Specializations and Generalizations of the Stackelberg Minimum Spanning Tree Game

Let be given a graph $G=(V,E)$ whose edge set is partitioned into a set $R$ of \emph{red} edges and a set $B$ of \emph{blue} edges, and assume that red edges are weighted and form a spanning tree of $G$. Then, the \emph{Stackelberg Minimum Spanning Tree} (\stack) problem is that of pricing (i.e., weighting) the blue edges in such a way that the total weight of the blue edges selected in a minimum spanning tree of the resulting graph is maximized. \stack \ is known to be \apx-hard already when the number of distinct red weights is 2. In this paper we analyze some meaningful specializations and generalizations of \stack, which shed some more light on the computational complexity of the problem. More precisely, we first show that if $G$ is restricted to be \emph{complete}, then the following holds: (i) if there are only 2 distinct red weights, then the problem can be solved optimally (this contrasts with the corresponding \apx-hardness of the general problem); (ii) otherwise, the problem can be approximated within $7/4 + \epsilon$, for any $\epsilon > 0$. Afterwards, we define a natural extension of \stack, namely that in which blue edges have a non-negative \emph{activation cost} associated, and it is given a global \emph{activation budget} that must not be exceeded when pricing blue edges. Here, after showing that the very same approximation ratio as that of the original problem can be achieved, we prove that if the spanning tree of red edges can be rooted so as that any root-leaf path contains at most $h$ edges, then the problem admits a $(2h+\epsilon)$-approximation algorithm, for any $\epsilon > 0$.Comment: 22 pages, 7 figure

Stackelberg Max Closure with Multiple Followers

In a Stackelberg max closure game, we are given a digraph whose vertices correspond to projects from which firms can choose and whose arcs represent precedence constraints. Some projects are under the control of a leader who sets prices in the first stage of the game, while in the second stage, the firms choose a feasible subset of projects of maximum value. For a single follower, the leader’s problem of finding revenue-maximizing prices can be solved in strongly polynomial time. In this paper, we focus on the setting with multiple followers and distinguish two situations. In the case in which only one copy of each project is available (limited supply), we show that the two-follower problem is solvable in strongly polynomial time, whereas the problem with three or more followers is NP-hard. In the case of unlimited supply, that is, when sufficient copies of each project are available, we show that the two-follower problem is already APX-hard. As a side result, we prove that Stackelberg min vertex cover on bipartite graphs with a single follower is APX-hard

Computational aspects of a 2-player Stackelberg shortest paths tree game

Let a communication network be modelled by a directed graph G = (V,E) of n nodes and m edges. We consider a one-round two-player network pricing game, the Stackelberg Shortest Paths Tree (StackSPT) game. This is played on G, by assuming that edges in E are partitioned into two sets: a set E F of edges with a fixed positive real weight, and a set E P of edges that should be priced by one of the two players (the leader). Given a distinguished node r ∈ V, the StackSPT game is then as follows: the leader prices the edges in E P in such a way that he will maximize his revenue, knowing that the other player (the follower) will build a shortest paths tree of G rooted at r, say S(r), by running a publicly available algorithm. Quite naturally, for each edge selected in the solution, the leader’s revenue is assumed to be equal to the loaded price of an edge, namely the product of the edge price times the number of paths from r in S(r) that use it. First, we show that the problem of maximizing the leader’s revenue is NP-hard as soon as |E P | = Θ(n). Then, in search of an effective method for solving the problem when the size of E P is constant, we focus on the basic case in which |E P | = 2, and we provide an efficient O(n 2 logn) time algorithm. Afterwards, we generalize the approach to the case |E P | = k, and we show that it can be solved in polynomial time whenever k = O(1)