We consider the following surveillance problem: Given a set P of n sites in a metric space and a set R of k robots with the same maximum speed, compute a patrol schedule of minimum latency for the robots. Here a patrol schedule specifies for each robot an infinite sequence of sites to visit (in the given order) and the latency L of a schedule is the maximum latency of any site, where the latency of a site s is the supremum of the lengths of the time intervals between consecutive visits to s. When k = 1 the problem is equivalent to the travelling salesman problem (TSP) and thus it is NP-hard. For k ? 2 (which is the version we are interested in) the problem becomes even more challenging; for example, it is not even clear if the decision version of the problem is decidable, in particular in the Euclidean case. We have two main results. We consider cyclic solutions in which the set of sites must be partitioned into l groups, for some l ? k, and each group is assigned a subset of the robots that move along the travelling salesman tour of the group at equal distance from each other. Our first main result is that approximating the optimal latency of the class of cyclic solutions can be reduced to approximating the optimal travelling salesman tour on some input, with only a 1 + e factor loss in the approximation factor and an O ((k/e) k) factor loss in the runtime, for any e > 0. Our second main result shows that an optimal cyclic solution is a 2(1 - 1/k)-approximation of the overall optimal solution. Note that for k = 2 this implies that an optimal cyclic solution is optimal overall. We conjecture that this is true for k ? 3 as well. The results have a number of consequences. For the Euclidean version of the problem, for instance, combining our results with known results on Euclidean TSP, yields a PTAS for approximating an optimal cyclic solution, and it yields a (2(1 - 1/k) + e)-approximation of the optimal unrestricted (not necessarily cyclic) solution. If the conjecture mentioned above is true, then our ...