2015
DOI: 10.1007/s10107-015-0912-3
|View full text |Cite
|
Sign up to set email alerts
|

Stable sets and graphs with no even holes

Abstract: We develop decomposition/composition tools for efficiently solving maximum weight stable sets problems as well as for describing them as polynomially sized linear programs (using "compact systems"). Some of these are well-known but need some extra work to yield polynomial "decomposition schemes".We apply the tools to graphs with no even hole and no cap. A hole is a chordless cycle of length greater than three and a cap is a hole together with an additional node that is adjacent to two adjacent nodes of the hol… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
2
1

Citation Types

1
22
0

Year Published

2017
2017
2021
2021

Publication Types

Select...
4
2

Relationship

1
5

Authors

Journals

citations
Cited by 12 publications
(23 citation statements)
references
References 16 publications
1
22
0
Order By: Relevance
“…We say that the type of is , according to the types and of the two holes with respect to the amalgam. Based on Lemma 7.1, and the property that these two holes share only one edges, there are 13 possible types for : 11,14,16,18,22,25,27,29,46,48,59,57,89. If belongs to type 89 then any vertex of 2 (this vertex is exist since 2 ≠ ∅) with hole type 8 induces a cycle with a long unichord in in 1 , a contradiction.…”
Section: Recognizing Long Unichord Free Graphsmentioning
confidence: 99%
See 2 more Smart Citations
“…We say that the type of is , according to the types and of the two holes with respect to the amalgam. Based on Lemma 7.1, and the property that these two holes share only one edges, there are 13 possible types for : 11,14,16,18,22,25,27,29,46,48,59,57,89. If belongs to type 89 then any vertex of 2 (this vertex is exist since 2 ≠ ∅) with hole type 8 induces a cycle with a long unichord in in 1 , a contradiction.…”
Section: Recognizing Long Unichord Free Graphsmentioning
confidence: 99%
“…This is the case because is not basic, so it is not chordal and contains a chordless cycle of length at least 4 (that provides at least two nonadjacent pairs). ■ The next lemma is implicitly proved in [11] (as Corollary 2.16), but the machinery there is much heavier and relies on many definitions, so we prefer to give our own simple proof. Note that in [12], it is claimed without proof that any graph with an amalgam should have an amalgam such that at least one block of decomposition has no amalgam.…”
Section: Lemma 73mentioning
confidence: 99%
See 1 more Smart Citation
“…On the other hand, it is easy to see how to use Theorem 8.1 to obtain a polynomial time algorithm to solve the maximum weight clique problem for cap-free graphs, see [51].…”
Section: Theorem 81 (Conforti Cornuéjols Kapoor and Vušković [43])mentioning
confidence: 99%
“…Conforti and Gerards [51] show how to obtain a polynomial time algorithm for solving maximum weight independent set problem, on any class of graphs that is decomposable by amalgams into basic graphs for which one can solve the maximum weight independent set problem in polynomial time. In particular, using Theorem 8.1, they obtain a polynomial time algorithm for solving the maximum weight independent set problem for (cap, odd-hole)-free graphs (i.e.…”
Section: Theorem 81 (Conforti Cornuéjols Kapoor and Vušković [43])mentioning
confidence: 99%