Large Independent Sets in Triangle-Free Planar Graphs

Abstract

Every triangle-free planar graph on n vertices has an independent set of size at least (n + 1)/3, and this lower bound is tight. We give an algorithm that, given a triangle-free planar graph G on n vertices and an integer k ≥ 0, decides whether G has an independent set of size at least (n + k)/3, in time 2 O( √ k) n. Thus, the problem is fixed-parameter tractable when parameterized by k. Furthermore, as a corollary of the result used to prove the correctness of the algorithm, we show that there exists ε > 0 such that every planar graph of girth at least five on n vertices has an independent set of size at least n/(3 − ε).

Linear Kernels and Linear-Time Algorithms for Finding Large Cuts

“…Ideally, we strive for algorithms that are linear in the input size, and optimal in the dependence on the parameter k assuming a standard hypothesis such as the Exponential Time Hypothesis [22]. New results in this direction include linear-time fixed-parameter algorithms for Graph Bipartization [23], Planar Subgraph Isomorphism [10], DAG Partitioning [36], Planar Independent Set [11] and Subset Feedback Vertex Set [27].…”

Triangle‐free planar graphs with the smallest independence number

“…Let C be the cycle in K v v v + 2 3 4 distinct from K and the boundary of f . Since C is not dangerous and ≥ v deg( ) 3 3 , we conclude that G C is either C c 6, or C v 6, . The former is excluded, since G is not isomorphic to C v 6, .…”

“…It is natural to ask whether the bound from Theorem can be improved, at the expense of having further families of exceptional graphs. Algorithmically, this question was answered by Dvořák and Mnich , who proved that if an ‐vertex planar triangle‐free graph does not have an independent set larger than , then its tree‐width is . Using their techique, a more detailed answer can be given, showing that all such graphs are created from graphs of bounded size by a construction similar to the one used to define the class below; we will give details in a followup paper.…”

“…Let p denote the number of internal vertices of degree three incident with f . If ≥ p 2, then G contains Conf (3) or Conf(5) not interfering with the outer face. If p = 0, then c f c f ( ) = ( ) = 0 0 .…”

“…, γ forms Conf(4) in G as well, unless G V f − ( ) contains a path u zz u 2 ′ 3 of length three. As in the previous paragraph, this is only possible if K zx x z y y = , and letting C be the 6-cycle u v v u zz3 3 2 2 ′ , we have ⊆ G G C K 1 , and since C separates x x , 1 2 from y y , 1 2 , it is dangerous. Considering the 4-cycles C zx x z =…”