Abstract
We study feedback vertex sets (FVS) in tournaments, which are orientations of complete graphs. As our main result, we show that any tournament on n nodes has at most 1.5949n minimal FVS. This significantly improves the previously best upper bound of 1.6667n by Fomin et al. [STOC 2016] and 1.6740n by Gaspers and Mnich [J. Graph Theory 72(1):72–89, 2013]. Our new upper bound almost matches the best‐known lower bound of 21n/7≈1.5448n, due to Gaspers and Mnich. Our proof is algorithmic, and shows that all minimal FVS of tournaments can be enumerated in time O(1.5949n).
Highlights
The Minimum Feedback Vertex Set (FVS) problem in directed graphs is a fundamental problem in combinatorial optimization: given a directed graph G, find a smallest set of vertices in G whose removal yields an acyclic digraph
The run time of this approach is within a polynomial factor of the number M (T ) of minimal FVS in T
The complexity of the Minimum FVS problem is within a polynomial factor of the maximum of M (T ) over all n-vertex tournaments, which we denote by M (n)
Summary
The Minimum Feedback Vertex Set (FVS) problem in directed graphs is a fundamental problem in combinatorial optimization: given a directed graph G, find a smallest set of vertices in G whose removal yields an acyclic digraph. We introduce a new function M (n, k) for the maximum number of maximal transitive vertex sets in a tournament of order n containing a fixed set of k vertices, and we will show that M (n, k) ≤ 1.5949n−k for all 0 ≤ k ≤ n. As Woeginger [14] showed that deciding whether a vertex is a Banks winner is NP-complete, a feasible approach to determine the Banks set is to enumerate all minimal FVS. For this purpose, Brandt et al [2] implemented the algorithm of Gaspers and Mnich. Our new algorithm in this paper can be used to compute the Banks set of a tournament asymptotically faster
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