# Kate commitments from the Lagrange basis without FFTs

Special thanks to Vitalik Buterin and Dan Boneh for rediscovering the barycentric formula, Alessandro Chiesa for discussions on the asymptotics of Pippenger, as well as Dankrad Feist and Dmitry Khovratovich for discussions on the Lagrange SRS FFT. Also thanks to various Ethereum researchers and members of the ZK Study Club for helpful discussions.

TLDR: We show how to commit, evaluate and open polynomials in the Lagrange basis without FFTs.

Notation

Let:

• \mathbb{F} be a prime field
• \lambda = \log(|\mathbb{F}|)
• \omega be a primitive root of unity of order n in \mathbb{F}
• L_i be the Lagrange polynomial for \omega^i, i.e. the minimal polynomial such that L_i(w^i) = 1 and L_i(w^j) = 0 if i \ne j
• e be a pairing over pairing-friendly groups \mathbb{G}_1, \mathbb{G}_2 with base field \mathbb{F}
• \mathbf{g} be a generator of \mathbb{G}_1 or \mathbb{G}_2
• \tau^j\textbf{g} (in the additive notation) for j=0, ..., n-1 be an SRS in the monomial basis (a âpowers of tauâ) where \tau\in\mathbb{F} is secret

Construction

preprocessingâDuring a one-time preprocessing step the prover transforms the SRS from the monomial basis to the Lagrange basis. That is, he computes and caches L_i(\tau)\mathbf{g} for i=0, ..., n-1. This transformation can be done efficiently using a single offline FFT, as presented in Appendix 1.

commitmentsâThe prover has a polynomial f with coefficients in the Lagrange basis, i.e. the prover has evaluations f(\omega^i) for i = 0, ..., n-1. The Kate commitment to f is the linear combination (also known as a multiexponentiation in the multiplicative notation) \sum_{i=0}^{n-1}f(\omega^i)L_i(\tau)\mathbf{g}. The commitment is readily computable from the evaluations f(\omega^i) and the Lagrange SRS elements L_i(\tau)\mathbf{g} without any FFT. When using Pippengerâs algorithm (see discussion in Appendix 3) the cost is O\big(\frac{n\lambda}{\log(n\lambda)}\big) group operations.

evaluationsâLet z be an evaluation point which is not a root of unity, e.g. a random evaluation point. We show how to compute f(z) using a linear number of field operations. When working with roots of unity the barycentric formula yields f(z) = \frac{1 - z^n}{n}\sum_{i=0}^{n-1}\frac{f(\omega^i)\omega^i}{\omega^i - z}. Computing the z^n term in the leading factor \frac{1 - z^n}{n} is negligeable (it costs roughly \log(n) multiplications). It therefore suffices to cheaply compute the inverses of \omega^i - z. We show in Appendix 2 how to do this using 1 inversion and 3(n - 1) multiplications using Montgomery batch inversion.

openingsâWe conclude by showing how to compute opening proofs given the evaluation f(z). The opening proof is the linear combination \sum_{i=0}^{n-1}\frac{f(\omega^i) - f(z)}{\omega^i - z}L_i(\tau)\mathbf{g} which can be computed using Pippenger. Notice that the inverses of \omega^i - z were computed in linear time for the evaluation of f(z) above.

#### Appendix 1âSRS transformation

The goal is to efficiently compute L_i(\tau)\mathbf{g} for i = 0, ..., n-1 from the powers of tau \tau^j\mathbf{g} for j = 0, ..., n-1. Notice that L_i(X) = \frac{1}{n}\sum_{j=0}^{n-1}(\omega^{-i}X)^j so that L_i(\tau)\mathbf{g} = \frac{1}{n}\sum_{j=0}^{n-1}\tau^j\mathbf{g}Y^j\rvert_{Y=\omega^{-i}}. To get the Lagrange SRS it therefore suffices to evaluate the polynomial \frac{1}{n}\sum_{j=0}^{n-1}\tau^j\mathbf{g}Y^j at every root of unity using a single FFT.

#### Appendix 2âbatch inversion

Montgomeryâs batch inversion trick inverts k field elements a_1, ..., a_{k} in four steps:

1. compute \prod_{i = 1}^j a_i for j = 2, ..., k
2. compute 1\big/\prod_{i = 1}^k a_i
3. compute 1\big/\prod_{i = 1}^{j-1} a_i for j = k, ..., 2
4. compute a_j^{-1} = \big(\prod_{i = 1}^j a_i\big)\big(1\big/ \prod_{i = 1}^{j-1} a_i\big) for j = 2, ..., k

Steps 1, 3, 4 each cost k - 1 field multiplications and step 2 costs 1 field inversion.

#### Appendix 3âasymptotics of Pippenger

Pippenger proved that his algorithm is assymptotically optimal. In the context of SNARKs we have \lambda > \log(n) and (except for tiny circuits) n > \lambda so that \frac{n\lambda}{\log(n\lambda)} > \frac{n\log(n)}{\log(n\lambda)} > \frac{n\log(n)}{\log(n^2)} = \frac{n}{2}. Therefore, contrary to folklore, linear combinations cost at least a linear number of group operations, even with Pippenger.

Technically n is at most polynomial in \lambda since provers are polynomially bounded, and therefore linear combinations with Pippenger are asymptotically superlinear. In practice, after setting \lambda to roughly 128 bits, a linear combination with n terms costs roughly 10n group operations.

1 Like

Hey Justin, this is a neat contribution! Thank you!

The f(z) = \frac{1 - z^n}{n}\sum_{i=1}^{n}\frac{f(\omega^i)}{(\omega^i - z)\omega^i} sum should have i from 0 to n-1 (not from 1 to n), no?

Also, Iâm trying to re-derive your formula, but Iâm getting something slightly different, with an \omega^{-i} in the denominator.

Hereâs my (probably flawed) thinkingâŚ

Let A(X)=X^n - 1 and recall that L_i(X) = \frac{A(X)}{A'(\omega^i)\cdot(X-\omega^i)}.
Let A'(X) be the derivative of X^n - 1 and let g(x)=A(X)/(X-\omega^i). Note that A'(\omega^i)=g(\omega^i).

Also, note that:

g(x) = (\omega^i)^0 x^{n-1} + (\omega^i)^1 x^{n-2} + (\omega^i)^2 x^{n-3} + \dots + (\omega^i)^{n-2} x^1 + (\omega^i)^{n-1} x^0

(You can verify this by carrying out the divison yourself.)

Now, we can derive what A'(\omega_i) is:

\begin{align*} A'(\omega^i) = g(\omega^i) &= (\omega^i)^0 \omega^{i(n-1)} + (\omega^i)^1 \omega^{i(n-2)} + (\omega^i)^2 \omega^{i(n-3)} + \dots + (\omega^i)^{n-2} \omega^{i\cdot1} + (\omega^i)^{n-1} \omega^{i\cdot0}\\ &= n\omega^{i(n-1)} = n(\omega^{i\cdot n-i})=n\omega^{-i} \end{align*}

Substituting in the Lagrange interpolation formula (e.g., BT04):

\begin{align*} f(X) &= \sum_{i=0}^{n-1} L_i(X)\cdot f(\omega^i)\Leftrightarrow\\ f(X) &= \sum_{i=0}^{n-1} \frac{A(X)}{A'(\omega^i)\cdot(X-\omega^i)}\cdot f(\omega^i)\Leftrightarrow\\ f(X) &= \sum_{i=0}^{n-1} \frac{(X^n - 1)}{n\cdot\omega^{-i}\cdot(X-\omega^i)}\cdot f(\omega^i)\Leftrightarrow\\ f(X) &= \frac{1}{n} \sum_{i=0}^{n-1} \frac{(X^n - 1)}{(X-\omega^i)\cdot\omega^{-i}}\cdot f(\omega^i)\Leftrightarrow\\ f(X) &= \frac{1-X^n}{n} \sum_{i=0}^{n-1} \frac{f(\omega^i)}{(\omega^i - X)\cdot \omega^{-i}} \end{align*}

Maybe I messed up somewhere? (I am very good at messing up on algebra/arithmetic.)

Since \omega^0 = \omega^n = 1 the two sums \sum_{i=1}^{n} and \sum_{i=0}^{n-1} are equivalent, right? I generally prefer \sum_{i=1}^{n} (no awkward n-1, fewer characters) but will edit my post for consistency.

I think youâre right. It suffices to show that the barycentric formula holds on all roots of unity. Hereâs a short informal argument:

\begin{align*}\lim_{X \to \omega^j}\frac{1 - X^n}{n}\sum_{i=0}^{n-1}\frac{f(\omega^i)\omega^i}{\omega^i - X} = \lim_{X \to \omega^j}\frac{1 - X^n}{n}\frac{f(\omega^j)\omega^j}{\omega^j - X} = f(\omega^j)\end{align*}

Ha, cool! Didnât notice that at all
(Quoting equations seems to not work wellâŚ)

I see. Whatâs the advantage of using limits here rather than normally evaluating \frac{1-X^n}{n}\sum_{i=0}^{n-1} \frac{f(\omega^i)\omega^i}{\omega^i - X} at X=\omega^j?

Oh itâs just because both the 1 - X^n numerator and the \omega^j - X denominator are 0 and you want to handle 0/0 carefully.

Oh, and use LâHĂ´pitalâs rules to compute the limit:

\begin{align*} \lim_{X \to \omega^j}\frac{1 - X^n}{n}\frac{f(\omega^j)\omega^j}{\omega^j - X} &= \frac{f(\omega^j)\omega^j}{n}\lim_{X \to \omega^j}\frac{1 - X^n}{\omega^j - X}\\ &= \frac{f(\omega^j)\omega^j}{n}\lim_{X \to \omega^j}\frac{- nX^{n-1}}{-1}\\ &= f(\omega^j)\omega^j (\omega^j)^{n-1}\\ &= f(\omega^j)\\ \end{align*}

Neat! Thank you.