Revolutionaries and spies: Spy-good and spy-bad graphs

We study a game on a graph G played by r revolutionaries and s spies. Initially, revolutionaries and then spies occupy vertices. In each subsequent round, each revolutionary may move to a neighboring vertex or not move, and then each spy has the same option. The revolutionaries win if m of them meet at some vertex having no spy (at the end of a round); the spies win if they can avoid this forever. Let σ(G, m, r) denote the minimum number of spies needed to win. To avoid degenerate cases, assume |V (G)| ≥ r −m + 1 ≥ ⌊r/m⌋ ≥ 1. The easy bounds are then ⌊r/m⌋ ≤ σ(G, m, r) ≤ r −m + 1. We prove that the lower bound is sharp when G has a rooted spanning tree T such that every edge of G not in T joins two vertices having the same parent in T . As a consequence, σ(G, m, r) ≤ γ(G) ⌊r/m⌋, where γ(G) is the domination number; this bound is nearly sharp when γ(G) ≤ m. For the random graph with constant edge-probability p, we obtain constants c and c′ (depending on m and p) such that σ(G, m, r) is near the trivial upper bound when r < c lnn and at most c′ times the trivial lower bound when r > c′ lnn. For the hypercube Qd with d ≥ r, we have σ(G, m, r) = r −m + 1 when m = 2, and for m ≥ 3 at least r − 39m spies are needed. For complete k-partite graphs with partite sets of size at least 2r, the leading term in σ(G, m, r) is approximately k k−1 r m when k ≥ m. For k = 2, we have σ(G, 2, r) = ⌈ ⌊7r/2⌋−3 5 ⌉ and σ(G, 3, r) = ⌊r/2⌋, and in general 3r 2m − 3 ≤ σ(G, m, r) ≤ (1+1/ √ 3)r m .