2010
DOI: 10.1007/978-3-642-13089-2_22
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Short Witnesses and Accepting Lassos in ω-Automata

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Cited by 13 publications
(14 citation statements)
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“…Proof: We prove by reduction from the problem of finding a shortest accepting lasso for a generalized Büchi automaton, which is known to be NP-hard [4], [8]. For Algorithm 2: MinimumCostAcceptingLasso Data: Product automaton P = (Q P , δ P , q 0,P , F P , w) Result: An accepting lasso for P minimizing f w 1 SCCs = STRONGLYCONNECTEDCOMPONENTS(P);…”
Section: B Trajectory Generationmentioning
confidence: 99%
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“…Proof: We prove by reduction from the problem of finding a shortest accepting lasso for a generalized Büchi automaton, which is known to be NP-hard [4], [8]. For Algorithm 2: MinimumCostAcceptingLasso Data: Product automaton P = (Q P , δ P , q 0,P , F P , w) Result: An accepting lasso for P minimizing f w 1 SCCs = STRONGLYCONNECTEDCOMPONENTS(P);…”
Section: B Trajectory Generationmentioning
confidence: 99%
“…As a result of this lemma, unless P=NP we cannot compute in a time polynomial to the size of P, a shortest lasso that is accepting for P and for which f w (r P ) is minimum. Unfortunately, it is also NP-hard to approximate within any constant factor, the length of such a lasso (the proof would utilize the same reduction in Lemma 4 along with the fact due to Ehlers [8], according to which it is NP-hard to approximate within any constant factor the length of a shortest accepting lasso for a generalized Büchi automaton). Consequently, we utilize a greedy algorithm to find a shortest such lasso which has the minimum cost.…”
Section: B Trajectory Generationmentioning
confidence: 99%
“…In this subsection we present our results for the certificate computation. Given a start vertex x that belongs to the winning set, a certificate or lasso [25] is a path from x that consists of a simple path and a not necessarily simple cycle such that the a play that traverses the cycle infinitely often satisfies the objective. From the certificate an example of an accepting run, i.e., an infinite path from x that satisfies the objective, can be constructed.…”
Section: Certificate Computationmentioning
confidence: 99%
“…Let S be a vertex set reachable from x that induces a strongly connected subgraph such that for all 1 ≤ j ≤ k we have either S ∩ L j = ∅ or S ∩ U j = ∅ (i.e., if S contains a vertex from L j then it also contains a vertex from U j ). A certificate is a "lasso-shaped" path that consists of a path to S and a (not necessarily simple) cycle between the vertices of S [25]. The basic algorithm [28,44] for the winning set problem has an asymptotic running time of O((m + b) min(n, k)) with b = k j=1 (|L j | + |U j |) ≤ 2nk.…”
Section: Introductionmentioning
confidence: 99%
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