-1} \cdots ~u_k^{-1},\\ |
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&~~~~~~u_1 ~~~ u_2^{-1} ~\cdots ~u_k^{-1},\\ |
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&~~~~~~u_1^{-1} ~~ u_2 ~~\cdots ~u_k^{-1},\\ |
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&~~~~~~~~~~~~~~\vdots\\ |
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&~~~~~~u_1 ~~~~ u_2 ~~\cdots ~u_k) |
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\end{align*} |
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And verifier checks: |
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$$[a]G + [r'] H + [ab] U == P' + \sum_{j=1}^k ( [u_j^2] L_j + [u_j^{-2}] R_j)$$ |
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where the synthetic blinding factor $r'$ is $r' = r + \sum_{j=1}^k (l_j u_j^2 + r_j u_j^{-2})$. |
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\vspace{1cm} |
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Unfold: |
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$$ |
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\textcolor{brown}{[a]G} + \textcolor{cyan}{[r'] H} + \textcolor{magenta}{[ab] U} |
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== |
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\textcolor{blue}{P'} + \sum_{j=1}^k ( \textcolor{violet}{[u_j^2] L_j} + \textcolor{orange}{[u_j^{-2}] R_j}) |
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$$ |
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\begin{align*} |
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&Right~side = \textcolor{blue}{P'} + \sum_{j=1}^k ( \textcolor{violet}{[u_j^2] L_j} + \textcolor{orange}{[u_j^{-2}] R_j})\\ |
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&= \textcolor{blue}{< \overrightarrow{a}, \overrightarrow{G}> + [r] H + [v] U}\\ |
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&+ \sum_{j=1}^k (\\ |
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&\textcolor{violet}{[u_j^2] \cdot <\overrightarrow{a}_{lo}, \overrightarrow{G}_{hi}> + [l_j] H + [<\overrightarrow{a}_{lo}, \overrightarrow{b}_{hi}>] U}\\ |
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&\textcolor{orange}{+ [u_j^{-2}] \cdot <\overrightarrow{a}_{hi}, \overrightarrow{G}_{lo}> + [r_j] H + [<\overrightarrow{a}_{hi}, \overrightarrow{b}_{lo}>] U} |
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) |
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\end{align*} |
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\begin{align*} |
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&Left~side = \textcolor{brown}{[a]G} + \textcolor{cyan}{[r'] H} + \textcolor{magenta}{[ab] U}\\ |
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& = \textcolor{brown}{< \overrightarrow{a}, \overrightarrow{G} >}\\ |
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&+ \textcolor{cyan}{[r + \sum_{j=1}^k (l_j \cdot u_j^2 + r_j u_j^{-2})] \cdot H}\\ |
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&+ \textcolor{magenta}{< \overrightarrow{a}, \overrightarrow{b} > U} |
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\end{align*} |
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\bibliography{paper-notes.bib} |
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\bibliographystyle{unsrt} |
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\end{document} |
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