On the Complexity of the Minimum-($k,\rho$)-Shortcut Problem

TL;DR

This paper investigates the computational complexity of the Minimum-$(k, ho)$-Shortcut problem, establishing NP-hardness for most parameter ranges and identifying a polynomial-time solvable case.

Contribution

It provides a simplified reduction proving NP-hardness for $k extgreater=2$ and $ ho extgreater= k+2$, and shows polynomial solvability for $ ho=k+1$ in undirected graphs.

Findings

NP-hardness for $k extgreater=2$ and $ ho extgreater= k+2$ in both directed and undirected graphs.

Polynomial-time solvability for $ ho=k+1$ in undirected graphs.

Remaining open case: $ ho=k+1$ in directed graphs.

Abstract

We consider the Minimum-$(k,\rho)$-$\mathrm{Shortcut}$ problem ($\min(k,\rho)\text{-}\mathrm{Shortcut}$), where the goal is to find the smallest set of shortcut edges such that every vertex in a given graph can reach its $\rho$ closest vertices using paths of at most $k$ edges. This is a fundamental graph optimization problem used to accelerate parallel shortest path algorithms. It is well-known that the problem is trivially solvable for the cases $k=1$ and $k\geq\rho$. While recent work by Leonhardt, Meyer, and Penschuck (ESA 2024) showed that in undirected graphs $\min(k,\rho)\text{-}\mathrm{Shortcut}$ is NP-hard for $k\geq 3$ if $\rho=\Theta(n^\epsilon)$, the boundary where the problem transitions from polynomial-time solvable to NP-hard remained open. In this paper, we narrow this gap significantly. We present a simpler and more direct reduction from the Hitting Set problem which establishes that $\min(k,\rho)\text{-}\mathrm{Shortcut}$ is NP-hard for $k\geq2$ and $\rho\geq k+2$ in both directed and undirected graphs. Complementing this, we use the symmetry of the undirected case to show that $\rho=k+1$ is solvable in polynomial time, a regime where the directed version remains a candidate for NP-hardness. Therefore, we obtain an almost complete characterization of the complexity of $\min(k,\rho)\text{-}\mathrm{Shortcut}$, with the sole remaining open case being $\rho = k+1$ in the directed setting.