Local Representatives and Shortest Completions for Next-to-Shortest Paths in Directed Graphs

Abstract

Given a directed graph with positive edge weights and two vertices s,t, a next-to-shortest s-t path is a shortest simple s-t path among those whose length is strictly larger than the shortest-path distance. The problem was introduced by Lalgudi, Papaefthymiou and Potkonjak in 1996; it is NP-hard when zero-weight edges are allowed, and its complexity on positively weighted digraphs remained open for almost three decades until Chen, Wein and Zhang recently gave a polynomial-time algorithm running in O(n^4 m^3 log n) time. We give a substantially faster algorithm within their optimal-middle-segment framework. The core idea is to split the problem into"choosing a prefix"and"completing it". Given a prefix P: s ->A made of shortest-path edges, delete the vertices used by P, forbid leaving A along shortest-path edges, and the best completion is one shortest-path computation. The difficulty lies in choosing P: even for a fixed A, deciding whether some shortest prefix admits a completion is NP-complete. We do not solve these fixed-A subproblems one by one. Fix any globally optimal next-to-shortest path; its middle segment induces a boundary edge x ->c in the shortest-path DAG. For the correct triple (A,B,x), the optimal path certifies c as a feasible next hop, and we prove that every feasible next hop that is not earlier than c in a topological order can be combined with the same middle segment into another globally optimal path. Hence only the feasible next hop of maximum topological index is kept per triple, giving O(n^3) representatives, all generated by a two-dimensional DAG dynamic program with a local reward. The total running time is O(n^3 (m + n log n)), and O(n^3 m) on unweighted graphs. The proof rests on an uncrossing lemma: the last intersection between a reference prefix and the candidate's partner suffix can always be moved strictly earlier, which cannot go on forever.