An Almost Quadratic Vertex Kernel for Subset Feedback Arc Set in Tournaments

TL;DR

This work delivers the first polynomial kernel for Subset-FAST by introducing regular orders and the rich-vertex concept, showing that Subset-FAST admits an $O((αk)^2)$-vertex kernel when an $α$-approximation exists. Leveraging the known $α = O(\log k \log \log k)$ approximation for Subset-FAS, the authors obtain an almost quadratic kernel. The approach hinges on transforming any solution into a regular order, applying a sequence of safe reductions, and replacing long maximal non-terminal intervals with controlled gadgets via Reduction Rule RICH REPLACE. Overall, the paper advances parameterized complexity for Subset-FAST, bridging kernelization with approximation to achieve near-quadratic kernels and outlining avenues for further improvement if stronger approximations become available.

Abstract

In the Feedback Arc Set in Tournaments (Subset-FAST) problem, we are given a tournament $D$ and a positive integer $k$, and the objective is to determine whether there exists an arc set $S \subseteq A(D)$ of size at most $k$ whose removal makes the graph acyclic. This problem is well-known to be equivalent to a natural tournament ranking problem, whose task is to rank players in a tournament such that the number of pairs in which the lower-ranked player defeats the higher-ranked player is no more than $k$. Using the PTAS for Subset-FAST [STOC 2007], Bessy et al. [JCSS 2011] present a $(2 + \varepsilon)k$-vertex kernel for this problem, given any fixed $\varepsilon > 0$. A generalization of Subset-FAST, called Subset-FAST, further includes an additional terminal subset $T \subseteq V(D)$ in the input. The goal of Subset-FAST is to determine whether there is an arc set $S \subseteq A(D)$ of size at most $k$ whose removal ensures that no directed cycle passes through any terminal in $T$. Prior to our work, no polynomial kernel for Subset-FAST was known. In our work, we show that Subset-FAST admits an $\mathcal{O}((αk)^{2})$-vertex kernel, provided that Subset-FAST has an approximation algorithm with an approximation ratio $α$. Consequently, based on the known $\mathcal{O}(\log k \log \log k)$-approximation algorithm, we obtain an almost quadratic kernel for Subset-FAST.

Figures (3)

• Figure 1: An order $\sigma$ of $15$ vertices with $\mathrm{cost}(\sigma) = 3$. Black vertices denote the terminals, and white vertices denote the non-terminals. In the order $\sigma$, black thick arcs present unaffected backward arcs, red arcs present affected arcs, and all forward arcs are omitted. The arc $\sigma_{5}\sigma_{3}$ is an affected arc above the terminal $\sigma_{5}$; the arc $\sigma_{11}\sigma_{1}$ is an affected arc above the terminals $\sigma_{1}$ and $\sigma_{11}$; the arc $\sigma_{15}\sigma_{6}$ is an affected arc above the terminals $\sigma_{11}$ and $\sigma_{12}$; The three intervals $[\sigma_{2}, \sigma_{4}]$, $[\sigma_{6}, \sigma_{10}]$, and $[\sigma_{13}, \sigma_{15}]$ are maximal non-terminal intervals w.r.t. the order $\sigma$.
• Figure 2: A regular order $\sigma$ and an optimal order ${\sigma}^{*}$ of $n$ vertices. Black vertices denote the terminals, and white vertices denote the non-terminals. In the order $\sigma$, $I = [\sigma_{l}, \sigma_{r}]$ is a maximal non-terminal interval, and $(L, I, R)$ is the $I$-partition, where $L = [\sigma_{1}, \sigma_{l - 1}]$ and $R = [\sigma_{r + 1}, \sigma_{n}]$. In the order ${\sigma}^{*}$, the interval $I^{*} = [{\sigma}^{*}_{l'}, {\sigma}^{*}_{r'}]$ is a maximal non-terminal interval w.r.t ${\sigma}^{*}$, and $(L^{*}, I^{*}, R^{*})$ is the $I^{*}$-partition, where $L = [{\sigma}^{*}_{1}, {\sigma}^{*}_{l' - 1}]$ and $R = [{\sigma}^{*}_{r' + 1}, {\sigma}^{*}_{n}]$. All out-rich vertices in $I$ are in the blue dotted boxes, and all in-rich vertices in $I$ are in the purple dotted boxes, where $d = (\alpha + 2)k + 1$.
• Figure 3: A maximal non-terminal interval $I$ w.r.t. an order $\sigma$ and the non-terminal interval obtained from $I$ by applying Reduction Rule \ref{['REDUCTION: RICH REPLACE']}. Black vertices denote the terminals, and white vertices denote the non-terminals. In the order $\sigma$, $I = [\sigma_{l}, \sigma_{r}]$ is a maximal non-terminal interval, black thick arcs present unaffected backward arcs, red arcs present affected arcs, and all forward arcs are omitted. In the interval $I$, assume that $[\sigma_{l}, \sigma_{l + 2d}]$ is the set of out-rich vertices (in the blue box), $W_{I} = [\sigma_{l + 2d + 1}, \sigma_{r - 2d - 1}]$ is the set of rich vertices, and $W^{-}_{I} = [\sigma_{r - 2d}, \sigma_{r}]$ is the set of in-rich vertices (in the purple box). In the order $\sigma$, $\sigma_{l + 2d + 1}$ is an affected rich vertex and $\sigma_{r - 2d}$ is an affected in-rich vertex. After applying Reduction Rule \ref{['REDUCTION: RICH REPLACE']}, the unaffected rich vertices $\sigma_{l + 2d + 2}, \sigma_{l + 2d +3}, \ldots, \sigma_{r - 2d - 1}$ are removed and $\ell$ non-terminals $v_{1}, v_{2}, \ldots, v_{\ell}$ are added.

Theorems & Definitions (29)

• proposition thmcounterproposition
• proof
• proposition thmcounterproposition
• proof
• definition thmcounterdefinition: regular orders
• definition thmcounterdefinition: regularization
• lemma thmcounterlemma
• proof
• lemma thmcounterlemma
• proof