Nearly Optimal Dynamic Set Cover: Breaking the Quadratic-in-$f$ Time Barrier

TL;DR

This paper breaks the $\Omega(f^2)$ update time barrier via the following results: $(1+\epsilon)f$-approximation can be maintained deterministically in O and the algorithm works against an adaptive adversary.

Abstract

The dynamic set cover problem has been subject to extensive research since the pioneering works of [Bhattacharya et al, 2015] and [Gupta et al, 2017]. The input is a set system $(U, S)$ on a fixed collection $S$ of sets and a dynamic universe of elements, where each element appears in a most $f$ sets and the cost of each set lies in the range $[1/C, 1]$, and the goal is to efficiently maintain an approximately-minimum set cover under insertions and deletions of elements. Most previous work considers the low-frequency regime, namely $f = O(\log n)$, and this line of work has culminated with a deterministic $(1+ε)f$-approximation algorithm with amortized update time $O(\frac{f^2}{ε^3} + \frac{f}{ε^2}\log C)$ [Bhattacharya et al, 2021]. In the high-frequency regime of $f = Ω(\log n)$, an $O(\log n)$-approximation algorithm with amortized update time $O(f\log n)$ was given by [Gupta et al, 2017]. Interestingly, at the intersection of the two regimes, i.e., $f = Θ(\log n)$, the state-of-the-art results coincide: approximation $Θ(f) = Θ(\log n)$ with amortized update time $O(f^2) = O(f \log n) = O(\log^2 n)$. Up to this date, no previous work achieved update time of $o(f^2)$. In this paper we break the $Ω(f^2)$ update time barrier via the following results: (1) $(1+ε)f$-approximation can be maintained in $O\left(\frac{f}{ε^3}\log^*f + \frac{f}{ε^3}\log C\right) = O_{ε,C}(f \log^* f)$ expected amortized update time; our algorithm works against an adaptive adversary. (2) $(1+ε)f$-approximation can be maintained deterministically in $O\left(\frac{1}εf\log f + \frac{f}{ε^3} + \frac{f\log C}{ε^2}\right) = O_{ε,C}(f \log f)$ amortized update time.

Figures (2)

• Figure 1: An illustration of a single iteration of the while loop of the $\mathsf{FixLevel}(e, l)$ subroutine, which raises set $s$ by one level. Depicted by black circles are three active elements $e_1, e_5, e_6$, and depicted by white circles are four passive elements $e, e_2, e_3, e_4$ (note that $e$ is passive). The small dashed rectangle represents $\mathsf{zlev}(e)$, which is equal to $\mathsf{lev}(e)$. There are three sets: $s = \{e, e_1, e_2, e_3, e_4, e_5\}$ colored red, $s_1 = \{e_1, e_2\}$ colored green, and $s_2 = \{e, e_3, e_4, e_5, e_6\}$ colored blue. The level of all three sets is $k$. The left and right parts of the figure illustrate the states right before and after the iteration, respectively. During the iteration, $e_1$ and $e_5$ were raised from level $k$ to level $k + 1$ and they remain active. Elements $e_2$ and $e_4$ became active, since they were passive elements at level $k + 1$, and since they belong to $s$. Since the lazy level of $e$ was $k$, both its lazy and intrinsic levels got raised by one, so the gap between them remains equal to $d$. The level of $s$ was raised to $k + 1$.
• Figure 2: In this illustration, the dirty elements are $e_1, e_2, e_3$, and the clean elements, which have been raised to level $k+1$, are $e_4, e_5, e_6, e_7, e_8$. The $\mathsf{Rebuild}(k)$ procedure reduces the gaps $\mathsf{ilev}(e_i) - \mathsf{zlev}(e_i), \forall i\in \{1, 2, 3\}$. In the deterministic algorithm, the gaps decrease by at least one; in the randomized algorithm, the gaps decrease exponentially (i.e., from $d$ to $\log d$).

Theorems & Definitions (97)

• Theorem 1.1
• Definition 2.1
• Lemma 2.1
• proof
• Definition 2.2
• Corollary 2.1
• proof
• Lemma 2.2: bhattacharya2021dynamic
• proof
• Lemma 2.3: bhattacharya2021dynamic