Maximum Centre-Disjoint Mergeable Disks

TL;DR

The paper studies the Maximum Centre-Disjoint Mergeable Disks (MCMD) problem, where one must select a largest centre-disjoint subset of disks while merging all others into nearby disks to enlarge radii. It establishes NP-hardness for MCMD and its variant with relaxed merge order (RMCMD) via reductions from Planar Monotone 3-SAT and Partition, respectively. To enable exact solutions, it provides an ILP formulation for MCMD and a polynomial-time dynamic programming algorithm for the collinear-centers special case. The work highlights practical applications to rotating-map labeling and spatial distributions, and outlines several directions for future research, including approximations and extensions to other distance measures or shapes.

Abstract

Given a set of disks in the plane, the goal of the problem studied in this paper is to choose a subset of these disks such that none of its members contains the centre of any other. Each disk not in this subset must be merged with one of its nearby disks that is, increasing the latter's radius. This problem has applications in labelling rotating maps and in visualizing the distribution of entities in static maps. We prove that this problem is NP-hard. We also present an ILP formulation for this problem, and a polynomial-time algorithm for the special case in which the centres of all disks are on a line.

Figures (7)

• Figure 1: An example rotating map with 4 labels. (a) The initial configuration in which two of the labels overlap during rotation (the circles show the area covered by the labels during rotation). (b) After rotating the map 45 degrees counterclockwise. (c) Two of the labels are merged so that none of the label overlap during rotation.
• Figure 2: The distribution of schools in Munich; disks corresponding to neighbouring schools were merged using the ILP of Section \ref{['silp']} to obtain larger, centre-disjoint disks.
• Figure 3: An example set of disks with two proper assignments: Either all disks are merged with $d_1$, which gives a proper assignment of cardinality 1, or just $d_3$ is merged with $d_2$, which gives a proper assignment of cardinality 4. The latter is a solution to MCMD, because it has the maximum cardinality.
• Figure 4: An example set of disks with no proper assignment. Either $d_3$ and $d_4$ are merged with $d_1$, after which $d_5$ cannot be merged with $d_2$ (but must be), or $d_3$ and $d_5$ are merged with $d_2$, after which $d_4$ cannot be merged with $d_1$ (but must be).
• Figure 5: A monotone rectilinear representation of a Planar Monotone 3-SAT instance with three clauses and four variables. Horizontal segments on the $x$-axis denote the variables, and horizontal segments above and below the $x$-axis denote positive and negative clauses, respectively: $c_1 = v_1 \lor v_2 \lor v_3$, $c_2 = v_1 \lor v_3 \lor v_4$, $c_3 = \neg v_1 \lor \neg v_2 \lor \neg v_4$.

Theorems & Definitions (21)

• Definition 2.1
• Definition 2.2
• Definition 2.3
• Definition 2.4
• Definition 3.1
• Definition 3.2
• Definition 3.3
• Definition 3.4
• Lemma 3.5
• Theorem 3.6