On the Hardness of the Drone Delivery Problem

TL;DR

This work analyzes the computational complexity of the Drone Delivery Problem (DDT) under restricted movement models. It proves NP-hardness for DDT-Line even when agents have only two speeds and initial positions are fixed, via a Partition-based reduction that ties delivery time to a exact partition sum. It also establishes strong inapproximability for DDT-GridR without initial positions on unit grids with rectangular movement areas, via a 2P1N-3SAT-based reduction, while offering a straightforward $O(n)$-time greedy approximation. The results sharpen the boundary between tractable and intractable collaborative routing problems and show that even simplified geometric settings with limited speed diversity resist efficient approximation, though practical greedy strategies remain viable in the grid setting. These insights inform design principles for multi-agent delivery systems operating under restricted movement licenses and speed classes, emphasizing careful instance-structure considerations when evaluating routing algorithms.

Abstract

Fast shipping and efficient routing are key problems of modern logistics. Building on previous studies that address package delivery from a source node to a destination within a graph using multiple agents (such as vehicles, drones, and ships), we investigate the complexity of this problem in specialized graphs and with restricted agent types, both with and without predefined initial positions. Particularly, in this paper, we aim to minimize the delivery time for delivering a package. To achieve this, we utilize a set of collaborative agents, each capable of traversing a specific subset of the graph and operating at varying speeds. This challenge is encapsulated in the recently introduced Drone Delivery Problem with respect to delivery time (DDT). In this work, we show that the DDT with predefined initial positions on a line is NP-hard, even when considering only agents with two distinct speeds. This refines the results presented by Erlebach, et al.[10], who demonstrated the NP-hardness of DDT on a line with agents of arbitrary speeds. Additionally, we examine DDT in grid graphs without predefined initial positions, where each drone can freely choose its starting position. We show that the problem is NP-hard to approximate within a factor of $O(n^{1-\varepsilon}$), where $n$ is the size of the grid, even when all agents are restricted to two different speeds as well as rectangular movement areas. We conclude by providing an easy $O(n)$ approximation algorithm.

Figures (6)

• Figure 1: An example of a DDT-Line instance and its optimal solution is depicted: The solid line from $s$ to $y$ forms the path graph, where each solid edge has a distance of $1$. The initial positions of four drones are $p_1=s, p_2, p_3, p_4$, with the solid lines above and below representing each drone's movement subgraph. Suppose drone $1$ and $2$ have a speed of $1$, drone $3$ and drone $4$ have a speed of $3$. The optimal solution for this example takes a total time of $\max\{1, 3/3\} + 3/3 + 3 = 5$ by sequentially using drones $1$, $3$ and $2$ to deliver the package from $s$ to $y$, as indicated by the bold solid arrows. The dashed arrows represent the respective empty phases.
• Figure 2: A sketch of the DDT-Line instance construction. There are $3n+2$ slow agents $\{e_i, b^l_i, b^r_i\}_{i\in [n]} \cup \{d, p\}$, $4n+1$ fast agents $\{f^l_i,f^r_i, h^l_i, h^r_i\}_{i\in [n]} \cup \{q\}$. The starting point $s$ is the leftmost node of agent $d$'s interval, and the destination $y$ is the rightmost point of agent $e_n$'s interval. In the middle part we have a gap of size $\varepsilon$ which can only be traversed by the $e_i$'s or $p$. Assigning $e_i$'s to the left side such that the respective $p_i$'s sum up to exactly $\frac{P}{2}$, the package meets $p$ at $l_0$ without waiting time.
• Figure 3: A close-up on $p_i$'s corresponding agents: slow agent $e_i$, $b_i^l$ and $b_i^r$; fast agents $f_i^l$, $f_i^r$, $h_i^l$, and $h_i^r$. From the construction, it is clear that $e_i$ can only help only on either the left or right side interval, utilizing the fast drones $h_i$ on that side. A helping agent $h_i$ is utilized by assigning $e_i$ to either pick up the package from or deliver it to $h_i$. Not assigning $e_i$ leaves $h_i$ ineffective, as the delivery time is only dependent on $b_i$ as it has to catch up and pick up the package again.
• Figure 4: Two DDT-Grid instances. On the left is an instance (a) where agents have rectangular movement areas and have two distinct speeds: there are four slow agents with speed $1$ and one fast agent with speed $5$, displayed black and red respectively. Each movement area is represented as a rectangle, indicated by shading. The optimal solution, with the respective trips indicated by bold solid arrows, takes a total time of $7.6$ by sequentially using agents $a_1$, $a_5$, $a_4$ and $a_3$ to deliver the package from $s$ to $y$. On the right is an instance (b) with two agents having unit speed, where the movement areas of the agents are not rectangular. Note that even though the nodes of the subgraph of the second agent have rectangular shape, it is not a rectangular movement area as some edges are missing. The optimal solution takes a total time of $8$ by first using agent $a_1$ and then $a_2$ as indicated by the bold solid arrows.
• Figure 5: A sketch of the DDT-GridR instance: The package is to be delivered from the bottom-right to the top-left. The grey bars represent the rectangular movement areas of the literal agents associated with the three occurrences of every variable (two positive $x_{i,1},x_{i,2}$, as well as one negative $x_{i,3}$ ). On the left side we have $m'$ clause gadgets. In this example clause $c_1$ contains literals $x_1$, $\neg x_2$, and $x_3$. Therefore we have the three agents $x'_{1,1},x'_{2,3},x'_{3,1}$, which work as counterparts to $x_{1,1},x_{2,3},x_{3,1}$ delivering the package between two clause gadgets. Agent $c_h$ (on the far left) serves as an auxiliary agent to assist with the delivery to $y$. On the right side we have $n'$ variable gadgets. Every optimal schedule (given $\phi$ is satisfiable) must traverse all variable gadgets up to point $B$, as delivering horizontally over a distance $a$ with a slow agent would require an excessive amount of time.

Theorems & Definitions (10)

• theorem 1
• theorem 2
• theorem 2
• lemma 1
• proof
• lemma 2
• proof
• theorem 2
• proof
• theorem 2