Maximum-Width Rainbow-Bisecting Empty Annulus
Sang Won Bae, Sandip Banerjee, Arpita Baral, Priya Ranjan Sinha Mahapatra, Sang Duk Yoon
TL;DR
This work addresses the problem of computing maximum-width rainbow-bisecting empty annuli across three basic shapes—axis-parallel square, axis-parallel rectangle, and circle—for a colored planar point set. It develops distinct algorithms tailored to each shape: RBSA is solved through three geometric configurations with overall $O(n^3)$ time and $O(n)$ space; RBRA is tackled via anchored-uniform strategies with a baseline $O(n^3)$-time approach improved to $O(k^2n^2\log n)$ time using minimal rainbow intervals and advanced data structures; RBCA is reduced to widest empty slabs in 3D via paraboloid lifting, achieving $O(n^3)$ time and $O(n^2)$ space, with a line-constrained variant at $O(n^2)$ time/space. The key innovations include a structured decomposition of RBSA into C1–C3 configurations, an optimization framework for RBRA leveraging DP$_{ij}(w)$ with improved decision procedures, and a 3D slab-based reduction for circular annuli. Collectively, the results advance the state of color-spanning and widest-empty-region problems, offering exact algorithmic solutions with rigorous time/space bounds and laying groundwork for further extensions (e.g., arbitrary orientations).
Abstract
Given a set of $n$ colored points with $k$ colors in the plane, we study the problem of computing a maximum-width rainbow-bisecting empty annulus (of objects specifically axis-parallel square, axis-parallel rectangle and circle) problem. We call a region rainbow if it contains at least one point of each color. The maximum-width rainbow-bisecting empty annulus problem asks to find an annulus $A$ of a particular shape with maximum possible width such that $A$ does not contain any input points and it bisects the input point set into two parts, each of which is a rainbow. We compute a maximum-width rainbow-bisecting empty axis-parallel square, axis-parallel rectangular and circular annulus in $O(n^3)$ time using $O(n)$ space, in $O(k^2n^2\log n)$ time using $O(n\log n)$ space and in $O(n^3)$ time using $O(n^2)$ space respectively.
