Centroidal Localization Game

One important problem in a network $G$ is to locate an (invisible) moving entity by using distance-detectors placed at strategical locations in $G$. For instance, the famous metric dimension of a graph $G$ is the minimum number $k$ of detectors placed in some vertices $\{v_1,\cdots,v_k\}$ such that the vector $(d_1,\cdots,d_k)$ of the distances $d(v_i,r)$ between the detectors and the entity's location $r$ allows to uniquely determine $r$ for every $r \in V(G)$. In a more realistic setting, each device does not get the exact distance to the entity's location. Rather, given locating devices placed in $\{v_1,\cdots,v_k\}$, we get only relative distances between the moving entity's location $r$ and the devices (roughly, for every $1\leq i,j\leq k$, it is provided whether $d(v_i,r) >$, $<$, or $=$ to $d(v_j,r)$). The centroidal dimension of a graph $G$ is the minimum number of devices required to locate the entity, in one step, in this setting.In this paper, we consider the natural generalization of the latter problem, where vertices may be probed sequentially (i.e., in several steps) until the moving entity is located. Roughly, at every turn, a set $\{v_1,\cdots,v_k\}$ of vertices are probed and then the relative order of the distances between the vertices $v_i$ and the current location $r$ of the moving entity is given. If it not located, the moving entity may move along one edge. Let $\zeta^* (G)$ be the minimum $k$ such that the entity is eventually located, whatever it does, in the graph $G$. We first prove that $\zeta^* (T)\leq 2$ for every tree $T$ and give an upper bound on $\zeta^*(G\square H)$ for the cartesian product of graphs $G$ and $H$. Our main result is that $\zeta^* (G)\leq 3$ for any outerplanar graph $G$. We then prove that $\zeta^* (G)$ is bounded by the pathwidth of $G$ plus 1 and that the optimization problem of determining $\zeta^* (G)$ is NP-hard in general graphs. Finally, we show that approximating (up to a small constant distance) the location of the robber in the Euclidean plane requires at most two vertices per turn.