Notes taken from \href{https://sites.google.com/site/vincenzoiovinoit/}{Vincenzo Iovino} explainations about FRI \cite{fri}, \cite{cryptoeprint:2022/1216}.
Notes taken from \href{https://sites.google.com/site/vincenzoiovinoit/}{Vincenzo Iovino}\cite{vincenzoiovino}explainations about FRI \cite{fri}, \cite{cryptoeprint:2022/1216}, \cite{cryptoeprint:2019/1020}.
These notes are for self-consumption, are not complete, don't include all the steps neither all the proofs.
An implementation of FRI can be found at \href{https://github.com/arnaucube/fri-commitment}{https://github.com/arnaucube/fri-commitment}.
An implementation of FRI can be found at\\\href{https://github.com/arnaucube/fri-commitment}{https://github.com/arnaucube/fri-commitment}\cite{fri-impl}.
\end{abstract}
\tableofcontents
\section{Preliminaries}
\subsection{Low degree testing}
V wants to ensure that $deg(f(x))\leq d$.
We are in the IOP setting, V asks on a point, P sends back the opening at that point.
TODO
\subsubsection{General degree d test}
\subsection{General degree d test}
Query at points $\{ x_i \}_0^{d+1},~z$ (with rand $z \overset{R}{\in}\mathbb{F}$).
Interpolate $p(x)$ at $\{f(x_i)\}_0^{d+1}$ to reconstruct the unique polynomial $p$ of degree $d$ such that $p(x_i)=f(x_i)~\forall i=1, \ldots, d+1$.
@ -91,7 +84,8 @@ Here, P proves $g,~h$ both have $deg \leq d$, but instead of doing $2 \cdot (d+2
So we halved the number of queries.
\subsection{FRI}
\subsection{FRI-LDT}\label{sec:fri-ldt}
FRI low degree testing.\\
Both P and V have oracle access to function $f$.
V wants to test if $f$ is polynomial with $deg(f)\leq d$.
@ -154,8 +148,9 @@ eg. for $f(x)=x^4+x^3+x^2+x+1$,
\vspace{30px}
\paragraph{Proof generation}
P starts from $f(x)$, and for $i=0$ sets $f_0(x)=f(x)$.
\emph{(Commitment phase)}
P starts from $f(x)$, and for $i=0$ sets $f_0(x)=f(x)$.
\begin{enumerate}
\item$\forall~i \in\{0, log(d)\}$, with $d = deg~f(x)$,\\
P computes $f_i^L(x),~ f_i^R(x)$ for which
@ -163,7 +158,7 @@ P starts from $f(x)$, and for $i=0$ sets $f_0(x)=f(x)$.
f_i(x) = f_i^L(x^2) + x f_i^R(x^2)
\end{equation}
holds.
\item V sends challenge $\alpha_i$
\item V sends challenge $\alpha_i\in\mathbb{F}$
\item P commits to the random linear combination $f_{i+1}$, for
\begin{equation}\tag{eq. $B_i$}
f_{i+1}(x) = f_i^L(x) + \alpha_i f_i^R(x)
@ -174,12 +169,16 @@ Notice that at each step, $deg(f_i)$ halves.
This is done until the last step, where $f_i^L(x),~ f_i^R(x)$ are constant (degree 0 polynomials). For which P does not commit but gives their values directly to V.
\emph{(Query phase)}
P would receive a challenge $z \in D$ set by V (where $D$ is the evaluation domain, $D \in\mathbb{F}$), and P would open the commitments at $\{z^{2^i}, -z^{2^i}\}$ for each step $i$.
(Recall, "opening" means that would provide a proof (MerkleProof) of it).
\item[] Constant values of last iteration: $\{f_k^L,~f_k^R\}$, for $k=log(d)$
\end{itemize}
@ -195,7 +194,8 @@ V receives:
\vspace{20px}
For all $i \in\{0, log(d)\}$, V knows the openings at $z^{2^i}$ and $-(z^{2^i})$ for $Comm(f_i(x))$, which are $o_i=f_i(z^{2^i})$ and $o_i'=f_i(-(z^{2^i}))$ respectively.
For all $i \in\{0, log(d)\}$, V knows the openings at $z^{2^i}$ and $-(z^{2^i})$ for\\
$Comm(f_i(x))$, which are $o_i=f_i(z^{2^i})$ and $o_i'=f_i(-(z^{2^i}))$ respectively.
V, from (eq. $A_i$), knows that
$$f_i(x)=f_i^L(x^2)+ x f_i^R(x^2)$$
@ -236,12 +236,12 @@ where V will find the values of $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$ being
Once, V has computed $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$, can use them to compute the linear combination of
obtaining then $f_{i+1}(z^2)$. This comes from (eq. $B_i$).
obtaining then $f_{i+1}(z^{2^i})$. This comes from (eq. $B_i$).
Now, V checks that the obtained $f_{i+1}(z^2)$ is equal to the received opening $o_{i+1}=f_{i+1}(z^2)$ from the commitment done by P.
V checks also the commitment of $Comm(f_{i+1}(x))$ for the opening $o_{i+1}=f_{i+1}(z^2)$.\\
Now, V checks that the obtained $f_{i+1}(z^{2^i})$ is equal to the received opening $o_{i+1}=f_{i+1}(z^{2^i})$ from the commitment done by P.
V checks also the commitment of $Comm(f_{i+1}(x))$ for the opening $o_{i+1}=f_{i+1}(z^{2^i})$.\\
If the checks pass, V is convinced that $f_1(x)$ was committed honestly.
Now, sets $i := i+1$ and starts a new iteration.
@ -251,8 +251,38 @@ For the last iteration, V checks that the obtained $f_i^L(z^{2^i}),~f_i^R(z^{2^i
\vspace{10px}
It needs $log(d)$ iterations, and the number of queries (commitments + openings sent and verified) needed is $2\cdot log(d)$.
\section{FRI as polynomial commitment}
\emph{[WIP. Unfinished document]}
\subsection{Parameters}
P commits to $f_i$ restricted to a subfield $F_0\subset\mathbb{F}$.
Let $0<\rho<1$ be the \emph{rate} of the code, such that
$$|F_0| =\rho^{-1}\cdot d$$
\begin{theorem}
For $\delta\in(0, 1-\sqrt{\rho})$, we have that if V accepts, then w.v.h.p. (with very high probability) $\Delta(f_0,~ p^d)\leq\delta$.
\end{theorem}
\section{FRI as polynomial commitment scheme}
This section overviews the trick from \cite{cryptoeprint:2019/1020} to convert FRI into a polynomial commitment.
Want to check that the evaluation of $f(x)$ at $r$ is $f(r)$, which is equivalent to proving that $\exists ~Q \in\mathbb{F}[x]$ with $deg(Q)=d-1$, such that
$$
f(x)-f(r) = Q(x) \cdot (x-r)
$$
note that $f(x)-f(r)$ evaluated at $r$ is $0$, so $(x-r) | (f(x)-f(r))$, in other words
$(f(x)-f(r))$ is a multiple of $(x-r)$ for a polynomial $Q(x)$.
Let us define $g(x)=\frac{f(x)-f(r)}{x-r}$.
Prover uses FRI-LDT \ref{sec:fri-ldt} to commit to $g(x)$, and then prove w.v.h.p that $deg(g)\leq d-1$ ($\Longleftrightarrow\Delta(g,~ p^{d-1}\leq\delta$).
Prover was already proving that $deg(f)\leq d$.
Now, the missing thing to prove is that $g(x)$ has the right shape. We can relate $g$ to $f$ as follows:
V does the normal FRI-LDT, but in addition, at the first iteration:
