Obedient Plane Drawings for Disk Intersection Graphs

  • Bahareh Banyassady
  • Michael Hoffmann
  • Boris Klemz
  • Maarten Löffler
  • Tillmann Miltzow
Conference paper
Part of the Lecture Notes in Computer Science book series (LNCS, volume 10389)

Abstract

Let \(\mathcal D\) be a set of disks and G be the intersection graph of \(\mathcal D\). A drawing of G is obedient to \(\mathcal D\) if every vertex is placed in its corresponding disk. We show that deciding whether a set of unit disks \(\mathcal D\) has an obedient plane straight-line drawing is \(\mathcal {NP}\)-hard regardless of whether a combinatorial embedding is prescribed or an arbitrary embedding is allowed. We thereby strengthen a result by Evans et al., who show \(\mathcal {NP}\)-hardness for disks with arbitrary radii in the arbitrary embedding case. Our result for the arbitrary embedding case holds true even if G is thinnish, that is, removing all triangles from G leaves only disjoint paths. This contrasts another result by Evans et al. stating that the decision problem can be solved in linear time if \(\mathcal D\) is a set of unit disks and G is thin, that is, (1) the (graph) distance between any two triangles is larger than 48 and (2) removal of all disks within (graph) distance 8 of a triangle leaves only isolated paths. A path in a disk intersection graph is isolated if for every pair AB of disks that are adjacent along the path, the convex hull of \(A\cup B\) is intersected only by disks adjacent to A or B. Our reduction can also guarantee the triangle separation property (1). This leaves only a small gap between tractability and \(\mathcal {NP}\)-hardness, tied to the path isolation property (2) in the neighborhood of triangles. It is therefore natural to study the impact of different restrictions on the structure of triangles. As a positive result, we show that an obedient plane straight-line drawing is always possible if all triangles in G are light and the disks are in general position (no three centers collinear). A triangle in a disk intersection graph is light if all its vertices have degree at most three or the common intersection of the three corresponding disks is empty. We also provide an efficient drawing algorithm for that scenario.

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Copyright information

© Springer International Publishing AG 2017

Authors and Affiliations

  • Bahareh Banyassady
    • 1
  • Michael Hoffmann
    • 2
  • Boris Klemz
    • 1
  • Maarten Löffler
    • 3
  • Tillmann Miltzow
    • 4
  1. 1.Institute of Computer ScienceFreie Universität BerlinBerlinGermany
  2. 2.Department of Computer ScienceETH ZürichZürichSwitzerland
  3. 3.Dept. of Computing and Information SciencesUtrecht UniversityUtrechtThe Netherlands
  4. 4.Université Libre de Bruxelles (ULB)BrusselsBelgium

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