From one {point,poly} to multi {point,poly} commitment schemes

6 minute read

This is my first entry here, so I would highly appreciate any feedback.

This is not an introduction to Polynomial Commitment Schemes (PCS). Instead, I will use some statements from vanilla KZG scheme and build upon these for multiple points and multiple polynomials. If you are new to PCS, I recommend the intro by Dankrad Feist or the original KZG paper.

Polynomial committment schemes

The primary focus of this article is to explore various flavours of KZG polynomial commitment scheme:

  • single polynomial, single opening (vanilla)
  • single polynomial, multiple openings
  • multiple polynomials, single opening
  • multiple polynomials, multiple openings

and show how the commitment changes, how are the proofs constructed, and what the verifier needs to do to get convinced of the proof’s validity.

Vanilla KZG

Here I just re-state the results from KZG and explain some notation:

  • secret $s$ from SRS: $ { G, sG, s^2G, …, s^{n-1}G } $
  • polynomial $p(X)$
  • commitment, $C = [p(s)]_1$
  • claim $p$ at $z$: $y = p(z)$
  • quotient polynomial $h(x) = \frac{p(x) - y}{x-z}$
  • proof, $\pi = [h(s)]$

One thing worth noting is that the quotient polynomial needs to be computed separately for each point $z$ (if we evaluate at multiple points).

Verifier’s job is to perform a pairing check:

$$ e(C - [y]_1, G_2) \stackrel{?}{=} e(\pi, [s-z]_2)$$

Single polynomial, multiple openings

Perhaps the easiest extension of the above scheme is verifying the polynomial $p$ at multiple points $z_1, …, z_t$. Let’s consider the case $t=2$, i.e. just two points. Of course the prover could provide two separate proofs $\pi_1$ & $\pi_2$ to prove that $p(z_1) = y_1$ & $p(z_2) = y_2$, but we can be more efficient than that.

Recall that for a single opening at $z$, we construct a polynomial $p(x) - p(z)$ which clearly evaluates to $0$ if $x=z$, i.e. it has a factor of $x-z$.

We would like to extend this to two points now, so we require that some polynomial $p(x) - r(x)$ evaluates to zero at both $z_1$ & $z_2$. To find $r(x)$, first rewrite the condition as:

$$p(z_1) = r(z_1), \quad p(z_2) = r(z_2) $$

$$ y_1 = r(z_1), \quad y_2 = r(z_2) $$

and notice that the function $r(x)$ needs to pass through two points $(z_1, y_1)$ and $(z_2, y_2)$ - it’s a line! We can use Lagrange interpolation to compute $r(x)$.

Ok, so we know that $(x-z_1)\cdot(x-z_2)$ is a factor of $p(x) - r(x)$. For the case of multiple openings, we define $h(x)$ slightly differently than before:

$$h(x) = \frac{p(x) - r(x)}{Z_s(x)}$$

where $Z_s(x) = (x-z_1)\cdot(x-z_2)$

Prover’s definitions of commitment $C$ and proof $\pi$ don’t change from the single-point version. The verifier’s check differs minimally (it can actually be generalised) to include the fact that we no longer have a single point but multiple ones:

$$ e(C - [r(s)]_1, G_2) \stackrel{?}{=} e(\pi, [Z_s(s)]_2)$$

The argument here can be easily extended to more than two points: $r(x)$ then becomes a quadratic, cubic, …, degree-(t-1)-polynomial instead of a line (degree-1-polynomial).

Multiple polynomials, one opening

This section serves well as an intermediate step on our way to opening multiple polynomials at multiple points. It is used in practice in Plonk, where seemingly when only two points are opened, it is more efficient to use the scheme described here twice, rather than applying the multi-openings optimisations.

Instead of dealing with a single polynomial $p(x)$, the prover now wants to commit to multiple polynomials (let’s assume they have the same degree) and later prove their openings at a single point $z$.

The committent phase changes, as now for each polynomial $p_i(x), i \in {1, …, k}$, we generate $C_i = [p_i(s)]$.

We define $h(x)$ as: $$h(x) = \sum_{i=1}^{k}\gamma^{i-1}\frac{p_i(x) - y}{x-z}$$

To see why, let’s digress slightly and define a sum of a bunch of functions $F_i$ defined over our field 𝔽: $$ G(x) = \sum_{i}^{k} \gamma^{i-1} \cdot F_i(x)$$

And also an arbitrary linear factor $Z(x) = (x-z)$. Notice that if even all functions $F_i$ were divisible by $Z(x)$ except one function, say, $F_j$, then the probability that $Z(x) | G(x)$ is $\frac{k}{𝔽}$, because: $$ F_j \mod Z(x) = R(x) \neq 0$$

$$ G(x) = \sum_{i \neq j} \gamma^{i-1} \cdot F_i(x) + \gamma^{j-1}\cdot R(x) \mod Z(x)$$

We fix some $\chi \in 𝔽$ s.t. $Z(\chi) = 0$ but $R(\chi) \neq 0$. If we wish $G(\chi) = 0$, then with a slight abuse of notation we have: $$ 0 = \sum_{i=1, i \neq j}^k \gamma^{i-1} \cdot F_i + \gamma^{j-1}\cdot R$$

which is just a polynomial in $\gamma$ of degree $k$, which has at most $k$ solutions.

The point of all that being: the verifier receives $[h(s)]_1$, which implicitly “commits” to all $p_i$’s and they can compute $Z(x) = (x-z)$ by themselves and so if at least one of the committed polynomials is not divisible by $(x-z)$, then the prover will get caught with high probability.

Let’s wrap up this section with the pairing check:

$$ \begin{aligned} e(\gamma^0 \cdot (C_i - [y]_1), G_2) \cdot … \cdot e(\gamma^{k-1} \cdot (C_k - [y]_1), G_2) \stackrel{?}{=} e(\pi, [s-z]_2) \end{aligned} $$

Which, thanks to the bilinearity property of pairings, can be simplified to: $$ \begin{aligned} e(\gamma^0 \cdot (C_i - [y]_1) + … + \gamma^{k-1} \cdot (C_k - [y]_1), G_2) \stackrel{?}{=} e(\pi, [s-z]_2) \end{aligned} $$

As it can be seen, having $k$ polynomials requires the prover to publish $k$ commitments, but the verifier still computes just one pairing check.

Multiple polynomials, multiple openings

Note: Multiple variants exist with different tradeoffs, here I summarise the first scheme from this paper. That same scheme is further improved in the same paper, optimising for verifier speed at the expense of an additional element required in the proof. I might write about it in the future.

As with the preceding scheme, the prover commits to each polynomial $p_i(x)$ separately: $C_1, …, C_k$.

We will need to define a set of all evaluation points $T$ as $T = \lbrace z_1, …, z_t \rbrace$.

The quotient polynomial $h(x)$ is a straightforward combination of the above two schemes: we take advantage of both the sum over $p_i(x)$ as well as the generalised $r_i(x)$ and $Z_{s_i}(x)$. You might notice that I’ve added an extra subscript to indicate that each polynomial might have a different set of evaluation points. This means that each $p_i$ gets evaluated not necessarily on the entire domain $T$, but its subset $S_i$, and that these subsets are not necessarily disjoint.

Example: $p_1(x)$ is opened at $S_1 = \lbrace z_1, z_2 \rbrace$, and $p_2(x)$ is opened at $S_2 = \lbrace z_1, z_3 \rbrace$.

Define the “combined” $h(x)$:

$$ \begin{aligned} h(x) = \sum_{i=1}^{k}\gamma^{i-1}\frac{p_i(x) - r_i(x)}{Z_{s_i}(x)} \end{aligned}\label{1}\tag{1} $$

Unfortunately, having distinct openings for each $p_i$ complicates the pairing check, since we cannot just plug in $Z_{s}$ into the RHS of $\eqref{1}$, because now $Z_{s_i}$ are distinct!

However, imagine for a minute that we allow ourselves to compute a product of pairings on the RHS (instead of a single pairing), where we define each $h_i$ as the fraction from the sum of $\eqref{1}$, and $\pi_i$ as its commitment.

$$ \begin{aligned} h_i = \frac{p_i(x) - r_i(x)}{Z_{s_i}(x)} \newline \newline \pi = [h_i(s)]_1 \end{aligned} $$

Then we have $$ \begin{aligned} e(C_1 - [r_1(s)]_1, G_2) \stackrel{?}{=} e(\pi_1, [Z_1(s)]_2) \newline e(C_2 - [r_2(s)]_1, G_2) \stackrel{?}{=} e(\pi_2, [Z_2(s)]_2) \newline … \newline e(C_k - [r_k(s)]_1, G_2) \stackrel{?}{=} e(\pi_k, [Z_k(s)]_2) \newline \end{aligned} $$

Which, in polynomial notation, checks the following equality: $$p_i(x) - r_i(x) = h_i(x)\cdot Z_{s_i}$$

We can use a trick to multiply both sides by $Z_T(x) = (x-z_1)\cdot…\cdot(x-z_t) = \prod_{i=1}^t{(x-z_i)}$:

$$(p_i(x) - r_i(x))\cdot Z_T(x) = h_i(x)\cdot Z_{s_i} \cdot Z_T(x)$$

and then divide by $Z_{s_i}$:

$$ \begin{aligned} (p_i(x) - r_i(x))\cdot Z_{T/S_i}(x) = h_i(x)\cdot Z_T(x) \end{aligned}\label{2}\tag{2} $$

and $Z_i = Z_{T/S_i}(x)$ is defined as $\prod_{i: z_i \notin S_i}{(x-z_i)}$

We rewrite Equation $\eqref{2}$ in pairing notation:

$$ \begin{aligned} e(C_i - [r_i(s)]_1, [Z_i(s)]_2 ) = e(\pi_i, [Z_T(s)]_2) \newline \end{aligned} $$

Finally, thanks to the properties of pairings (summation in $G_1$ corresponds to multiplication in the target group $G_t$), we can combine the RHS for all $i \in \lbrace 1, …, k\rbrace$, including the “weight” $\gamma$, as:

$$ e(\gamma^0 \cdot C_1 - [r_1(s)]_1, [Z_1(s)]_2) \cdot … \cdot e(\gamma^{k-1} \cdot C_k - [r_k(s)]_1, [Z_k(s)]_2 ) = e(\pi, [Z_T(s)]_2) $$

For a full proof, refer to the Plonk paper, Claim 4.6.

Notice that the described protocol is not optimal for the verifier - I might explore optimisations in a future post.

Author: Marcin Górny