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 10
 11\begin{filecontents}[overwrite]{pinocchio-groth16.bib}
 12@misc{cryptoeprint:2013/279,
 13  author = {Bryan Parno and Craig Gentry and Jon Howell and Mariana Raykova},
 14  title = {Pinocchio: Nearly Practical Verifiable Computation},
 15  howpublished = {Cryptology {ePrint} Archive, Paper 2013/279},
 16  year = {2013},
 17  url = {https://eprint.iacr.org/2013/279}
 18}
 19@misc{cryptoeprint:2016/260,
 20  author = {Jens Groth},
 21  title = {On the Size of Pairing-based Non-interactive Arguments},
 22  howpublished = {Cryptology {ePrint} Archive, Paper 2016/260},
 23  year = {2016},
 24  url = {https://eprint.iacr.org/2016/260}
 25}
 26
 27@misc{pairings,
 28  note = {\url{https://github.com/arnaucube/math/blob/master/weil-pairing.pdf}},
 29  url = {https://github.com/arnaucube/math/blob/master/weil-pairing.pdf}
 30}
 31\end{filecontents}
 32\nocite{*} % TODO review
 33
 34
 35\newtheorem{thm}{Theorem}[section]
 36\newtheorem{lemma}[thm]{Lemma}
 37\newtheorem{prop}[thm]{Proposition}
 38\newtheorem{cor}[thm]{Corollary}
 39\newtheorem{defn}[thm]{Definition}
 40
 41
 42\title{Pinocchio and Groth16 notes}
 43\author{arnaucube}
 44\date{2026}
 45
 46\begin{document}
 47
 48\maketitle
 49
 50\begin{abstract}
 51    Notes taken while studying the schemes Pinocchio (\cite{2013/279}) and Groth16 (\cite{2016/260}).
 52
 53    Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
 54
 55    The proofs may slightly differ from the ones from the books or papers, since I try to extend them for a deeper understanding.
 56\end{abstract}
 57
 58\tableofcontents
 59\newpage
 60
 61
 62\section{Introduction}
 63
 64% TODO.
 65% what we want to prove, how it connects with R1CS & QAP
 66
 67
 68\section{R1CS}
 69% TODO Add R_{R1CS} relation definition.
 70
 71Rank-One Constraint System (R1CS) is the representation of the constraints of the circuit by 3 linear combinations, where there is one constraint per operation.
 72
 73$$
 74\begin{aligned}
 75(a_{11}w_1 + a_{12}w_2 + ... + a_{1n}w_n) \cdot (b_{11}w_1 + b_{12}w_2 + ... + b_{1n}w_n)\\- (c_{11}w_1 + c_{12}w_2 + ... + c_{1n}w_n) = 0
 76\\
 77(a_{21}w_1 + a_{22}w_2 + ... + a_{2n}w_n) \cdot (b_{21}w_1 + b_{22}w_2 + ... + b_{2n}w_n)\\- (c_{21}w_1 + c_{22}w_2 + ... + c_{2n}w_n) = 0
 78\\
 79(a_{31}w_1 + a_{32}w_2 + ... + a_{3n}w_n) \cdot (b_{31}w_1 + b_{32}w_2 + ... + b_{3n}w_n)\\- (c_{31}w_1 + c_{32}w_2 + ... + c_{3n}w_n) = 0
 80\\
 81\ldots
 82\\
 83(a_{m1}w_1 + a_{m2}w_2 + ... + a_{mn}w_n) \cdot (b_{m1}w_1 + b_{m2}w_2 + ... + b_{mn}w_n)\\- (c_{m1}w_1 + c_{m2}w_2 + ... + c_{mn}w_n) = 0
 84\end{aligned}
 85$$
 86
 87Where the R1CS linear combinations can be represented by three matrices $A$, $B$ and $C$:
 88$$
 89\begin{aligned}
 90A =
 91\begin{pmatrix}
 92a_{11} & a_{12} & ... & a_{1n}\\
 93a_{21} & a_{22} & ... & a_{2n}\\
 94a_{31} & a_{32} & ... & a_{3n}\\
 95...\\
 96a_{m1} & a_{m2} & ... & a_{mn}
 97\end{pmatrix}
 98B =
 99\begin{pmatrix}
100b_{11} & b_{12} & ... & b_{1n}\\
101b_{21} & b_{22} & ... & b_{2n}\\
102b_{31} & b_{32} & ... & b_{3n}\\
103...\\
104b_{m1} & b_{m2} & ... & b_{mn}
105\end{pmatrix}
106\\
107C =
108\begin{pmatrix}
109c_{11} & c_{12} & ... & c_{1n}\\
110c_{21} & c_{22} & ... & c_{2n}\\
111c_{31} & c_{32} & ... & c_{3n}\\
112...\\
113c_{m1} & c_{m2} & ... & c_{mn}
114\end{pmatrix}
115\end{aligned}
116$$
117
118Where
119$(A, B, C) = A w \circ B w - C w = 0$
120
121The process that converts the \emph{flat constraints} into \emph{R1CS} is the following:
122
123Lets assume that we have a constraint $5 w_3 = w_1 \cdot 2 w_2$ (being $w_1$, $w_2$ and $w_3$ the signals 1, 2 and 3 respectively)). We can represent this constraint in the matrices form by:
124
125$$
126\begin{aligned}
127A =
128\begin{pmatrix}
1291 & 0 & 0 & ... & 0\\
130\end{pmatrix}
131\\
132B =
133\begin{pmatrix}
1340 & 2 & 0 & ... & 0\\
135\end{pmatrix}
136\\
137C =
138\begin{pmatrix}
1390 & 0 & 5 & ... & 0\\
140\end{pmatrix}
141\end{aligned}
142$$
143
144
145Which in the R1CS equation form would look as:
146$$
147\begin{aligned}
148(1 \cdot w_1 + 0 \cdot w_2 + 0 \cdot w_3 + ... + 0 \cdot w_n) \cdot
149(0 \cdot w_1 + 2 \cdot w_2 + 0 \cdot w_3 + ... + 0 \cdot w_n) -\\
150(0 \cdot w_1 + 0 \cdot w_2 + 5 \cdot w_3 + ... + 0 \cdot w_n) = 0
151\end{aligned}
152$$
153
154If we simplify the previous liniar combination we get:
155$$
156\begin{aligned}
157(1 \cdot w_1) \cdot (2 \cdot w_2) - (5 \cdot w_3) = 0
158\\
159w_1 \cdot 2w_1 - 5 w_3 = 0
160\\
1615 w_3 = w_1 \cdot 2w_1
162\end{aligned}
163$$
164Where $5 w_3 = w_1 \cdot 2w_1$ is the constraint that we wanted to represent in the R1CS format.
165
166The nice thing from R1CS is that allows us to standarize the constraints in an equation fashion, where all the signals are involved and we 'activate' these ones that are used in each constraint.
167Since we have now the constraints in a polynomial format, we can do work with it in the polynomial way.
168
169
170\section{QAP (Quadratic Arithmetic Programs)} \label{qap}
171
172Once we have the constraints of a circuit represented in \emph{R1CS}, we can do one more iteration and represent it in \emph{Quadratic Arithmetic Programs (QAP)}.
173%% TODO %%
174\emph{QAP} is a representation of the XXX in a linear combination of the \emph{R1CS}, represented by 3 polynomials $a(X)$, $b(X)$ and $c(X)$.
175The \emph{QAP} consists of the 3 polynomials $a(X)$, $b(X)$ and $c(X)$, plus a divisor polynomial $h(X) \in \mathbb{F}[X]^{<d}$.
176
177The main idea behind the \emph{QAP} is that they 'contain' all the constraints of the circuit, meaning that these polynomials need to be able to generate all the constraints.
178
179In order to 'collapse' the \emph{R1CS} polynomials into the 3 polynomials of the \emph{QAP}, we make use of \emph{polynomial interpolation}, also named \emph{Lagrange interpolation}. \emph{Lagrange interpolation} allows for a group of points, to find the smallest degree and unique polynomial that goes through all those points. It can be calculated by computing:
180
181$$L(X) = \sum_{j=0}^{n} y_j l_j(X)$$
182
183where $l_j(X)$ is:
184$$
185l_j(X) = \prod_{0\leq m \leq k ; m\neq j}
186\frac{X-x_m}{x_j - x_m}
187...
188\frac{(X-x_{j-1})}{(x_j - x_{j-1})}
189\frac{(X-x_{j+1})}{(x_j - x_{j+1})}
190...
191\frac{X-x_k}{x_j - x_k}
192$$
193
194The intuition behind the \emph{Lagrange interpolation} is quite simple. If we have two points (on a 2 dimensional plane) we can find a line that goes through both points (this can be done with a pen and a ruler on a paper), if we have 3 points we can find a curved line that goes through all the three points, if we do this through math trying to get the lowest degree curve we obtain a  2nd degree polynomial. This can be generalized using the *Lagrange interpolation*, allowing us for a group of $n$ points to find a $n-1$ degree polynomial that goes through all of them.
195
196In our case, \emph{Lagrange interpolation} is used to go from the \emph{R1CS} polynomials into the \emph{QAP} polynomials, which are defined by:
197
198$$
199\begin{aligned}
200    a(X) = (\alpha_1(X)w_1 + \alpha_2(X)w_2 + ... + \alpha_n(X)w_n) = \sum_{i=1}^n \alpha_i(X) w_i\\
201    b(X) = (\beta_1(X)w_1 + \beta_2(X)w_2 + ... + \beta_n(X)w_n) = \sum_{i=1}^n \beta_i(X) w_i\\
202    c(X) = (\gamma_1(X)w_1 + \gamma_2(X)w_2 + ... + \gamma_n(X)w_n) = \sum_{i=1}^n \gamma_i(X) w_i
203\end{aligned}
204$$
205
206The polynomials $a(X), b(X), c(X)$ are the QAP, such that
207$$a(X) \cdot b(X) = c(X)$$
208so that
209$$p(X) = a(X) \cdot b(X) - c(X) = 0$$
210
211that is,
212$$
213\begin{aligned}
214p(X) = (\alpha_1(X)w_1 + \alpha_2(X)w_2 + ... + \alpha_n(X)w_n)
215\\ \cdot (\beta_1(X)w_1 + \beta_2(X)w_2 + ... + \beta_n(X)w_n)
216\\ - (\gamma_1(X)w_1 + \gamma_2(X)w_2 + ... + \gamma_n(X)w_n)
217\\
218= (\sum_{i=1}^n \alpha_i(X) w_i) \cdot (\sum_{i=1}^n \beta_i(X) w_i) - (\sum_{i=1}^n \gamma_i(X) w_i)
219\end{aligned}
220$$
221
222The main idea, is that $p(X)$ defines the constraints of our circuit, so that for a valid witness evaluated at $p(X)$, it will be equal to zero.
223
224
225\vspace{0.5cm}
226
227
228\subsection{Participant contribution}
229In the context of SRS (Structured Reference String), the powers of tau consist of an array containing the scalar multiplication of the generator point by each power of tau, eg. $\tau^0 \cdot G,~\tau^1 \cdot G,~\tau^2 \cdot G,~\tau^3 \cdot G \ldots$
230
231For our case, we'll generate $n$ powers of tau on $\mathbb{G}_1$, and $m$ on $\mathbb{G}_2$. We will denote the combination of both $\mathbb{G}_1$ and $\mathbb{G}_2$ powers of tau by SRS.
232
233Each participant will obtain the previous-participant generated SRS, which is formed by
234
235\begin{align*}
236        \{ [\tau^0]_1, [\tau^1]_1, [\tau^2]_2, \ldots, [\tau^{n-1}]_1 \},\\
237        \{ [\tau^0]_2, [\tau^1]_2, [\tau^2]_2, \ldots, [\tau^{m-1}]_2 \}
238\end{align*}
239
240for simplicity, we will represent it as
241
242$$\{~[~\tau^i]_1~\}^{n-1}_{i=0},~\{~[~\tau^i]_2~\}^{m-1}_{j=0}$$
243
244
245Each participant generates their secret random tau $\tau_p$, which to avoid confusion we will denote as $p \in \mathbb{F}_r$. From it, the participant can compute $[p]_2 \in \mathbb{G}_2$.
246
247Then the participant proceeds to update the reference string (SRS), by multiplying each element by $p^i$:
248$$\{~ p^i \cdot [~ \tau^i]_1~\}^{n-1}_{i=0},~\{~ p^i \cdot [~\tau^i]_2~\}^{m-1}_{j=0}$$
249which, by the properties of scalar multiplication of a point, is equivalent to
250$$\{~ [~ (p \cdot \tau)^i]_1~\}^{n-1}_{i=0},~\{~ [~ (p \cdot \tau)^i]_2~\}^{m-1}_{j=0}$$
251and we denote $p \cdot \tau$ as $\tau'$, so we can write the previous expression as
252$$\{~[~\tau'^i]_1~\}^{n-1}_{i=0},~\{~[~\tau'^i]_2~\}^{m-1}_{j=0}$$
253
254which is
255\begin{align*}
256        \{ [(\tau \cdot p)^0]_1, [(\tau \cdot p)^1]_1, [(\tau \cdot p)^2]_2, \ldots, [(\tau \cdot p)^{n-1}]_1 \},\\
257        \{ [(\tau \cdot p)^0]_2, [(\tau \cdot p)^1]_2, [(\tau \cdot p)^2]_2, \ldots, [(\tau \cdot p)^{m-1}]_2 \}
258\end{align*}
259
260Notice that the contributor does not know the previous $\tau$, but can obtain $(\tau \cdot p)^i$.
261
262
263\subsection{Proof of correct computation}
264The participant will compute the proof, which consists of
265$$\pi = ([\tau']_1,~[p]_2)$$
266which is equivalent to $\pi = (\tau' \cdot G,~p \cdot H) = ((p \cdot \tau) \cdot G,~p \cdot H)$
267
268In the next section we will see how this is used to verify the contribution correct computation.
269
270
271\subsection{Contribution verification}
272This is the interesting part. We want to check that the new SRS is well computed from the previous SRS together with the new tau. We want to avoid accepting contributions that do not follow the powers of tau structure, or that contain empty elements or that do not derive from the previous SRS.
273
274Following the powers of tau structure means that
275$$\tau'^{i+1} = \tau' \cdot \tau'^i ~~\forall~i \in [1, n-1]$$
276
277And the relation between the new SRS and the previous SRS is
278$$\tau' = \tau \cdot p$$
279where $\tau$ is the secret element from the previous SRS and $p$ is the secret one from the new SRS, resulting in the combined $\tau'$.
280
281Now, we're interested into how we can check these relations in the context of the SRS elements. A new contribution can be verified as follows:
282
283\begin{enumerate}
284        \item Check that elements of the new SRS are non-empty, non-zero and in the correct prime order subgroups.
285\item Get the proof ($\pi = ([\tau']_1, [p]_2)$), and confirm that the 1st element is equal to the element in position 1 $[\tau'^1]_1$ of the new SRS (which corresponds to $\tau'$ powered to $1$).
286
287\item Check that $\tau'$ (the new SRS) is correctly related to $\tau$ (the previous SRS).
288
289    To check this, we rely on the pairing properties and we use the proof value generated by the contributor, which consists of $[p]_2$
290    $$e([\tau]_1, [p]_2) \stackrel{?}{=} e([\tau']_1, [1]_2)$$
291    
292    We can see how this holds, as $\tau' = p \cdot \tau$, and by the bilinear property of the pairing (see \cite{pairings}) we have $e(a \cdot G, b \cdot H) = e((a \cdot b) \cdot G, H)$, which in our notation is represented as
293    $$e([a]_1, [b]_2) = e([a \cdot b]_1, [1]_2)$$
294    by applying this to our case we can see that:
295    $$e([\tau]_1, [p]_2) = e([\tau \cdot p]_1, [1]_2) = e([\tau']_1, [1]_2)$$
296
297\item Check if the new SRS follows the powers of tau structure:
298    $$e([\tau'^i]_1, [\tau']_2) \stackrel{?}{=} e([\tau'^{i+1}]_1, [1]_2) ~~ \forall ~i \in [1, n-1]$$
299    $$e([\tau']_1, [\tau'^j]_2) \stackrel{?}{=} e([1]_1, [\tau'^{j+1}]_2) ~~ \forall ~i \in [1, m-1]$$
300    
301    Which, again, by the bilinarity property, we can see that
302    $$e([\tau'^i]_1, [\tau']_2) = e([\tau'^{i+1}]_1, [1]_2)$$
303    is enforcing that $\tau'^i \cdot \tau'^1 = \tau'^{i+1}$.
304    
305    And the same for
306    $$e([\tau']_1, [\tau'^j]_2) = e([1]_1, [\tau'^{j+1}]_2)$$
307    which checks that $\tau'^1 \cdot \tau'^j = \tau'^{j+1}$.
308    
309    With this check, we're ensuring that each power of tau of the SRS is consistent with the previous one.
310\end{enumerate}
311
312
313% TODO
314% - uniformize x & X at polynomials notation (currently using p(x) and p(X), uniformize it)
315
316
317\bibliographystyle{unsrt}
318\bibliography{pinocchio-groth16.bib}
319
320\end{document}