c 2009 Society for Industrial and Applied Mathematics

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1 SIAM J. DISCRETE MATH. Vol., No., pp c 009 Society for Indstrial and Applied Mathematics THE SURVIVING RATE OF A GRAPH FOR THE FIREFIGHTER PROBLEM CAI LEIZHEN AND WANG WEIFAN Abstract. We consider the following firefighter problem on a graph G =(V, E). Initially, a fire breaks ot at a ertex of G. In each sbseqent time nit, a firefighter protects one ertex, and then the fire spreads to all nprotected neighbors of the ertices on fire. The objectie of the firefighter is to sae as many ertices as possible. Let sn() denote the maximm nmber of ertices the firefighter can sae when a fire breaks ot at ertex of G. We define the sriing rate ρ(g) of G to be the aerage percentage of ertices that can be saed when a fire randomly breaks ot at a ertex of G, i.e., ρ(g) = V sn()/n. In this paper, we proe that for eery tree T on n ertices, ρ(t ) > /n. Frthermore, we show that ρ(g) > /6 for eery oterplanar graph G, and ρ(h) > /0 for eery Halin graph H with at least ertices. Key words. firefighter problem, sriing rate, tree, oterplanar graph, Halin graph AMS sbject classifications. 0C, 0C90 DOI. 0.7/ Introdction. The firefighter problem was introdced by Hartnell [9] in 99 in an attempt to model the spread of fire, diseases, compter irses, and similar phenomena at a macro-control leel. A fire breaks ot at a ertex of a graph G = (V,E) at time 0. For each time interal (i, i+], i 0, a firefighter protects one ertex not yet on fire (the ertex remains protected afterwards), and the fire spreads from brning ertices (i.e., ertices on fire) to all nprotected neighbors of these ertices. The process ends when the fire can no longer spread, and all ertices that are not brning are saed. The objectie of the firefighter is to sae as many ertices as possible. It is not srprising that the firefighter problem is ery difficlt. Indeed, it has been shown by Finbow et al. [] that the problem is NP-complete een for trees of maximm degree. On the other hand, Hartnell and Li [0] hae proed that a simple greedy method for trees is a /-approximation algorithm, and MacGilliray and Wang [8] hae gien a 0- integer programming formlation of the problem for trees and soled the problem in polynomial time for some sbclass of trees. Recently, Cai, Verbin, and Yang [] hae obtained a ( /e)-approximation algorithm for trees based on a linear programming relaxation and randomized ronding, and they hae also considered fixed parameter tractability of the problem on trees. Frthermore, seeral athors (Deelin and Hartke [], Fogarty [7], and Wang and Moeller []) hae considered arios aspects of the problem for d-dimensional grids, especially the nmber of firefighters reqired to contain the fire. We refer the reader to a recent srey of Finbow and MacGilliray [6] for more information on the firefighter problem. Receied by the editors Agst 7, 007; accepted for pblication (in reised form) Jly, 009; pblished electronically December 9, Department of Compter Science and Engineering, The Chinese Uniersity of Hong Kong, Shatin, Hong Kong SAR, China (lcai@cse.chk.ed.hk). This athor was partially spported by grant CUHK6/0E from the Research Grants Concil of Hong Kong. Institte of Mathematics, Zhejiang Normal Uniersity, Jinha, Zhejiang 00, China (wwf@ zjn.cn). This athor was partially spported by grant NSFC of China. 8

2 THE SURVIVING RATE OF A GRAPH 8 In this paper, we stdy the fire defending ability of a graph as a whole by considering the aerage percentage of ertices the firefighter can sae. Let sn() denotethe maximm nmber of ertices in G that the firefighter can sae when a fire breaks ot at ertex, which will be referred to as the sriing nmber for. We define the sriing rate ρ(g) ofg to be the aerage percentage of ertices that can be saed when a fire randomly breaks ot at a ertex of the graph, i.e., V ρ(g) = sn() n. For example, the sriing rates of paths and cycles are ρ(p n )= n + n for n and ρ(c n )= n. We note that the concept of sriing rates is closely related to the notion of expected damage introdced by Finbow et al. [], who inestigated graphs of minimm expected damage. To be precise, ρ(g) = ed(g)/n, where ed(g) = n V (n sn()) is the expected damage ed(g) ofg. Or main reslts are the following lower bonds on sriing rates of trees, oterplanar graphs, and Halin graphs : /n if G is a tree with n ertices, ρ(g) > /6 if G is an oterplanar graph, /0 if G is a Halin graph with at least ertices. We will proceed as follows. We fix notation and gie definitions in section. We then establish lower bonds on the sriing rates of trees (section ), oterplanar graphs (section ), and Halin graphs (section ). We will also discss ftre directions and propose some open problems in section 6.. Notation and definitions. All graphs in this paper are connected ndirected simple graphs with m edges and n ertices. For a graph G =(V,E) and aertex V, d() denotes the degree of in G, andn() thesetofneighbors of, i.e., ertices adjacent to. A d-ertex is a ertex of degree d, and we se V d and n d, respectiely, to denote the set and the nmber of d-ertices in G. We se Δ to denote the maximm degree of G. For a sbgraph H of G, we se sn(h) = V (H) sn() todenotethesriing nmber for H, andsn(h) = sn(h)/ V (H) to denote the aerage sriing nmber for H. We will also se the aboe notation and terms for a sbset V of ertices; e.g., sn(v ) denotes the sriing nmber for V. A rooted tree is a tree T with one ertex r chosen as the root. For each ertex in T, its neighbor on the niqe (r, )-path in T is the parent of, and each of the other neighbors of is a child of. A ertex is a descendant of if is on the niqe (r, )-path in T.Arooted sbtree T of T is the sbtree of T consisting of, whichis the root of T, and all other descendants of. The height of a rooted tree T r is the length of a longest root-to-leaf path, and the height of a ertex in T r is the height of the rooted sbtree T.. Trees. Let s start with the following simple obseration: if the firefighter protects ertices,,..., t sccessiely to sae k ertices in a graph G, thenfor Agraphisanoterplanar graph if it has a planar embedding with all ertices on the bondary of the oter face. AgraphisaHalin graph if it is formed from a planar embedding of a tree withot ertices of degree by connecting all its leaes by a cycle that crosses no edges.

3 86 CAI LEIZHEN AND WANG WEIFAN any spanning sbgraph H of G, he can protect these ertices in the same order to sae at least k ertices in H. This implies the following connection between sriing rates of a graph and its spanning sbgraphs. Fact.. For any spanning sbgraph H of a graph G, we hae ρ(g) ρ(h). The aboe fact indicates that, among all graphs, trees hae the highest sriing rates, as eery connected graph contains a spanning tree. Since we can sae at most n d() ertices when a fire breaks ot at a ertex, eerytreet satisfies ρ(t ) n V (n d()) = m n = n + n. We note that trees attaining this pper bond are exactly caterpillars where the distance between eery pair of ertices of degree at least is not eqal to, which was shown by Finbow et al. [] in their stdy of graphs with minimm expected damage. Table. gies sriing rates of small trees. Table. Sriing rates of trees with at most 9 ertices. n ρ(t ) , , , 6 8, 6 8 In this section, we show that the sriing rate of eery tree on n ertices is greater than /n, which tends to as n. To establish this lower bond, we first gie two lemmas abot sriing nmbers for paths and sbtrees in a tree. Lemma.. For eery path P on k ertices in a tree T, sn(p ) > (k )n. Proof. Let P = k. Let T i denote the connected component of T { i, i+ } containing ertex i,andt i the nmber of ertices in T i. When a fire breaksotatertex i, i k, we protect i first and then i+ (do nothing if i or i+ does not exist) to sae all ertices in T T i T i+. Therefore sn( i ) n t i t i+ and sn(p ) k (n t i t i+ ) i= = kn k (t i + t i+ ) i= > (k )n. Lemma.. For eery rooted sbtree S of height h 0 in a rooted tree T, sn(s) (n h ) S, where S is the nmber of ertices in S. Proof. When a fire breaks ot at a ertex of S, wesethesimplestrategyof protecting its parent in T, which will sae all ertices in T T (note that T = S ). If S = T and the fire breaks ot at the root, we protect a child of the root. A caterpillar is a tree that has a path sch that eery ertex not in the path is adjacent to some ertex in the path.

4 THE SURVIVING RATE OF A GRAPH 87 Let H i,0 i h, be the set of ertices of height i in S. Thenforeach0 i h, H i n S, wheren denotes the nmber of ertices in T. Therefore sn(h i )= (n n ) H i n S, H i and ths sn(s) = = n h sn(h i ) i=0 h ( H i n S ) i=0 h H i i=0 h S i=0 =(n h ) S. The aboe lemma implies that the sriing rate of a tree of small diameter is close to, where the diameter diam(g) of a graph G eqals the maximm pairwise distance in G. diam(t )/ + n. Corollary.. For eery n-ertex tree T, ρ(t ) Proof. LetP = 0,,..., t be a longest path in T and h = t/ = diam(t )/. We can make T arootedtreeofheighth by choosing h as the root, and the bond follows directly from Lemma.. Now we establish the /n lower bond on the sriing rate of a tree. For arootedtreeonn ertices, a long path is a path that contains at least n ertices, and a short tree is a rooted sbtree whose height is at most n. Clearly, if a rooted sbtree does not contain a long path from the root to a leaf, then it is a short tree. The main idea in or proof is to partition a rooted tree into long paths and short trees, and then se Lemmas. and. to obtain a lower bond on the sriing rate of the tree. We choose n to define long paths in order to balance the lower bonds from these two lemmas. Theorem.. For eery n-ertex tree T, ρ(t ) > /n. Proof. It sffices to proe the theorem for n. We first make T arootedtree by arbitrarily choosing a ertex as its root, and then se indction on the height of T to show that T can be partitioned into long paths and short trees. Let P be a longest root-to-leaf path in T.IfPhas less than n ertices, then the height of T is at most n. Therefore T itself is a short tree, and we are done. Otherwise, we delete all ertices of P from T to obtain a forest F.NotethatF is a collection of rooted trees, and the choice of P garantees that eery rooted sbtree of any rooted tree in F is also a rooted sbtree of T. Frthermore, the height of any rooted tree in F is one less than that of T. By the indction hypothesis, each rooted tree in F can be partitioned into long paths and short trees. These partitions and P together gie s a reqired partition of T. Let P be the set of long paths and S the set of short trees in the partition. Denote by P the nmber of ertices in a path P,by S the nmber of ertices in a short tree S, andbyh(s) the height of S. If T itself is a short tree, the theorem follows directly from Lemma.. Otherwise P, and it follows from Lemmas. and.

5 88 CAI LEIZHEN AND WANG WEIFAN that sn(t )= sn(p )+ sn(s) P P S S > ( P )n + h(s) ) S P P S S(n ( = P + ) ( S n n + ) (h(s)+) S P P S S P P S S ( ) n n S P P n + S S as P P P + S S S = n and h(s) n. Let p denote P P P. Then P p/ n p/ n, and it follows that ( p sn(t ) >n n + ) n(n p) = n nn, n which yields ρ(t ) > /n.. Oterplanar graphs. In this section we proe that the sriing rate of eery oterplanar graph with at least ertices is greater than /6. For this prpose, we first gie a lower bond on the sriing nmber for a ertex in a maximal oterplanar graph. Lemma.. For eery ertex of a maximal oterplanar graph G with n n (d() ). ertices, we hae sn() Proof. Let G be an oterplanar embedding of G. Then the bondary of the oter face of G forms an n-cycle C. For two ertices x and y in C, letc[x, y] denote the section of C from x to y, and let C[x, y) =C[x, y] \{y}. Let be a d-ertex and x,x,...,x d its neighbors ordered according to their orders in C. Since {C[x i,x i+ ): i<d} partitions C {, x d } into d sections, one section, say C[x t,x t+ ), contains at least (n )/(d ) ertices, and ths C[x t,x t+ ] contains at least (n )/(d ) + ertices. If C[x t,x t+ ] =,thend = n and the lemma is clearly tre. Otherwise C[x t,x t+ ], and by the maximality of G, there is a niqe ertex y t C(x t,x t+ ) that forms a -face with ertices x t,x t+ (see Figre.). y t x t x t+ Fig... Inside section C[x t,x t+ ]. An oterplanar graph is maximal if no edge can be added withot losing oterplanarity.

6 THE SURVIVING RATE OF A GRAPH 89 Clearly, one of C[x t,y t ]andc[y t,x t+ ], say C[x t,y t ], contains at least (n )/(d ) + ertices. When a fire breaks ot at, we protect x t first and then y t to sae all ertices in C[x t,y t ]. Therefore sn() n n +d += (d ) (d ) n (d ) as d. We now se the aboe lemma to establish the following lower bond on sriing rates of oterplanar graphs. Theorem.. The sriing rate of eery oterplanar graph with ertices is greater than /6. Proof. In light of Fact., it sffices to proe the theorem for eery maximal oterplanar graph G since eery oterplanar graph is a spanning sbgraph of some maximal oterplanar graph. Also we assme n, as the theorem is clearly tre for n =,. Let V d be the set of d-ertices, and n d the nmber of d-ertices. By Lemma., we hae sn(g) = sn(v d ) d= d= n (d ) n d = 6 nβ for β = Δ d= d n d. To calclate the ale of β, wenotethatg contains n edges, which implies V d() =n 6. Therefore n +n + +Δn Δ =(n + n + + n Δ ) 6, which yields (.) (d )n d +6=n + n. d= We now se (.) to show that β>n: β = d n d d= =n + n + n + d= d= d n d n + n + n + (n + n )+ d n d d= ( = n + n + n + Δ ) (d )n d +6 + d n d (by (.)) = n + n + n ++ ( d d= + d d= ) n d. Since d + d > ford, we hae β>n+ and therefore ρ(g) =sn(g)/n > 6.

7 80 CAI LEIZHEN AND WANG WEIFAN. Halin graphs. In this section we show that the sriing rate of eery Halin graph with n ertices is greater than /0. Recall that a Halin graph is formed from a planar embedding of a tree T withot -ertices by connecting all its leaes by acyclec that crosses no edges. We note that a Halin graph contains at least ertices and at most n edges. Table. gies sriing rates of small Halin graphs. Table. Sriing rates of Halin graphs with at most 8 ertices. n ρ(g) 9 6, 6 9 9, , 9 6, 0 6 We assme that a planar embedding of a Halin graph G is gien where all ertices in the cycle C lie on the oter face. All ertices not on C, i.e., nonleaf ertices of the tree T,areinternal ertices, and neighbors of a ertex are ordered in a conterclockwise direction. We also se the following legend for the figres in this section: the sorce of a fire is indicated by a black ertex, a ertex on fire is shaded, a protected ertex is enclosed by a sqare box, and the time a ertex is on fire or protected is gien by a nmber beside the ertex. To obtain a lower bond on the sriing rate of a Halin graph, we start with two lemmas on the sriing nmber for a ertex. The first lemma shows that some ertices on the cycle C case ery little damage to the graph when they are sorces of fire, and the second lemma gies a lower bond on the sriing nmber for a ertex in terms of its degree. Lemma.. The cycle C of eery Halin graph G contains at least two ertices sch that the sriing nmber for each ertex is at least n. Proof. The lemma is triially tre for n 6, and ths we may assme n 7. Let T be the tree obtained from the tree T of G by remoing all leaes of T, i.e., ertices of the cycle C of G. Case. T has a leaf with d G (). If d G () =, then, as illstrated in Figre.(a), the leftmost and rightmost neighbors of on C both hae sriing nmbers at least n. Otherwise, d G () and, as illstrated in Figre.(b), the second leftmost and second rightmost neighbors of on C both hae sriing nmbers at least n. (a) (b) Fig... (a)the sitation for d G () =p to symmetry, and (b) the sitation for d G () p to symmetry (the rightmost ertex may or may not be a neighbor of ).

8 THE SURVIVING RATE OF A GRAPH 8 Case. All leaes of T hae degree in G. Let P be a longest path in T. Since G has at least 7 ertices, T contains at least ertices. We claim that for each end of P there is a distinct ertex on C whose sriing nmber is at least n. Let be an arbitrary end of P,and the niqe ertex in T that is adjacent to. Order the neighbors N G () of in G conterclockwise. Let x N G () betheertex immediately preceding, andy N G () the ertex immediately following. If one of x, y, sayx, isonc, then we hae the sitation shown in Figre.(a), and the sriing nmber for ertex z in the figre is at least n. Otherwise, both x and y are ertices of T. Since is an end of a longest path in T,oneofx, y, sayx, is aleafoft, and we hae the sitation shown in Figre.(b). As illstrated in the figre, the sriing nmber for ertex z is at least n. x x z z (a) (b) Fig... (a) Vertex x is on the cycle C, and(b) ertex x is a leaf of T. Edges of T are indicated by thick lines. Lemma.. Let G be a Halin graph with n 7 ertices. Then for eery ertex in G, sn() n d(). Proof. We consider two cases, depending on whether ertex is in the cycle C. Case. is a ertex in C. Let be the niqe neighbor of in T, and regard T as a tree rooted at. Order the sbtrees rooted at the children of conterclockwise. Call the first one the leftmost sbtree, and the last one the rightmost sbtree. Note that d() = and we need to show that sn() (n )/. Case.. One of the leftmost and rightmost sbtrees of contains at least (n )/ ertices. Withot loss of generality, we may assme that the rightmost sbtree T of contains at least (n )/ ertices. If the cycle C has at least two more ertices from to the leftmost ertex of T, we can sae all ertices in T by protecting its rightmost ertex, root, and leftmost ertex in this order (see Figre.(a)), which implies sn() (n )/. Otherwise, C has only one ertex from to the leftmost ertex of T, and ths T contains n ertices. We first protect the rightmost ertex of T and then the root of T. At this point, the leftmost ertex of T is on fire. From now on, we contain the fire to spread along the cycle C conterclockwise by protecting the niqe internal neighbor w of the ertex x most recently on fire ntil we hae the sitation that w has been protected already. When this happens, we protect the right neighbor x of x on C (do nothing if x has been protected already), and the fire can no longer spread (see Figre.(b)). Therefore we sae more than half of the ertices in T, and ths sn() > (n )/ (n )/ forn 7.

9 8 CAI LEIZHEN AND WANG WEIFAN T 6 T (a) (b) Fig... The sitation for the case that the rightmost sbtree T contains at least (n )/ ertices: (a) C contains at least two ertices from to the leftmost ertex of T,and(b) C contains only one ertex from to the leftmost ertex of T. leftmost sbtree rightmost sbtree S 6 S (a) (b) Fig... The sitation for the case that both the leftmost and rightmost sbtrees contain fewer than (n )/ ertices: (a) one of the leftmost and rightmost sbtrees contains more than one ertex, and (b) both the leftmost and rightmost sbtrees contain one ertex. Case.. Both the leftmost and rightmost sbtrees of contain fewer than (n )/ ertices. In this case, let S denote the tree rooted at after deleting both the leftmost and rightmost sbtrees of from T. Then S contains at least (n )/ ertices as n 7. Withot loss of generality, we may assme that the leftmost sbtree of contains at least as many ertices as the rightmost sbtree of. If the leftmost sbtree of contains at least two ertices, then C contains at least two more ertices from to the leftmost ertex of S, and we can sae all ertices in S by protecting ertices in the order shown in Figre.(a), implying sn() (n )/. Otherwise, both the leftmost and rightmost sbtrees of contain one ertex, and ths S has n ertices. We protect ertex first, then the rightmost ertex in S. At

10 THE SURVIVING RATE OF A GRAPH 8 T l() T r() T T l() T T r() 6 (a) (b) Fig... Case. is an internal ertex: (a) one of the leftmost and rightmost sbtrees contains more than one ertex, and (b) both the leftmost and rightmost sbtrees contain one ertex. this point, the leftmost ertex of S is on fire, and we protect ertices in S as we did for T in Case. to contain the fire to spread along the cycle C conterclockwise (see Figre.(b)). Therefore we sae more than half of the ertices in S, and ths sn() > (n )/ (n )/ forn 7. Case. is not a ertex in C. In this case, is an internal ertex of T. Regard T as a tree rooted at, and order sbtrees rooted at the children of conterclockwise arond. By pairing p each sbtree with the sbtree on its right, we get d() pairs of sbtrees with a total nmber of (n ) ertices. Therefore there is a pair T,T r() of sbtrees with at least (n )/d() ertices. Withot loss of generality, we may assme that T contains at least as many ertices as T r(). Then T contains at least (n )/d() ertices. Let T l() denote the sbtree to the left of T. If one of T l() and T r(),say T l(), contains at least two ertices, we can sae all ertices in T by first protecting, then the rightmost ertex of T, followed by the leftmost ertex of T (see Figre.(a)). Otherwise, both T l() and T r() contain one ertex, and ths T contains at least (n )/d() ertices since T T r() has at least (n )/d() ertices. We first protect ertex and then the rightmost ertex of T. At this point, the leftmost ertex in T is on fire, and we deal with T as what we did for T in Case. to contain the fire to spread along the cycle C, conterclockwise (see Figre.(b)). Therefore we sae at least + (n )/ ertices in T,wheren is the nmber of ertices in T, which yields sn() (n )/d(). We now se Lemmas. and. to obtain a lower bond on the sriing rate of a Halin graph. Theorem.. For eery Halin graph G with n ertices, its sriing rate ρ(g) > 0. Proof. From Table., we see that the theorem holds for n 8, and ths we assme n 9. We also note that the nmber of leaes in an n-ertex tree eqals (d )n d +, d=

11 8 CAI LEIZHEN AND WANG WEIFAN where n d is the nmber of d-ertices in the tree. This is easily deried from the following relation: ( Δ ) dn d =(n ) = n d. d= We are now ready to establish the lower bond in the theorem. By Lemma., the cycle C contains two ertices x, y sch that each of them has sriing nmber at least n. Let V = V (C) {x, y}. Since the ertices in the cycle C of G are exactly the leaes of the tree T of G, wehae V = V (C) = d= (d )n d. Therefore each internal ertex can be matched with d() distinct ertices V () in V, and we consider the sriing nmber for {} V () collectiely. Note that ertices in V are -ertices, and eery internal ertex has degree at least. By Lemma., we hae sn() (n )/d() for each ertex. Therefore the aerage sriing nmber sn({} V ()) d() = n d= ( n d() + n ( + d() d() ) (d() ) ). In the interal [, ), fnction f(x) =+/(x ) /x monotonically decreases in [, + 6) and monotonically increases in ( + 6, ). Since + 6., f(d()) reaches its minimm at d() = or 6, which is 9/0 for both and 6. It follows that sn({} V ()) (n ). 0 Combining this with the fact that sn(x) n andsn(y) n, we hae sn(g) 0 (n )(n ) + (n ) ( 0 n 7 ) = 0 n + > 0 n (as n 9), and hence ρ(g) > 0 for n. 6. Conclding remarks. In this paper, we hae introdced the notion of the sriing rate of a graph for the firefighter problem to measre the fire defending ability of a graph as a whole, and stdied sriing rates of trees, oterplanar graphs, and Halin graphs. Trees, oterplanar graphs, and Halin graphs all hae constant lower bonds on their sriing rates. In other words, for eery graph in these graph classes, we can sae, on aerage, at least a constant percentage of ertices when afirebreaksotata

12 THE SURVIVING RATE OF A GRAPH 8 ertex. It is natral to ask whether other graph classes hae this property as well. In particlar, one may wish to generalize this property to planar graphs. Unfortnately, for planar graphs K,n, lim n ρ(k,n ) = 0. Neertheless, we feel that this property holds for planar graphs of maximm degree. Conjectre 6.. There is a positie constant c sch that eery nontriial planar graph G of maximm degree satisfies ρ(g) c. On the other hand, we can also generalize the sriing rate of a graph to the k-sriing rate by allowing the firefighter to protect k ertices each time, which brings s the following problem for planar graphs. Problem 6.. Determine the minimm k for which planar graphs hae a constant lower bond on their k-sriing rates. The lower bonds of sriing rates that we hae obtained for trees, oterplanar graphs, and Halin graphs are not tight. It wold be interesting to improe these lower bonds. Howeer, this may reqire new techniqes, instead of detailed analysis. For oterplanar graphs and Halin graphs, it is nclear whether their asymptotic sriing rates are. We note that there are infinitely many oterplanar graphs (n-fans, for instance) and Halin graphs (n-wheels, for instance) G with ρ(g) approaching as n. Problem 6.. For oterplanar graphs (Halin graphs, respectiely) G on n ertices, determine whether lim n ρ(g) =. We hae shown that the sriing rate of an n-ertex tree T is at least /n, which approaches as n tends to infinity. We conjectre that actally ρ(t ) approaches mch more qickly as follows. Conjectre 6.. For eery n-ertex tree T, ρ(t ) Θ( log n n ). In terms of algorithmic and complexity isses, we certainly expect that it is ery hard to determine the sriing rate of a graph, bt it seems also difficlt to obtain an NP-completeness proof. Conjectre 6.. It is NP-complete to determine the sriing rate of a tree. We also note that the proofs of or theorems imply polynomial-time approximation algorithms for sriing rates of trees, oterplanar graphs, and Halin graphs. In particlar, or discssions in section imply a polynomial-time ( /n)- approximation algorithm for the sriing rate of a tree. On the other hand, it follows directly from a reslt of Hartnell and Li [0] that the following greedy method for trees is a /-approximation algorithm for the sriing rate of a tree: eery time the firefighter protects a ertex that cts off the maximm nmber of ertices from the fire. We feel that this greedy algorithm achiees a mch better approximation ratio for sriing rates, and leae its performance analysis as the last open problem of the paper. Problem 6.6. Determine the approximation ratio of the greedy algorithm for the sriing rate of a tree. Is it actally eqal to Θ( log n n )? Note added in proof. Recently, Cai et al. [] hae proed Conjectre 6., and soled Problem 6. for oterplanar graphs and Problem 6.6. REFERENCES [] L. Cai, Y. Cheng, E. Verbin, and Y. Zho, Sriing rates of trees and oterplanar graphs for the firefighter problem, manscript, 009. [] L. Cai, E. Verbin, and L. Yang, Firefighting on trees: ( /e)-approximation, fixed parameter tractability and a sbexponential algorithm, in Proceedings of ISAAC 008, S.-H. Hong, H. Hagamochi, and T. Fknaga, eds., Lectre Notes in Compt. Sci. 69, Springer, New York, 008, pp

13 86 CAI LEIZHEN AND WANG WEIFAN [] M. Deelin and S. Hartke, Fire containment in grids of dimension three and higher, Disc. Appl. Math., (007), pp [] S. Finbow, B. Hartnell, Q. Li, and K. Schmeisser, On minimizing the effects of fire or a irs on a network, J. Combin. Math. Combin. Compt., (000), pp.. [] S. Finbow, A. King, G. MacGilliray, and R. Rizzi, The firefighter problem for graphs of maximm degree three, Disc. Math., 07 (007), pp [6] S. Finbow and G. MacGilliray, The firefighter problem: A srey of reslts, directions, and qestions, Astralas. J. Combin., (009), pp [7] P. Fogarty, Catching the Fire on Grids, M.Sc. Thesis, Department of Mathematics, Uniersity of Vermont, Brlington, VT, 00. [8] G. MacGilliray and P. Wang, On the firefighter problem, J. Combin. Math. Combin. Compt., 7 (00), pp [9] B. Hartnell, Firefighter! An application of domination, presentation at the th Manitoba Conference on Combinatorial Mathematics and Compting, Uniersity of Manitoba, Winnipeg, Canada, 99. [0] B. Hartnell and Q. Li, Firefighting on trees: How bad is the greedy algorithm?, in Proceedings of the Thirty-first Sotheastern International Conference on Combinatorics, Graph Theory and Compting (Boca Raton, FL, 000), Congr. Nmer., Utilitas Math., Winnipeg, MB, 000, pp [] P. Wang and S. Moeller, Fire control on graphs, J. Combin. Math. Combin. Compt., (00), pp. 9.