KZH polynomial commitments

tl;dr: KZG + Hyrax = KZH¹. This name makes me happy: not only it stands on its own but it also coincides with the first three authors’ initials!

Preliminaries

Notation

• $[m]=\{1,2,3,\ldots,m\}$
• $[m)=\{0,1,2,\ldots,m-1\}$
• $\bin^s$ denotes the boolean hypercube of size $2^s$.
• Let $\F$ denote a finite field of prime order $p$
• Let $\Gr_1,\Gr_2,\Gr_T$ denote cyclic groups of prime order $p$ with a bilinear map $e : \Gr_1\times\Gr_2\rightarrow\Gr_T$ where computing discrete logs is hard
  • We use additive group notation
• Denote $\MLE{s}$ as the space of all multilinear extensions (MLEs) $f(X_1,\ldots,X_s)$ of size $2^s$ with entries in $\F$
  • We also use $\MLE{s_1,s_2,\ldots,s_\ell} \bydef \MLE{\sum_{i\in[\ell]} s_i}$
• Denote $i$’s binary representation as $\vect{i} = (i_0, i_1, \ldots, i_{s-1})\in \bin^s$, s.t. $i=\sum_{k=0}^{s-1} 2^k \cdot i_k$
  • We often naturally interchange between these two, when it is clear from context
• $(v_0, v_2, \ldots, v_{n-1})^\top$ denotes the transpose of a row vector
• We typically use bolded variables to indicate vectors and matrices
  • e.g., a matrix $\mat{A}$ consists of rows $\mat{A}_i,\forall i\in[n)$, where each row $\mat{A}_i$ consists of entries $A_{i,j},\forall j\in[m)$
  • e.g., vectors $\A$ are typically italicized while matrices $\mat{M}$ are not
• We use $\vect{a}\cdot G\bydef (a_0\cdot G,a_1\cdot G,\ldots, a_{n-1}\cdot G)$
• We use $a\cdot \G\bydef (a\cdot G_0,a\cdot G_1,\ldots, a\cdot G_{n-1})$
• We use $\vect{a}\cdot \G \bydef \sum_{i\in[n)} a_i\cdot G_i$
• Recall the definition of $\eq(\x,\boldsymbol{b})$ Lagrange polynomials from here
• For time complexities, we use:
  • $\Fadd{n}$ for $n$ field additions in $\F$

  • $\Fmul{n}$ for $n$ field multiplications in $\F$

  • $\msm{n}_b$ for a single multi-scalar multiplication (MSM) in $\Gr_b$ of size $n$
  • $\Gadd{n}_b$ for $n$ additions in $\Gr_b$
  • $\Gmul{n}_b$ for $n$ individual scalar multiplications in $\Gr_b$
  • $\multipair{n}$ for a multipairing of size $n$ and $\pair$ for one pairing
• It is useful to understand Hyrax, which KZH is highly-related to.

$\mathsf{KZH}_2$ construction

This construction can be parameterized to commit to any MLE $f(\X,\Y)\in \MLE{\term{\nu},\term{\mu}}$ representing a matrix of $\term{n} = 2^\nu$ rows and $\term{m}=2^\mu$ columns, where $\X\in \bin^\nu$ indicates the row and $\Y\in\bin^\mu$ indicates the column.

$\mathsf{KZH}_2.\mathsf{Setup}(1^\lambda, \nu,\mu) \rightarrow (\mathsf{vk},\mathsf{ck})$²

Notation:

• $n \gets 2^\nu$ denotes the # of matrix rows
• $m \gets 2^\mu$ denotes the # of matrix columns
• $N = n\cdot m\bydef 2^{\nu + \mu}$ denotes the total # of entries in the matrix

Pick trapdoors and generators:

• $\term{\alpha}\randget\F$
• $\term{\btau} \bydef (\tau_0, \tau_1,\ldots,\tau_{n-1})\randget \F^n$
• $\term{\G}\bydef(G_0,\ldots, G_{m-1})\randget \Gr_1^m$
• $\term{\V}\randget \Gr_2$

Compute $\H\in\Gr_1^{n \times m}$: \begin{align} H_{i,j} &\gets \tau_i \cdot G_j\in \Gr_1 , \forall i\in[n),j\in[m) \\
\H_i &\gets \tau_i\cdot \G %\\
\bydef (\tau_i \cdot G_0,\tau_i\cdot G_1,\dots,\tau_i\cdot G_{m-1})\in\Gr_1^m , \forall i\in[n) \\
%&\bydef (H_{i,0},\ldots,H_{i,m-1})\\
\term{\H} &\bydef \begin{pmatrix} \text{ —} & \H_0 & \text{— }\\\ \text{ —} & \H_1 & \text{— }\\\ & \vdots & \\
\text{ —} & \H_{n-1} & \text{— }\\
\end{pmatrix} %\\
\bydef \begin{pmatrix} \text{ —} & \tau_0 \cdot \G & \text{— }\\
\text{ —} & \tau_1\cdot\G & \text{— }\\
&\vdots &\\
\text{ —} &\tau_{n-1}\cdot\G & \text{— }\\
\end{pmatrix} %\bydef\begin{pmatrix} % \tau_0 \cdot G_0 &\tau_0 \cdot G_1 & \dots & \tau_0\cdot G_{m-1}\\
% \tau_1 \cdot G_0 &\tau_1 \cdot G_1 & \dots & \tau_1\cdot G_{m-1}\\
% \vdots & & & \vdots\\
% \tau_{n-1} \cdot G_0 & \tau_{n-1}\cdot G_1 & \dots & \tau_{n-1}\cdot G_{m-1}\\
%\end{pmatrix} %\\
%\bydef \begin{pmatrix} % | & | & & | \\
% \btau\cdot G_0 & % \btau\cdot G_1 & % \cdots & % \btau\cdot G_{m-1}\\
% | & | & & | \\
%\end{pmatrix} \end{align}

Compute $\A\in\Gr_1^m$, $\VV\in\Gr_2^n$ and $\Vp\in\Gr_2$: \begin{align} \term{\A} &\gets (\alpha\cdot\G) %\\
\bydef (\alpha\cdot G_0, \alpha\cdot G_1,\ldots,\alpha\cdot G_{m-1})\in\Gr_1^m\\
%&\bydef (A_0,\ldots,A_{m-1})\\
\term{\VV} &\gets (\btau\cdot \V) %\\
\bydef (\tau_0\cdot \V, \tau_1\cdot \V,\ldots,\tau_{n-1}\cdot \V)\in\Gr_2^n\\
%&\bydef (V_0,\ldots,V_{n-1})\\
\term{\Vp} &\gets \alpha\cdot \V\in\Gr_2\\
\end{align}

Return the VK and proving key:

• $\vk\gets (\Vp,\VV,\A)$
• $\ck\gets (\A, \H)$³

Interestingly, the $G_i$’s and $\V$ generators are not needed in the $\ck$ (when proving) nor in the $\vk$ (when verifying), although the KZH paper does include them. They would indeed be useful when trying to verify correctness of the $\ck$ and $\vk$.

$\mathsf{KZH}_2.\mathsf{Commit}(\mathsf{ck}, f(\boldsymbol{X},\boldsymbol{Y})) \rightarrow (C, \mathsf{aux})$

Parse the $\ck$ as: \begin{align} ((\cdot,\cdot, \A), \H) &\parse \ck,\ \text{where:}\\
\A &= %(A_j)_{j\in[m)} = \alpha\cdot \G %=(\alpha\cdot G_j)_{j\in[m)} \\
\H &= %(H_{i,j})_{i\in[n),j\in[m)} = (\tau_i \cdot \G)_{i\in[n)} \end{align}

Let $\term{\vec{f_i}\bydef(f(\i,\j))_{\j \in\bin^\mu}}$ denote the $i$th row of the matrix encoded by $f$. Compute the full commitment to $f$ (via a single $\msm{N}_1$): \begin{align} \term{C} \gets \sum_{i \in [n)} \sum_{j\in [m)} f(\i, \j)\cdot H_{i,j} \bydef \emph{\sum_{i\in [n)} \vec{f_i} \cdot \mat{H}_i} \in \Gr_1 \end{align}

Compute the $n$ row commitments of $f$ (via $n$ $\msm{m}_1$): \begin{align} \term{D_i} \gets \sum_{j\in[m)} f(\i, \j) \cdot A_j \bydef \emph{\vec{f_i} \cdot \A}\in\Gr_1 , \forall i\in[n) \end{align}

Set the auxiliary info to be these $n$ row commitments:

• $\term{\aux}\gets (D_i)_{i\in[n)}\in\Gr_1^n$

$\mathsf{KZH}_2.\mathsf{Open}(f(\boldsymbol{X},\boldsymbol{Y}), (\boldsymbol{x}, \boldsymbol{y}), z; \mathsf{aux})\rightarrow \pi$

Partially-evaluate $f\in \MLE{\nu,\mu}$: \begin{align} \term{f_\x(\Y)} &\gets f(\x, \Y) \in \MLE{\mu} \\
\label{eq:fxY} &\bydef \sum_{i\in[n)} \eq(\x, \i) f(\i,\Y) \end{align}

Return the proof⁴:

• $\pi \gets (f_\x, \aux) \in \F^m \times \Gr_1^n$

When $\x\in\bin^\nu$ and $\y\in{\bin^\mu}$, the step above involves zero work: $f_\x(\Y)$ is just the $x$th column in the matrix encoded by $f$. Furthermore, $z=f(\x,\y)$ is simply the entry at location $(x,y)$ in the matrix. When $\x$ is an arbitrary, point, computing all the $\eq(\x, \i)$’s requires $\Fmul{2n}$ (see here). Then, assuming a Lagrange-basis representation for all $f(\i,\Y)$ rows, combining them together as per Eq. \ref{eq:fxY} will require (1) $\Fmul{m}$ for each row $i$ to multiply $\eq(\x, \i)$ by $f(\i,\Y)$ and (2) $\Fadd{(n-1)m}$ to add all multiplied rows together. So, $\Fmul{n(m + 2)} + \Fadd{(n-1)m}$ in total.

$\mathsf{KZH}_2.\mathsf{Verify}(\mathsf{vk}, C, (\boldsymbol{x}, \boldsymbol{y}), z; \pi)\rightarrow \{0,1\}$

Parse the VK and the proof: \begin{align} (\Vp,\VV,\A) &\parse \vk\\
(f_\x,\aux) &\parse\pi\\
(D_i)_{i\in[n)} &\parse \aux\\
\end{align}

Check the row commitments are consistent with the full commitment (via a multipairing $\multipair{n+1}$): \begin{align} e(C, \Vp) \equals \sum_{i\in[n)} e(D_i, V_i)\Leftrightarrow\\
\end{align}

Check the auxiliary data: \begin{align} \label{eq:kzh2-verify-aux} \sum_{j\in[m)} f_\x(\j) \cdot A_j \equals \sum_{i\in[n)}\eq(\x, \i) \cdot D_i\Leftrightarrow \end{align}

Check $z$ against the partially-evaluated $f_\x$: \begin{align} z\equals f_\x(\y) \end{align}

Assuming $f_\x$ is received in Lagrange basis, computing all $f_\x(\j)$ is just fetching entries. Therefore, the LHS of the auxiliary check from Eq. \ref{eq:kzh2-verify-aux} always involves an $\msm{m}_1$. When $(\x,\y)$ are on the hypercube (1) the RHS is a single $\Gr_1$ scalar multiplication which can be absorbed into the MSM on the LHS and (2) the last check on $z$ involves simply fetching the $y$th entry in $f_\x$. When $(\x,\y)$ are arbitrary, the RHS involves $\Fmul{2n}$ to evaluate all $\eq(\x,\i)$ Lagrange polynomials (see here) and then an $\msm{n}_1$ which can be absorbed into the LHS. The final check involves evaluating the $f_\x$ MLE at an arbitrary $\y$. This involves evaluating all $\eq(\y,\j)$ Lagrange polynomials in $\Fmul{2m}$ time and then taking a dot product in $\Fmul{m} + \Fadd{m}$ time.

Correctness

The first check is correct because: \begin{align} e(C, \Vp) &\equals \sum_{i\in[n)} e(D_i, V_i)\Leftrightarrow\\
e\left(\sum_{i\in[n)} \vec{f_i} \cdot \mat{H}_i, \alpha\cdot \V\right) &\equals \sum_{i\in[n)} e\left(\vec{f_i} \cdot \A, \tau_i \cdot \V\right) \Leftrightarrow \\
\sum_{i\in[n)} e\left(\vec{f_i} \cdot \mat{H}_i, \alpha\cdot \V\right) &\equals \sum_{i\in[n)} e\left(\vec{f_i} \cdot \A, \tau_i \cdot \V\right) \Leftrightarrow \\
\sum_{i\in[n)} e\left((\vec{f_i} \cdot \tau_i) \cdot \G, \alpha\cdot \V\right) &\equals \sum_{i\in[n)} e\left((\vec{f_i} \cdot \alpha)\cdot \G, \tau_i \cdot \V\right) \Leftrightarrow \\
\sum_{i\in[n)} e\left(\vec{f_i} \cdot \G, (\alpha\cdot \tau_i)\cdot \V\right) &\goddamnequals \sum_{i\in[n)} e\left(\vec{f_i} \cdot \G, (\alpha\cdot \tau_i)\cdot \V\right) \end{align}

The second check is correct because: \begin{align} \sum_{j\in[m)} f_\x(\j) \cdot A_j &\equals \sum_{i\in[n)}\eq(\x, \i) \cdot D_i\Leftrightarrow\\
\alpha \sum_{j\in[m)} f(\x, \j) \cdot G_j &\equals \sum_{i\in[n)}\eq(\x, \i) \cdot \left( \sum_{j\in [m)} f(\i,\j) \cdot A_j \right)\Leftrightarrow\\
\alpha \sum_{j\in[m)} \sum_{i\in[n)} \eq(\x,\i) f(\i, \j) \cdot G_j &\equals \sum_{i\in[n)}\eq(\x, \i) \cdot \alpha \left( \sum_{j\in [m)} f(\i,\j) \cdot G_j \right)\Leftrightarrow\\
\alpha \sum_{i\in[n)} \sum_{j\in[m)} \eq(\x,\i) f(\i, \j) \cdot G_j &\equals \alpha \sum_{i\in[n)}\eq(\x, \i) \cdot \left( \sum_{j\in [m)} f(\i,\j) \cdot G_j \right)\Leftrightarrow\\
\alpha \sum_{i\in[n)} \sum_{j\in[m)} \eq(\x,\i) f(\i, \j) \cdot G_j &\goddamnequals \alpha \sum_{i\in[n)} \sum_{j\in [m)} \eq(\x, \i) f(\i,\j) \cdot G_j \end{align}

Efficient instantiation

Typically, when commiting to a size-$N$ MLE, the scheme is most-efficiently set up with $n = m = \sqrt{N} = 2^s$ via $\kzhSetup{2}(1^\lambda, s, s)$. (Assuming $\sqrt{N}$ is a power of two, for simplicity here; otherwise, must pad.)

Performance

Include vanilla $\kzhK(d)$, explaining eval proofs for hypercube and for non-hypercube points and how $\kzh{\log_2{N}}$ yields $2\log_2{N}$-sized proofs. Include the optimized variant of $\kzhK(d)$.

We use $\kzhTwo^{n,m}$ to refer to the $\kzhTwo$ scheme set up with $\kzhSetup{2}(1^\lambda, \log_2{n},\log_2{m})$ We use $\kzhTwoSqr$ to refer to $\kzhTwo^{\sqN,\sqN}$.

Setup, commitments and proof sizes:

Openings at arbitry points:

Openings at points on the hypercube:

For “Open time (random)” the time should technically have $\Fmul{n(m+2)} + \Fadd{(n-1)m}$ instead, but it’s peanuts, so ignoring.

References

For cited works, see below 👇👇