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Setup. Let $\mathbb{F}$ be a finite field with a multiplicative subgroup $E = \{e_1, \dots, e_k\}$ of order $k$. Given a list $y = y_1, \dots, y_k\in \mathbb{F}$ let $p$ be the unique polynomial of degree $k-1$ s.t. $p(e_i) = y_i$ for all $i\in [k]$. Let $E'=\{e'_1, \dots, e'_t\}\not\subset E$ be a additive coset of a subgroup of $E$ (i.e. $t < k$).

Problem. Given $y$ as input, can we compute $p(e'_i)$ for all $i\in [t]$ in time $O_{\log|\mathbb{F}|}(k + t\;\mathrm{polylog}\;k)$? That is, if $t = k / \mathrm{polylog}(k)$, the computation would run in time $O_{\log|\mathbb{F}|}(k)$. Note that if $t = O(k)$, FFT solves this problem in time $O_{\log|\mathbb{F}|}(k + k \;\mathrm{polylog}\; k) = O(k \;\mathrm{polylog}\; k)$.

Relaxation. It would be still be interesting to solve this problem over any finite field $\mathbb{F}$ and any (disjoint) subsets $E,E'\subset \mathbb{F}$ of size $k,t$, respectively, in time $O_{\log|\mathbb{F}|}(k\log\log k + t\;\mathrm{polylog}\;k)$.

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  • $\begingroup$ What is $n$? Is it the size of $𝔽$? $\endgroup$
    – Bruno
    Commented Jul 12 at 14:02
  • $\begingroup$ @Bruno Sorry, $n$ was supposed to be $k$. I updated the question to reflect that -- thanks! $\endgroup$ Commented Jul 13 at 20:08
  • $\begingroup$ Perhaps you can compute $p \pmod{(x-e_1')(x-e_2')\cdots}$? $\endgroup$ Commented Jul 14 at 11:27
  • $\begingroup$ Thanks for the suggestion @CommandMaster! I agree finding $r = p \mod \prod_{i\in t}(x - e_i')$, and evaluating $r$ on $E'$ (in time $t\mathrm{polylog} \;t)$ is a valid approach. However, it is not clear to me how to find $r$ within the requested time bound. One can find the remainder of a degree $k$ polynomial by a degree $O(k)$ polynomial in time $O(k\mathrm{polylog} k)$ using [Sieveking-Kung division])(people.csail.mit.edu/madhu/ST12/scribe/lect06.pdf), but that does not meet our time bound. Not clear how to optimize for reducing by a smaller degree $t$ polynomial. $\endgroup$ Commented Jul 15 at 2:27

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