Complexity of the path avoiding forbidden pairs problem revisited

TL;DR

This paper revisits the complexity of the path avoiding forbidden pairs problem in directed acyclic graphs, providing new complexity results and an improved algorithm for special cases based on the arrangement of forbidden pairs.

Contribution

The paper proves NP-hardness for cases with no nested pairs and introduces an O(M(n)) time algorithm for cases with no halving pairs, improving previous solutions.

Findings

NP-hardness when no two pairs are nested

Polynomial-time solvability when no two pairs are halving

An improved algorithm with O(M(n)) complexity for certain cases

Abstract

Let G = (V, E) be a directed acyclic graph with two distinguished vertices s, t and let F be a set of forbidden pairs of vertices. We say that a path in G is safe, if it contains at most one vertex from each pair {u, v} in F. Given G and F, the path avoiding forbidden pairs (PAFP) problem is to find a safe s-t path in G. We systematically study the complexity of different special cases of the PAFP problem defined according to the mutual positions of forbidden pairs. Fix one topological ordering of vertices; we say that pairs {u, v} and {x, y} are disjoint, if u, v < x, y, nested, if u < x, y < v, and halving, if u < x < v < y. The PAFP problem is known to be NP-hard in general or if no two pairs are disjoint; we prove that it remains NP-hard even when no two forbidden pairs are nested. On the other hand, if no two pairs are halving, the problem is known to be solvable in cubic time. We simplify and improve this result by showing an O(M(n)) time algorithm, where M(n) is the time to multiply two n \times n boolean matrices.