To buy or not to buy: deterministic rent-or-buy problems on node-weighted graphs

TL;DR

This paper introduces improved deterministic and randomized algorithms for the rent-or-buy problem on node- and edge-weighted graphs, achieving better competitive ratios than previous methods through a novel charging scheme.

Contribution

It presents new deterministic and randomized algorithms with improved competitive ratios for the rent-or-buy problem on node-weighted graphs, extending previous results and techniques.

Findings

Deterministic algorithm with $O(\log n \log ar{n})$-competitiveness.

Deterministic algorithm with $O(ar{n}\log ilde{k})$-competitiveness.

Randomized algorithm with $O(\log ilde{k} \\log ar{n})$-competitiveness.

Abstract

We study the rent-or-buy variant of the online Steiner forest problem on node- and edge-weighted graphs. For $n$-node graphs with at most $\bar{n}$ non-zero node-weights, and at most $\tilde{k}$ different arriving terminal pairs, we obtain a deterministic, $O(\log n \log \bar{n})$-competitive algorithm. This improves on the previous best, $O(\log^4 n)$-competitive algorithm obtained by the black-box reduction from (Bartal et al. 2021) combined with the previously best deterministic algorithms for the simpler 'buy-only' setting. We also obtain a deterministic, $O(\bar{n}\log \tilde{k})$-competitive algorithm. This generalizes the $O(\log \tilde{k})$-competitive algorithm for the purely edge-weighted setting from (Umboh 2015). We also obtain a randomized, $O(\log \tilde{k} \log \bar{n})$-competitive algorithm. All previous approaches were based on the randomized, black-box reduction from~\cite{AwerbuchAzarBartal96} that achieves a $O(\log \tilde{k} \log n)$-competitive ratio when combined with an algorithm for the 'buy-only' setting. Our key technical ingredient is a novel charging scheme to an instance of \emph{online prize-collecting set cover}. This allows us to extend the witness-technique of (Umboh 2015) to the node-weighted setting and obtain refined guarantees with respect to $\bar{n}$, already in the much simpler 'buy-only' setting.