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 70      <h1>Notes on ProtoGalaxy</h1>
 71
 72<p><em>2023-08-20</em></p>
 73
 74<blockquote>
 75<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>
 76</blockquote>
 77
 78<p>The idea of these notes is to provide some extra intuition on <a href="https://eprint.iacr.org/2023/1106">ProtoGalaxy paper</a>.
 79The 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>
 80
 81<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>
 82
 83<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>
 84
 85<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>
 86
 87<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.
 88Also thanks to <a href="https://twitter.com/kiliconu">Onur</a> for comments and corrections.</p>
 89
 90<div style="margin-left:20%;" >
 91
 92<img src="img/posts/protogalaxy/protogalaxies-colliding.jpg"/>
 93
 94<br>
 95(img: protogalaxies colliding, <a href="https://en.wikipedia.org/wiki/File:Stellar_Fireworks_Finale.jpg">from Wikipedia</a>)
 96
 97</div>
 98
 99<h2>Preliminaries</h2>
100
101<p>Before jumping into the protocol itself, let&rsquo;s overview some building blocks first.</p>
102
103<h4><span class="math inline">\(pow_i\)</span></h4>
104
105<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>
106
107<p><span class="math inline">\(pow_i\)</span> is a <span class="math inline">\(t\)</span>-variate polynomial such that</p>
108<p><span class="math display">\[
109% pow_i(X_1, \ldots, X_t) = \prod_{i \in [n]} ((1-b_j) + b_j \cdot X^{2^j})
110pow_i(X) = \prod_{i \in [n]} ((1-b_j) + b_j \cdot X^{2^j})
111\]</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>
112
113<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>
114<p><span class="math display">\[
115pow_i(\overrightarrow{\beta}) = \prod_{j \in [n]} ((1-b_j) + b_j \cdot \beta_j)
116\]</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>
117
118<p>So, for example,</p>
119<p><span class="math display">\[
120    i=2_d=10_b;~
121    pow_2(\overrightarrow{\beta}) =
122    \underbrace{((1-0) + 0 \cdot \beta)}_{1}
123    \cdot \underbrace{((1-1) + 1 \cdot \beta^2)}_{\beta^2}
124    = \beta^2
125    \\
126    i=3_d=11_b;~
127    pow_3(\overrightarrow{\beta}) =
128    \underbrace{((1-1) + 1 \cdot \beta)}_{\beta}
129    \cdot \underbrace{((1-1) + 1 \cdot \beta^2)}_{\beta^2}
130    = \beta^3
131    \\
132    i=4_d=100_b;~
133    pow_4(\overrightarrow{\beta}) =
134    \underbrace{((1-0) + 0 \cdot \beta)}_{1}
135    \cdot \underbrace{((1-0) + 0 \cdot \beta^2)}_{1}
136    \cdot \underbrace{((1-1) + 1 \cdot \beta^4)}_{\beta^4}
137    = \beta^4
138    \\
139    i=5_d=101_b;~
140    pow_5(\overrightarrow{\beta}) =
141    \underbrace{((1-1) + 1 \cdot \beta)}_{\beta}
142    \cdot \underbrace{((1-0) + 0 \cdot \beta^2)}_{1}
143    \cdot \underbrace{((1-1) + 1 \cdot \beta^4)}_{\beta^4}
144    = \beta^5
145\]</span></p>
146<h4>Lagrange bases</h4>
147
148<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>
149
150<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>
151<p><span class="math display">\[L_j(\omega^i)=
152\begin{cases}
153	&amp;1 ~\text{if}~ i=j\\
154	&amp;0 ~\text{otherwise}
155\end{cases}
156\]</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>
157
158<p>To get a bit of more intuition here, let&rsquo;s represent this visually:</p>
159
160<div>
161
162<img src="img/posts/protogalaxy/lagrange-basis-intuition.png" style="width:40%;display:inline-block;" />
163
164<div style="width:50%; display:inline-block;">
165The 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 $L_2(w^2) \cdot v_k = 1 \cdot v_k$.
166
167While, when we evaluate $L_2(X)$ at for example $\omega^1$, we will obtain a $0$, thus the result will be a $0$.
168
169</div>
170</div>
171
172<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>
173
174<h4>Function <span class="math inline">\(f(X)\)</span></h4>
175
176<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>
177
178<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>
179
180<!--$$f(X)= \sum_{i=0}^n A(X_i) \cdot B(X_i) - C(X_i)$$-->
181<p><span class="math display">\[f(X)= A \cdot X \circ B \cdot X - C \cdot X \]</span></p>
182<!-- $$f(X)= \sum_{i=0}^n A(X_i) \cdot B(X_i) - C(X_i)$$ -->
183
184<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>
185
186<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>
187
188<h4>Lemma 4.2</h4>
189
190<p>The following lemma is from the ProtoGalaxy paper:</p>
191
192<blockquote>
193<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>
194<p><span class="math display">\[
195	f \left( \sum_{i=0}^k a_i L_i(X) \right) = \sum_{i=0}^k f(a_i) L_i(X) + Z(X) Q(X)
196\]</span></p></blockquote>
197
198<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>
199
200<p>Recall from the <a href="https://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor#Euclidean_division">Euclidean polynomial division</a>:</p>
201
202<blockquote>
203<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>
204</blockquote>
205
206<p>Thus,</p>
207<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>
208
209<p>So, when evaluating at <span class="math inline">\(a_j, ~\forall j=0, \ldots, k\)</span>,</p>
210<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>
211<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>
212<div style="float:right;">
213
214$\square$
215
216</div>
217
218<h2>ProtoGalaxy protocol</h2>
219
220<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>
221
222<p><img src="img/posts/protogalaxy/folding-diagram.png" style="width:80%;" /></p>
223
224<h4>Instances</h4>
225
226<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.
227Additionally, we will have <span class="math inline">\(k\)</span> randomized instance pairs <span class="math inline">\(( \{ \phi_i \}^k; \{ w_i \}^k )\)</span>.</p>
228
229<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>
230
231<p>As we start folding instances, the folded instance must satisfy the relation</p>
232<p><span class="math display">\[
233\sum_{i \in [n]} pow_i(\overrightarrow{\beta}) f_i(w) = e
234\]</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>
235
236<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>
237
238<h4>Protocol</h4>
239
240<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>
241
242<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>
243
244<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>
245
246<p><br>
247<em>(Let P denote the Prover, and V the Verifier)</em></p>
248
249<ol>
250<li><p>V sends a challenge <span class="math inline">\(\delta \in^R \mathbb{F}\)</span></p></li>
251
252<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>
253
254<li><p>P computes the polynomial</p>
255<p><span class="math display">\[
256F(X) := \sum_{i \in [n]} pow_i(
257\underbrace{
258	\textcolor{orange}{\overrightarrow{\beta}} + X \cdot
259	\textcolor{orange}{\overrightarrow{\delta} }
260}_\text{represent it by $\textcolor{orange}{\overrightarrow{\eta}}$}
261) \textcolor{ForestGreen}{f_i(w)}
262\\
263\text{\scriptsize{where $\textcolor{orange}{\overrightarrow{\eta}}$ is just an intermediate}}
264\\
265\text{\scriptsize{representation of the new random vector over $X$}}
266\\
267\text{\scriptsize{where $\overrightarrow{\eta}=\{ \beta+X \delta, (\beta + X \delta)^2, (\beta +X \delta)^4, \ldots, (\beta + X \delta)^{2^{t-1}} \}$}}
268\\
269= \sum_{i \in [n]}
270	pow_i(\textcolor{orange}{\overrightarrow{\eta}})
271\textcolor{ForestGreen}{f_i(w)}
272\]</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>
273
274<div style="font-size:85%;">
275
276
277In order to see this, let's simplify $\overrightarrow{\eta}$ and instead let's get take $\overrightarrow{\eta}$ getting rid of $\beta,~\delta$. Let $\overrightarrow{X} = \{X, X, X, \ldots, X \}$,
278recall that 
279
280
281$$
282pow_i(\overrightarrow{X}) = \prod_{j \in [n]} ((1-b_j) + b_j \cdot X_j)
283$$
284
285
286then we have
287
288
289$$
290i=2_d=10_b;~
291pow_2(\overrightarrow{X}) = 1 \cdot X = X;~ deg=1
292\\
293i=3_d=11_b;~
294pow_3(\overrightarrow{X}) = X \cdot X = X^2;~ deg=2
295\\
296i=4_d=100_b;~
297pow_4(\overrightarrow{X}) = 1 \cdot 1 \cdot X = X;~ deg=1
298\\
299i=5_d=101_b;~
300pow_5(\overrightarrow{X}) = X \cdot 1 \cdot X = X^2;~ deg=2
301\\
302i=6_d=110_b;~
303pow_6(\overrightarrow{X}) = 1 \cdot X \cdot X = X^2;~ deg=2
304\\
305i=7_d=111_b;~
306pow_7(\overrightarrow{X}) = X \cdot X \cdot X = X^3;~ deg=3
307\\
308i=8_d=1000_b;~
309pow_8(\overrightarrow{X}) = 1 \cdot 1 \cdot 1 \cdot X = X;~ deg=1
310$$
311
312
313</div>
314
315<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>.
316<br></p>
317
318<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>
319
320<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>
321
322<li><p>V sends a challenge <span class="math inline">\(\textcolor{magenta}{\alpha} \in^R \mathbb{F}\)</span></p></li>
323
324<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>
325
326<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>
327
328<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>
329
330<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>
331<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>
332
333<p>The following image tries to show how the previous steps relate:
334<img src="img/posts/protogalaxy/steps-3-7.png" style="width:80%;" /></p></li>
335
336<li><p>P computes the polynomial</p>
337<p><span class="math display">\[
338G(X) := \sum_{i \in [n]}
339pow_i(\overrightarrow{\beta^*}) \cdot f_i(L_0(X) w + \sum_{j \in [k]} L_j(X) w_j)
340\]</span></p>
341<div style="font-size:85%;">Here we make use of the <a href="#lemma-4.2">Lemma 4.2</a> to see that:</div>
342<p><span class="math display">\[
343G(X) := \sum_{i \in [n]}
344pow_i(\overrightarrow{\beta^*}) \cdot
345\underbrace{ f_i(\textcolor{SkyBlue}{L_0(X)} w }_{\textcolor{SkyBlue}{L_0(X)} \cdot f_i(w)} + \underbrace{ \sum_{j \in [k]} \textcolor{SkyBlue}{L_j(X)} w_j) }_{0}
346\\
347\text{\small{ notice that $f_i(w_j)=0 ~\forall j \in [k], i \in [n]$, thus $\sum_{j \in [k]} \textcolor{SkyBlue}{L_j(X)} \cdot f_i(w_j) = 0$ }}
348\\
349= \sum_{i \in [n]}
350\underbrace{ pow_i(\overrightarrow{\beta^*}) \cdot f_i(w) }_{F(\alpha) ~\text{from step 6}} \cdot \textcolor{SkyBlue}{L_0(X)} \pmod{Z(X)}
351\\
352= F(\alpha) \cdot \textcolor{SkyBlue}{L_0(X)} \pmod{Z(X)}
353\]</span></p>
354<div style="font-size:85%;">(where $Z(X)$ is the vanishing polynomial from the <a href="#lagrange-bases">preliminaries section</a>)</div>
355
356<p>which leads to the next step.</p></li>
357
358<li><p>P computes the polynomial <span class="math inline">\(K(X)\)</span> such that</p>
359<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>
360
361<p><br>
362What 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>).
363We&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>.
364Here 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>
365
366<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>
367
368<li><p>V sends a challenge <span class="math inline">\(\gamma \in^R \mathbb{F}\)</span></p></li>
369
370<li><p>P, V compute</p>
371<p><span class="math display">\[e^* := F(\alpha) L_0(\gamma) + Z(\gamma) K(\gamma)\]</span></p></li>
372
373<li><p>P, V outputs the new folded committed instance</p>
374<p><span class="math display">\[\Phi^* = (\phi^*, \overrightarrow{\beta^*}, e^*)\]</span></p><p>where</p>
375<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>
376
377<li><p>Additionally, P outputs the folded witness</p>
378<p><span class="math display">\[w^* = L_0(\gamma) w + \sum_{i \in [k]} L_i(\gamma) \cdot w_i\]</span></p></li>
379</ol>
380
381<p>To check that <span class="math inline">\((\Phi^*; w^*)\)</span> satisfies the relation we check that</p>
382<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>
383
384<!--
385## Wrapup
386
387#### Folding a large amount of instances
388In the previous protocol, prover's costs are $O(k \log k)$, and verifier's costs are $O(k)$.
389ProtoGalaxy paper proposes an alternative construction of the previous scheme, for When we have a larger number of instances being fold.
390
391The main idea will be to replace the vanishing check by a Sumcheck over the boolean hypercube.
392
393
394We will put the running instance $w$ from the previous section together with the other witnesses, so 
395
396In step 8, we had
397
398$$
399\textcolor{gray}{
400G(X) := \sum_{i \in [n]} pow_i(\overrightarrow{\beta^*}) \cdot f_i \left( \sum_{j=0}^k \textcolor{SkyBlue}{L_j(X)} w_j \right)
401}
402$$
403
404Instead of the Lagrange bases, we will use the polynomial $eq(x, y)$.
405
406TODO explain eq poly, maybe in the preliminaries section
407
408
409so we will have
410
411
412$$
413G(t) :=
414eq(t, \overrightarrow{\delta^n})
415\sum_{i \in [n]} pow_i(\overrightarrow{\beta^*})
416\cdot f_i \left( \sum_{j'=0} \textcolor{SkyBlue}{eq(t, j')} w_j' \right)
417$$
418
419P and V run the SumCheck protocol over $G$.
420-->
421
422<h3>Conclusions</h3>
423
424<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>
425
426<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>
427
428<p>Which for example, the equation that we had on the step 8:</p>
429<p><span class="math display">\[
430\textcolor{gray}{
431G(X) := \sum_{i \in [n]} pow_i(\overrightarrow{\beta^*}) \cdot f_i \left( \sum_{j=0}^k \textcolor{SkyBlue}{L_j(X)} w_j \right)
432}
433\]</span></p><p>is now replaced by</p>
434<p><span class="math display">\[
435G(t) :=
436eq(t, \overrightarrow{\delta^n})
437\sum_{i \in [n]} pow_i(\overrightarrow{\beta^*})
438\cdot f_i \left( \sum_{j'=0} \textcolor{SkyBlue}{eq(t, j')} w_j' \right)
439\]</span></p><p>Further details can be found at section 5 of the paper.</p>
440
441<p><br></p>
442
443<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>
444
445<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>
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