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dc.contributor.authorFomin, Fedor
dc.contributor.authorGolovach, Petr
dc.contributor.authorLokshtanov, Daniel
dc.contributor.authorPanolan, Fahad
dc.contributor.authorSaurabh, Saket
dc.contributor.authorZehavi, Meirav
dc.date.accessioned2021-05-19T12:08:05Z
dc.date.available2021-05-19T12:08:05Z
dc.date.created2021-01-04T11:40:57Z
dc.date.issued2020
dc.PublishedSIAM Journal on Discrete Mathematics. 2020, 34 (3), 1578-1601.
dc.identifier.issn0895-4801
dc.identifier.urihttps://hdl.handle.net/11250/2755693
dc.description.abstractAn undirected graph $G$ is $d$-degenerate if every subgraph of $G$ has a vertex of degree at most $d$. By the classical theorem of Erdös and Gallai from 1959, every graph of degeneracy $d>1$ contains a cycle of length at least $d+1$. The proof of Erdös and Gallai is constructive and can be turned into a polynomial time algorithm constructing a cycle of length at least $d+1$. But can we decide in polynomial time whether a graph contains a cycle of length at least $d+2$? An easy reduction from Hamiltonian Cycle provides a negative answer to this question: Deciding whether a graph has a cycle of length at least $d+2$ is NP-complete. Surprisingly, the complexity of the problem changes drastically when the input graph is 2-connected. In this case we prove that deciding whether $G$ contains a cycle of length at least $d+k$ can be done in time $2^{\mathcal{O}(k)}\cdot|V(G)|^{\mathcal{O}(1)}$. In other words, deciding whether a 2-connected $n$-vertex $G$ contains a cycle of length at least $d+\log{n}$ can be done in polynomial time. Similar algorithmic results hold for long paths in graphs. We observe that deciding whether a graph has a path of length at least $d+1$ is NP-complete. However, we prove that if graph $G$ is connected, then deciding whether $G$ contains a path of length at least $d+k$ can be done in time $2^{\mathcal{O}(k)}\cdot n^{\mathcal{O}(1)}$. We complement these results by showing that the choice of degeneracy as the “above guarantee parameterization” is optimal in the following sense: For any $\varepsilon>0$ it is NP-complete to decide whether a connected (2-connected) graph of degeneracy $d$ has a path (cycle) of length at least $(1+\varepsilon)d$.en_US
dc.language.isoengen_US
dc.publisherSIAMen_US
dc.titleGoing Far from Degeneracyen_US
dc.typeJournal articleen_US
dc.typePeer revieweden_US
dc.description.versionpublishedVersionen_US
dc.rights.holderCopyright 2020 Society for Industrial and Applied Mathematicsen_US
cristin.ispublishedtrue
cristin.fulltextpostprint
cristin.qualitycode1
dc.identifier.doihttps://doi.org/10.1137/19M1290577
dc.identifier.cristin1864692
dc.source.journalSIAM Journal on Discrete Mathematicsen_US
dc.source.4034
dc.source.143
dc.source.pagenumber1578-1601en_US
dc.relation.projectNorges forskningsråd: 263317en_US
dc.relation.projectERC-European Research Council: 819416en_US
dc.identifier.citationSIAM Journal on Discrete Mathematics. 2020, 34(3), 1587–1601en_US
dc.source.volume34en_US
dc.source.issue3en_US


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