arnaucube
arnaucube/arnaucube.comPublic Self-hosted repository
arnaucube.com/blog/protogalaxy.html
blog/protogalaxy.htmlRaw
  1<!DOCTYPE html>
  2<html lang="en">
  3
  4  <head>
  5    <meta name="description" content="The idea of these notes is to provide some extra intuition and explainations on ProtoGalaxy's paper. ProtoGalaxy is a folding scheme which iterates on ideas from ProtoStar paper." />
  6    <meta charset="utf-8">
  7    <title> Notes on ProtoGalaxy - arnaucube - blog</title>
  8    <meta name="title" content=" Notes on ProtoGalaxy - arnaucube - blog">
  9    <meta name="description" content="The idea of these notes is to provide some extra intuition and explainations on ProtoGalaxy's paper. ProtoGalaxy is a folding scheme which iterates on ideas from ProtoStar paper.">
 10
 11    <meta property="og:title" content=" Notes on ProtoGalaxy - arnaucube - blog" />
 12    <meta property="og:description" content="The idea of these notes is to provide some extra intuition and explainations on ProtoGalaxy's paper. ProtoGalaxy is a folding scheme which iterates on ideas from ProtoStar paper." />
 13    <meta property="og:url" content="https://arnaucube.com/blog/protogalaxy.html" />
 14    <meta property="og:type" content="article" />
 15    <meta property="og:image" content="https://arnaucube.com/blog/img/posts/protogalaxy/protogalaxies-colliding.jpg" />
 16    <meta name="twitter:title" content=" Notes on ProtoGalaxy - arnaucube - blog">
 17    <meta name="twitter:description" content="The idea of these notes is to provide some extra intuition and explainations on ProtoGalaxy's paper. ProtoGalaxy is a folding scheme which iterates on ideas from ProtoStar paper.">
 18    <meta name="twitter:image" content="https://arnaucube.com/blog/img/posts/protogalaxy/protogalaxies-colliding.jpg">
 19    <meta name="twitter:card" content="summary_large_image">
 20    <meta name="author" content="arnaucube">
 21
 22    <link rel="icon" type="image/png" href="img/logoArnauCubeFavicon.png">
 23
 24    <meta name="viewport" content="width=device-width, initial-scale=1">
 25
 26    <link href="css/bootstrap.min.css" rel="stylesheet">
 27    <link rel="stylesheet" href="css/style.css">
 28
 29    <!-- highlightjs -->
 30    <!-- <link rel="stylesheet" href="js/highlightjs/atom-one-dark.css"> -->
 31    <link rel="stylesheet" href="js/highlightjs/atom-one-light.css">
 32    <!-- <link rel="stylesheet" href="js/highlightjs/gruvbox-dark.css"> -->
 33    <script src="js/highlightjs/highlight.pack.js"></script>
 34
 35    <!-- katex -->
 36    <link rel="stylesheet" href="js/katex/katex.min.css">
 37  </head>
 38
 39  <body>
 40
 41    <!-- o_gradient_background" -->
 42    <nav id="mainNav" class="navbar navbar-default navbar-fixed-top"
 43                      style="height:50px;font-size:100%;">
 44      <div class="container">
 45        <div style="float:left;">
 46        <a href="/blog" style="color:#000;display:inline-block;">blog</a>
 47        <span style="margin-right:20px; margin-left:20px;">|</span>
 48        <a href="/blog/notes.html" style="font-size:90%;color:#000;display:inline-block;">notes</a>
 49        </div>
 50        <div style="float:right;">
 51          <a href="/" style="color:#000;display:inline-block;margin-right:10px;">arnaucube.com</a>
 52          <div class="onoffswitch" style="margin-left:0px;display:inline-block;" title="change theme">
 53            <input onclick="switchTheme()" type="checkbox" name="onoffswitch" class="onoffswitch-checkbox"
 54              id="themeSwitcher">
 55            <label class="onoffswitch-label" for="themeSwitcher"></label>
 56          </div>
 57        </div>
 58      </div>
 59      <!-- <img style="height:5px; width:100%; margin-top:8px;" src="img/gradient-line.jpg" /> -->
 60    </nav>
 61
 62
 63
 64    <div class="container" style="margin-top:0px;max-width:800px;">
 65      <h1>Notes on ProtoGalaxy</h1>
 66
 67<p><em>2023-08-20</em></p>
 68
 69<blockquote>
 70<p><strong>Note</strong>: I&rsquo;m not a mathematician, just a math enthusiast studying in my free time, and this article is just an attempt to try to sort the notes that I took while studying the ProtoGalaxy scheme.</p>
 71</blockquote>
 72
 73<p>The idea of these notes is to provide some extra intuition on <a href="https://eprint.iacr.org/2023/1106">ProtoGalaxy paper</a>.
 74The paper is very well written and exposes the ideas very clearly, so it&rsquo;s worth to check directly the paper instead of these notes.</p>
 75
 76<p>I make use of <span class="math inline">\(\textcolor{blue}{c}\textcolor{magenta}{o}\textcolor{orange}{l}\textcolor{cyan}{o}\textcolor{green}{u}\textcolor{red}{r}\textcolor{pink}{s}\)</span> to try to relate the different parameters across the scheme.</p>
 77
 78<p>ProtoGalaxy is a <em>folding scheme</em> which iterates on ideas from <a href="https://eprint.iacr.org/2023/620">ProtoStar paper</a> (here you can find <a href="https://geometry.xyz/notebook/paper-speedrun-protostar">Geometry&rsquo;s post</a> overviewing its main results and techniques).</p>
 79
 80<p>For an introduction to <em>folding schemes</em> I highly recommend <a href="https://youtu.be/IzLTpKWt-yg?t=6367">this talk (at 1:46)</a> by <a href="https://twitter.com/CPerezz19">Carlos Pérez</a>.</p>
 81
 82<p>I would like to thank <a href="https://twitter.com/LiamEagen">Liam Eagen</a> and <a href="https://twitter.com/rel_zeta_tech">Ariel Gabizon</a> for their kind explanations on the scheme.
 83Also thanks to <a href="https://twitter.com/kiliconu">Onur</a> for comments and corrections.</p>
 84
 85<div style="margin-left:20%;" >
 86
 87<img src="img/posts/protogalaxy/protogalaxies-colliding.jpg"/>
 88
 89<br>
 90(img: protogalaxies colliding, <a href="https://en.wikipedia.org/wiki/File:Stellar_Fireworks_Finale.jpg">from Wikipedia</a>)
 91
 92</div>
 93
 94<h2>Preliminaries</h2>
 95
 96<p>Before jumping into the protocol itself, let&rsquo;s overview some building blocks first.</p>
 97
 98<h4><span class="math inline">\(pow_i\)</span></h4>
 99
100<p>In the protocol we need the powers of a random value. We use the <span class="math inline">\(pow_i\)</span> function to store and operate only <span class="math inline">\(log(n)\)</span> entries instead of <span class="math inline">\(n\)</span>.</p>
101
102<p><span class="math inline">\(pow_i\)</span> is a <span class="math inline">\(t\)</span>-variate polynomial such that</p>
103<p><span class="math display">\[
104% pow_i(X_1, \ldots, X_t) = \prod_{i \in [n]} ((1-b_j) + b_j \cdot X^{2^j})
105pow_i(X) = \prod_{i \in [n]} ((1-b_j) + b_j \cdot X^{2^j})
106\]</span></p><p>where <span class="math inline">\(b_j\)</span> is the binary decomposition of <span class="math inline">\(n\)</span>, such that <span class="math inline">\(n = \sum 2^j\)</span>.</p>
107
108<p>In our case, we will have <span class="math inline">\(\overrightarrow{X}\)</span> being <span class="math inline">\(\overrightarrow{\beta} = \{\beta^{2^i} \}_0^{t-1} = (\beta, \beta^2, \beta^4, \ldots, \beta^{2^{t-1}})\)</span>, thus</p>
109<p><span class="math display">\[
110pow_i(\overrightarrow{\beta}) = \prod_{j \in [n]} ((1-b_j) + b_j \cdot \beta_j)
111\]</span></p><p>which, basically is a way to have <span class="math inline">\(pow_i(\overrightarrow{\beta}) = \beta^i\)</span>, but storing just <span class="math inline">\(t\)</span> betas, to actually represent <span class="math inline">\(2^t\)</span> of them, so we work with <span class="math inline">\(log\)</span> of the total amount actually used.</p>
112
113<p>So, for example,</p>
114<p><span class="math display">\[
115    i=2_d=10_b;~
116    pow_2(\overrightarrow{\beta}) =
117    \underbrace{((1-0) + 0 \cdot \beta)}_{1}
118    \cdot \underbrace{((1-1) + 1 \cdot \beta^2)}_{\beta^2}
119    = \beta^2
120    \\
121    i=3_d=11_b;~
122    pow_3(\overrightarrow{\beta}) =
123    \underbrace{((1-1) + 1 \cdot \beta)}_{\beta}
124    \cdot \underbrace{((1-1) + 1 \cdot \beta^2)}_{\beta^2}
125    = \beta^3
126    \\
127    i=4_d=100_b;~
128    pow_4(\overrightarrow{\beta}) =
129    \underbrace{((1-0) + 0 \cdot \beta)}_{1}
130    \cdot \underbrace{((1-0) + 0 \cdot \beta^2)}_{1}
131    \cdot \underbrace{((1-1) + 1 \cdot \beta^4)}_{\beta^4}
132    = \beta^4
133    \\
134    i=5_d=101_b;~
135    pow_5(\overrightarrow{\beta}) =
136    \underbrace{((1-1) + 1 \cdot \beta)}_{\beta}
137    \cdot \underbrace{((1-0) + 0 \cdot \beta^2)}_{1}
138    \cdot \underbrace{((1-1) + 1 \cdot \beta^4)}_{\beta^4}
139    = \beta^5
140\]</span></p>
141<h4>Lagrange bases</h4>
142
143<p>Let <span class="math inline">\(\omega\)</span> denote a root of unity, such that <span class="math inline">\(\omega^N=1\)</span>, and set <span class="math inline">\(\mathbb{H}=\{1, \omega, \omega^2, \ldots, \omega^{N-1}\}\)</span>.</p>
144
145<p>Lagrange bases are a whole topic by itself, but for following the next sections, we just need to have a clear intuition on how they operate. The main idea is that</p>
146<p><span class="math display">\[L_j(\omega^i)=
147\begin{cases}
148	&amp;1 ~\text{if}~ i=j\\
149	&amp;0 ~\text{otherwise}
150\end{cases}
151\]</span></p><p>In other words, when we&rsquo;re evaluating the Lagrange base <span class="math inline">\(j\)</span> at <span class="math inline">\(w^i\)</span>, the evaluation is going to be <span class="math inline">\(1\)</span> when <span class="math inline">\(i=j\)</span>, and <span class="math inline">\(0\)</span> in all the other cases.</p>
152
153<p>To get a bit of more intuition here, let&rsquo;s represent this visually:</p>
154
155<div>
156
157<img src="img/posts/protogalaxy/lagrange-basis-intuition.png" style="width:40%;display:inline-block;" />
158
159<div style="width:50%; display:inline-block;">
160The intuition here is that when we evaluate $L_2(X)$ at $\omega_2=\omega^2$, it will give us $1$. In the blue equation of the image we're doing $L_2(w^2) \cdot v_k = 1 \cdot v_k$.
161
162While, when we evaluate $L_2(X)$ at for example $\omega^1$, we will obtain a $0$, thus the result will be a $0$.
163
164</div>
165</div>
166
167<p>Let the vanishing polynomial of <span class="math inline">\(\mathbb{H}\)</span> be <span class="math inline">\(Z(X)=\prod_{i=0}^{N-1} (X - \omega^i) = X^N -1\)</span>.</p>
168
169<h4>Function <span class="math inline">\(f(X)\)</span></h4>
170
171<p>Assume we have a function <span class="math inline">\(f(X)\)</span> which when evaluated at a valid witness is equal to 0.</p>
172
173<p>For example, in the R1CS case, this would be (for given matrices <span class="math inline">\(A, B, C \in \mathbb{F}^{m \times m}\)</span> representing our constraints),</p>
174
175<!--$$f(X)= \sum_{i=0}^n A(X_i) \cdot B(X_i) - C(X_i)$$-->
176<p><span class="math display">\[f(X)= A \cdot X \circ B \cdot X - C \cdot X \]</span></p>
177<!-- $$f(X)= \sum_{i=0}^n A(X_i) \cdot B(X_i) - C(X_i)$$ -->
178
179<p>where <span class="math inline">\(f(w)=0^n\)</span> if satisfied for a valid witness <span class="math inline">\(w \in \mathbb{F}^n\)</span>.</p>
180
181<p>Our goal will be to prove that we have folded various instantiations of valid witnesses and that they fulfill the check of <span class="math inline">\(f(w)=0^n\)</span>.</p>
182
183<h4>Lemma 4.2</h4>
184
185<p>The following lemma is from the ProtoGalaxy paper:</p>
186
187<blockquote>
188<p><strong>Lemma 4.2:</strong> Fix any polynomial <span class="math inline">\(f(X) \in \mathbb{F}[X]\)</span> and <span class="math inline">\(a_0, \ldots, a_k \in \mathbb{F}\)</span>. There exists <span class="math inline">\(Q(X) \in \mathbb{F}[X]\)</span> such that</p>
189<p><span class="math display">\[
190	f \left( \sum_{i=0}^k a_i L_i(X) \right) = \sum_{i=0}^k f(a_i) L_i(X) + Z(X) Q(X)
191\]</span></p></blockquote>
192
193<p>The way to check that the lemma is true for me was to implement it with code and check that it is satisfied. This is not a proper way, so luckily later <a href="https://hecmas.github.io">Héctor Masip</a> showed me an actual proof of this lemma, which goes as follows:</p>
194
195<p>Recall from the <a href="https://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor#Euclidean_division">Euclidean polynomial division</a>:</p>
196
197<blockquote>
198<p>For <span class="math inline">\(f(X), g(X) \in \mathbb{F}[X]\)</span> with <span class="math inline">\(\deg f \geq \deg g\)</span>, <span class="math inline">\(\exists\)</span> unique polynomials <span class="math inline">\(q(X), r(X) \in \mathbb{F}[X]\)</span> such that <span class="math inline">\(f(X) = g(X) q(X) + r(X)\)</span>, with <span class="math inline">\(0 \leq \deg r &lt; \deg g\)</span>.</p>
199</blockquote>
200
201<p>Thus,</p>
202<p><span class="math display">\[f(\sum_{i=0}^k a_i \cdot L_i(X)) = Q(X) \cdot Z(X) + r(X)\]</span></p><p>with <span class="math inline">\(0 \leq \deg r &lt; \deg Z = k+1\)</span>.</p>
203
204<p>So, when evaluating at <span class="math inline">\(a_j, ~\forall j=0, \ldots, k\)</span>,</p>
205<p><span class="math display">\[f(\sum_{i=0}^k a_i \cdot L_i(a_j)) = f(a_j) = \underbrace{Q(a_j) \cdot Z(a_j)}_{0} + r(a_j)\]</span></p><p>so <span class="math inline">\(f(a_j)=r(a_j)\)</span>, therefore</p>
206<p><span class="math display">\[r(X) = \sum_{i=0}^k r(a_i) \cdot L_i(X) = \sum_{i=0}^k f(a_i) \cdot L_i(X)\]</span></p>
207<div style="float:right;">
208
209$\square$
210
211</div>
212
213<h2>ProtoGalaxy protocol</h2>
214
215<p>The main idea of this scheme, is to be able to fold <span class="math inline">\(k+1\)</span> instances that satisfy the relation, producing a single <em>folded instance</em> which still satisfies the relation.</p>
216
217<p><img src="img/posts/protogalaxy/folding-diagram.png" style="width:80%;" /></p>
218
219<h4>Instances</h4>
220
221<p>Let <span class="math inline">\((\phi; w)\)</span> be our running instance, the witness is represented by <span class="math inline">\(w\)</span>, and <span class="math inline">\(\phi = cm(w)\)</span> represents the committed witness.
222Additionally, we will have <span class="math inline">\(k\)</span> randomized instance pairs <span class="math inline">\(( \{ \phi_i \}^k; \{ w_i \}^k )\)</span>.</p>
223
224<p>The running instance is paired with two more parameters, so the complete running instance consists of <span class="math inline">\(((\phi, \overrightarrow{\beta}, e); w)\)</span>, where <span class="math inline">\(e\)</span> is the &lsquo;error&rsquo; term and <span class="math inline">\(\overrightarrow{\beta} \in^R \mathbb{F}^t\)</span> (where <span class="math inline">\(t=\log(n)\)</span>).</p>
225
226<p>As we start folding instances, the folded instance must satisfy the relation</p>
227<p><span class="math display">\[
228\sum_{i \in [n]} pow_i(\overrightarrow{\beta}) f_i(w) = e
229\]</span></p><p>where <span class="math inline">\(pow_i\)</span> is the function explained in the <a href="#preliminaries">preliminaries section</a>, and <span class="math inline">\(f_i\)</span> is the <span class="math inline">\(i\)</span>-th value from evaluating <span class="math inline">\(f(w)\)</span>.</p>
230
231<p>For the first instance (before folding) <span class="math inline">\(e=0\)</span>, which we can see that matches <span class="math inline">\(\sum_{i \in [n]} pow_i(\overrightarrow{\beta}) f_i(w) = 0\)</span>, since <span class="math inline">\(f(w)=0^n\)</span> for a valid <span class="math inline">\(w\)</span>.</p>
232
233<h4>Protocol</h4>
234
235<p>We want a way for the Prover to fold <span class="math inline">\(k+1\)</span> instances into a single one, combining them while ensuring that they satisfy the relation by showing that <span class="math inline">\(f(w_i)=0\)</span> for all <span class="math inline">\(k+1\)</span> instances.</p>
236
237<p>For doing so, we could just add all the instances naively, but they could cancel each other giving <span class="math inline">\(0\)</span> when evaluated at the different <span class="math inline">\(w_i\)</span>, even when individually don&rsquo;t evaluate at <span class="math inline">\(0\)</span>.</p>
238
239<p>One way to combine the instances is through making use of a random linear combination (as done in Nova and ProtoStar), but when combining <span class="math inline">\(k\)</span> instances the degree of the cross-terms would grow exponentially. Thus ProtoGalaxy uses the Lagrange bases, by associating each instance to a different Lagrange coefficient, thus keeping the instances &lsquo;separated&rsquo; while being combined in the same expression.</p>
240
241<p><br>
242<em>(Let P denote the Prover, and V the Verifier)</em></p>
243
244<ol>
245<li><p>V sends a challenge <span class="math inline">\(\delta \in^R \mathbb{F}\)</span></p></li>
246
247<li><p>P, V compute <span class="math inline">\(\overrightarrow{\delta} = \{\delta^{2^i} \}_0^{t-1} = (\delta, \delta^2, \delta^4, \ldots, \delta^{2^{t-1}}) \in \mathbb{F}^t\)</span></p></li>
248
249<li><p>P computes the polynomial</p>
250<p><span class="math display">\[
251F(X) := \sum_{i \in [n]} pow_i(
252\underbrace{
253	\textcolor{orange}{\overrightarrow{\beta}} + X \cdot
254	\textcolor{orange}{\overrightarrow{\delta} }
255}_\text{represent it by $\textcolor{orange}{\overrightarrow{\eta}}$}
256) \textcolor{ForestGreen}{f_i(w)}
257\\
258\text{\scriptsize{where $\textcolor{orange}{\overrightarrow{\eta}}$ is just an intermediate}}
259\\
260\text{\scriptsize{representation of the new random vector over $X$}}
261\\
262\text{\scriptsize{where $\overrightarrow{\eta}=\{ \beta+X \delta, (\beta + X \delta)^2, (\beta +X \delta)^4, \ldots, (\beta + X \delta)^{2^{t-1}} \}$}}
263\\
264= \sum_{i \in [n]}
265	pow_i(\textcolor{orange}{\overrightarrow{\eta}})
266\textcolor{ForestGreen}{f_i(w)}
267\]</span></p><p>Notice that each <span class="math inline">\(pow_i(\textcolor{orange}{\overrightarrow{\eta}})\)</span> will be a polynomial of degree equal to the number of positive bits in the binary representation of <span class="math inline">\(i\)</span>.</p>
268
269<div style="font-size:85%;">
270
271
272In order to see this, let's simplify $\overrightarrow{\eta}$ and instead let's get take $\overrightarrow{\eta}$ getting rid of $\beta,~\delta$. Let $\overrightarrow{X} = \{X, X, X, \ldots, X \}$,
273recall that 
274
275
276$$
277pow_i(\overrightarrow{X}) = \prod_{j \in [n]} ((1-b_j) + b_j \cdot X_j)
278$$
279
280
281then we have
282
283
284$$
285i=2_d=10_b;~
286pow_2(\overrightarrow{X}) = 1 \cdot X = X;~ deg=1
287\\
288i=3_d=11_b;~
289pow_3(\overrightarrow{X}) = X \cdot X = X^2;~ deg=2
290\\
291i=4_d=100_b;~
292pow_4(\overrightarrow{X}) = 1 \cdot 1 \cdot X = X;~ deg=1
293\\
294i=5_d=101_b;~
295pow_5(\overrightarrow{X}) = X \cdot 1 \cdot X = X^2;~ deg=2
296\\
297i=6_d=110_b;~
298pow_6(\overrightarrow{X}) = 1 \cdot X \cdot X = X^2;~ deg=2
299\\
300i=7_d=111_b;~
301pow_7(\overrightarrow{X}) = X \cdot X \cdot X = X^3;~ deg=3
302\\
303i=8_d=1000_b;~
304pow_8(\overrightarrow{X}) = 1 \cdot 1 \cdot 1 \cdot X = X;~ deg=1
305$$
306
307
308</div>
309
310<p>So, for example <span class="math inline">\(pow_{255}\)</span>, since <span class="math inline">\(i=255_d = 11111111_b\)</span>, will have degree <span class="math inline">\(8\)</span>. But <span class="math inline">\(pow_{256}\)</span>, since <span class="math inline">\(i=256_d= 100000000_b\)</span>, will have degree <span class="math inline">\(1\)</span>.
311<br></p>
312
313<p>Notice also that when <span class="math inline">\(X=0\)</span>, <span class="math inline">\(\overrightarrow{\eta}=\overrightarrow{\beta}\)</span>, so we&rsquo;re left with the original equation <span class="math inline">\(F(0)=\sum_{i\in[n]} pow_i(\textcolor{orange}{\overrightarrow{\beta}}) \textcolor{ForestGreen}{f_i(w)} = e\)</span>.</p></li>
314
315<li><p>P sends non constant coefficients <span class="math inline">\(F_1, \ldots, F_t\)</span> of <span class="math inline">\(F(X)\)</span> to V</p></li>
316
317<li><p>V sends a challenge <span class="math inline">\(\textcolor{magenta}{\alpha} \in^R \mathbb{F}\)</span></p></li>
318
319<li><p>P, V compute <span class="math inline">\(F(\textcolor{magenta}{\alpha}) = e + \sum_{i \in [t]} \textcolor{ForestGreen}{F_i} \textcolor{magenta}{\alpha^i}\)</span></p></li>
320
321<li><p>P, V compute <span class="math inline">\(\textcolor{orange}{\overrightarrow{\beta}}\textcolor{magenta}{^*} \in \mathbb{F}^t\)</span> where <span class="math inline">\(\textcolor{orange}{\beta_i \textcolor{magenta}{^*}} := \textcolor{orange}{\beta_i} + \textcolor{magenta}{\alpha} \cdot \textcolor{orange}{\delta^{2^{i-1}}}\)</span></p>
322
323<p>Notice that <span class="math inline">\(\textcolor{orange}{\overrightarrow{\beta}}\textcolor{magenta}{^*} = \textcolor{orange}{\overrightarrow{\beta}} + \textcolor{magenta}{\alpha} \cdot \textcolor{orange}{\overrightarrow{\delta}}\)</span>, is the same as <span class="math inline">\(\textcolor{orange}{\overrightarrow{\eta}}\)</span> with <span class="math inline">\(X=\textcolor{magenta}{\alpha}\)</span>.</p>
324
325<p>We have now a new randomized instance <span class="math inline">\((\textcolor{orange}{\overrightarrow{\beta}}\textcolor{magenta}{^*}, w, F(\alpha))\)</span>, for which the relation from the equation (1) holds:</p>
326<p><span class="math display">\[F(\textcolor{magenta}{\alpha}) = \sum_{i \in [n]} pow_i(\textcolor{orange}{\overrightarrow{\beta}}\textcolor{magenta}{^*}) \textcolor{ForestGreen}{f(w)}\]</span></p><p>since both sides of the equality represent some <span class="math inline">\(\textcolor{magenta}{re-randomized}\)</span> (eg. <span class="math inline">\(\textcolor{magenta}{\alpha}\)</span>) <span class="math inline">\(\textcolor{orange}{random~value}\)</span> (eg. <span class="math inline">\(\textcolor{orange}{\overrightarrow{\beta}, ~\overrightarrow{\delta}}\)</span>) combined with the <span class="math inline">\(\textcolor{ForestGreen}{evaluated~function}\)</span> (eg. <span class="math inline">\(\textcolor{ForestGreen}{f(w)}\)</span>).</p>
327
328<p>The following image tries to show how the previous steps relate:
329<img src="img/posts/protogalaxy/steps-3-7.png" style="width:80%;" /></p></li>
330
331<li><p>P computes the polynomial</p>
332<p><span class="math display">\[
333G(X) := \sum_{i \in [n]}
334pow_i(\overrightarrow{\beta^*}) \cdot f_i(L_0(X) w + \sum_{j \in [k]} L_j(X) w_j)
335\]</span></p>
336<div style="font-size:85%;">Here we make use of the <a href="#lemma-4.2">Lemma 4.2</a> to see that:</div>
337<p><span class="math display">\[
338G(X) := \sum_{i \in [n]}
339pow_i(\overrightarrow{\beta^*}) \cdot
340\underbrace{ f_i(\textcolor{SkyBlue}{L_0(X)} w }_{\textcolor{SkyBlue}{L_0(X)} \cdot f_i(w)} + \underbrace{ \sum_{j \in [k]} \textcolor{SkyBlue}{L_j(X)} w_j) }_{0}
341\\
342\text{\small{ notice that $f_i(w_j)=0 ~\forall j \in [k], i \in [n]$, thus $\sum_{j \in [k]} \textcolor{SkyBlue}{L_j(X)} \cdot f_i(w_j) = 0$ }}
343\\
344= \sum_{i \in [n]}
345\underbrace{ pow_i(\overrightarrow{\beta^*}) \cdot f_i(w) }_{F(\alpha) ~\text{from step 6}} \cdot \textcolor{SkyBlue}{L_0(X)} \pmod{Z(X)}
346\\
347= F(\alpha) \cdot \textcolor{SkyBlue}{L_0(X)} \pmod{Z(X)}
348\]</span></p>
349<div style="font-size:85%;">(where $Z(X)$ is the vanishing polynomial from the <a href="#lagrange-bases">preliminaries section</a>)</div>
350
351<p>which leads to the next step.</p></li>
352
353<li><p>P computes the polynomial <span class="math inline">\(K(X)\)</span> such that</p>
354<p><span class="math display">\[G(X) = F(\alpha) L_0(X) + Z(X) K(X)\]</span></p><p>(Recall, <span class="math inline">\(F(\alpha)\)</span> <em>&lsquo;contains&rsquo;</em> the <em>&rsquo;re-randomized&rsquo;</em> expression comming from <span class="math inline">\(\textcolor{magenta}{\alpha}\)</span> combined with <span class="math inline">\(\textcolor{orange}{\overrightarrow{\beta},~ \overrightarrow{\delta}}\)</span>, combined with <span class="math inline">\(\textcolor{ForestGreen}{f(w)}\)</span> from step 7).</p>
355
356<p><br>
357What we&rsquo;re using here is the same trick that we can find in Plonk, KZG commitments, and other known protocols (which, in fact was proposed by <a href="https://sci-hub.hkvisa.net/10.1137/050646445">Eli Ben-Sasson &amp; Madhu Sudan in 2008</a>).
358We&rsquo;re using the fact that if <span class="math inline">\(P(X)/Z(X)\)</span> leads to a polynomial (and not to a rational function) means that <span class="math inline">\(P(X)\)</span> is divisible by <span class="math inline">\(Z(X)\)</span> (the <em>vanishing polynomial</em>), which means that <span class="math inline">\(P(X)\)</span> vanishes over the evaluation domain. From another perspective, there exists a polynomial <span class="math inline">\(K(X)\)</span> such that <span class="math inline">\(P(X) = Z(X) \cdot K(X)\)</span>.
359Here we&rsquo;re using this fact with <span class="math inline">\(K(X)\)</span>, where if exists a <span class="math inline">\(K(X)\)</span> such that <span class="math inline">\(G(X)- F(\alpha)L_0(X) = Z(X)K(X)\)</span>, means that <span class="math inline">\(G(X)-F(\alpha)L_0(X)\)</span> vanishes over the evaluation domain.</p></li>
360
361<li><p>P sends the coefficients <span class="math inline">\(K_1, \ldots, K_t\)</span> of <span class="math inline">\(K(X)\)</span> to V</p></li>
362
363<li><p>V sends a challenge <span class="math inline">\(\gamma \in^R \mathbb{F}\)</span></p></li>
364
365<li><p>P, V compute</p>
366<p><span class="math display">\[e^* := F(\alpha) L_0(\gamma) + Z(\gamma) K(\gamma)\]</span></p></li>
367
368<li><p>P, V outputs the new folded committed instance</p>
369<p><span class="math display">\[\Phi^* = (\phi^*, \overrightarrow{\beta^*}, e^*)\]</span></p><p>where</p>
370<p><span class="math display">\[\phi^* = L_0(\gamma) \phi + \sum_{i \in [k]} L_i(\gamma) \cdot \phi_i\]</span></p><p>Recall from <a href="#lagrange-bases"><em>Lagrange bases</em> section</a>, that the intuition behind Lagrange bases usage is to <em>&lsquo;select&rsquo;</em> the different values associated to each <span class="math inline">\(L_i\)</span>. The idea here is that we are combinning the different instances into a new instance, while keeping them &lsquo;separated&rsquo; by the different <span class="math inline">\(L_i\)</span>.</p></li>
371
372<li><p>Additionally, P outputs the folded witness</p>
373<p><span class="math display">\[w^* = L_0(\gamma) w + \sum_{i \in [k]} L_i(\gamma) \cdot w_i\]</span></p></li>
374</ol>
375
376<p>To check that <span class="math inline">\((\Phi^*; w^*)\)</span> satisfies the relation we check that</p>
377<p><span class="math display">\[\sum_{i \in [n]} pow_i(\overrightarrow{\beta^*}) \cdot f_i(w^*) = e^*\]</span></p><p>As usual, this interactive protocol is converted into a non-interactive one through the Fiat-Shamir transformation.</p>
378
379<!--
380## Wrapup
381
382#### Folding a large amount of instances
383In the previous protocol, prover's costs are $O(k \log k)$, and verifier's costs are $O(k)$.
384ProtoGalaxy paper proposes an alternative construction of the previous scheme, for When we have a larger number of instances being fold.
385
386The main idea will be to replace the vanishing check by a Sumcheck over the boolean hypercube.
387
388
389We will put the running instance $w$ from the previous section together with the other witnesses, so 
390
391In step 8, we had
392
393$$
394\textcolor{gray}{
395G(X) := \sum_{i \in [n]} pow_i(\overrightarrow{\beta^*}) \cdot f_i \left( \sum_{j=0}^k \textcolor{SkyBlue}{L_j(X)} w_j \right)
396}
397$$
398
399Instead of the Lagrange bases, we will use the polynomial $eq(x, y)$.
400
401TODO explain eq poly, maybe in the preliminaries section
402
403
404so we will have
405
406
407$$
408G(t) :=
409eq(t, \overrightarrow{\delta^n})
410\sum_{i \in [n]} pow_i(\overrightarrow{\beta^*})
411\cdot f_i \left( \sum_{j'=0} \textcolor{SkyBlue}{eq(t, j')} w_j' \right)
412$$
413
414P and V run the SumCheck protocol over $G$.
415-->
416
417<h3>Conclusions</h3>
418
419<p>As mentioned initially, the <a href="https://eprint.iacr.org/2023/1106">ProtoGalaxy paper</a> is very well written and clear, so I highly recommend reading it to get the full explaination and proofs. Hopefully these notes might have provided some intuition behind the steps helping to follow the paper descriptions.</p>
420
421<p>One thing to mention is that we&rsquo;ve covered the version of the protocol presented in the section 4 of the paper, which has prover costs <span class="math inline">\(O(k \log k)\)</span> and verifier costs <span class="math inline">\(O(k)\)</span>. In section 5, the protocol is modified to obtain prover costs <span class="math inline">\(O(k)\)</span> and verifier costs <span class="math inline">\(O(\log k)\)</span>, at the expense of worse constant costs, which makes it more suitable for larger <span class="math inline">\(k\)</span> (more instances folded in each step). The main idea is by replacing the vanishing polynomial check with the SumCheck protocol, which internally works with similar equations that we had before, but replacing the Lagrange bases <span class="math inline">\(\textcolor{SkyBlue}{L_j(X)}\)</span> by the equivalent over the multilinar extension world <span class="math inline">\(\textcolor{SkyBlue}{eq(x, t)}\)</span>.</p>
422
423<p>Which for example, the equation that we had on the step 8:</p>
424<p><span class="math display">\[
425\textcolor{gray}{
426G(X) := \sum_{i \in [n]} pow_i(\overrightarrow{\beta^*}) \cdot f_i \left( \sum_{j=0}^k \textcolor{SkyBlue}{L_j(X)} w_j \right)
427}
428\]</span></p><p>is now replaced by</p>
429<p><span class="math display">\[
430G(t) :=
431eq(t, \overrightarrow{\delta^n})
432\sum_{i \in [n]} pow_i(\overrightarrow{\beta^*})
433\cdot f_i \left( \sum_{j'=0} \textcolor{SkyBlue}{eq(t, j')} w_j' \right)
434\]</span></p><p>Further details can be found at section 5 of the paper.</p>
435
436<p><br></p>
437
438<p>I&rsquo;m still discovering all the marvelous world of folding schemes, <a href="https://raw.githubusercontent.com/arnaucube/math/master/notes_nova.pdf">trying</a> to <a href="https://raw.githubusercontent.com/arnaucube/math/master/notes_hypernova.pdf">grasp</a> intuition across the different designs. One thing that will be interesting is to be able to benchmark on the practical side all the different folding schemes to see how they work in practice, and for example depending on the use case, evaluate for which threshold of <span class="math inline">\(k\)</span> makes sense to switch between them, and which <span class="math inline">\(k\)</span> makes sense for different use cases in the context of <a href="https://hackmd.io/@9MPvD9czTQS18zEjAERVAg/Hy-vKK04h">parallelizing folding</a>.</p>
439
440<p>In the repo <a href="https://github.com/arnaucube/protogalaxy-poc">https://github.com/arnaucube/protogalaxy-poc</a> you can find a proof of concept implementation of the ProtoGalaxy scheme described in this post.</p>
441
442    </div>
443
444    <footer style="text-align:center; margin-top:100px;margin-bottom:50px;">
445      <div class="container">
446        <br>
447        <a href="/blog">Go to main</a>
448        <br><br>
449        <div class="row">
450          <ul class="list-inline">
451            <li><a href="https://twitter.com/arnaucube"
452                   style="color:gray;text-decoration:none;"
453                   target="_blank">twitter.com/arnaucube</a>
454            </li>
455            <li><a href="https://github.com/arnaucube"
456                   style="color:gray;text-decoration:none;"
457                   target="_blank">github.com/arnaucube</a>
458            </li>
459          </ul>
460        </div>
461        <div class="row" style="display:inline-block;">
462          Blog made with <a href="http://github.com/arnaucube/blogo/"
463                            target="_blank" style="color: gray;text-decoration:none;">Blogo</a>
464        </div>
465      </div>
466    </footer>
467
468    <script>
469    </script>
470    <script src="js/external-links.js"></script>
471    <script>hljs.initHighlightingOnLoad();</script>
472    <script defer src="js/katex/katex.min.js"></script>
473    <script defer src="js/katex/auto-render.min.js"></script>
474    <script>
475      document.addEventListener("DOMContentLoaded", function() {
476        renderMathInElement(document.body, {
477          displayMode: false,
478          // customised options
479          // • auto-render specific keys, e.g.:
480          delimiters: [
481            {left: '$$', right: '$$', display: true},
482            {left: '$', right: '$', display: false},
483            {left: "\\[", right: "\\]", display: true},
484            {left: "\\(", right: "\\)", display: false},
485            {left: "\\begin{equation}", right: "\\end{equation}", display: true}
486          ],
487          // • rendering keys, e.g.:
488          throwOnError : true
489        });
490      });
491
492      ///
493      let theme = localStorage.getItem("theme");
494      if ((theme === "light-theme")||(theme==null)) {
495          theme = "light-theme";
496          document.getElementById("themeSwitcher").checked = false;
497      } else if (theme === "dark-theme") {
498          theme = "dark-theme";
499          document.getElementById("themeSwitcher").checked = true;
500      }
501      document.body.className = theme;
502      localStorage.setItem("theme", theme);
503
504      function switchTheme() {
505        theme = localStorage.getItem("theme");
506        if (theme === "light-theme") {
507          theme = "dark-theme";
508          document.getElementById("themeSwitcher").checked = true;
509        } else {
510          theme = "light-theme";
511          document.getElementById("themeSwitcher").checked = false;
512        }
513        document.body.className = theme;
514        localStorage.setItem("theme", theme);
515
516        console.log(theme);
517      }
518    </script>
519    <script>
520    function tagLinks(tagName) {
521      var tags = document.getElementsByTagName(tagName);
522      for (var i=0, hElem; hElem =  tags[i]; i++) {
523        if (hElem.parentNode.className=="row postThumb") {
524          continue;
525        }
526        hElem.id = hElem.innerHTML.toLowerCase().replace(" ", "-");
527        hElem.innerHTML = "<a style='color:black!important;' href='#"+hElem.id+"'>"+hElem.innerHTML+"</a>";
528      }
529    }
530    tagLinks("h2");
531    tagLinks("h3");
532    tagLinks("h4");
533    tagLinks("h5");
534    </script>
535    <script>
536function generateTOC(tocContainerId = "toc") {
537    const tocContainer = document.getElementById(tocContainerId);
538    if (!tocContainer) return;
539
540    const headings = document.querySelectorAll("h2, h3, h4, h5, h6");
541    if (!headings.length) return;
542
543    // Generate IDs if missing
544    headings.forEach((heading, index) => {
545        if (!heading.id) {
546            heading.id =
547                heading.textContent
548                    .trim()
549                    .toLowerCase()
550                    .replace(/[^\w\s-]/g, "")
551                    .replace(/\s+/g, "-") +
552                "-" +
553                index;
554        }
555    });
556
557    const toc = document.createElement("ul");
558    let currentLevel = 1;
559    let currentList = toc;
560    const stack = [toc];
561
562    headings.forEach((heading) => {
563        const level = parseInt(heading.tagName.substring(1), 10);
564
565        const li = document.createElement("li");
566        const a = document.createElement("a");
567
568        a.href = "#" + heading.id;
569        a.textContent = heading.textContent;
570
571        li.appendChild(a);
572
573        if (level > currentLevel) {
574            for (let i = currentLevel; i < level; i++) {
575                const nestedList = document.createElement("ul");
576
577                const lastLi =
578                    currentList.lastElementChild ||
579                    currentList.appendChild(document.createElement("ul"));
580
581                lastLi.appendChild(nestedList);
582
583                stack.push(nestedList);
584                currentList = nestedList;
585            }
586        } else if (level < currentLevel) {
587            for (let i = currentLevel; i > level; i--) {
588                stack.pop();
589            }
590            currentList = stack[stack.length - 1];
591        }
592
593        currentList.appendChild(li);
594        currentLevel = level;
595    });
596
597    tocContainer.innerHTML = "";
598    tocContainer.appendChild(toc);
599}
600
601document.addEventListener("DOMContentLoaded", () => {
602    generateTOC();
603});
604    </script>
605    <script src="js/mermaid.min.js"></script>
606
607  </body>
608</html>
609