Subscribe to the gold package and get unlimited access to Shamra Academy
Register a new userStructure and algorithms for (cap, even hole)-free graphs
230
0
0.0
(
0
)
Added by
Shenwei Huang
Publication date
2016
fields
Informatics Engineering
and research's language is
English
Ask ChatGPT about the research
No Arabic abstract
A graph is even-hole-free if it has no induced even cycles of length 4 or more. A cap is a cycle of length at least 5 with exactly one chord and that chord creates a triangle with the cycle. In this paper, we consider (cap, even hole)-free graphs, and more generally, (cap, 4-hole)-free odd-signable graphs. We give an explicit construction of these graphs. We prove that every such graph $G$ has a vertex of degree at most $frac{3}{2}omega (G) -1$, and hence $chi(G)leq frac{3}{2}omega (G)$, where $omega(G)$ denotes the size of a largest clique in $G$ and $chi(G)$ denotes the chromatic number of $G$. We give an $O(nm)$ algorithm for $q$-coloring these graphs for fixed $q$ and an $O(nm)$ algorithm for maximum weight stable set. We also give a polynomial-time algorithm for minimum coloring. Our algorithms are based on our results that triangle-free odd-signable graphs have treewidth at most 5 and thus have clique-width at most 48, and that (cap, 4-hole)-free odd-signable graphs $G$ without clique cutsets have treewidth at most $6omega(G)-1$ and clique-width at most 48.
rate research
Read More
A hole is a chordless cycle with at least four vertices. A pan is a graph which consists of a hole and a single vertex with precisely one neighbor on the hole. An even hole is a hole with an even number of vertices. We prove that a (pan, even hole)-free graph can be decomposed by clique cutsets into essentially unit circular-arc graphs. This structure theorem is the basis of our $O(nm)$-time certifying algorithm for recognizing (pan, even hole)-free graphs and for our $O(n^{2.5}+nm)$-time algorithm to optimally color them. Using this structure theorem, we show that the tree-width of a (pan, even hole)-free graph is at most 1.5 times the clique number minus 1, and thus the chromatic number is at most 1.5 times the clique number.
The class of all even-hole-free graphs has unbounded tree-width, as it contains all complete graphs. Recently, a class of (even-hole, $K_4$)-free graphs was constructed, that still has unbounded tree-width [Sintiari and Trotignon, 2019]. The class has unbounded degree and contains arbitrarily large clique-minors. We ask whether this is necessary. We prove that for every graph $G$, if $G$ excludes a fixed graph $H$ as a minor, then $G$ either has small tree-width, or $G$ contains a large wall or the line graph of a large wall as induced subgraph. This can be seen as a strengthening of Robertson and Seymours excluded grid theorem for the case of minor-free graphs. Our theorem implies that every class of even-hole-free graphs excluding a fixed graph as a minor has bounded tree-width. In fact, our theorem applies to a more general class: (theta, prism)-free graphs. This implies the known result that planar even hole-free graph have bounded tree-width [da Silva and Linhares Sales, Discrete Applied Mathematics 2010]. We conjecture that even-hole-free graphs of bounded degree have bounded tree-width. If true, this would mean that even-hole-freeness is testable in the bounded-degree graph model of property testing. We prove the conjecture for subcubic graphs and we give a bound on the tree-width of the class of (even hole, pyramid)-free graphs of degree at most 4.
A vertex of a graph is bisimplicial if the set of its neighbors is the union of two cliques; a graph is quasi-line if every vertex is bisimplicial. A recent result of Chudnovsky and Seymour asserts that every non-empty even-hole-free graph has a bisimplicial vertex. Both Hadwigers conjecture and the ErdH{o}s-Lovasz Tihany conjecture have been shown to be true for quasi-line graphs, but are open for even-hole-free graphs. In this note, we prove that for all $kge7$, every even-hole-free graph with no $K_k$ minor is $(2k-5)$-colorable; every even-hole-free graph $G$ with $omega(G)<chi(G)=s+t-1$ satisfies the ErdH{o}s-Lovasz Tihany conjecture provided that $ tge s> chi(G)/3$. Furthermore, we prove that every $9$-chromatic graph $G$ with $omega(G)le 8$ has a $K_4cup K_6$ minor. Our proofs rely heavily on the structural result of Chudnovsky and Seymour on even-hole-free graphs.
The class of even-hole-free graphs is very similar to the class of perfect graphs, and was indeed a cornerstone in the tools leading to the proof of the Strong Perfect Graph Theorem. However, the complexity of computing a maximum independent set (MIS) is a long-standing open question in even-hole-free graphs. From the hardness point of view, MIS is W[1]-hard in the class of graphs without induced 4-cycle (when parameterized by the solution size). Halfway of these, we show in this paper that MIS is FPT when parameterized by the solution size in the class of even-hole-free graphs. The main idea is to apply twice the well-known technique of augmenting graphs to extend some initial independent set.
Given two independent sets $I, J$ of a graph $G$, and imagine that a token (coin) is placed at each vertex of $I$. The Sliding Token problem asks if one could transform $I$ to $J$ via a sequence of elementary steps, where each step requires sliding a token from one vertex to one of its neighbors so that the resulting set of vertices where tokens are placed remains independent. This problem is $mathsf{PSPACE}$-complete even for planar graphs of maximum degree $3$ and bounded-treewidth. In this paper, we show that Sliding Token can be solved efficiently for cactus graphs and block graphs, and give upper bounds on the length of a transformation sequence between any two independent sets of these graph classes. Our algorithms are designed based on two main observations. First, all structures that forbid the existence of a sequence of token slidings between $I$ and $J$, if exist, can be found in polynomial time. A sufficient condition for determining no-instances can be easily derived using this characterization. Second, without such forbidden structures, a sequence of token slidings between $I$ and $J$ does exist. In this case, one can indeed transform $I$ to $J$ (and vice versa) using a polynomial number of token-slides.
Log in to be able to interact and post comments
comments
Fetching comments
Sign in to be able to follow your search criteria
