I am going through Dan Boneh's video on PLONK - https://www.youtube.com/watch?v=LbpPCN-f_XA&t=952s

At around 19 minutes, he gets to the Prod Check Gadget.


$\omega \in F_p$ is the primitive $k$th root of unity (i.e. $\omega^{k} = 1$)

$\Omega = \{1, \omega, \omega^{2}, ..., \omega^{k-1}\}$

Then he sets $t \in {F_p}^{(\le k)}[X]$ to be the degree-$k$ polynomial such that

$t(1) = f(1)$ and $t(\omega^s) = \prod_{i=0}^s f(\omega^i)$ for $s = 1, ..., k-1$.

So $t(\omega) = f(1)\cdot f(\omega)$

$t(\omega^2) = f(1)\cdot f(\omega)\cdot f(\omega^2)$


$t(\omega^{k-1})= f(1)\cdot f(\omega)\cdot f(\omega^2)\dots f(\omega^{k-2})\cdot f(\omega^{k-1})$

In the next slide, he says that the Prover should construct the polynomial $t(X) \in {F_p}^{(\le k)}$ & he will use this polynomial $t(X)$ for the proof.

I am confused as to how do I construct a recursive polynomial like $t(X)$.

I know how to calculate the value of $t(X)$ for any value of $X \in \Omega$ - that's pretty simple.

i.e. if I want to calculate $t(\omega^3)$, it can be calculated as

$t(\omega^3) = f(1) \cdot f(\omega) \cdot f(\omega^2) \cdot f(\omega^3)$

However, how do I construct a univariate polynomial $t$ in a way that the Prover can send a commitment of this polynomial to the Verifier?

Let's say my $f$ polynomial is

$f(X) = 3X^2 + 4X + 7$

How do I go from here to constructing $t(X)$?


1 Answer 1


To construct $t(X)$, you can use the Lagrange interpolation method.

Suppose $F=GF(17)$ and $w=4$ is an $4^{th}$ roots of unity in $F$. (I'm taking your example from this post )

Now $f(X)=3X^2+4X+7$ and $\Omega=\{1,4,16,13\}$.

Then $t(X)$ will be $6X^3 + 13X^2 + 11X + 1$.

Here is the sage code for this:



6*x^3 + 13*x^2 + 11*x + 1
  • $\begingroup$ Thank you very much for the answer. If you get time, could you also look at this question of mine - crypto.stackexchange.com/questions/105901/… - my question was only about $\prod_{a \in \Omega} f(a) = 1$ & the answer seems satisfactory however, in the comments, I have also discussed $\prod_{a \in \Omega} f(a) = m$ - I am not so convinced about the answer for that. If you have any thoughts, do let me know. Let me know if I should ask a separate question for $\prod_{a \in \Omega} f(a) = m$? $\endgroup$
    – user93353
    Apr 8, 2023 at 8:40

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