Two-person Positive Shortest Path Games Have Nash Equilibria in Pure Stationary Strategies

TL;DR

This paper proves the existence of Nash equilibria in pure stationary strategies for finite two-person shortest path games with positive local costs, extending results to infinite graphs and interdiction variants, with polynomial-time algorithms.

Contribution

It establishes the existence of pure stationary Nash equilibria in positive shortest path games and provides polynomial-time algorithms for their computation, including interdiction game variants.

Findings

Nash equilibria exist in finite positive shortest path games.

Pure stationary NE can be computed in polynomial time.

Extension of results to infinite graphs and interdiction games.

Abstract

We prove that every finite two-person shortest path game, where the local cost of every move is positive for each player, has a Nash equilibrium (NE) in pure stationary strategies, which can be computed in polynomial time. We also extend the existence result to infinite graphs with finite out-degrees. Moreover, our proof gives that a terminal NE (in which the play is a path from the initial position to a terminal) exists provided at least one of the two players can guarantee reaching a terminal. If none of the players can do it, in other words, if each of the two players has a strategy that separates all terminals from the initial position $s$, then, obviously, a cyclic NE exists, although its cost is infinite for both players, since we restrict ourselves to positive games. We conjecture that a terminal NE exists too, provided there exists a directed path from $s$ to a terminal. However, this is open. We extend our result to short paths interdiction games, where at each vertex, we allow one player to block some of the arcs and the other player to choose one of the non-blocked arcs. Assuming that blocking sets are chosen from an independence system given by an oracle, we give an algorithm for computing a NE in time $O(|E|(\log|V|+\tau))$, where $V$ is the set of vertices, $E$ is the set of arcs, and $\tau$ is the maximum time taken by the oracle on any input.