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    On the complexity of kings

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    On the complexity of kings

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    AbstractA king in a directed graph is a vertex from which each vertex in the graph can be reached through paths of length at most two. There is a broad literature on tournaments (completely oriented digraphs), and it has been known for more than half a century that all tournaments have at least one king. Recently, kings have proven useful in theoretical computer science, in particular in the study of the complexity of reachability problems and semifeasible sets.In this article, we study the complexity of recognizing kings. For each succinctly specified family of tournaments, the king problem is already known to belong to Π2p. We prove that the complexity of kingship problems is a rich enough vocabulary to pinpoint every nontrivial many-one degree in Π2p. That is, we show that every set in Π2p other than 0̸ and Σ∗ is equivalent to a king problem under ≤mp-reductions. Indeed, we show that the equivalence can even be realized by relatively simple padding, and holds even if the notion of kings is redefined to refer to k-kings (for any fixed k≥2)—vertices from which all vertices can be reached through paths of length at most k. In contrast, we prove that for each succinctly specified family of tournaments the source problem (the problem of deciding whether a given vertex v has the property that there exists a k such that v is a k-king) also falls within Π2p, yet cannot be Π2p-complete—or even NP-hard—unless P=NP.Using these and related techniques, we obtain a broad range of additional results about the complexity of king problems, diameter problems, and radius problems. It follows easily from our proof approach that the problem of testing kingship in succinctly specified graphs (which need not be tournaments) is Π2p-complete. We show that the radius problem for arbitrary succinctly represented graphs is Σ3p-complete, but that in contrast the diameter problem for arbitrary succinctly represented graphs (or even tournaments) is Π2p-complete

    On the Complexity of Kings

    No full text
    A king in a directed graph is a node from which each node in the graph can be reached via paths of length at most two. There is a broad literature on tournaments (completely oriented digraphs), and it has been known for more than half a century that all tournaments have at least one king [Lan53]. Recently, kings have proven useful in theoretical computer science, in particular in the study of the complexity of reachability problems [Tan01,NT05]. and semifeasible sets [HNP98,HT06,HOZZ06]. In this paper, we study the complexity of recognizing kings. For each succinctly specified family of tournaments, the king problem is already known to belong to Π2p\Pi_2^p [HOZZ06]. We prove that the complexity of kingship problems is a rich enough vocabulary to pinpoint every nontrivial many-one degree in Π2p\Pi_2^p. That is, we show that \emph{every} set in Π2p\Pi_2^p other than ∅\emptyset and Σ∗\Sigma^* is equivalent to a king problem under ≤mp\leq_m^p-reductions. Indeed, we show that the equivalence can even be instantiated via relatively simple padding, and holds even if the notion of kings is redefined to refer to k-kings (for any fixed k≥2k \geq 2)---nodes from which the all nodes can be reached via paths of length at most k. Using these and related techniques, we obtain a broad range of additional results about the complexity of king problems, diameter problems, radius problems, and initial component problems. It follows easily from our proof approach that the problem of testing kingship in succinctly specified graphs (which need not be tournaments) is Π2p\Pi_2^p-complete. We show that the radius problem for arbitrary succinctly represented graphs is Σ3p\Sigma_3^p-complete, but that in contrast the diameter problem for arbitrary succinctly represented graphs (or even tournaments) is Π2p\Pi_2^p-complete. Yet again in contrast, we prove that initial component languages (which ask whether a given node can reach all other nodes in its tournament) all fall within Π2p\Pi_2^p, yet cannot be Π2p\Pi_2^p-complete---or even NP-hard---unless P=NP
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