## Going Far From Degeneracy

##### Peer reviewed, Journal article

##### Published version

##### View/Open

##### Date

2019-09-06##### Metadata

Show full item record##### Collections

##### Original version

Leibniz International Proceedings in Informatics. 2019, 47. https://doi.org/10.4230/lipics.esa.2019.47##### Abstract

An undirected graph G is d-degenerate if every subgraph of G has a vertex of degree at most d. By the classical theorem of Erd\H{o}s and Gallai from 1959, every graph of degeneracy d>1 contains a cycle of length at least d+1. The proof of Erd\H{o}s 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^{O(k)}|V(G)|^{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^{O(k)}n^{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 \epsilon>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+\epsilon)d.