Finding Efficient Region in The Plane with Line segments

TL;DR

This paper introduces three algorithms to efficiently find a region in the plane that contains all points reachable within a maximum path length from a start point amid line-segment obstacles, improving computational complexity.

Contribution

The paper presents three algorithms, progressively optimizing the process of computing the achievable region with respect to obstacle complexity and computational efficiency.

Findings

First algorithm runs in O(n^3) time.

Second algorithm improves to O(n^2 log n) using the short path map.

Third algorithm achieves O(n log n) complexity by direct boundary tracing.

Abstract

Let $\mathscr O$ be a set of $n$ disjoint obstacles in $\mathbb{R}^2$, $\mathscr M$ be a moving object. Let $s$ and $l$ denote the starting point and maximum path length of the moving object $\mathscr M$, respectively. Given a point $p$ in ${R}^2$, we say the point $p$ is achievable for $\mathscr M$ such that $\pi(s,p)\leq l$, where $\pi(\cdot)$ denotes the shortest path length in the presence of obstacles. One is to find a region $\mathscr R$ such that, for any point $p\in \mathbb{R}^2$, if it is achievable for $\mathscr M$, then $p\in \mathscr R$; otherwise, $p\notin \mathscr R$. In this paper, we restrict our attention to the case of line-segment obstacles. To tackle this problem, we develop three algorithms. We first present a simpler-version algorithm for the sake of intuition. Its basic idea is to reduce our problem to computing the union of a set of circular visibility regions (CVRs). This algorithm takes $O(n^3)$ time. By analysing its dominant steps, we break through its bottleneck by using the short path map (SPM) technique to obtain those circles (unavailable beforehand), yielding an $O(n^2\log n)$ algorithm. Owing to the finding above, the third algorithm also uses the SPM technique. It however, does not continue to construct the CVRs. Instead, it directly traverses each region of the SPM to trace the boundaries, the final algorithm obtains $O(n\log n)$ complexity.