2017
DOI: 10.1016/j.dam.2016.04.012
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From matchings to independent sets

Vadim Lozin

Abstract: In 1965, Jack Edmonds proposed his celebrated polynomial-time algorithm to find a maximum matching in a graph. It is well-known that finding a maximum matching in G is equivalent to finding a maximum independent set in the line graph of G. For general graphs, the maximum independent set problem is NP-hard. What makes this problem easy in the class of line graphs and what other restrictions can lead to an efficient solution of the problem? In the present paper, we analyze these and related questions. We also re… Show more

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Cited by 11 publications
(6 citation statements)
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“…• 6 does not belong to C . Indeed, if 6 belongs to C , then it must be of distance at most 2 from a (Claim 1), in which case we obtain an induced cycle of length at least p − 2 containing four edges (0, 3), (3,6), (0, 2), (2,5) • u is adjacent to 2. Indeed, assume u is not adjacent to 2.…”
Section: Lemmamentioning
confidence: 99%
See 2 more Smart Citations
“…• 6 does not belong to C . Indeed, if 6 belongs to C , then it must be of distance at most 2 from a (Claim 1), in which case we obtain an induced cycle of length at least p − 2 containing four edges (0, 3), (3,6), (0, 2), (2,5) • u is adjacent to 2. Indeed, assume u is not adjacent to 2.…”
Section: Lemmamentioning
confidence: 99%
“…If a is not adjacent to 3 and is adjacent to 6, then -if 6 ∈ C , i.e. if 6 = b, then 3 has no other neighbours on C by Claim 1, in which case we obtain an induced cycle of length p − 1 containing four edges (0, 3), (3,6), (0, 2), (2, 5) of H . -if 6 / ∈ C , then any other neighbour of 6 on C (if any) must be of distance at most 3 from a by Claim 1, in which case we obtain an induced cycle of length at least p − 2 containing four edges (0, 3), (3,6), (0, 2), (2,5) of H .…”
Section: Lemmamentioning
confidence: 99%
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“…Theorem 1 implies that (unless P = NP) for any graph F, if WIS can be solved for F-free graphs in polynomial time, then each component of F is S i;j;k for some indices i, j, k. Then Lozin [25] conjectured that WIS can be solved in polynomial time for S i;j;k -graphs for any fixed indices i, j, k. The above allows one to focus on possible open problems, i.e., on possible graph classes for which WIS may be solved in polynomial time.…”
Section: Introductionmentioning
confidence: 99%
“…Theorem 1 implies that (unless P = NP) for any graph F , if WIS can be solved for F -free graphs in polynomial time, then each connected component of F is S i,j,k for some indices i, j, k. Then Lozin [23] conjectured that WIS can be solved in polynomial time for S i,j,k -graphs for any fixed indices i, j, k. The above allows one to focus on possible open problems, i.e., on possible graph classes for which WIS may be solved in polynomial time. This manuscript shows that (i) WIS can be solved for (P 4 + P 4 , Triangle)-free graphs in polynomial time, and that in particular it turns out that (ii) for every (P 4 + P 4 , Triangle)-free graph G there is a family S of subsets of V (G) inducing (complete) bipartite subgraphs of G, which contains polynomially many members and can be computed in polynomial time, such that every maximal independent set of G is contained in some member of S.…”
Section: Introductionmentioning
confidence: 99%