Revolutionaries and spies on trees and unicyclic graphs

Rev olutionaries and spies on trees and unicyclic graph s Daniel W. Cranston ∗ , Clifford D. Sm yth † , Douglas B. W est ‡ No v emb er 7, 2018 Abstract A team of r r evolutionaries and a team of s spies pla y a game on a graph G . Initially , rev olutionaries a n d then spies tak e p ositions at v ertices. In eac h subsequen t round, eac h rev olutionary m a y mo ve to an adjacen t verte x or n ot mov e, and then eac h sp y has the same op tion. The rev olutionaries wa nt to h old an unguar d e d me eting , meaning m rev olutionaries at some v ertex having no sp y at the end of a roun d. T o prev ent this forev er, trivially at least min {| V ( G ) | , ⌊ r /m ⌋} spies are needed. When G is a tree, this man y spies s uffices. When G is a unicyclic graph, min {| V ( G ) | , ⌈ r /m ⌉} spies suffice, and we c h aracterize those unicyclic graphs where ⌊ r /m ⌋ + 1 sp ies are needed. 1 In tro ductio n Man y pursuit games ha ve b een studied on graphs. W e study suc h a game that can b e in terpreted as mo deling a problem of net work securit y . One team consists of r r evolutionaries ; the o t her consists of s spies. The rev olutionaries w an t to arrange a one-time meeting o f m rev olutionaries free of ov ersigh t b y spies. Initially , the rev olutionaries tak e p ositions at v ertices, a nd then the spies do the same. In each subsequen t round, each rev olutiona ry may mo ve to an adjacen t v ertex or not mov e, and t hen eac h sp y has the same option. Ev ery o ne kno ws where ev eryone else is. The rev o lutionaries w in if at t he end of a round there is an unguar de d me eting , where a me eting is a set of (at least) m rev olutionaries on one v ertex, and a meeting is unguar de d if there is no sp y a t that v ertex. The spies win if they can prev ent this forev er. Let RS( G, m, r , s ) denote this game play ed on the graph G b y s spies a nd r revolutionaries seeking a n unguarded meeting of size m . The rev olutionaries can form min {| V ( G ) | , ⌊ r /m ⌋} meetings initially; if s is smaller than this, then the spies immediately lose. On the ot her hand, the spies win if s ≥ r − m + 1; they follo w r − m + 1 distinct revolutionaries, and the ot her m − 1 r ev olutionaries cannot form a ∗ Virginia Commonw ealth Univ er sity , dcranston@vcu.edu. † Univ ersity of North Car olina – Greensb or o , cdsmyth@uncg.edu. ‡ Univ ersity of Illinois, west@math.uiuc.edu, par tially s upp o rted by NSA gra nt H982 30-10 -1-03 63. 1 meeting. F or fix ed G, r , m w e study the minimu m s suc h that the spies win R S( G, m, r , s ); let σ ( G, m, r ) denote this threshold. The t r ivial b ounds are min {| V ( G ) | , ⌊ r/m ⌋ } ≤ σ ( G, m, r ) ≤ min {| V ( G ) | , r − m + 1 } . The game was inv en ted b y Jozef Bec k in the mid-199 0 s. Shortly thereafter, Sm yth pro ved that the trivial lo w er b ound on σ ( G, m, r ) is sufficien t when G is a tree. This was not publis hed; w e include a proo f here and use the result in solving the game for unicyclic graphs. Ho ward and Sm yth [2] studied RS( G, 2 , r , s ) when G is the infinite 2-dimensional in teger grid with o ne- step horizontal, v ertical, and diagonal edges. They pro ve d 6 ⌊ r / 8 ⌋ ≤ σ ( G, 2 , r ) ≤ r − 2 . F or the upp er b ound, whic h they conjectured is sharp, they sho wed that r − 2 spies can win b y ha ving all but one follow rev o lutionaries; the one remaining spy can prev en t the remaining t hree revolutionaries from making a n unguarded me eting. In this pap er, w e determine σ ( G, m, r ) for all trees and unicyclic gra phs. F or a graph G with r /m < | V ( G ) | , w e sho w that σ ( G, m, r ) ≤ ⌈ r /m ⌉ if G has at most one cycle, by summing o v er comp onen ts. When G is a tr ee, ⌊ r /m ⌋ spies s uffice, and ⌈ r /m ⌉ suffice when G is connected and unicyclic. Our final result is tha t if G is unicyclic with a cycle of length ℓ and t v ertices not o n the cycle (connected or not), and m ∤ r , then σ ( G, m, r ) = ⌊ r /m ⌋ if and only if ℓ ≤ max {⌊ r / m ⌋ − t + 2 , 3 } . Butterfield, Cranston, Puleo, W est, and Zamani [1] study σ ( G, m, r ) on a v ariet y of graphs. With r /m < | V ( G ) | , they sho w that σ ( G , m, r ) = ⌊ r /m ⌋ when G ha s a ro oted spanning tree T such that eve ry edge of G not in T joins ve r tices ha ving the same paren t in T ; this includes graphs with a dominating v ertex and in terv al graph. F or ev ery graph G , they pro v e σ ( G, m, r ) ≤ γ ( G ) ⌊ r /m ⌋ , where γ ( G ) is the domination n um b er of G , and this is nearly sharp: fo r t, m, r ∈ N with t ≤ m , there is a graph G with domination n um b er t suc h that σ ( G, m, r ) > t ( r / m − 1). Also, there are c hordal graphs (and bipartite gra phs) for giv en r and m such that σ ( G, m, r ) = r − m + 1. If G is the d -dimensional hypercub e Q d with d ≥ r , then σ ( G, 2 , r ) = r − 1; in general, if d ≥ r ≥ m ≥ 3, then σ ( Q d , m, r ) > r − 3 4 m 2 . They also consider r -larg e complete k - partite graphs, where “ r - la rge” means that each partite set ha s size at least 2 r . When k ≥ m and G is suc h a g r a ph, k · k ⌊ r /k ⌋ ( k − 1) m + 1 − k ≤ σ ( G, m, r ) ≤  k k − 1 r m  + k . When G is a la r g e bipartite graph, they sho w that σ ( G, 2 , r ) = 7 r / 10 a nd σ ( G, 3 , r ) = 3 r / 2 (within a dditive constants ) and that in general ⌊ r / 2 ⌋ ⌈ m/ 3 ⌉ − 2 ≤ σ ( G, m, r ) ≤  1 + 1 √ 3  r m + 1 . The upp er b ound co efficien t on r m is ab out 1 . 58; the low er one is ab out 1 . 5 when m is larg e and r > m ; they conjecture that the low er o ne is the true co efficien t, at least when 3 | m . 2 2 T rees and Cycles If r / m < | V ( G ) | , then the spies lose whenev er s < ⌊ r /m ⌋ . Hence the first chance for the spies is when s = ⌊ r / m ⌋ . W e prov e that this suffices when G is a tree, yielding σ ( G, m, r ) = ⌊ r /m ⌋ . Since the spies alwa ys win when s ≥ | V ( G ) | , the statemen t of our first theorem remains true regardless o f the relationship b et wee n r /m and | V ( G ) | . Nev ertheless, to a void t rivial statemen ts, w e henceforth assume alw a ys that r /m ≤ | V ( G ) | . Theorem 1. If G is a tr e e and s ≥ ⌊ r /m ⌋ , t hen the spies win RS( G, m, r , s ) . Pr o o f. It suffices to sho w that the spies win when s = ⌊ r /m ⌋ < | V ( G ) | . Cho ose a ro ot z ∈ V ( G ). The p ar ent o f a no n-ro o t v ertex v , denoted v + , is its neighbor on the path f r o m v to z . The o t her neigh b ors o f v are its ch ildr en ; let C ( v ) b e this set of c hildren of V . The desc endants of v are all t he vertice s ( including v ) whose path to z con tains v ; let D ( v ) b e the set of descendan ts of v . F or eac h v ertex v , let r ( v ) and s ( v ) denote the n um b er of rev olutionar ies and spies on v at t he current time, resp ectiv ely , and let w ( v ) = P u ∈ D ( v ) r ( u ). The spies maintain the follo wing inv ariant giving the num ber of spies on eac h vertex a t the end of an y ro und: s ( v ) =  w ( v ) m  − X u ∈ C ( v )  w ( u ) m  for v ∈ V ( G ) . (1) Since P u ∈ C ( v ) w ( u ) = w ( v ) − r ( v ), the formula is alw ays nonnegat iv e. Also, if r ( v ) ≥ m , then s ( v ) ≥ j w ( v ) m k − j w ( v ) − r ( v ) m k ≥ 1. Hence (1) guara ntees that ev ery meeting is guarded. T o show that the spies can establish (1 ) after the first round, it suffices that all the form ulas sum t o ⌊ r /m ⌋ . More generally , summing ov er the descendan ts of an y ve r tex v , X u ∈ D ( v ) s ( u ) =  w ( v ) m  , (2) since ⌊ w ( u ) /m ⌋ o ccurs p ositiv ely in the term fo r u and negative ly in the term for u + , except that ⌊ w ( v ) /m ⌋ o ccurs only p ositive ly . When v = z , the total is ⌊ r /m ⌋ , since w ( z ) = r . T o sho w that the spies can maintain (1), let r ′ ( v ) denote the new num b er of rev olutionaries at v aft er the revolutionaries mov e, and let w ′ ( v ) = P u ∈ D ( v ) r ′ ( v ). The spies mo ve to ac hiev e the new v alues required b y (1), starting from the lea v es; w e will pro duce s ′ ( v ) at v after adjusting at a ll the c hildren (a nd lo we r descendan ts) of v . W e pro cess siblings sim ultaneously . That is, having updated all c hildren of C ( v ), w e adjust at all of C ( v ) sim ultaneously . In doing this, excess spies mov e to v , and needed spies come from v ; no c hanges are made in v olving c hildren of C ( v ). Similarly , fixing C ( v + ) later includes an exc hange b et wee n v and v + but do es not disturb the spies on C ( v ). After 3 success fully up da t ing C ( v ) for all v , the ro ot v ertex z will ha v e exactly s ′ ( z ) spies, since alw ays P s ′ ( v ) = P s ( v ) = ⌊ r /m ⌋ . W e can now pro cess C ( v ). Let D ∗ ( u ) = D ( u ) − { u } for all u . F or u ∈ C ( v ) , all ve rtices in D ∗ ( u ) ha v e b een adjusted. Spies previously on D ∗ ( u ) remained in D ( u ), and those no w on D ∗ ( u ) came fro m D ( u ). Hence u no w has P t ∈ D ( u ) s ( t ) − P t ∈ D ∗ ( u ) s ′ ( t ) spies. Let ∂ ( u ) = s ′ ( u ) − X t ∈ D ( u ) s ( t ) + X t ∈ D ∗ ( u ) s ′ ( t ) =  w ′ ( u ) m  −  w ( u ) m  . (3) As defined, c hang ing by ∂ ( u ) ac hiev es s ′ ( u ) spies at u ; the second eq ua lity uses (2). When ∂ ( u ) is p ositiv e, ∂ ( u ) spies mov e fr o m v to u ; when it is negativ e, u sends − ∂ ( u ) spies to v . Let C + = { u ∈ C ( v ) : w ′ ( u ) > w ( u ) } and C − = C − C + . Note that b oth C + and C − ma y con tain v ertices f o r whom t he adjustmen t f r om or to v is 0. T o av oid making spies take t wo steps to reac h C + , w e must ensure P u ∈ C + ∂ ( u ) ≤ s ( v ). T o a void f orcing spies from C − to take a second step, w e m ust ensure P u ∈ C − | ∂ ( u ) | ≤ s ′ ( v ). F or the first inequalit y , note that P u ∈ C + [ w ′ ( u ) − w ( u )] ≤ r ( v ), since rev o lutionaries who en tered subtrees root ed at C ( v ) on this round we re previously at v . Th us X u ∈ C +  w ′ ( u ) m  ≤ $ X u ∈ C + w ′ ( u ) m % ≤  r ( v ) + P u ∈ C + w ( u ) m  =  w ( v ) − P u ∈ C − w ( u ) m  ≤  w ( v ) m  − X u ∈ C −  w ( u ) m  = s ( v ) + X u ∈ C +  w ( u ) m  . By (3), this yields the desired inequalit y . F or the second inequalit y , P u ∈ C − [ w ( u ) − w ′ ( u )] ≤ r ′ ( v ), since rev olutionaries who left subtrees ro ot ed at C − on this round a re no w at v . Th us X u ∈ C −  w ( u ) m  ≤ $ X u ∈ C − w ( u ) m % ≤  r ′ ( v ) + P u ∈ C − w ′ ( u ) m  =  w ′ ( v ) − P u ∈ C + w ′ ( u ) m  ≤  w ′ ( v ) m  − X u ∈ C +  w ′ ( u ) m  = s ′ ( v ) + X u ∈ C −  w ′ ( u ) m  . Again (3) yie lds the desired inequalit y . These inequalities ensure that the adjustmen ts restore (1) by legal mo v es. The adjust- men t t o C ( v ) is allow ed, after which v has s ( v ) − P u ∈ C ( v ) ∂ ( u ) spies. Using (1) and (3), s ( v ) − X u ∈ C ( v ) ∂ ( u ) = s ′ ( v ) −  w ′ ( v ) m  +  w ( v ) m  = s ′ ( v ) − ∂ ( v ) . Th us, when C ( v + ) is later pro cessed, the adjustmen t needed at v is exactly what w e hav e said will b e made. F urthermore, spies mo ving to v from C − do not come from b elo w C − 4 and do no t con tinu e on to v + . The argumen t for the latter applies also to the former. If ∂ ( v ) < 0 , requiring − ∂ ( v ) spies to mov e to v + , then the adjustmen t at C ( v ) left more than s ′ ( v ) spies at v . Ho w ev er, since P u ∈ C − | ∂ ( u ) | ≤ s ′ ( v ), the extra spies b ey ond s ′ ( v ) w ere among the s ( v ) spies on v at the b eginning of the round, so they can mo ve to v + . Corollary 2. I f G is a fo r est and s ≥ ⌊ r /m ⌋ , then the spies win RS( G, m, r , s ) . W e next show that ⌈ r /m ⌉ spies suffice to win on a cycle. Lemma 3. If G is a cycle, then σ ( G, m, r ) ≤ ⌈ r /m ⌉ . Equality holds when m | r and r /m ≤ | V ( G ) | . Pr o o f. Ex tra rev olutionaries cannot mak e the game easie r for the spies, so w e may assume that r = sm . Giv en the initial lo cations o f rev olutionaries, index the rev olutionaries in order from 0 to s m − 1 around the cycle. Place spies on the ve rtices o ccupied by the r evolutionaries whose index is a multiple of m (this ma y put more t han one sp y o n a v ertex). Because rev olutionaries a r e iden tical, w e ma y assume t ha t the rev olutionaries alwa ys remain indexed in order around the cycle as they mov e (equiv alen tly , if rev olutionaries switc h p ositions, then they t r ade indices). Since indices mov e alo ng at most one edge in eac h round, the i th sp y can contin ue to follow the r evolutionary with index im . Th us after eve ry r o und, an y v ertex o ccupied b y at least m r evolutionaries is guarded by at least one sp y . When the cycle is short enough, the threshold for a spy win improv es to the trivial lo wer b ound ⌊ r /m ⌋ . F o r longer cycles, the strategy for the rev olutionaries to defeat ⌊ r / m ⌋ spies ma y tak e many rounds to pro duce an unguarded meeting. Note t ha t when r | m , the upp er b ound from Lemma 3 coincides with the trivial low er bo und to yield σ ( C n , m, r ) = r /m when n ≥ r /m . Theorem 4. I f G is a cycle of le ngth ℓ , an d m ∤ r with r /m ≤ ℓ , then σ ( G, m, r ) = ⌊ r /m ⌋ if and only if ℓ ≤ ⌊ r /m ⌋ + 2 or r < m . Pr o o f. Let s = ⌊ r /m ⌋ . When r < m , no meetings can b e for med, so no spies are needed. When ℓ ≤ s + 2, the spies can guard all initial meetings and leav e at most t wo v ertices unguarded. After a mo v e b y the rev olutionaries, at least tw o v ertices u and v fail to host a meeting. The spies can mov e to lea ve only u and v unguarded, b y shifting one step along paths f rom u and v to the previously unguarded vertice s. F or the con vers e, consider ℓ ≥ s + 3. It suffices to show that the rev olutionaries win when r = sm + 1. They will first distract one spy S , arranging for S to guard a vertex o ccupied b y at most one revolutionary . They then win by making s meetings on the remaining path, guarded b y at most s − 1 spies. The rev olutionaries mo v e so that no mor e than m of them ev er o ccup y o ne v ertex. Sub ject to this, they start with a ny initial distribution. The spies tak e initia l p ositions, a nd 5 the rev olutionaries designate one sp y as S . T o reduce what S g uards, on eac h round the rev olutionaries guarded by S mo v e aw a y , half in one direction and half in the other (rounded to in tegers). The rev olutionaries that were on the neigh b oring v ertices mov e farther aw a y , but still each vertex has at most m revolutionaries; this is p ossible since ℓ ≥ s + 3. No matter ho w S mo ves , the n um b er of rev olutionaries guarded by S is at most half (rounded up) of what it was b efor e. After at most ⌈ lo g 2 m ⌉ ro unds, S guards at most one rev olutiona r y . No w the revolutionaries shorten the path containing the other r evolutionaries by mo ving those nearest t o S a w ay from S , maintaining that each v ertex has at most m revolutionaries (again p o ssible since ℓ ≥ s + 3). They contin ue until the path consists of s consecutiv e v ertices with meetings, whic h cannot all b e guar ded by the s − 1 spies other than S . 3 Unicyclic Graphs A connected unicyclic graph has a cycle and trees attac hed to it. The cycle and trees in teract, since r evolutionaries may mo ve on and off the cycle. The spies must resp ond appropriately . Theorem 5. If G is a unicyclic gr aph, then σ ( G , m, r ) ≤ ⌈ r /m ⌉ . Pr o o f. It suffices to sho w that the spies win RS( G, m, r , s ) when s = ⌈ r / m ⌉ ≤ | V ( G ) | . If there a re r i rev olutionaries in a comp onen t G i of G , and G i is a tree, then only ⌊ r i /m ⌋ spies are needed in G i . Since at most one comp onen t contains a cycle, a nd ⌊ a ⌋ + ⌈ b ⌉ ≤ ⌈ a + b ⌉ for all a, b ∈ R , w e ma y assume tha t G is connected and contains a cycle C . View G − V ( C ) as disjoin t trees ro oted at v ertices neighboring C . In order to use the strategies of Theorem 1 and Lemma 3, a spy m ust b e av ailable when needed to mo v e from C to a tr ee (or vice v ersa). Sa y that the curren t state satisfies the cycle c ondition if the n umber of r ev olutionaries on C is m times the num b er of spies o n C and t here is a sp y g ua rding ev ery m th rev olutiona ry as described in the pro of of Lemma 3. The k ey fact needed is that adding one sp y and m rev olutionaries t o an y vertex of C (or remov ing them) preserv es the cycle condition. As in Lemma 3, w e may a ssume m | r . W e may also assum that a ll rev olutionaries s tart on the cycle. T o play against another initial p osition, the spies imagine a n initial p osition on the cyc le and follo w their winning strategy as the rev olutionaries mov e t o the actual start . With a ll rev olutionaries initially on the cycle, the cycle condition holds at the start. The att ache d tr e es are the components o f G − V ( C ). The ro o t of an attac hed tree T is the v ertex z adjacen t to a v ertex of C . The neighbor of z in V ( C ), denoted z ∗ , is the ma te of T . As rev olutionaries disp erse to or return from the atta ched trees, main taining the cycle condition requires k eeping a buffer of “ fak e” rev olutio na ries for eac h tree at its ma t e on C (a ve rtex may b e the mate of man y tr ees). When a r evolutionary mov es from C to an attache d tree T , it mov es to the ro ot z of T from the mate z ∗ on C . Un til m rev olutionaries mo v e to T , no sp y needs to follo w, since 6 m − 1 rev olutionaries cannot make a meeting o n T . F or eac h rev olutiona r y that mov es from z ∗ to z , we add a fak e revolutionary at z ∗ ; this main ta ins the cycle condition. The fak e rev olutionaries a re mark ers maintained by the spies and do no t mo v e. When m actual rev olutiona ries hav e mov ed to T , a sp y is needed. Before the final mo v e of j rev olutio naries from z ∗ to z , there w ere also m − j fak e rev olut io naries at z ∗ , so by the cycle condition there w as a sp y at z ∗ . This sp y mov es to z follo wing the j rev olutionar ies and m − j fa k e rev olutio naries disapp ear fro m z ∗ . This preserv es the cycle condition. In the strategy on T given in Theorem 1, when rev olutionaries are added at the ro ot to increase the n umber of spies needed, the v ertex needing the extra sp y is the r o ot. Hence t he arriv al of the spy at z fro m z ∗ p ermits the sp y strategy on T to op erate locally . Similarly , when rev olutiona r ies leav e T to reduce the num b er of spies needed, t hey lea ve the ro ot. The n umber of spies computed for other vertice s of T do es not c hange, so t he lo cation o f the extra sp y is z . It can return to z ∗ and reestablish the appropriate num b er of fake rev olutionaries. As spies follow rev olutionaries on to and o ff the cycle, the f ak e rev olutionaries enable the spies to main tain the cycle condition, and the strategies for spies on the cycle and o n the attac hed trees can op erate indep enden tly as previously giv en. By Theorem 5, σ ( G, m, r ) ∈ {⌊ r /m ⌋ , ⌈ r /m ⌉} when G is unicyclic and | V ( G ) | ≥ r /m . Theorem 6 determines whic h is the answe r. The role of vertice s outside the cycle is sho wn b y the disjoint union of C 5 and P 2 . Three spies defeat sev en rev olutiona ries on C 5 when m = 2, by Theorem 4. How ev er, four rev olutiona r ies can sit on P 2 forev er, o ccup ying tw o spies, a nd the remaining three rev olutionaries defeat the remaining sp y on C 5 . Adding a n edge to join t he tw o comp o nen ts do es not affect the rev olutiona r ies’s strategy; it do es not matter whether the graph is connected. Theorem 6. L et G b e a unicyclic g r aph having a cycle C of length l and exactly t vertic es not on C . If | V ( G ) | ≥ r /m , then σ ( G, m, r ) ∈ {⌊ r /m ⌋ , ⌈ r /m ⌉} , e qual ling ⌊ r /m ⌋ when m ∤ r if and only if ℓ ≤ max( ⌊ r /m ⌋ − t + 2 , 3) or r < m . Pr o o f. By Theorem 5, σ ( G, m, r ) ≤ ⌈ r /m ⌉ , whether or not G is connected. Assume m ∤ r , and let s = ⌊ r / m ⌋ ≤ | V ( G ) | . It suffic es to show that the rev olutionaries win RS( G , m, r, s ) if ℓ ≥ max { s − t + 3 , 4 } , and the spies win R S( G, m, r , s ) if ℓ ≤ max { s − t + 2 , 3 } . When ℓ ≥ max { s − t + 3 , 4 } , the rev olutionaries first mak e meetings a t k v ertices outside C , where k = min { t, s − 1 } . Thes e meetings m ust b e guarded by k spies. The se k m rev olutionaries (and th us also k spies) will not mov e (it do es not matt er whether their comp onen t contains C ). On C , the remaining r − k m rev olutionaries play against s − k spies. Since r /m > s , also ( r − k m ) /m > s − k . Since s − k = max { s − t, 1 } , w e ha ve ℓ ≥ s − k + 3. By Theorem 4, the remaining rev o lutionaries win on the cycle. No w suppo se ℓ ≤ max { s − t + 2 , 3 } . If G has a comp onent H with ˆ t v ertices a nd ˆ r rev olutionaries outside the comp onen t con taining C , then H needs only min { ˆ t, ⌊ ˆ r /m ⌋} spies. 7 On G − V ( H ), since ⌊ a ⌋ + ⌊ b ⌋ ≤ ⌊ a + b ⌋ , the sp ecified conditions hold for the remaining spies to win. Hence to consider a minimal coun t erexample we ma y assum e that G is connected. As in Theorem 4 , w e ma y also assume that a ll rev olutionaries start on C . Case 1: s > t . In this case, ℓ ≤ s − t + 2. S ince | V ( G ) | > ⌊ r /m ⌋ = s , in fact ℓ ∈ { s − t + 1 , s − t + 2 } . Eac h attached tree T reserv es | V ( T ) | spies, lo cated initia lly a t the mate of T . These “ t r ee spies” mostly remain at the mate of T except to mov e into T as needed to pla y the sp y strategy on T from Theorem 5. When r T rev olutionaries are in T , exactly ⌊ r T /m ⌋ tree spies will hav e follow ed them. By Theorem 1, the r ev olutionaries nev er mak e a n unguarded meeting outside C . The s − t unreserv ed spies alw ays o ccup y distinct vertice s of C ; call them “cycle s pies”. T o draw t ′ tree spies off C , at least mt ′ rev olutionaries m ust ha v e left C . Since s = ⌊ r /m ⌋ , few er than m ( s − t ′ + 1) rev olutionaries remain on C . Hence at most s − t ′ meetings on C need to b e guarded. Since ℓ ≤ s − t + 2, w e hav e seen in Theorem 4 that the s − t cycle spies can mov e to guard an y des ired set of s − t v ertices on C . If at least ℓ − ( s − t ) v ertices of C retain tree spies, then the cycle spies can g uard the remaining v ertices of C . If no v ertices of C retain tree spies, then t ′ = t . In this case few er than m ( s − t + 1) rev olutionaries a re on C , t hey make at most s − t meetings, and the cycle spies can g ua rd those meetings. Hence w e ma y a ssume that ℓ = s − t + 2 and that all tree spies on C are at one v ertex, v . The cycle spies can guard a ll the other meetings on C unless there are s − t + 1 suc h meetings. If there is a lso a meeting at v , then there are at least m ( s − t + 2) rev o lut io naries on C . Hence m ( s − t + 2) < m ( s − t ′ + 1), whic h yields t ′ < t − 1. Th us at least t w o tree spies remain at v , and one of them can mo ve to gua r d a meeting at a neigh b or u of v on C . If there is no meeting at v , then a g ain a tree sp y from v can guard a meeting a t u . This is the only wa y a t ree sp y leav es its reserv ed subgraph; when the conditio n ends the sp y mo v es back. It cannot b e pulled in to T (t w o steps) at the same time, b ecause hav ing m ( s − t + 1) rev o lutionaries on V ( C ) − { v } leav es few er than mt revolutionaries in the union of the trees a nd v : not enough to pull t he la st t r ee spy bac k in to its tree. Case 2: s ≤ t . In this case, ℓ = 3, and the ve r t ices of C are pairwise adjacent. Deleting the edges of C leav es three disjoin t trees, ro oted at the ve rt ices of C . As usual, the spies ma y assume that the rev olutionaries all start on V ( C ). Alwa ys, an initial p osition can b e defended by ⌊ r /m ⌋ spies, and in this case they a re all on t he cycle. A t a giv en time, let r i b e the num b er of rev olutionaries on the comp onen t T i of G − E ( C ) (ro oted at v i ∈ V ( C )), for i ∈ { 1 , 2 , 3 } . The spies maintain that at the end of eac h round there ar e at least ⌊ r i /m ⌋ spies on tree T i , arrang ed a ccording to the strategy of Theorem 1, with a ny extra spies lo cated at the ro ot v i . Since each v ertex appears in some T i , abilit y to main tain this inv ariant completes t he pro of. The inv ariant holds af ter the initial placemen t, since r evolutionaries app ear o nly a t the ro ots. 8 After the revolutionaries mo ve, the spies up date their p osition on eac h T i via the strategy in the pro of of Theorem 1. The up date starts from the lea ve s a nd w orks to ward the ro ot. The rev o lutionaries no w a t non- r o ot v ertices of T i w ere in T i at the end of the previous round, sin ce rev olutionaries en ter or lea ve T i only via edges of C . As sho wn in Theorem 1 , the up dates to all v ertices except the ro o t can b e completed using spies that we r e in T i b efore the ro und, since t he in v arian t held at that time. Also, spies who mo v ed to v i during this pro cess ar e cov ering rev olutio na ries who came there fr o m T i − v i . Let s i = ⌊ r i /m ⌋ . If not enough revolutionaries arriv e at v i from the rest of C on this round to push the num b er of revolutionaries on T i up to ms i + m , then t he in v arian t already holds on T i . Ho w ev er, if T i no w con ta ins at least ms i + mk rev olutio na ries (for some p o sitiv e k), then the num b er o f rev olutionaries remaining in the other trees is at most r − m ( s i + k ) , so the n umber of spies needed on those trees is at most s − s i − k . That is, k s pies are freed to mo v e to v i . F urthermore, since the new rev olutionaries in T i came from the other ro ots and the tree strategy w as fo llo wed using ⌊ r j /m ⌋ spies on each T j , the freed spies w ere at the other ro ots and are no w av ailable to mo ve to v i . Doing so restores the desired in v ariant. The tec hnique of Case 2 a b o ve do es not work in Case 1, since the rev olutionaries can mak e the verte x on C needing extra spies b e fa r from the v ertex with freed spies. W e hav e now determined the winner for ev ery ga me RS( G, m, r , s ) suc h that G is uni- cyclic, a nd we hav e pro vided a constructiv e strat egy fo r the winner in each case. References [1] J.V. Butterfield, D.W. Cranston, G.J. Puleo, D.B. W est, and R. Zamani, R evolutio naries and spies: S p y -go o d and spy-bad graphs, subm itted. [2] D. How ard and C.D. Smyth, Rev olutionaries and spies on grid-lik e graphs, to app ear in Discrete Math. 9