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

Transcription

1 Revolutionaries and spies: Spy-good and spy-bad graphs Jane V. Butterfield, Daniel W. Cranston, Gregory J. Puleo, Douglas B. West, and Reza Zaani May 6, 01 Abstract We study a gae on a graph G played by r revolutionaries and s spies. Initially, revolutionaries and then spies occupy vertices. In each subsequent round, each revolutionary ay ove to a neighboring vertex or not ove, and then each spy has the sae option. The revolutionaries win if of the eet at soe vertex having no spy (at the end of a round); the spies win if they can avoid this forever. Let σ(g,,r) denote the iniu nuber of spies needed to win. To avoid degenerate cases, assue V (G) r + 1 r/ 1. The easy bounds are then r/ σ(g,,r) r + 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 sae parent in T. As a consequence, σ(g,,r) γ(g) r/, where γ(g) is the doination nuber; this bound is nearly sharp when γ(g). For the rando graph with constant edge-probability p, we obtain constants c and c (depending on and p) such that σ(g,,r) is near the trivial upper bound when r < cln n and at ost c ties the trivial lower bound when r > c ln n. For the hypercube Q d with d r, we have σ(g,,r) = r + 1 when =, and for 3 at least r 39 spies are needed. For coplete k-partite graphs with partite sets of size at least r, the leading ter in σ(g,,r) is approxiately 7r/ 3 5 k r k 1 when k. For k =, we have σ(g,,r) = and σ(g,3,r) = r/, and in general 3r 3 σ(g,,r) (1+1/ 3)r. Matheatics Departent, University of Illinois, jbutter@illinois.edu, partially supported by NSF grant DMS , EMSW1-MCTP: Research Experience for Graduate Students. Matheatics Departent, Virginia Coonwealth University, dcranston@vcu.edu. Matheatics Departent, University of Illinois, puleo@illinois.edu, partially supported by NSF grant DMS , EMSW1-MCTP: Research Experience for Graduate Students. Matheatics Departent, University of Illinois, west@ath.uiuc.edu, partially supported by NSA grant H Coputer Science Departent, University of Illinois, zaani@uiuc.edu. 1

2 1 Introduction We study a pursuit gae involving two teas on a graph. The first tea consists of r revolutionaries; the second consists of s spies. The revolutionaries want to arrange a onetie eeting of revolutionaries free of oversight by spies. Initially, the revolutionaries take positions at vertices, and then the spies do the sae. In each subsequent round, each revolutionary ay ove to a neighboring vertex or not ove, and then each spy has the sae option. All positions are known by all players at all ties. The revolutionaries win if at the end of a round there is an unguarded eeting, where a eeting is a set of (at least) revolutionaries on one vertex, and a eeting is unguarded if there is no spy at that vertex. The spies win if they can prevent this forever. Let RS(G,, r, s) denote this gae played on the graph G by s spies and r revolutionaries seeking an unguarded eeting of size. The spies trivially win if s V (G) or r <. If r/ < V (G), then the revolutionaries can for r/ eetings initially, and hence at least r/ spies are needed to avoid losing iediately. On the other hand, the spies win if s r +1; they follow r +1 distinct revolutionaries, and the other 1 revolutionaries cannot for a eeting. To avoid degenerate or trivial gaes, henceforth in this paper we always assue V (G) r + 1 r/ 1. Let σ(g,, r) denote the iniu s such that the spies win the gae RS(G,, r, s). The gae of Revolutionaries and Spies was invented by Jozef Beck in the id-1990s (unpublished). Syth proptly showed that σ(g,, r) = r/ when G is a tree, achieving the trivial lower bound (a later proof appears in []). Howard and Syth [4] studied the gae when G is the infinite -diensional integer grid with one-step horizontal, vertical, and diagonal edges. They observed that the spy wins RS(G,, 1, 1) (the spy stays at the edian position), and hence σ(g,, r) r + when r 1 (note that always σ(g,, r) σ(g,, r 1) + 1). For =, they proved that 6 r/8 σ(g,, r) r when r 3; they conjectured that the upper bound is the correct answer. Cranston, Syth, and West [] showed that σ(g,, r) r/ when G has at ost one cycle. Furtherore, let G be a unicyclic graph consisting of a cycle of length l and t vertices not on the cycle. They showed that if r (and as usual V (G) > r/ to avoid degeneracies), then σ(g,, r) = r/ if and only if l ax{ r/ t +, 3}. Our objective in this paper is to advance the systeatic study of this gae. We show that the trivial lower and upper bounds on σ(g,, r) each ay be sharp on various classes of graphs. Furtherore, we obtain classes where neither bound is asyptotically sharp and yet still σ(g,, r) can be deterined or closely approxiated. Say that G is spy-good if σ(g,, r) equals the trivial lower bound r/ for all and r such that r/ < V (G). In Section, we prove that every webbed tree is spy-good, where

3 a webbed tree is a graph G containing a rooted spanning tree T such that every edge of G not in T joins vertices having the sae parent in T. For exaple, every graph having a doinating vertex u is a webbed tree (rooted at u). Section 3 considers general bounds. Always σ(g,, r) γ(g) r/, where γ(g) is the doination nuber of G (the iniu size of a set S such that every vertex outside S has a neighbor in S). Since always r/ (r + 1)/, this upper bound is nontrivial only when γ(g) <. In that case, it is nearly sharp: for t,, r N with t <, we construct a graph with doination nuber t such that σ(g,, r) > t(r/ 1). In contrast to spy-good graphs, a graph G is spy-bad for r revolutionaries and eeting size if σ(g,, r) equals the trivial upper bound r + 1. Section 3 constructs chordal graphs (and bipartite graphs) that are spy-bad (for given r and ). In Section 4 we study hypercubes, showing first that the d-diensional hypercube Q d is spy-bad when d r and =. Also, the winning strategy for the revolutionaries uses only vertices near a fixed vertex. By splitting the revolutionaries into disjoint groups who play this strategy around vertices far apart, it follows that when d < r d /d 8, the revolutionaries win against (d 1) r/d spies on Q d (for = ). For general, we show that hypercubes are nearly spy-bad by proving σ(q d,, r) r 39 for d r. (For sall, the bound σ(q d,, r) r 3 4 when d r is better.) In these exaples of spy-bad graphs, there are few revolutionaries copared to the nuber of vertices. Siilar behavior holds for the rando graph with constant edge-probability (Section 5); the threshold for spies to win depends on the relationship between r and the nuber of vertices, n. Via fairly siple arguents, we obtain constants c and c (depending on ) such that alost always r + 1 spies are needed when r < c lnn, while a ultiple of r/ spies are enough when r > c ln n. Using ore intricate structural characteristics of the rando graph and a ore coplex strategy for the spies, Mitsche and Pra lat [5] independently proved that σ(g,, r) = (1+o(1))r/ spies suffice when r grows faster than (log n)/p (here also p ay depend on n). A coplete k-partite graph is r-large if each part has at least r vertices, which is as any vertices as the players ight want to use. In Section 6, we prove σ(g,, r) k r + k. k 1 Also σ(g,, r) k r 1 k when k and c =. k 1 +c k 1 Section 7 focuses on coplete bipartite graphs and contains our ost delicate results. When G is an r-large coplete bipartite graph, we obtain σ(g,, r) = 7r/ 3 5 and σ(g, 3, r) = r/. For larger we do not have the coplete answer; we prove ( ) ( 3 r o(1) σ(g,, r) ) r 3 < 1.58 r, where the upper bound requires r 1 1 1/ 3. We conjecture that σ(g,, r) is approxiately 3r when 3 divides, but in other cases the revolutionaries do a bit better. That advantage should fade as grows, with σ(g,, r) 3r. 3

4 Upper bounds for σ(g,, r) are proved using strategies for the spies. We define a notion of stable position in the gae. Proving that a particular nuber of spies can win involves showing that in a stable position all eetings are guarded and that for any ove by the revolutionaries fro a stable position, the spies can reestablish stability. This technique is used for graphs with doinating vertices and for webbed trees in Section, for rando graphs in Section 5, and for coplete ultipartite and coplete bipartite graphs in Sections 6 and 7. Each setting uses its own definition of stability tailored to the graphs under study. Lower bounds are proved by strategies for the revolutionaries, which usually are uch sipler. Most of our winning strategies for revolutionaries take at ost two rounds, but on hypercubes they take 1 rounds. In [], strategies for revolutionaries proving that σ(c n,, r) = r/ (when r/ < n) ay take any rounds. Many questions reain open, such as a characterization of spy-good graphs. In all known spy-good graphs, the spies can ensure that at the end of each round the nuber of spies at any vertex v is at least r(v)/, where r(v) is the nuber of revolutionaries at v. Existence of such a strategy is preserved when vertices expand into a coplete subgraph. Also, Howard and Syth [4] observed that σ(g,, r) is preserved by taking the distance power of a graph. Hence every graph obtained fro soe webbed tree via soe sequence of distance powers or vertex expansions is spy-good, but these are not the only spy-good graphs. It would also be interesting to bound σ(g,, r) in ters of other graph paraeters, such as treewidth. Generalizations of the gae are also possible, such as by allowing players to travel farther in a ove or by requiring ore spies to guard a eeting. One can also consider analogous gaes on directed graphs. Doinating Vertices and Webbed Trees We begin with graphs having a doinating vertex (a vertex adjacent to all others); we then apply this result to webbed trees. Let N(v) denote the neighborhood of a vertex v. Also N[v] = N(v) {v}, and N(S) = v S N(v). Definition.1. For a graph G having a doinating vertex u, a position in the gae RS(G,, r, s) is stable if, for each vertex v other than u, the nuber of spies at v is exactly r(v)/, where r(v) is the nuber of revolutionaries at v. The other spies, if any, are at u. Theore.. If a graph G has a doinating vertex, then σ(g,, r) = r/. Proof. Let u be a doinating vertex in G, and let s = r/. Since s = r/, a stable position will have a spy at u if there is a eeting at u. Hence a stable position has no unguarded eeting. When s = r/, there are enough spies to establish a stable position after the initial round. We show that the spies can reestablish a stable position at the end of each round. 4

5 Consider a stable position at the start of round t. Let X be a axial faily of disjoint sets of revolutionaries such that each set is located at one vertex other than u. Let Y be such a axial faily after the revolutionaries ove in round t. In X or Y, ore than one set ay be located at a single vertex in G. For exaple, a vertex v having p + q revolutionaries at the start of round t (where 0 q < ) corresponds to p eleents of X, and there are p spies at v at that tie. Let X = {x 1,..., x k } and Y = {y 1,...,y k }. Let X = {x k+1,...,x s }, representing the excess spies waiting at u after round t. Define an auxiliary bipartite graph H with partite sets X X and Y. For x i X and y j Y, put x i y j E(H) if soe revolutionary fro eeting x i is in eeting y j (note that x i and y j ay be the sae set). Also ake all of X adjacent to all of Y. If soe atching in H covers Y, then the spies can ove so that every vertex other than u having p +q revolutionaries at the end of round t (where 0 q < ) has exactly p spies on it (and the reaining spies are at u). The existence of such a atching follows fro Hall s Theore. For S Y, always X N(S), so N(S) = X + N(S) X. Consider the S revolutionaries in the eetings corresponding to S. Such revolutionaries cae fro eetings in N(S) X or were not in any of the k eetings indexed by X. Hence S N(S) X + (r k). Since X = s k and s = r/, N(S) X + S ( r/ k) = s k + S ( r/ k) = S, so Hall s Condition holds. Corollary.3. Fix n,, r with n r/. For 0 k ( n ), there is an n-vertex graph G with k edges such that σ(g,, r) = r/. Proof. For k n, for G by adding the desired nuber of edges joining leaves of an n- vertex star; Theore. applies. For k n 1, let G be a star plus isolated vertices; use Theore. and a + b a + b. Definition.4. For any vertex v in a rooted tree, the parent of a non-root vertex v (written v + ) is the first vertex after v on the path fro v to the root. The set of children of v (written C(v)) is the set of neighbors of v other than its parent, and the set of descendants of v (written D(v)) is the set of vertices whose path to the root contains v. A webbed tree is a graph G having a rooted spanning tree T such that every edge of G outside T joins two vertices having the sae parent (called siblings). Figure 1 shows a webbed tree, with the rooted spanning tree T in bold. Trivially, every tree is a webbed tree, as is every graph having a doinating vertex. In fact, a -connected graph is a webbed tree if and only if it has a doinating vertex. Every webbed tree is a graph whose blocks have doinating vertices, but the converse does not 5

6 hold. Consider the graph obtained fro two 4-cycles with a coon vertex by adding chords of the 4-cycles to create four vertices of degree 3; every block has a doinating vertex, but the graph is not a webbed tree. Our ain result in this section is that all webbed trees are spy-good. This conclusion is proved for trees in []. In that paper, an invariant defined in ters of the positions of the revolutionaries specifies how any spies should be placed on each vertex. The invariant guarantees that all eetings are covered, and a direct proof is given to show that the spies can restore the invariant after each round. Here we use the sae invariant to generalize the tree result to the class of webbed trees. Our ethod of proving that the invariant has the desired properties is different fro that in []. Here we decopose the spies response into independent responses in iagined gaes on subgraphs having a doinating vertex. After the revolutionaries ove, the spies restore the invariant by applying the strategy in Theore. independently to each graph induced by a vertex and its children in the spanning tree. Because we will apply Theore., we don t use stable for positions satisfying the invariant in a webbed tree; instead, we reserve that ter for positions in the auxiliary local gaes, whose graphs have doinating vertices. In [], the result on trees is extended in a different direction to deterine the winner in RS(G,, r, s) whenever G has at ost one cycle. A siilar extension is possible here for graphs obtained by adding a cycle through the roots of disjoint webbed trees, but the resulting faily is not as natural as the faily of unicyclic graphs. Theore.5. If G is a webbed tree, then σ(g,, r) = r/. Proof. Let T be a rooted spanning tree in G such that every edge of G not in T joins sibling vertices in T. Let z be the root of T, and let s = r/. The notation for children and descendants is as in Definition.4 with respect to T. For each vertex v, let r(v) and s(v) denote the nuber of revolutionaries and spies on v at the current tie, respectively, and let w(v) = u D(v) r(u). The spies aintain the following invariant specifying the nuber of spies on each vertex at the end of any round: w(v) s(v) = w(x) for v V (G). (1) x C(v) Since x C(v) w(x) = w(v) r(v), the forula is always nonnegative. Also, if r(v), then s(v) w(v) r(v) 1. Hence (1) guarantees that every eeting is guarded. w(v) To show that the spies can establish (1) after the first round, it suffices that all the forulas su to r/. More generally, suing over the descendants of any vertex v, w(v) s(u) =, () u D(v) 6

7 since w(u)/ occurs positively in the ter for u and negatively in the ter for u +, except that w(v)/ occurs only positively. When v = z, the total is r/, since w(z) = r. To show that the spies can aintain (1), let r(v) and s(v) refer to the start of round t, let r (v) denote the nuber of revolutionaries at v after the revolutionaries ove in round t, and let w (v) = u D(v) r (v). The spies will ove in round t to achieve the new values required by (1). To deterine these oves, we will use Theore. to obtain a stable position in each subgraph induced by a vertex and its children, independently. Let G(v) denote the subgraph induced by C(v) {v}; note that v is a doinating vertex in G(v). We will play a round in an iagined local gae on G(v) for each vertex v. v + G(v + ) v G(v) Figure 1: Decoposition of a webbed tree To set up the local gaes, we partition the s(v) spies at each vertex v into a set of š(v) spies to be used in the local gae on G(v) and a set of ŝ(v) spies to be used in the local gae on G(v + ), where š(v) and ŝ(v) su to s(v) (when the tree is drawn with the root z at the top, the accent indicates the direction of the relevant subgraph). Let D (v) = D(v) {v}. Let w (v) be the nuber of revolutionaries that are in D (v) at the start of round t or are there after the revolutionaries ove in round t. Every revolutionary counted by w (v) is also counted by w(v), and every revolutionary counted by x C(v) w(x) is also counted by w (v). These stateents also hold with w in place of w. Hence w(v) w (v) and w (v) w(x). (3) x C(v) By (3), ŝ(v) and š(v) are nonnegative when we define w(v) w (v) w (v) ŝ(v) = and š(v) = x C(v) w(x). (4) By (1), ŝ(v) + š(v) = s(v). Note also that if v is a leaf of T, then š(v) = 0 and ŝ(v) = s(v). For each non-leaf vertex v, the spies first iagine positions of revolutionaries in a gae on the graph G(v) that together with (4) for the spies for a stable position. After viewing 7

8 the actual oves by revolutionaries within G(v) as oves in this gae, the spies reestablish stability as in Theore.. We will show that the resulting positions satisfy the global invariant. The spies iagine ˆr(v) spies at v in G(v + ) and ř(v) spies at v in G(v), where w (v) ˆr(v) = w(v) and ř(v) = w (v) x C(v) w(x). (5) By (3), the values of ř(v) and ˆr(v) are nonnegative. Furtherore, we clai that if (4) and (5) hold at each vertex v, then the position on each subgraph induced by one parent and its children is stable. In G(v) we use š(v) and ř(v), and we use ŝ(x) and ˆr(x) for x C(v). By definition, ŝ(x) = ˆr(x)/. It reains only to check the su. We copute the total nuber of revolutionaries in the local gae: ř(v) + ˆr(x) = w (v) w(x) + w(x) w (x) x C(v) x C(v) w (x) x C(v) x C(v), whose floor is š(v) + x C(v) ŝ(x), as desired. Dividing by yields w (v) x C(v) The spies next view the actual oves by revolutionaries in the global gae as oves by the revolutionaries in the iagined local gaes. Each such ove occurs within the subgraph G(v) for one vertex v. The local gae can odel these oves if the relevant value of ˆr or ř is at least the nuber of real revolutionaries leaving this vertex and staying within this subgraph. The revolutionaries leaving v by edges in G(v + ) are those that were in D(v) and now are not; there are at ost w(v) w (v) of the. By (5), ˆr(v) is at least this large. Siilarly, revolutionaries leaving v via G(v) wind up in D (v) but were not there previously, so the nuber of the is at ost w (v) x C(v) w(x), which equals ř(v). The net change in the actual nuber of revolutionaries at v is r (v) r(v). Soe of this change is due to oves in G(v) and the rest to oves in G(v + ). Moves in G(v + ) enter or leave D(v). Hence the net change in the nuber of revolutionaries at v due to such oves is w (v) w(v). The reaining net change, due to oves between v and its children (in G(v)), is (r (v) r(v)) (w (v) w(v)). Therefore, after executing the actual oves in the iagined local gaes, the new iagined distributions for the revolutionaries are given by ˆr (v) = ˆr(v) + w (v) w(v) and ř (v) = ř(v) + (r (v) r(v)) (w (v) w(v)). (6) The specification of ˆr(v) in (5) and the change fro ˆr(v) to ˆr (v) in (6) iediately yield the forula for ˆr (v) in (7). To obtain ř (v), start with the forula for ř (v) in (5) and adjust by the definitions of r(v) r(v) and w (v) r (v), as indicated in (6). We copute ř (v) = ř(v) + (w(v) r(v)) (w (v) r (v)) = w (v) w(x) + w(x) x C(v) x C(v) x C(v) w (x) = w (v) x C(v) w (x). 8

9 Thus w ˆr (v) = w (v) (v) and ř (v) = w (v) x C(v) w (x). (7) The spies now respond in the local gaes. By Theore., these positions are stable, so ŝ (x) = ˆr (x)/ for x C(v), and š (v) is the leftover aount for v in the local gae on G(v). By the sae coputation that earlier showed š(v) was the correct needed aount of spies left for v in G(v), also š (v) = w (v) w (x) x C(v). Because each spy participated in exactly one local gae, playing the local gaes independently ensures autoatically that each spy oves at ost once in round t. Hence the spy oves we have described are feasible. It reains only to show that (1) holds for the resulting distribution of spies; that is w ŝ (v) + š (v) (v) = w (x) for v V (G). x C(v) Since the ters involving w again cancel, we use (7) to show that ŝ (v) + š (v) equals the desired value s (v) in the sae way we used (5) to show that the invented values ŝ(v) and š(v) su to s(v). 3 Spy-good vs. Spy-bad It is not true that all spy-good graphs are webbed trees. Given G, let G k denote the graph defined by V (G k ) = V (G) and E(G k ) = {uv: d G (u, v) k}. The spies can siulate one round of the gae on G k by playing k rounds on G. Thus σ(g k,, r) σ(g,, r), as noted by Howard and Syth [4]. This akes the square of a webbed tree spy-good, even though it is not generally a webbed tree (consider G = P n, for exaple). Say that a spy strategy is conforal if at the end of each round the nuber of spies at each vertex v is at least r(v)/, where r(v) is the nuber of revolutionaries there. For any conforal spy strategy on G, the strategy described above for G k is also conforal. Another graph operation also preserves the existence of conforal strategies. Proposition 3.1. Obtain G fro a graph G by expanding a vertex of G into a clique. If r/ spies win RS(G,, r, s) by a conforal strategy, then the sae holds for G. Proof. Let Q be the clique into which vertex v of G is expanded to for G. The spies play on G by iagining a gae on G. At each round, the revolutionaries on Q in G are collected onto v in G, with r(v) there after the previous round and r (v) after the revolutionaries ove. For other vertices, the aounts before and after are as in the real gae on G. Since a i a i, the spies on v at the end of the round in G suffice to cover the r (v) revolutionaries on Q in G and can ove there, since all vertices of Q have the sae 9

10 neighbors outside Q that v has in G. Extra spies ove to any vertex of Q. Moveents of spies fro v in G can also be atched by oves in the gae on G. Other oveents are the sae in G and G. This produces a conforal strategy on G. Proposition 3.. On a webbed tree G, the winning strategy in Theore.5 is conforal. Proof. Let T be a rooted spanning tree such that edges outside T join siblings in T. After each round, the nuber of spies on vertex v is given by r(v) + x C(v) w(x) w(x). Since a i a i, the strategy is conforal. x C(v) These results iply that graphs obtained fro webbed trees by vertex expansions and distance powers are spy-good. For exaple, the square of a path is spy-good. This graph is not a webbed tree, since it is -connected but has no doinating vertex (when it has at least six vertices). On the other hand, it is an interval graph, where an interval graph is a graph representable by assigning each vertex v an interval on the real line so that vertices are adjacent if and only if their intervals intersect. An interval graph that is not a distance power and has no two vertices with the sae closed neighborhood is obtained fro the square of an 8-vertex path by adding an edge joining the third and sixth vertices. Question 3.3. Which graphs are spy-good? We believe that all interval graphs are spy-good, even though the class is not contained in the spy-good classes obtained above. Although not all graphs are spy-good, Theore. yields good upper bounds on σ(g,, r) for graphs with sall doinating sets. A doinating set in a graph G is a set S V (G) such that every vertex outside S has a neighbor in S; the doination nuber γ(g) is the iniu size of a doinating set in G. Corollary 3.4. σ(g,, r) γ(g) r/ for any graph G. Proof. Let S be a sallest doinating set. With each vertex u S, associate r/ spies. Let G u be the subgraph of G induced by N[u]; it has u as a doinating vertex. The spies associated with u stay in G u, following the strategy of Theore. on G u. When there are fewer than r revolutionaries in G u, the spies iagine that the issing ones are at u. When a real revolutionary coes to vertex v in G u fro outside G u, a revolutionary in the iagined gae oves fro u to v to perfor its oves. When the real revolutionary leaves G u, the revolutionary tracking it in the gae on G u returns to u. These oves are possible, since u is a doinating vertex in G u. Since the spies win each iagined gae, the revolutionaries in the real gae never ake an unguarded eeting at the end of a round. 10

11 As rearked in the introduction, Corollary 3.4 is of interest only when γ(g), because otherwise the trivial upper bound r + 1 is stronger. When γ(g), the bound in Corollary 3.4 cannot be iproved. To otivate the proof, we first present a siple construction of spy-bad graphs. A split graph is a graph whose vertices can be partitioned into a clique and an independent set. A chordal graph is a graph in which every cycle of length at least 4 has a chord; split graphs clearly have this property. Recall that for fixed r and a graph is spy-bad if the revolutionaries can beat r spies (r + 1 spies trivially win). Proposition 3.5. Given r, N, there is a chordal graph G (in fact a split graph) such that σ(g,, r) = r + 1. Proof. Let G,r be the split graph consisting of a clique Q of size r and an independent set S of size ( r ), with the neighborhoods of the vertices in S being distinct -sets in Q. We show that r spies cannot win. The revolutionaries initially occupy each vertex of Q. Let s be the nuber of vertices of Q initially occupied by spies. The nuber of threatened eetings that spies on Q are not adjacent to is ( ) r s. Protecting against such threats requires putting spies initially on the ( r s ) vertices of S corresponding to these -sets, but only r s reaining spies are available, and ( ) r s > r s when r s. Note that r +1 r/ can be ade arbitrarily large. When r =, the ratio exceeds /. Letting also grow, we observe that σ(g,, r) cannot be bounded by a constant ultiple of r/, even on split graphs. Furtherore, the strategy for revolutionaries in Proposition 3.5 does not use any edges within the clique, so the stateent reains true also for the bipartite graph obtained by deleting those edges. When grows, the degrees of all vertices in G,r also grow. If the degrees in the independent set are bounded, then the spies can do better. We state the next result without proof, because the proof is a bit technical and the class of graphs is soewhat specialized. The technique is as usual for upper bounds: defining stable positions and showing that the spies can reestablish a stable position after each round. The proof will appear in the thesis of the third author. Theore 3.6. Let G be a split graph with clique Q and independent set S in which each vertex of S has degree at ost d. If is a ultiple of d, then σ(g,, r) d r/. A construction like that of Proposition 3.5 enables us to show that Corollary 3.4 is nearly sharp. When t =, the upper and lower bounds in this result are equal; when r, the difference between the is t 1. Theore 3.7. Given t,, r N such that t r, there is a graph G with doination nuber t such that σ(g,, r) > t(r/ 1). 11

12 Proof. First we construct a graph G. Begin with a copy of K t,r having partite sets T of size t and R of size r. Add an independent set U of size t ( r ), grouped into sets of size t. With each -set A in R, associate one t-set A in U. Make all of A adjacent to all of A, and add a atching joining A to T (see Figure ). Note that T is a doinating set. T = t T A R R = r A = U t A t t ( ) r Figure : Sharpness of the doination bound To show that γ(g) = t, let S be a sallest doinating set. For each -set A in R, the t vertices in A are adjacent only to A in R. Thus if S R < t r, then soe t-set A in U is undoinated by S R. Outside of R, the closed neighborhoods of the vertices in A are pairwise disjoint, so S needs t additional vertices to doinate the. Hence S t. Now, we give a strategy for the revolutionaries to win against t(r/ 1) spies on G. Let s = t(r/ 1). The revolutionaries initially occupy R, one on each vertex. A spy on a vertex u of U can protect all the sae threats (and ore) by locating at the neighbor of u in T instead. Hence we ay assue (at least for the purpose of trying to survive the next round) that no spies locate initially in U. Let v be a vertex of T having the fewest initial spies, and let s(v) be the nuber of spies there. The revolutionaries will win by attacking the neighbors of v. Let s be the nuber of spies initially in R, so s(v) (s s )/t. The revolutionaries want to for eetings at s(v) + 1 neighbors of v that are neighbors of no other vertices with spies. Let R be the set of vertices in R that do not have spies; note that R r s. If R (s(v) + 1), then the revolutionaries win as follows. First, group vertices in R into s(v) + 1 sets of size. For each such set A, the revolutionaries on A ove to the unique vertex u A,v in the associated subset A of U that is adjacent to v in T. For each such vertex, the only neighbor having a spy is v, so the eetings cannot all be guarded and the revolutionaries win. It thus suffices to show that r s (s(v) + 1). Since v has the fewest spies aong vertices of T, we have ts(v) s s t(r/ 1) s. Multiplying by /t and adding yields (s(v) + 1) r s (/t) r s, as desired, using t at the end. 1

13 Although the construction in Theore 3.7 depends heavily on, it does not depend uch on r. Indeed, the construction works equally well whenever the nuber of revolutionaries is at ost r, because the revolutionaries can use the strategy for a saller nuber of revolutionaries on the appropriate subgraph of the graph constructed for r revolutionaries. The sae coent applies to Proposition Hypercubes and Retracts For d N, let [d] = {1,..., d}. The d-diensional hypercube Q d is the graph with vertex set {v S : S [d]} such that v S and v T are adjacent when the syetric difference of S and T has size 1. The weight of the vertex v S is S. For vertices of sall weight, we write the subscripts without set brackets. We show first that Q d is spy-bad for = when d r. For larger, we will later obtain a lower bound on σ(q d,, r) using the sae basic idea. Theore 4.1. If G = Q d and d r, then σ(g,, r) = r 1. Proof. The upper bound is trivial; we show that r spies cannot win. The revolutionaries begin by occupying v 1,...,v r, threatening eetings of size at and at ( r ) vertices of weight. Let t be the nuber of revolutionaries left uncovered by the initial placeent of the spies. Threats at ( t ) vertices ust be watched by spies not on vertices of weight 1. A spy at a vertex of weight can watch one such threat; spies at vertices of weight 3 can watch three of the. Hence s (r t) + 3( 1 t ) if the spies stop the revolutionaries fro winning on the first round. This yields s r 1 if t 5 or t. If t = 4 and s = r, then the spies need to watch six threats at weight using two spies at vertices of weight 3. A spy at a vertex of weight 3 watches the three pairs in its nae. The four uncovered revolutionaries threaten eetings at six vertices of weight 3 corresponding to the edges of the coplete graph K 4. A spy at weight 3 can watch three pairs corresponding to a triangle. Since the edges of K 4 cannot be covered with two triangles, r spies are not enough when t = 4. If t = 3, then the counting bound yields s r for spies to avoid losing on the first round. If the initial placeent of r spies can watch all iediate threats, then they ust cover r 3 revolutionaries at vertices of weight 1 and occupy one vertex at weight 3. By syetry, we ay assue the spies locate at v 13 and v 4,...,v r. In the first round, revolutionaries at v 1 and v ove to v ; the others wait where they are. To guard the eeting at v, a spy at soe vertex of weight 1 ust ove there; let v j be the vertex fro which a spy oves to v. In the second round, the revolutionaries at v 3 and v j ove to v 3j, winning. The distance fro each spy to v 3j after round 1 is at least 3, except for the spy at v j, so no other spy could have oved after round 1 to watch that threat. 13

14 Extra spies on vertices of weight at least 5 cannot prevent the revolutionaries fro winning with the strategy given in the proof of Theore 4.1. This enables the revolutionaries to win against soewhat fewer spies when r is larger than the diension. A code with length d and distance k is a set of vertices in Q d such that the distance between any two of the is at least k. Let A(d, k) denote the axiu size of a code with distance k in Q d, and let B(d, k) be the nuber of vertices with distance less than k fro a fixed vertex in Q d. Note that B(d, k) = k 1 ( d ) i=0 i < d k 1 when k >. If M < d /B(d, k), then any code of size M having distance k can be extended by adding soe vertex, so A(d, k) d /d k 1 when k >. Corollary 4.. If d < r d /d 7, then σ(q d,, r) (d 1) r/d. Proof. Let X be a code in Q d with distance 9 and size at least d /d 8. The revolutionaries devote d revolutionaries to playing the strategy in the proof of Theore 4.1 at each of r/d vertices of X. If the ball of radius 4 at any such vertex has fewer than d 1 spies in the initial configuration, then the revolutionaries win in that ball in two rounds, since any spy initially outside that ball is too far away to guard a eeting fored at distance fro the central point in round. Since the code has distance 9, the balls of radius 4 are disjoint. Hence (d 1) r/d spies are needed to keep the revolutionaries fro winning within two rounds. Theore 4.1 and Corollary 4. together iply that at least (d 1) r/d spies are needed to win against r revolutionaries on Q d unless d < log r +7log log r. That any spies ay not be enough, since three revolutionaries easily defeat one spy on Q by starting initially at distinct vertices. Although four revolutionaries can threaten eetings at all eight vertices of Q 3, two spies can watch all those eetings and survive the next round. It appears that σ(q 3,, 4) =, though we have not worked out a coplete strategy for two spies against four revolutionaries. We have no nontrivial general upper bounds on σ(q d,, r) when r > d. Next we consider the gae on hypercubes when >. Again we use the threats ade by revolutionaries placed initially at vertices of weight 1. However, for larger we use a probabilistic arguent instead of explicit counting. The probabilistic arguents are sipler and yield a stronger lower bound on σ(q d,, r) than the counting arguents would, but we no longer copletely deterine the threshold (and hence we separate this fro the case = ). Again V (Q d ) = {v S : S [d]}, as specified as before Theore 4.1. Lea 4.3. For v V (Q d ), a vertex u of weight is within distance 1 of v if and only if u v v +1. Proof. The distance between any two vertices is their syetric difference. Always the size of the syetric difference is u + v u v. When u =, it follows that d Qd (u, v) 1 is equivalent to u v v

15 Our ain tool for the gae on Q d is a lea about failies of sets. Lea 4.4. Let S be a set of at ost t vertices in Q t, all having weight at least. If t 38.73, then Q t has a vertex w of weight such that d Qt (v, w) for all v S. Proof. Fix p (0, 1), to be deterined later. Construct a rando index set I [t] by independently including each eleent of [t] with probability p. In light of Lea 4.3, for v S we say that I avoids v if v I < v +1. Our goal is to show that with p chosen appropriately, with positive probability I avoids all of S and has size at least. The desired vertex w can then be any vertex of weight contained in such a set I. Our first task is to obtain a lower bound on P[A v ], where A v is the event that I avoids v. Let Bin(n, p) denote a rando variable having the binoial distribution with n trials and success probability p. Let B be the event that k+1 trials yield k successes in the first k 1 trials plus two failures at the end. Let B be the event that k+1 trials yield k 1 successes in the first k 1 trials plus two successes at the end. Canceling coon factors yields P[B] > P[B ] if and only if p < 1/. As a consequence, P[Bin(k+1, p) < k+1] > P[Bin(k 1, p) < k] when p < 1/. Note also that P[Bin(k, p) < k] P[Bin(k 1, p) < k]. v +1, so k and v {k, k 1}. For the event that I has fewer Now let k = than k eleents of v, our observations about the binoial distribution yield P[A v ] P[Bin(k 1, p) < k] P[Bin(3, p) < ] = (1 p) (1 + p). Let q = in v P[A v ]. Events of the for A v are down-sets in the subset lattice. By the FKG inequality (see Theore 6..1 of Alon and Spencer [1]), such events are positively correlated when p < 1/, so [ ] P A v q t = e tln q. Now let X = I. For αtp with α < 1, Chernoff s Inequality yields v S P[X < ] = P[X tp < tp] e ( tp) /(tp) = e (1 α) tp/. Our goal is to show P [ v S A v] > P[X < ], which follows fro ln[(1 p) (1 + p)] > (1 α) p/. With α =.347 and p =.07953, the strict inequality holds, and we obtain αp Hence when d /(αp) 38.73, soe -set avoids all vertices in S. Before we apply this lea to the gae on the hypercube, we prove a general result that relates the gae on a graph and its retracts. The notion of retract appeared as early as Hell [3], as a hooorphis fixing a subgraph. The variation fro [6] that we use becoes the hooorphis version when loops are available at all vertices. 15

16 Definition 4.5. An induced subgraph H of a graph G is a retract of G if there is a ap f : V (G) V (H) such that (1) f(v) = v for v V (H), and () uv E(G) iplies that f(u) and f(v) are equal or adjacent. Nowakowski and Winkler [6] proved a theore for the classical cop-and-robber pursuit gae that is analogous to our next result. Theore 4.6. Let H be a retract of a graph G. If the revolutionaries win RS(H,, r, s), then the revolutionaries win RS(G,, r, s). Equivalently, σ(g,, r) σ(h,, r). Proof. Let f : G H be as guaranteed in Definition 4.5. The revolutionaries play in G by playing exclusively on H, using the ap f to play as if the spies in V (G) V (H) were actually in V (H). The revolutionaries take initial positions as specified by their winning strategy on H. They siulate a spy on v V (G) by a spy on f(v) V (H). Whenever a spy can legally ove fro u to v in G, the definition of retract guarantees that the siulated spy can ove fro f(u) to f(v) in H. Therefore, the siulated spies always play legal oves in the iagined gae. The revolutionaries play their winning strategy against the siulated spies in H and eventually for an uncovered eeting at soe vertex w. Since f(w) = w, the absence of a siulated spy on w eans that there is no real spy on w, and the revolutionaries have won the real gae in G. Theore 4.7. If s r and d r, then the revolutionaries win RS(Q d,, r, s). Proof. The revolutionaries initially occupy v 1,...,v r. The revolutionaries threaten eetings after 1 steps at ( r ) vertices of weight. The vertices of weight protected by a spy at v i are precisely those whose corresponding sets contain i. Let t be the nuber of revolutionaries uncovered after the initial placeent of spies. By syetry, we ay assue that the uncovered revolutionaries are at v 1,..., v t. Let S be the set of spies initially on vertices having weight at least ; only such spies can protect vertices in the set of ( t ) vertices of weight above uncovered revolutionaries. Note that 0 S s (r t) t 38.73, and hence t Every subcube of Q d is a retract of Q d, by projection. Hence by Theore 4.6, we ay assue that the spies in S are all in Q t. We can therefore apply Lea 4.4. With t and S t < t, soe vertex of weight in Q t is too far fro S to be reached by any spy within 1 rounds, and the revolutionaries win. Although S t in Theore 4.7 while Lea 4.4 allows S t, generalizing the lea to vary S in ters of t does not noticeably strengthen the application. When t, an explicit counting bound on the nuber of vertices of weight in Q t that are within distance 1 of a given vertex of S leads to the following theore. 16

17 Theore 4.8. If d r 3 and s r 3 4, then the revolutionaries win RS(Q d,, r, s), so σ(q d,, r) > r 3 4. Theore 4.8 is stronger than Theore 4.7 when 5. We oit the proof, because the proofs of this counting lea and theore are longer and ore technical than those of Lea 4.4 and Theore 4.7, and because we believe that the revolutionaries ay win against as any as r spies. As in Theore 4.1, the revolutionaries in Theore 4.7 play locally, winning by staying within distance of a fixed vertex. Hence with general eeting size we can apply the sae coding theory arguent as in Corollary 4.. Given a code with distance 4 1, the balls of radius 1 are disjoint. Any vertex with distance ore than 1 fro the central point has distance ore than 1 fro the threatened eetings and cannot reach the in 1 turns, which is the nuber of rounds the revolutionaries need to win in the strategy of Theore 4.7. We thus have the following. Corollary 4.9. If d < r d /d 4, then σ(q d,, r) > (d 38) r/d. Finally, the hypercube result applies to ore general cartesian products via the notion of retract. For U V (G), we use G[U] to denote the subgraph of G induced by U. Corollary Let G = G 1 G d, where G 1,...,G d are graphs with at least one edge. If the revolutionaries win RS(Q d,, r, s), then the revolutionaries win RS(G,, r, s). Proof. By Theore 4.6, it suffices to show that G contains a retract isoorphic to Q d. Select v i w i E(G i ) for each i, and let U = {v 1, w 1 } {v d, w d }. Note that G[U] = Q d. To define f : V (G) U, first define g i : V (G i ) {v i, w i } by setting g i (x) = v i if x = v i and g i (x) = w i otherwise. Now let f(x 1,..., x d ) = (g 1 (x 1 ),...,g d (x d )). Clearly f fixes U. If xy E(G), then there exists exactly one i such that x i y i ; without loss of generality, x i v i. If also y i v i, then g i (x i ) = g i (y i ) = w i, so f(x) = f(y). On the other hand, if y i = v i, then g i (x i ) = w i and g i (y i ) = v i while g j (x j ) = g j (y j ) for all j i, so f(x)f(y) E(G[U]) since w i v i E(G). Therefore f satisfies the conditions in Definition 4.5, and G[U] is a retract of G isoorphic to Q d. 5 Rando Graphs In the Erdős Renyi binoial odel G(n, p), the vertex set is [n], pairs of vertices occur as edges independently with probability p, and we say that an event occurs alost surely if its probability tends to 1 as n. When the graph is randoly generated and there are not too any revolutionaries, the revolutionaries can play a strategy like that in Proposition 3.5 to defeat r spies: the revolutionaries occupy vertices so that no atter where the spies are placed, any uncovered 17

18 vertices can eet at soe vertex adjacent to no spy. When the nuber of revolutionaries is larger, also the allowed nuber of spies is larger; the revolutionaries no longer can find such a placeent, and the nuber of spies needed is only a fraction of r. Our ain task in this section is to show that for constant edge-probability p, these two situations for the nuber of revolutionaries are surprisingly close together, differing only by a constant factor. In particular, when r < ln ln n the revolutionaries alost always win agains r spies, and when r > c ln n alost always cr/ spies can win, where c is any constant greater than 4. The arguent in the first setting also yields results when p depends on n. Independently, Mitsche and Pra lat [5] have proved that for G in G(n, p), alost surely σ(g,, r) r +(+ +ǫ) log 1/(1 p) n; here p can depend on n (they also obtain conditions under which r + 1 spies are needed). Their upper bound is sharp within an additive constant, but also they require r to grow faster than (log n)/p. In coparison to our ethod, they use ore intricate structural characteristics of the rando graph and a ore coplex strategy for the spies. Our strategy for the spies is like that used elsewhere in this paper: introduce a notion of stable position that keeps the eetings covered, and show that the spies can aintain a stable position. First we consider the range where r + 1 spies are needed. Motivated by Alon and Spencer [1], we say that G has the r-extension property if for any disjoint T, U V (G) with T + U r, there is a vertex x V (G) adjacent to all of T and none of U. We first show why this property akes the gae easy for the revolutionaries. Proposition 5.1. If a graph G satisfies the r-extension property, and r r, then G is spy-bad for r revolutionaries and eeting size. Proof. The r revolutionaries initially occupy any set of r vertices in G. To see that r spies cannot prevent the fro winning on the first round, let U be the set occupied by the spies, and let T be the set occupied by uncovered revolutionaries. The revolutionaries on T win by oving to the vertex x guaranteed by the r-extension property. Alon and Spencer [1, Theore ] present the result below for constant r, but the proof holds ore generally. Theore 5.. Let ǫ = in{p, 1 p}, where p is a probability that depends on n. If r = o ( nǫ r ln n) and nǫ r, then G(n, p) alost surely has the r-extension property (and hence is spy-bad for all and r with r r). Proof. Let G be distributed as G(n, p). Given T, U V (G) with T + U r, write t = T and u = U. For x V (G) (T U), let A T,U,x be the event that x is adjacent to all of T and none of U; note that P[A T,U,x ] = p t (1 p) u ǫ r. 18

19 Let A T,U be the event that A T,U,x fails for all x V (G) (T U). The events A T,U,x for different x are deterined by disjoint sets of vertex pairs, so P[A T,U ] (1 ǫ r ) n r e ǫr (n r). The r-extension property fails if and only if soe event of the for A T,U occurs. Hence it suffices to show that the probability of their union tends to 0. There are 3 r ways to for T and U within a fixed r-set of vertices, since a vertex can be in either set or be oitted, and there are ( n r) sets of size r. Hence the union consists of at ost (3n) r events, each of whose probability is at ost e ǫr (n r). We copute (3n) r e ǫr (n r) = e r ln(3n) ǫr (n r) = e r ln3+r ln n ǫr (n r). Since ǫ 1/, the condition r = o ( nǫ r ln n) iplies r = o(n), so the exponent is doinated by nǫ r and tends to. Thus the bound on the probability of lacking the r-extension property tends to 0, and G(n, p) alost surely satisfies this property. In particular, when p is constant, G(n, p) is alost surely spy-bad for r when r c ln n, where c < ln(1/ǫ). Siilarly, when r is constant, G(n, p) is alost surely spy-bad when p tends to 0 ore slowly than 1/n 1/r. With p 1/, the key condition is np r. Now we confine our attention to the real of constant edge-probability p and consider well-known properties of the rando graph that enable the spies to do well. For every vertex, the expected degree is p(n 1), and for any two vertices the expected size of their coon neighborhood is p (n ). Moreover, these rando variables are so highly concentrated at their expectations that alost always the degrees of all vertices and the sizes of coon neighborhoods of all pairs are within constant factors of their expected values. We begin by stating this forally; the proofs are standard and straightforward using the Chernoff Bound. We treat G as a saple fro the odel G(n, p). Lea 5.3. Fix p and γ with 0 < γ < p < 1. In the rando graph odel G(n, p), alost surely (p γ)n < d(v) < (p + γ)n and (p γ )n < N(v) N(w) < (p + γ )n for all v, w V (G). Lea 5.4. Fix p and γ with 0 < γ < p < 1. In the rando graph odel G(n, p), alost p γ for all v, w V (G). surely N(v) N(w) N(v) Proof. Using the lower bound on coon neighborhood size and the upper bound on degree fro Lea 5.3, alost surely N(v) N(w) (p γ )n = p γ for all v, w V (G). N(v) (p+γ)n Definition 5.5. For q (0, 1), a graph G is q-coon if N(v) N(w) N(v) q for all v, w G. We develop a strategy for spies that will be successful on q-coon graphs under certain conditions. In a gae position, we need to distinguish players occupied in foring or covering eetings fro those who are not. These notions will also be iportant for spy strategies on coplete ultipartite or bipartite graphs. 19

20 Definition 5.6. Given a gae position, say that specified revolutionaries in a eeting and one spy covering the are bound. After designating the bound players for all vertices hosting eetings, the reaining spies and revolutionaries are free. A vertex having at least revolutionaries has exactly bound revolutionaries. For a vertex subset U, let r U and ˆr U denote the total nuber of revolutionaries and nuber of free revolutionaries on U. Siilarly, let s U and ŝ U denote the total nuber of spies and nuber of free spies on U. Write ˆr and ŝ for ˆr V (G) and ŝ V (G). A gae position is stable if (1) all eetings are covered, and () ŝ N[v] ˆr/ for all v V (G). As in Section, the nae stable is otivated by peritting the gae to continue. Lea 5.7. On any graph G, if the position at the beginning of a round is stable, then the spies can respond to cover all eetings at the end of the round. Proof. Let the notation in Definition 5.6 refer to the counts at the beginning of round t, in a stable position. Let X be the set of distinct vertices hosting eetings after the revolutionaries ove in round t. Let Y be the set of spies. Define an auxiliary bipartite graph H with partite sets X and Y. For x X and y Y, put xy E(H) if spy y can reach x fro its position at the start of round t, being adjacent to x or already there. If soe atching in H covers X, then the spies can ove in round t to cover all the eetings. It suffices to show that H satisfies Hall s Condition for a atching that covers X. Consider S X. If N G [S] contains b vertices that hosted eetings at the start of round t, then S ˆr+b eetings, because revolutionaries who were in eetings not in N G[S] cannot reach S in one ove. On the other hand, every free spy at a vertex of N G [S] can reach S in one ove, as can every spy bound to a eeting in S. Choosing x S, we have N H (S) ŝ N[v] + b ˆr/ + b S. Hence Hall s Condition is satisfied and the atching exists. The next lea provides the second half of what the spies need to do. Lea 5.8. Let G be a q-coon graph with n vertices, and fix ǫ > 0. Given a position in RS(G,, r, s) such that (1) all eetings are covered, () ŝ 1+ǫ ˆr, and (3) q lnn ŝ, the free spies can ove to produce a stable postion. (1 1/(1+ǫ)) q Proof. We prove that if each free spy oves to a uniforly rando vertex in the neighborhood of its current position, then with positive probability a stable position is produced. For v V (G), let X v be the nuber of spies in N[v] after the frees spies ove. Since G is q-coon, each free spy lands in N[v] with probability at least q. Also, these events for individual spies are independent, so X v is a su of ŝ independent indicator variables, each 0