Budgeted Out-tree Maximization with Submodular Prizes
April 26, 2022 Β· Declared Dead Β· π International Symposium on Algorithms and Computation
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Authors
Gianlorenzo D'Angelo, Esmaeil Delfaraz, Hugo Gilbert
arXiv ID
2204.12162
Category
cs.DS: Data Structures & Algorithms
Citations
5
Venue
International Symposium on Algorithms and Computation
Last Checked
4 months ago
Abstract
We consider a variant of the prize collecting Steiner tree problem in which we are given a \emph{directed graph} $D=(V,A)$, a monotone submodular prize function $p:2^V \rightarrow \mathbb{R}^+ \cup \{0\}$, a cost function $c:V \rightarrow \mathbb{Z}^{+}$, a root vertex $r \in V$, and a budget $B$. The aim is to find an out-subtree $T$ of $D$ rooted at $r$ that costs at most $B$ and maximizes the prize function. We call this problem \emph{Directed Rooted Submodular Tree} (\textbf{DRSO}). Very recently, Ghuge and Nagarajan [SODA\ 2020] gave an optimal quasi-polynomial-time $O\left(\frac{\log n'}{\log \log n'}\right)$-approximation algorithm, where $n'$ is the number of vertices in an optimal solution, for the case in which the costs are associated to the edges. In this paper, we give a polynomial-time algorithm for \textbf{DRSO} that guarantees an approximation factor of $O(\sqrt{B}/Ξ΅^3)$ at the cost of a budget violation of a factor $1+Ξ΅$, for any $Ξ΅\in (0,1]$. The same result holds for the edge-cost case, to the best of our knowledge this is the first polynomial-time approximation algorithm for this case. We further show that the unrooted version of \textbf{DRSO} can be approximated to a factor of $O(\sqrt{B})$ without budget violation, which is an improvement over the factor $O(Ξ\sqrt{B})$ given in~[Kuo et al.\ IEEE/ACM\ Trans.\ Netw.\ 2015] for the undirected and unrooted case, where $Ξ$ is the maximum degree of the graph. Finally, we provide some new/improved approximation bounds for several related problems, including the additive-prize version of \textbf{DRSO}, the maximum budgeted connected set cover problem, and the budgeted sensor cover problem.
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