Negative result: Non-ZK DeKART range proof

This blog post describes a flawed, non-ZK range range proof based on univariate polynomials. (For an actually-ZK scheme, see this post.)

Introduction

In a short blog post¹, Borgeaud describes a very simple range proof for a single value $z$, which we summarize here. In the summer of 2024, we observed that Borgeaud’s elegant protocol very efficiently extends to batch-proving many values². This ultimately led to a ZK range proof scheme based on multilinear polynomials and sumcheck, which is fully described in our academic paper²,

However, initially, we started with a ZK range proof based on univariate polynomials and KZG commitments.

This blog post describes that range proof and explains why it cannot be ZK in Type III bilinear groups³.

Preliminaries

• Univariate polynomials
• KZG polynomial commitments
• MLEs
• $[0,n]\bydef\{0,1,\ldots,n\}$
• $[\ell)\bydef\{0,1,\ldots,\ell-1\}$
• $\F$ is a finite field of prime order $p$
• $r\randget S$ denotes randomly sampling from a set $S$
• We use $\one{a}\bydef a\cdot G_1$ and $\two{b}\bydef b\cdot G_2$ and $\three{c}\bydef c\cdot G_\top$ to denote scalar multiplication in bilinear groups $(\Gr_1,\Gr_2,\Gr_\top)$ with generators $G_1,G_2,G_\top$, respectively (i.e., additive group notation).
• We use “small-MSM” to refer to multi-scalar multiplications (MSMs) where the scalars are small; we use “L-MSM” to refer to ones where the scalars are large
• We use $a\fsget S$ to denote sampling from a set $S$ in a deterministic manner using the Fiat-Shamir transcript $\FS$ derived so far

• We use $\mathcal{A}^\FSo$ to indicate that an algorithm $\mathcal{A}$ has oracle access to the Fiat-Shamir transcript $\FS$ and can read and append to it.

• We will often interpolate polynomials over the FFT basis $\term{\H}\bydef\{\omega^0,\omega^1,\ldots,\omega^n\}$, where $\term{\omega}$ is a primitive $(n+1)$th root of unity.

Borgeaud’s unbatched range proof

The verifier has a homomorphic commitment (e.g., a KZG commitment) to a polynomial $\term{f(X)}$ that encodes a value $\term{z}$ as: \begin{align} f(X) = \term{r}X + z \end{align} where $\emph{r}\randget \F$ is a blinder. Note that $f(0)=z$. The prover wants to prove that $z$ is an $\term{\ell}$-bit value: i.e., that $z\in[0,2^\ell)$.

Prover commits to each bit $\term{z_j}$ of $z\bydef \sum_{j\in[\ell)} z_j 2^j$ via $\ell$ blinded polynomials: \begin{align} \term{f_j(X)} = \term{r_j} X + z_j \end{align} where $\emph{r_j}\randget \F$ is a blinder. Note that $f_j(0)=z_j$. As a result, we have: \begin{align} \label{eq:f} f(X) = \sum_{j\in[\ell)} f_j(X) 2^j \end{align}

For Eq. \ref{eq:f} to hold, the prover picks $r_j$’s randomly but correlates them such that: \begin{align} r = \sum_{j\in[\ell)} r_j 2^j \end{align} We denote this by $(r_j)_{j\in[\ell)}\randget \term{\correlate{r, \ell}}$.

Note that, if the prover gives the $f_j$ commitments to the verifier, then the verifier can combine them homomorphically and obtain’s $f$’s commitment. This assures the verifier that $z=\sum_{j\in[\ell)} z_j 2^j$. Thus the prover’s remaing work is to prove that $z_j\in\{0,1\},\forall j\in[\ell)$.

The key observation is that $z_j\in\{0,1\}$ can be reduced to a claim about the $f_j$’s: \begin{align} z_j\in\{0,1\} \Leftrightarrow\\
f_j(0) \in \{0,1\} \Leftrightarrow\\
\begin{cases} f_j(0) = 0 \lor{}\\
f_j(0) = 1\\
\end{cases}\Leftrightarrow\\
\begin{cases} (X - 0) \mid f_j(X) - 0 \lor {}\\
(X - 0) \mid f_j(X) - 1\\
\end{cases}\Leftrightarrow\\
\emph{X \mid f_j(X) (f_j(X) - 1)} %\Leftrightarrow\\\ \end{align} The last claim can be proved by showing there exists a quotient polynomial $\term{h_j(X)}$ such that: \begin{align} X\cdot h_j(X) = f_j(X)\left(f_j(X) - 1\right) \end{align}

For example, if the verifier has KZG commitments to the $f_j$’s, the verifier can be given a KZG commitment to $h_j$ and use a pairing-check to enforce the above relation: \begin{align} \pair{\one{h_j(\tau)}}{\two{\tau}} &= \pair{\one{f_j(\tau)}}{\two{f_j(\tau)}-\two{1}},\forall j\in[\ell) \end{align} (Note that $\term{\tau}$ denotes the KZG trapdoor here.) For this to work, the verifier must verify “duality” of the $\Gr_1$ and $\Gr_2$ commitments to $f_j$: \begin{align} \label{eq:duality} \pair{\one{f_j(\tau)}}{\two{1}} &= \pair{\one{1}}{\two{f_j(\tau)}},\forall j\in[\ell) \end{align}

(Likely-not-ZK) univariate batched range proof

We observe that the $f$ and $f_j$ polynomials could be re-defined to store the bits of $n$ different values $\term{z_0, \ldots, z_{n-1}}$ without affecting the proof size and verifier time too much!

Firt, we store all the $n$ values in the polynomial $\emph{f}$, now of degree $n$: \begin{align} \label{eq:f-batched} f(\omega^i) &= z_i,\forall i\in[n)\\
f(\omega^n) &= r \end{align} where $r\randget \F$ is a blinding factor as before.

Let $\term{z_{i,j}}$ denote the $j$th bit of the $i$th value $z_i \bydef \sum_{i\in[\ell)} z_{i,j} 2^j$. Second, we store the $j$th bit of all the values in the $\emph{f_j}$ polynomials, now of degree $n$ too: \begin{align} f_j(\omega^i) &= z_{i,j},\forall i\in[n)\\
f_j(\omega^n) &= r_j \end{align} where $(r_j)_{j\in[\ell)} \randget \correlate{r, \ell}$ are randomly picked as before and $\term{\omega}$ is a $(n+1)$th primitive root of unity in $\F$.

We will typically (and a bit awkwardly) require that $n \gets 2^k-1$, since many fields $\F$ of interest have prime order $p$ where $p-1$ is divisible by $2^k\bydef n+1$ and thus will admit an $(n+1)$th primitive root of unity. e.g., BLS12-381 admits a root of unity for $k=32$.

Importantly, note that Eq. \ref{eq:f} still holds for these (redefined) $f$ and $f_j$ polynomials! \begin{align} f(X) = \sum_{j\in[\ell)} f_j(X) 2^j \end{align}

The key observation is that proving that $f_j$ stores bits is equivalent to: \begin{align} f_j(X) \in \{0,1\}, \forall X\in \H\setminus\{\omega^n\}\Leftrightarrow \\
\left. \vanish\ \middle|\ f_j(X)\left(f_j(X) - 1\right) \right. \end{align} This, in turn, is equivalent to proving there exists a quotient polynomial $\term{h_j(X)}$ of degree $2n - n = n$ such that: \begin{align} \label{eq:binarity-check} \vanish\cdot h_j(X) = f_j(X)\left(f_j(X) - 1\right) \end{align}

To verify Eq. \ref{eq:binarity-check} holds, the verifier could check, for each $j\in[\ell)$, that: \begin{align} \label{eq:hj-inefficient} \pair{\one{h_j(\tau)}}{\two{\frac{\tau^{n+1} - 1}{\tau - \omega^n}}} &= \pair{\one{f_j(\tau)}}{\two{f_j(\tau)}-\two{1}},\forall j\in[\ell) \end{align} For this to work, the verifier would also verify “duality” of the $\Gr_1$ and $\Gr_2$ commitments to $f_j$ as per Eq. \ref{eq:duality}.

For performance, the verifier could pick random challenges $\term{\beta_j}\randget\F$ and combine all the checks from Eq. \ref{eq:hj-inefficient} into one: \begin{align} \label{eq:hj} \pair{\underbrace{\sum_{j\in[\ell)} \beta_j \cdot \one{h_j(\tau)}}_{\term{D}}}{\two{\frac{\tau^{n+1} - 1}{\tau - \omega^n}}} &= \sum_{j\in[\ell)} \pair{\beta_j \cdot \one{f_j(\tau)}}{\two{f_j(\tau)}-\two{1}} \end{align} (A similar trick can be applied for the duality check as well from Eq. \ref{eq:duality}. Furthermore, everything can be combined into a single multi-pairing.)

Next, instead of asking for the individual $h_j(X)$ commitments, the verifier will send the $\beta_j$’s to the prover and expect to receive just the commitment $\emph{D}$ to the random linear combination of the $h_j$’s. This reduces proof size and makes the check in Eq. \ref{eq:hj} slightly faster: \begin{align} \label{eq:verification} \pair{D}{\two{\frac{\tau^{n+1} - 1}{\tau - \omega^n}}} &= \sum_{j\in[\ell)} \pair{\beta_j \cdot \one{f_j(\tau)}}{\two{f_j(\tau)}-\two{1}} \end{align}

We describe the scheme in detail below.

$\widetilde{\mathsf{Dekart}}^\mathsf{FFT}.\mathsf{Setup}(1^\lambda, n)\rightarrow \mathsf{prk},\mathsf{vk}$

Generate powers of $\tau$ up to and including $\tau^n$:

• $\term{\tau}\randget\F$
• $\term{\omega} \gets$ a primitive $(n+1)$th root of unity in $\F$
• $\term{\H}\bydef\{\omega^0,\omega^1,\ldots,\omega^n\}$

Let $\term{\lagr_i(X)} \bydef \prod_{j\in\H, j\ne i} \frac{X - \omega^j}{\omega^i - \omega^j}$ denote the $i$th Lagrange polynomial, for $i\in[0, n]$.

Return the public parameters:

• $\vk\gets \left(\tauTwo,\vanishTwo\right)$
• $\prk\gets \left(\vk, \left(\ellOne{i}\right)_{i\in[0,n]}, {\left(\ellTwo{i}\right)_{i\in[0,n]}}\right)$

$\widetilde{\mathsf{Dekart}}^\mathsf{FFT}.\mathsf{Commit}(\mathsf{prk},z_0,\ldots,z_{n-1}; r)\rightarrow C$

The same as $\dekartUni.\mathsf{Commit}$, but ignores the extra $\ellTwo{i}$ parameters in the $\prk$, of course.

$\widetilde{\mathsf{Dekart}}^\mathsf{FFT}.\mathsf{Prove}^{\mathcal{FS}(\cdot)}(\mathsf{prk}, C, \ell; z_0,\ldots,z_{n-1}, r)\rightarrow \pi$

Recall $\emph{z_{i,j}}$ denotes the $j$th bit of each $z_i\in[0,2^\ell)$.

• $\left(\vk, \left(\ellOne{i}\right)_{i\in[0,n]}, \left(\ellTwo{i}\right)_{i\in[0,n]}\right)\parse\prk$
• $(r_j)_{j\in[n)} \randget \correlate{r, \ell}$
• $C_j \gets r_j\cdot \ellOne{n} + \sum_{i\in[n)} z_{i,j}\cdot \ellOne{i} \bydef \one{\emph{f_j(\tau)}},\forall j\in[\ell)$
• ${\tilde{C}_j \gets r_j \cdot \ellTwo{n} + \sum_{i\in[n)} z_{i,j}\cdot \ellTwo{i} \bydef \two{\emph{f_j(\tau)}},\forall j\in[\ell)}$
• add $(\vk, C, \ell, (C_j, \tilde{C}_j)_{j\in[\ell})$ to the $\FS$ transcript
• $h_j(X)\gets \frac{f_j(X)(f_j(X) - 1)}{(X^{n+1} - 1) / (X-\omega^n)} = \frac{(X-\omega^n)f_j(X)(f_j(X) - 1)}{X^{n+1} - 1},\forall j \in[\ell)$
• $(\term{\beta_j})_{j\in[\ell)} \fsget \{0,1\}^\lambda$
• $\term{h(X)}\gets \sum_{j\in[\ell)} \beta_j \cdot h_j(X) = \frac{\sum_{j\in[\ell)}\beta_j (X-\omega^n)f_j(X)(f_j(X) - 1)}{X^{n+1} - 1}$
• $D \gets \sum_{i\in[0,n]} h(\omega^i) \cdot \ellOne{i} \bydef \one{\emph{h(\tau)}}$
  • Note: We discuss above how to interpolate these efficiently!
• $\term{\pi}\gets {\left(D, (C_j,\tilde{C}_j)_{j\in[\ell)}\right)}$

Proof size and prover time

Proof size is trivial: $(\ell+1)\Gr_1 + \ell \Gr_2$ group elements $\Rightarrow$ independent of the batch size $n$, but linear in the bit-width $\ell$ of the values.

Prover time is:

• $\ell n$ $\Gr_1$ $\textcolor{green}{\text{additions}}$ for each $c_j, j\in[\ell)$
• $\ell$ $\Gr_1$ scalar multiplications to blind each $c_j$ with $r_j$
• $\ell n$ $\Gr_2$ $\textcolor{green}{\text{additions}}$ for each $\tilde{c}_j, j\in[\ell)$
• $\ell$ $\Gr_2$ scalar multiplications to blind each $\tilde{c}_j$ with $r_j$
• $O(\ell n\log{n})$ $\F$ multiplications to interpolate $h(X)$
  • See break down here.
• 1 size-$(n+1)$ L-MSM for committing to $h(X)$

$\widetilde{\mathsf{Dekart}}^\mathsf{FFT}.\mathsf{Verify}^{\mathcal{FS}(\cdot)}(\mathsf{vk}, C, \ell; \pi)\rightarrow \{0,1\}$

Step 1: Parse the $\vk$ and the proof $\pi$:

• $\left(\tauTwo,\vanishTwo\right) \parse \vk$
• $\left(D, (C_j,\tilde{C}_j)_{j\in[\ell)}\right) \parse \pi$

Step 2: Make sure the radix-2 decomposition is correct:

• assert $C \equals \sum_{j=0}^{\ell-1} 2^j \cdot C_j$

Step 3: Make sure the $C_j$’s are bit commitments:

• add $(\vk, C, \ell, (C_j, \tilde{C}_j)_{j\in[\ell})$ to the $\FS$ transcript
• $(\term{\beta_j})_{j\in[\ell)} \fsget \{0,1\}^\lambda$
• assert ${\pair{D}{\vanishTwo} \equals \sum_{j\in[\ell)}\pair{\beta_j\cdot C_j}{\tilde{C}_j - \two{1}}}$

Step 4: Ensure duality of the $C_j$ and $\tilde{C}_j$ bit commitments:

• ${(\term{\alpha_j})_{j\in[\ell)} \fsget \{0,1\}^\lambda}$
• assert ${\pair{\sum_{j\in[\ell)} \alpha_j \cdot C_j}{\two{1}} \stackrel{?}{=} \pair{\one{1}}{\sum_{j\in[\ell)} \alpha_j \cdot \tilde{C}_j}}$

The two pairing equation checks above can be combined into a single size $\ell+3$ multipairing by picking a random $\term{\gamma}\in\F$ and checking: \begin{align} \pair{\sum_{j\in[\ell)} \alpha_j\cdot C_j}{-\two{\emph{\gamma}}} + \pair{\one{\emph{\gamma}}}{\sum_{j\in[\ell)} \alpha_j\cdot \tilde{C}_j} + {} \\\ {} + \pair{-D}{\vanishTwo} + \sum_{j\in[\ell)} \pair{\beta_j \cdot C_j}{\tilde{C}_j - \two{1}} \equals \three{0} \end{align} (Unclear what will be faster: computing $(\one{\gamma},\two{\gamma})$, or multiplying $\gamma$ by the $\alpha_j$’s, which will increase the MSM scalar length from $\lambda$ bits to $2\lambda$. My sense is that the 1st option, which we take above, should be faster.)

Verifier time

The verifier must do:

• size-$\ell$ $\mathbb{G}_1$ small-MSM (i.e., small $2^0, 2^1, 2^2, \ldots, 2^{\ell-1}$ scalars)
• size-$\ell$ $\mathbb{G}_1$ L-MSM for the $C_j$’s (i.e., 128-bit $\alpha_j$ scalars)
• size-$\ell$ $\mathbb{G}_2$ L-MSM for the $\tilde{C}_j$’s (i.e., 128-bit $\alpha_j$ scalars)
• size-$(\ell+3)$ multipairing

Concrete performance

We implemented $\tildeDekartUni$ in Rust over BLS12-381 using blstrs in this pull request⁴. Our code stands to be further optimized. Here are a few benchmarks, for 16-bit ranges, comparing it with Bulletproofs over the faster Ristretto255 curve:

For 32-bit ranges:

Why the odd 4064 and 2032 numbers? From $16\times254$ and $8\times 254$, which arises when chunking 254 ElGamal ciphertexts into into 16 chunks, each encrypting 16-bits.

Some sample MSM times for BLS12-381 here may be useful to make sense of the numbers below. (e.g., a size-4096 MSM in $\Gr_1$ was 6.04 ms on that machine).

In terms where most of the time is spent in $\tildeDekartUni$, here’s a breakdown for $\ell=16$ and $n=2^{12}-1=4095$ (run on a different machine) via ELL=16 N=4095 cargo bench -- range_proof inside aptos-core/crates/aptos-dkg/ in the pull request⁴:

 39.47 ms: All 16 deg-4095 f_j G_1 commitments
           + Each C_j took: 2.467122ms
101.62 ms: All 16 deg-4095 f_j G_2 commitments
           + Each \hat{C}_j took: 6.351356ms
 37.03 ms: All 16 deg-4095 h_j(X) coeffs
  1.07 ms: h(X) as a size-16 linear combination
  6.77 ms: deg-4095 h(X) commitment

The time to commit to the $f_j(X)$’s currently $\bad{dominates}$ but this can be sped up significantly using pre-computation since the scalars are small and the bases $\ellOne{i}$ and $\ellTwo{i}$ are fixed.

Cannot simulate in the Type 3 setting

Suppose, there exists a simulator $\sim$ for the scheme above such that $\sim(\tau, C, \ell)$ outputs a valid proof $\pi \bydef (D, (C_j,\tilde{C}_j)_{j\in[\ell)})$.

Then, the verifier’s last check will pass: \begin{align} \pair{\sum_{j\in[\ell)} \alpha_j \cdot C_j}{\two{1}} \equals \pair{\one{1}}{\sum_{j\in[\ell)} \alpha_j \cdot \tilde{C}_j} \end{align}

In particular, for $\ell = 1$, recall that the verifier ensures that $C \equals \sum_{j\in[\ell)} 2^j\cdot C_j = 2^0 C_0 = C_0$. Furthermore, the pairing check from above will imply: \begin{align} \pair{\alpha_0 \cdot C_0}{\two{1}} &\equals \pair{\one{1}}{\alpha_0 \cdot \tilde{C}_0}\Leftrightarrow\\
\pair{C_0}{\two{1}} &\equals \pair{\one{1}}{\tilde{C}_0}\Leftrightarrow\\
\pair{C}{\two{1}} &\equals \pair{\one{1}}{\tilde{C}_0} \Leftrightarrow\\
\pair{\underbrace{\one{x}}_{\bydef C}}{\two{1}} &\equals \pair{\one{1}}{\underbrace{\two{x}}_{\bydef \tilde{C}_0}} \end{align}

Let $\term{\phi_\tau} : \Gr_1 \rightarrow \Gr_2$, be defined as: \begin{align} \forall G \in \Gr_1, \emph{\phi_\tau(G)} \bydef \tilde{C}_0,\ \text{where}\ (\cdot, (\cdot,\tilde{C}_0))\gets \sim(\tau, G, 1) \end{align}

When $\tau \randget \F$ is randomly picked, $\phi_\tau$ is a homomorphism, since $\phi_\tau(\one{x}) = \two{x}$ for all inputs $\one{x}$. Therefore, the simulator $\sim$ yields a homomorphism $\Rightarrow$ symmetric external Diffie-Hellman (SXDH) would be broken in $(\Gr_1,\Gr_2)$.

Conclusion

Your thoughts or comments are welcome on this thread.

Appendix: Computing $h(X)$

We borrow differentiation tricks from Groth16 to ensure we only do size-$(n+1)$ FFTs. (Otherwise, we’d have to use size-$2(n+1)$ FFTs to compute the $\ell$ different $f_j(X)(f_j(X) - 1)$ multiplications.)

Our goal will be to obtain all $(h(\omega^i))_{i\in[0,n]}$ evaluations and then do a size-$(n+1)$ L-MSM to commit to it and obtain $\emph{D}$.

Recall that: \begin{align} h(X) &= \frac{\sum_{j\in[\ell)}\beta_j \cdot \overbrace{(X-\omega^n)f_j(X)(f_j(X) - 1)}^{\term{N_j(X)}}}{X^{n+1} - 1} \\
&\bydef \frac{\sum_{j\in[\ell)} \beta_j \cdot \emph{N_j(X)}}{X^{n+1} - 1} \Leftrightarrow\\
\Leftrightarrow h(X) (X^{n+1} - 1) &= \sum_{j\in[\ell)} \beta_j \cdot N_j(X) \end{align} Differentiating the above expression: \begin{align} h’(X)(X^{n+1} - 1) + h(X) (n+1)X^n &= \sum_{j\in[\ell)} \beta_j \cdot N_j’(X)\Leftrightarrow\\
\Leftrightarrow h(X) &= \frac{\sum_{j\in[\ell)} \beta_j \cdot N_j’(X) - h’(X)(X^{n+1} - 1)}{(n+1)X^n} \end{align} This reduces computing all $h(\omega^i)$’s to computing all $N_j’(\omega^i)$’s: \begin{align} \label{eq:h} \emph{h(\omega^i)} &= \frac{\sum_{j\in[\ell)} \beta_j \cdot N_j’(\omega^i)}{(n+1)\omega^{in}} \end{align} Our challenge is to compute all the $N_j’(\omega^i)$’s efficiently. Recall that: \begin{align} N_j(X) &\bydef (X-\omega^n)f_j(X)(f_j(X) - 1) \end{align} Differentiating it yields: \begin{align} N_j’(X) &= (X-\omega^n)’ \cdot \left(f_j(X)(f_j(X) - 1)\right) + (X-\omega^n) \left(f_j(X)(f_j(X) - 1)\right)’\Leftrightarrow\\
N_j’(X) &= f_j(X)(f_j(X) - 1) + (X-\omega^n) \left(f_j(X)^2 - f_j(X)\right)’\Leftrightarrow\\
N_j’(X) &= f_j(X)(f_j(X) - 1) + (X-\omega^n) \left(2f_j(X)f’_j(X) - f_j’(X)\right)\Leftrightarrow\\
N_j’(X) &= f_j(X)(f_j(X) - 1) + (X-\omega^n) f_j’(X)\left(2f_j(X)-1\right) \end{align} This reduces computing all $N_j’(\omega^i)$’s to computing all $f_j’(\omega^i)$’s: \begin{align} N_j’(\omega^i) = f_j(\omega^i)(f_j(\omega^i) - 1) + (\omega^i-\omega^n) f_j’(\omega^i)\left(2f_j(\omega^i)-1\right)\Rightarrow\\
\label{eq:nj-prime} \Rightarrow\begin{cases} \emph{N_j’(\omega^i)} &= (\omega^i-\omega^n) f_j’(\omega^i)\left(2f_j(\omega^i)-1\right) = \pm(\omega^i-\omega^n) f_j’(\omega^i), i\in[0,n)\\
\emph{N_j’(\omega^n)} &= f_j(\omega^n)(f_j(\omega^n) - 1) + 0 = r_j (r_j - 1) \end{cases} \end{align}

Time complexity

To compute all $f_j’(\omega^i)$’s for a single $j$:

1. 1 size-$(n+1)$ inverse FFT, for $f_j$’s coefficients in monomial basis
2. $n$ $\F$ multiplications, for the coefficients of the derivative $f_j’$
3. 1 size-$(n+1)$ FFT, for all $f_j’(\omega^i)$’s.

Then, to compute all $N_j’(\omega^i)$’s for a single $j$:

1. $2n+1$ $\F$ multiplications, for all $N’_j(\omega^i)$’s as per Eq. \ref{eq:nj-prime}
   • (Assuming all $\pm (\omega^i - \omega^n)$ are precomputed.)

All the numbers above get multipled by $\ell$, since we are doing this for every $j\in[\ell)$.

Lastly, to compute all the $h(\omega^i)$’s, we do:

1. $\ell n$ $\F$ multiplications to compute the $n$ different numerators from Eq. \ref{eq:h}, one for each evaluation $h(\omega^i)$
2. $n$ $\F$ multiplications, to divide the $n$ numerators by (the precomputed) $(n+1)\omega^{-in}$’s

Doing this $h(X)$ interpolation faster is an open problem, which is why in the paper we explore a multinear variant of DeKART².

References

For cited works, see below 👇👇