A Lossless Deamortization for Dynamic Greedy Set Cover

TL;DR

This work addresses dynamic set cover with a dynamic universe and a fixed set family, achieving a near-optimal approximation while providing strong worst-case update guarantees. The authors introduce a lossless deamortization framework that converts amortized efficiency into worst-case performance, culminating in a $(1+\varepsilon)\ln n$-approximation with update time $O\left(\frac{f\log n}{\varepsilon^2}\right)$ under frequency bound $f$. The approach relies on a fully global algorithm that uses a global notion of dirt and level-based resets, plus a unified deamortization technique that applies to both high- and low-frequency regimes; they further show how to obtain a $(1+\varepsilon)f$-approximation with the same worst-case bound via a primal–dual extension. The results extend to practical implications, including improved bounds for dynamic dominating set and vertex cover via standard reductions, highlighting the broad applicability of lossless deamortization to dynamic covering problems.

Abstract

The dynamic set cover problem has been subject to growing research attention in recent years. In this problem, we are given as input a dynamic universe of at most $n$ elements and a fixed collection of $m$ sets, where each element appears in a most $f$ sets and the cost of each set is in $[1/C, 1]$, and the goal is to efficiently maintain an approximate minimum set cover under element updates. Two algorithms that dynamize the classic greedy algorithm are known, providing $O(\log n)$ and $((1+ε)\ln n)$-approximation with amortized update times $O(f \log n)$ and $O(\frac{f \log n}{ε^5})$, respectively [GKKP (STOC'17); SU (STOC'23)]. The question of whether one can get approximation $O(\log n)$ (or even worse) with low worst-case update time has remained open -- only the naive $O(f \cdot n)$ time bound is known, even for unweighted instances. In this work we devise the first amortized greedy algorithm that is amenable to an efficient deamortization, and also develop a lossless deamortization approach suitable for the set cover problem, the combination of which yields a $((1+ε)\ln n)$-approximation algorithm with a worst-case update time of $O(\frac{f\log n}{ε^2})$. Our worst-case time bound -- the first to break the naive $O(f \cdot n)$ bound -- matches the previous best amortized bound, and actually improves its $ε$-dependence. Further, to demonstrate the applicability of our deamortization approach, we employ it, in conjunction with the primal-dual amortized algorithm of [BHN (FOCS'19)], to obtain a $((1+ε)f)$-approximation algorithm with a worst-case update time of $O(\frac{f\log n}{ε^2})$, improving over the previous best bound of $O(\frac{f \cdot \log^2(Cn)}{ε^3})$ [BHNW (SODA'21)]. Finally, as direct implications of our results for set cover, we [...]

Figures (2)

• Figure 1: On the left: Each black circle represents a single element with level and passive level up to $4$, and is assigned to different collections $A_k$ and $P_k$. Recall that for each element $e$, $\mathsf{plev}(e) \geq \mathsf{lev}(e)$. We can see the following properties: If an element is in $P_k$ it is also in $P_{k'}$ for any $k'>k$. If an element is in $A_k$ it is also in $A_{k'}$ for any $k'<k$ if its level is up to $k'$. For each $k$ we have that $A_k \cup P_k$ is the collection of all elements up to level $k$, and for each $k$$A_k \cap P_k = \emptyset$. On the right: Following a reset up to level $3$, we have that $P_i = \emptyset$ for any $i \leq 3$, and the level of all elements that were at level up to $3$ are now at a level up to $4$, but the passive level of each such element is now at least $4$. The black circles here can represent several elements. Notice that ten elements participated in this reset, and these elements are represented by one of the five black circles. Any element that was in $P_4$ before the reset is still there after.
• Figure 3: Element $e$ is in $\mathcal{U}_l$ and $\mathcal{U}_{l+1}$. Thus it must be covered by a set in $\mathcal{S}_l$, at level $\mathsf{lev}_l(e) \in I_l$, and it must be covered by a set in $\mathcal{S}_{l+1}$, at level $\mathsf{lev}_{l+1}(e) \in I_{l+1}$. Notice the overlap between the different intervals.

Theorems & Definitions (47)

• Theorem 1.1: High-frequency set cover
• Theorem 1.2: Low-frequency set cover
• Theorem 1.3: Dominating set
• Theorem 3.1
• Definition 3.1
• Lemma 3.1
• proof
• Corollary 3.1
• Definition 3.2
• Definition 3.3