32,268 research outputs found
Graphs of Edge-Intersecting Non-Splitting Paths in a Tree: Representations of Holes-Part II
Given a tree and a set P of non-trivial simple paths on it, VPT(P) is the VPT
graph (i.e. the vertex intersection graph) of the paths P, and EPT(P) is the
EPT graph (i.e. the edge intersection graph) of P. These graphs have been
extensively studied in the literature. Given two (edge) intersecting paths in a
graph, their split vertices is the set of vertices having degree at least 3 in
their union. A pair of (edge) intersecting paths is termed non-splitting if
they do not have split vertices (namely if their union is a path). We define
the graph ENPT(P) of edge intersecting non-splitting paths of a tree, termed
the ENPT graph, as the graph having a vertex for each path in P, and an edge
between every pair of vertices representing two paths that are both
edge-intersecting and non-splitting. A graph G is an ENPT graph if there is a
tree T and a set of paths P of T such that G=ENPT(P), and we say that is
a representation of G.
Our goal is to characterize the representation of chordless ENPT cycles
(holes). To achieve this goal, we first assume that the EPT graph induced by
the vertices of an ENPT hole is given. In [2] we introduce three assumptions
(P1), (P2), (P3) defined on EPT, ENPT pairs of graphs. In the same study, we
define two problems HamiltonianPairRec, P3-HamiltonianPairRec and characterize
the representations of ENPT holes that satisfy (P1), (P2), (P3).
In this work, we continue our work by relaxing these three assumptions one by
one. We characterize the representations of ENPT holes satisfying (P3) by
providing a polynomial-time algorithm to solve P3-HamiltonianPairRec. We also
show that there does not exist a polynomial-time algorithm to solve
HamiltonianPairRec, unless P=NP
The Clique Problem in Ray Intersection Graphs
Ray intersection graphs are intersection graphs of rays, or halflines, in the
plane. We show that any planar graph has an even subdivision whose complement
is a ray intersection graph. The construction can be done in polynomial time
and implies that finding a maximum clique in a segment intersection graph is
NP-hard. This solves a 21-year old open problem posed by Kratochv\'il and
Ne\v{s}et\v{r}il.Comment: 12 pages, 7 figure
- …