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small updates to notes_fri_stir.tex

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arnaucube
2025-04-05 02:32:54 +02:00
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@@ -66,7 +66,7 @@ V wants to check that two functions $g,~h$ are both polynomials of degree $\leq
Consider the following protocol:
\begin{enumerate}
\item V sends $\alpha \in \mathbb{F}$ to P. P sends $f(x) = g(x) + \alpha h(x)$ to V.
\item V sends $\alpha \in \mathbb{F}$ to P.
\item P sends $f(x)=g(x) + \alpha h(x)$ to V.
\item V queries $f(r), ~g(r), ~h(r)$ for rand $r \in \mathbb{F}$.
\item V checks $f(r)=g(r) + \alpha h(r)$. (Schwartz-Zippel lema).
@@ -169,12 +169,19 @@ P starts from $f(x)$, and for $i=0$ sets $f_0(x)=f(x)$.
\end{equation}
\item P sets $f_i(x) := f_{i+1}(x)$ and starts again the iteration.
\end{enumerate}
Notice that at each step, $deg(f_i)$ halves.
Note on step 3: when we say "commits", this means that the prover P evaluates $f_{i+1}(x)$ at the $(\rho^{-1} \cdot d)$-sized evaluation
domain $D$ (ie. $f_{i+1}(x) |_D$), and constructs a merkle tree with the
evaluations as leaves.
\vspace{0.5cm}
Notice that at each step, $deg(f_i)$ halves with respect to $deg(f_{i-1}$).
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 $i$.
P would receive a challenge $z \in D$ set by V (where $D$ is the evaluation
domain, $D \subset \mathbb{F}$), 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).
\paragraph{Data sent from P to V}
@@ -284,7 +291,8 @@ $$|F_0| = \rho^{-1} \cdot d$$
\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
Want to check that the evaluation of $f(x)$ at $r$ is $f(r)$, for $r \notin D, r
\in^R \mathbb{F}$; 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)