Simple strategies versus optimal schedules in multi-agent patrolling

Suppose that we want to patrol a fence (line segment) using k mobile agents with given speeds v₁,..., v_k so that every point on the fence is visited by an agent at least once in every unit time period. A simple strategy where the ith agent moves back and forth in a segment of length v_i/2 patrols the length (v₁ +· · ·+v_k)/2, but it has been shown recently that this is not always optimal. Thus a natural question is to determine the smallest c such that a fence of length c(v₁ + · · · + v_k)/2 cannot be patrolled. We give an example showing c ≥ 4/3 (and conjecture that this is the best possible). We also consider a variant of this problem where we want to patrol a circle and the agents can move only clockwise. We can patrol a circle of perimeter rv_r by a simple strategy where the r fastest agents move at the same speed, but it has been shown recently that this is not always optimal. We conjecture that this is not even a constant-approximation strategy. To tackle this conjecture, we relate it to what we call constant gap families. Using this relation, we give another example where the simple strategy is not optimal. We propose another variant where we want to patrol a single point under the constraint that for each i = 1,..., k, the time between two consecutive visits of agent i should be a_i or longer. This problem can be reduced to the discretized version where the ai are integers and the goal is to visit the point at every integer time. It is easy to see that this discretized patrolling is impossible if 1/a₁+· · ·+1/a_k < 1, and that there is a simple strategy if 1/a₁+· · ·+1/a_k ≥ 2. Thus we are interested in the smallest c such that patrolling is always possible if 1/a₁ + · · · + 1/a_k ≥ c. We prove that α ≤ c < 1.546, where α = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.