The n-vehicle exploration problem is NP-complete

TL;DR

The paper investigates the n-vehicle exploration problem (NVEP), a nonlinear optimization with mid-trip refueling where a permutation $\pi$ of $n$ vehicles yields a total distance $D_{\pi}$ that is maximized under the convoy's constraints and ends with all vehicles at the start. It proves NP-completeness by constructing a polynomial-time reduction from the Hamiltonian path problem: each graph vertex becomes a vehicle with $a_i=i$ and $b_i=\tfrac{1}{2}$, edges induce positive segment distances $d_i$ so that $V_i \Rightarrow V_j$ is feasible, and terminal nodes enforce a path of length $n$. The reduction shows that a Hamiltonian path exists iff NVEP admits a feasible sequence of length at least $n$, establishing $HP \le_P NVEP$ and, since NVEP is in $\mathcal{NP}$, NVEP is NP-complete. This work clarifies the source of hardness in NVEP and relates it to jeep- and aircraft-refueling variants, indicating a broader class of intractable problems arising from fractional objectives and sequencing constraints.

Abstract

The $n$-vehicle exploration problem (NVEP) is a nonlinear unconstrained optimization problem. Given a fleet of $n$ vehicles with mid-trip refueling technique, the NVEP tries to find a sequence of $n$ vehicles to make one of the vehicles travel the farthest, and at last all the vehicles return to the start point. NVEP has a fractional form of objective function, and its computational complexity of general case remains open. Given a directed graph $G$, it can be reduced in polynomial time to an instance of NVEP. We prove that the graph $G$ has a hamiltonian path if and only if the reduced NVEP instance has a feasible sequence of length at least $n$. Therefore we show that Hamiltonian path $\leq_P$ NVEP, and consequently prove that NVEP is NP-complete.

Theorems & Definitions (2)

• Theorem 1
• proof