math/notes_fri.tex
notes_fri.texRaw
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27
28\title{Notes on FRI}
29\author{arnaucube}
30\date{February 2023}
31
32\begin{document}
33
34\maketitle
35
36\begin{abstract}
37 Notes taken from \href{https://sites.google.com/site/vincenzoiovinoit/}{Vincenzo Iovino} \cite{vincenzoiovino} explanations about FRI \cite{fri}, \cite{cryptoeprint:2022/1216}, \cite{cryptoeprint:2019/1020}.
38
39 These notes are for self-consumption, are not complete, don't include all the steps neither all the proofs.
40
41 An implementation of FRI can be found at\\ \href{https://github.com/arnaucube/fri-commitment}{https://github.com/arnaucube/fri-commitment} \cite{fri-impl}.
42\end{abstract}
43
44\tableofcontents
45
46\section{Preliminaries}
47\subsection{General degree d test}
48
49Query at points $\{ x_i \}_0^{d+1},~z$ (with rand $z \overset{R}{\in} \mathbb{F}$).
50Interpolate $p(x)$ at $\{f(x_i)\}_0^{d+1}$ to reconstruct the unique polynomial $p$ of degree $d$ such that $p(x_i)=f(x_i)~\forall i=1, \ldots, d+1$.
51
52V checks $p(z)=f(z)$, if the check passes, then V is convinced with high probability.
53
54This needs $d+2$ queries, is linear, $\mathcal{O}(n)$. With FRI we will have the test in $\mathcal{O}(\log{}d)$.
55
56\section{FRI protocol}
57Allows to test if a function $f$ is a poly of degree $\leq d$ in $\mathcal{O}(\log{}d)$.
58
59Note: "P \emph{sends} $f(x)$ to V", "\emph{sends}", in the ideal IOP model means that all the table of $f(x)$ is sent, in practice is sent a commitment to $f(x)$.
60
61\subsection{Intuition}
62V wants to check that two functions $g,~h$ are both polynomials of degree $\leq d$.
63
64Consider the following protocol:
65
66\begin{enumerate}
67 \item V sends $\alpha \in \mathbb{F}$ to P. P sends $f(x) = g(x) + \alpha h(x)$ to V.
68 \item P sends $f(x)=g(x) + \alpha h(x)$ to V.
69 \item V queries $f(r), ~g(r), ~h(r)$ for rand $r \in \mathbb{F}$.
70 \item V checks $f(r)=g(r) + \alpha h(r)$. (Schwartz-Zippel lema).
71 If holds, V can be certain that $f(x)=g(x)+ \alpha h(x)$.
72 \item P proves that $deg(f) \leq d$.
73 \item If V is convinced that $deg(f) \leq d$, V believes that both $g, h$ have $deg \leq d$.
74\end{enumerate}
75
76%/// TODO tabulate this next lines
77With high probablility, $\alpha$ will not cancel the coeffs with $deg \geq d+1$. % TODO check which is the name of this theorem or why this is true
78
79Let $g(x)=a \cdot x^{d+1}, ~~ h(x)=b \cdot x^{d+1}$, and set $f(x) = g(x) + \alpha h(x)$.
80Imagine that P can chose $\alpha$ such that $a x^{d+1} + \alpha \cdot b x^{d+1} = 0$, then, in $f(x)$ the coefficients of degree $d+1$ would cancel.
81%///
82
83\quad
84
85Here, P proves $g,~h$ both have $deg \leq d$, but instead of doing $2 \cdot (d+2)$ queries ($d+2$ for $g$, and $d+2$ for $h$), it is done in $d+2$ queries (for $f$).
86So we halved the number of queries.
87
88
89\subsection{FRI-LDT}\label{sec:fri-ldt}
90FRI low degree testing.\\
91Both P and V have oracle access to function $f$.
92
93V wants to test if $f$ is polynomial with $deg(f) \leq d$.
94
95Let $f_0(x)=f(x)$.
96
97Each polynomial $f(x)$ of degree that is a power of $2$, can be written as
98$$f(x) = f^L(x^2) + x f^R(x^2)$$
99for some polynomials $f^L,~f^R$ of degree $\frac{deg(f)}{2}$, each one containing the even and odd degree coefficients as follows:
100
101% $f^L(x)$ is built from the even degree coefficients divided by $x$, and $f^R(x)$ from the odd degree coefficients divided by $x$.
102
103$$f^L(x)= \sum_0^{\frac{d+1}{2}-1} c_{2i} x^i ,~~ f^R(x)= \sum_0^{\frac{d+1}{2}-1} c_{2i+1} x^i$$
104
105eg. for $f(x)=x^4+x^3+x^2+x+1$,
106\begin{align*}
107 \begin{rcases}
108 f^L(x)=x^2+x+1\\
109 f^R(x)=x+1
110 \end{rcases}
111 ~f(x) = f^L(x^2) &+ x \cdot f^R(x^2)\\
112 = (x^2)^2 + (x^2) + 1 &+ x \cdot ((x^2) + 1)\\
113 = x^4 + x^2 + 1 &+ x^3 + x
114\end{align*}
115
116% \begin{enumerate}
117% \item V sends to P some $\alpha_0 \in \mathbb{F}$.
118% Let
119% \begin{equation}\tag{$A_0$}
120% f_0(x) = f_0^L(x^2) + x f_0^R(x^2)
121% \end{equation}
122% \item P sends
123% \begin{equation}\tag{$B_0$}
124% f_1(x) = f_0^L(x) + \alpha_0 f_0^R(x)
125% \end{equation}
126% to V.
127%
128% (remember that "sends" in IOP model is that P commits to it)
129% \item V sends to P some $\alpha_1 \in \mathbb{F}$.
130% Let
131% \begin{equation}\tag{$A_1$}
132% f_1(x) = f_1^L(x^2) + x f_1^R(x^2)
133% \end{equation}
134% \item P sends
135% \begin{equation}\tag{$B_1$}
136% f_2(x) = f_1^L(x) + \alpha_1 f_1^R(x)
137% \end{equation}
138% to V.
139% \item Keep repeating the process, eg. let
140% \begin{equation}\tag{$A_2$}
141% f_2(x) = f_2^L(x^2) + x f_2^R(x^2)
142% \end{equation}
143% until $f_i^L,~ f_i^R$ are constant (degree 0 polynomials).
144% \item Once $f_i^L,~ f_i^R$ are constant, P sends them to V.
145% \end{enumerate}
146%
147% Notice that at each step, $deg(f_i)$ halves.
148
149
150\vspace{30px}
151
152\paragraph{Proof generation}
153
154\emph{(Commitment phase)}
155P starts from $f(x)$, and for $i=0$ sets $f_0(x)=f(x)$.
156\begin{enumerate}
157 \item $\forall~i \in \{0, log(d)\}$, with $d = deg~f(x)$,\\
158 P computes $f_i^L(x),~ f_i^R(x)$ for which
159 \begin{equation}\tag{eq. $A_i$}
160 f_i(x) = f_i^L(x^2) + x f_i^R(x^2)
161 \end{equation}
162 holds.
163 \item V sends challenge $\alpha_i \in \mathbb{F}$
164 \item P commits to the random linear combination $f_{i+1}$, for
165 \begin{equation}\tag{eq. $B_i$}
166 f_{i+1}(x) = f_i^L(x) + \alpha_i f_i^R(x)
167 \end{equation}
168 \item P sets $f_i(x) := f_{i+1}(x)$ and starts again the iteration.
169\end{enumerate}
170Notice that at each step, $deg(f_i)$ halves.
171
172This 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.
173
174\emph{(Query phase)}
175P 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 $i$.
176(Recall, "opening" means that would provide a proof (MerkleProof) of it).
177
178\paragraph{Data sent from P to V}
179\begin{itemize}
180 \item[] Commitments: $\{Comm(f_i)\}_0^{log(d)}$\\
181 {\scriptsize eg. $\{Comm(f_0),~ Comm(f_1),~ Comm(f_2),~ ...,~ Comm(f_{log(d)})\}$ }
182 \item[] Openings: $\{ f_i(z^{2^i}),~f_i(-(z^{2^i})) \}_0^{log(d)}$\\
183 for a challenge $z \in D$ set by V\\
184 {\scriptsize eg. $f_0(z),~ f_0(-z),~ f_1(z^2),~ f_1(-z^2),~ f_2(z^4),~ f_2(-z^4),~ f_3(z^8),~ f_3(-z^8),~ \ldots$}
185 \item[] Constant values of last iteration: $\{f_k^L,~f_k^R\}$, for $k=log(d)$
186\end{itemize}
187
188% \begin{figure}[htp]
189% \centering
190% \begin{footnotesize}
191% \begin{sequencediagram}
192% \newinst[0]{p}{Prover}
193% \newinst[5]{v}{Verifier}
194%
195% \mess{p}{$\{Comm(f_i)\}_0^{log(d)},~ \{f_i(z^{2^i}),~f_i(-(z^{2^i})) \}_0^{log(d)},~ \{f_k^L,~ f_k^R\}$}{v}
196%
197% \end{sequencediagram}
198% \end{footnotesize}
199% \caption[FRI-LDT]{sketch of the FRI-LDT flow}
200% \label{fig:fri-ldt}
201% \end{figure}
202
203
204\paragraph{Verification}
205
206V receives:
207\begin{align*}
208 \text{Commitments:}~ &Comm(f_i),~ \forall i \in \{0, log(d)\}\\
209 \text{Openings:}~ &\{o_i, o_i'\}=\{ f_i(z^{2^i}),~f_i(-(z^{2^i})) \},~ \forall i \in \{0, log(d)\}\\
210 \text{Constant vals:}~ &\{f_k^L,~f_k^R\}
211\end{align*}
212
213\vspace{20px}
214
215For all $i \in \{0, log(d)\}$, V knows the openings at $z^{2^i}$ and $-(z^{2^i})$ for\\
216$Comm(f_i(x))$, which are $o_i=f_i(z^{2^i})$ and $o_i'=f_i(-(z^{2^i}))$ respectively.
217
218V, from (eq. $A_i$), knows that
219$$f_i(x)=f_i^L(x^2) + x f_i^R(x^2)$$
220should hold, thus
221$$f_i(z)=f_i^L(z^2) + z f_i^R(z^2)$$
222where $f_i(z)$ is known, but $f_i^L(z^2),~f_i^R(z^2)$ are unknown.
223But, V also knows the value for $f_i(-z)$, which can be represented as
224$$f_i(-z)=f_i^L(z^2) - z f_i^R(z^2)$$
225(note that when replacing $x$ by $-z$, it loses the negative in the power, not in the linear combination).
226
227Thus, we have the system of independent linear equations
228\begin{align*} % TODO add braces on left
229 f_i(z)&=f_i^L(z^2) + z f_i^R(z^2)\\
230 f_i(-z)&=f_i^L(z^2) - z f_i^R(z^2)
231\end{align*}
232for which V will find the value of $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$.
233Equivalently it can be represented by
234$$
235\begin{pmatrix}
236 1 & z\\
237 1 & -z
238\end{pmatrix}
239\begin{pmatrix}
240 f_i^L(z^2)\\
241 f_i^R(z^2)
242\end{pmatrix}
243=
244\begin{pmatrix}
245 f_i(z)\\
246 f_i(-z)
247\end{pmatrix}
248$$
249where V will find the values of $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$ being
250\begin{align*}
251 f_i^L(z^{2^i})=\frac{f_i(z) + f_i(-z)}{2}\\
252 f_i^R(z^{2^i})=\frac{f_i(z) - f_i(-z)}{2z}\\
253\end{align*}
254
255Once, V has computed $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$, can use them to compute the linear combination of
256$$
257f_{i+1}(z^{2^i}) = f_i^L(z^{2^i}) + \alpha_i f_i^R(z^{2^i})
258$$
259obtaining then $f_{i+1}(z^{2^i})$. This comes from (eq. $B_i$).
260
261Now, 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.
262V checks also the commitment of $Comm(f_{i+1}(x))$ for the opening $o_{i+1}=f_{i+1}(z^{2^i})$.\\
263If the checks pass, V is convinced that $f_1(x)$ was committed honestly.
264
265Now, sets $i := i+1$ and starts a new iteration.
266
267For the last iteration, V checks that the obtained $f_i^L(z^{2^i}),~f_i^R(z^{2^i})$ are equal to the constant values $\{f_k^L,~f_k^R\}$ received from P.
268
269\vspace{10px}
270It needs $log(d)$ iterations, and the number of queries (commitments + openings sent and verified) needed is $2 \cdot log(d)$.
271
272\subsection{Parameters}
273
274P commits to $f_i$ restricted to a subfield $F_0 \subset \mathbb{F}$.
275Let $0<\rho<1$ be the \emph{rate} of the code, such that
276$$|F_0| = \rho^{-1} \cdot d$$
277
278\begin{theorem}
279 For $\delta \in (0, 1-\sqrt{\rho})$, we have that if V accepts, then w.v.h.p. (with very high probability) $\Delta(f_0,~ p^d) \leq \delta$.
280\end{theorem}
281
282\section{FRI as polynomial commitment scheme}
283This section overviews the trick from \cite{cryptoeprint:2019/1020} to convert FRI into a polynomial commitment.
284
285Want 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
286
287$$
288f(x)-f(r) = Q(x) \cdot (x-r)
289$$
290
291note that $f(x)-f(r)$ evaluated at $r$ is $0$, so $(x-r) | (f(x)-f(r))$, in other words
292$(f(x)-f(r))$ is a multiple of $(x-r)$ for a polynomial $Q(x)$.
293
294Let us define $g(x) = \frac{f(x)-f(r)}{x-r}$.
295
296Prover uses FRI-LDT \ref{sec:fri-ldt} to commit to $g(x)$, and then prove w.v.h.p that $deg(g) \leq d-1$ ($\Longleftrightarrow \Delta(g,~ p^{d-1} \leq \delta$).
297
298Prover was already proving that $deg(f) \leq d$.
299
300Now, the missing thing to prove is that $g(x)$ has the right shape. We can relate $g$ to $f$ as follows:
301V does the normal FRI-LDT, but in addition, at the first iteration:
302V has $f(z)$ and $g(z)$ openings, so can verify
303$$g(z) = (f(z)-f(r))\cdot (z-r)^{-1}$$
304
305
306\bibliography{paper-notes.bib}
307\bibliographystyle{unsrt}
308
309\end{document}