On the tree cover number and the positive semidefinite maximum nullity of a graph

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1 arxi: [math.co] Oct 018 On the tree coer number and the positie semidefinite maximum nullity of a graph Chassidy Bozeman Abstract For a simple graph G = (V,E), let S + (G) denote the set of real positie semidefinite matrices A = (a ij ) such that a ij 0 if {i,j} E and a ij = 0 if {i,j} / E. The maximum positie semidefinite nullity of G, denoted M + (G), is max{null(a) A S + (G)}. A tree coer of G is a collection of ertex-disjoint simple trees occurring as induced subgraphs of G that coer all the ertices of G. The tree coer number of G, denoted T(G), is the cardinality of a minimum tree coer. It is known that the tree coer number of a graph and the maximum positie semidefinite nullity of a graph are equal for outerplanar graphs, and it was conjectured in 011 that T(G) M + (G) for all graphs [Barioli et al., Minimum semidefinite rank of outerplanar graphs and the tree coer number, Elec. J. Lin. Alg., 011]. We show that the conjecture is true for certain graph families. Furthermore, we proe bounds on T(G) to show that if G is a connected outerplanar graph on n ertices, thenm + (G) = T(G) n, andif Gis aconnected outerplanar graphon n 6ertices with no three or four cycle, then M + (G) = T(G) n. We characterize connected outerplanar graphs with M + (G) = T(G) = n, and for each cactus graph G, we gie a formula for computing T(G) (and therefore M + (G))). 1 Introduction A graph is a pair G = (V,E) where V is the ertex set and E is the set of edges (two element subsets of the ertices). All graphs discussed are simple (no loops or multiple edges) and finite. For a graph G = (V,E) on n ertices, we use S + (G) to denote the set of real n n positie semidefinite matrices A = (a ij ) satisfying a ij 0 if and only if {i,j} E, for i j, and a ii is any non negatie real number. The maximum positie semidefinite nullity of G, denoted M + (G), is defined as max{null(a) A S + (G)}. The minimum positie semidefinite rank of G, denoted mr + (G), is defined as min{rank(a) A S + (G)}, and it follows from the Rank-Nullity Theorem that M + (G)+mr + (G) = n. Barioli et al. [] define a tree coer of G to be a collection of ertex-disjoint simple trees occurring as induced subgraphs of G that Department of Mathematics, Iowa State Uniersity, Ames, IA 50011, USA (cbozeman@iastate.edu) 1

2 coer all the ertices of G. The tree coer number of G, denoted T(G), is the cardinality of a minimum tree coer, and it is used as a tool for studying the positie semidefinite maximum nullity of G. (In their paper [], G is allowed to be a multigraph, but we restrict ourseles to simple graphs.) It was conjectured in [] that T(G) M + (G) for all graphs, and it is shown there that T(G) = M + (G) for outerplanar graphs. We show that T(G) M + (G) for certain families of graphs in Section. In Section, we study T(G) for connected graphs with girth at least 5 and deduce bounds on M + (G) for connected outerplanar graphs with girth at least 5. In section 4, we characterize connected outerplanar graphs on n ertices haing positie semidefinite maximum nullity and tree coer number equal to the upper bound of n, and we gie a formula for computing T(G) and M + (G) for cactus graphs. 1.1 Graph theory terminology For a graph G = (V,E) and V, the neighborhood of, denoted N(), is the set of ertices adjacent to. The degree of is the cardinality of N() and is denoted by deg(). A ertex of degree one is called a leaf. A set S V is independent if no two of the ertices of S are adjacent. The path P n is the graph with ertex set { 1,..., n } and edge set {{ i, i+1 } i {1,...,n 1}}. The cycle C n is formed by adding the edge { n, 1 } to P n. The girth of a graph is the size of the smallest cycle in the graph. We denote the graph on n ertices containing eery edge possible by K n, and we use K s,t to denote the complete bipartite graph, the graph whose ertex set may be partitioned into two independent sets X and Y such that X = s, Y = t, for each x X and y Y, {x,y} is an edge, and each edge has one endpoint in X and one endpoint in Y. The graph K 1, is referred to as a claw and more generally, the graph K 1,t is called a star. For a graph G = (V,E), a graph G = (V,E ) is a subgraph of G if V V and E E. A subgraph G is an induced subgraph of G if V(G ) V(G) and E(G ) = {{u,} {u,} E(G) and u, V(G )}. If S V(G), then we use G[S] to denote the subgraph induced by S. For S V(G), we use G S to denote G[V(G) \S], and for e E, G e denotes the graph obtained by deleting e. For a graph G and an induced subgraph H, G H denotes the graph that results from G by deleting V(H). A graph H = (V(H),E(H)) is a clique if for each u, V(H),{u,} E(H). The clique number of G, denoted ω(g), is ω(g) = max{ V(H) : H is a subgraph of G and H is a clique}. The independence number of G, denoted α(g), is the cardinality of a maximum independent set. A graph is connected if there is a path from any ertex to any other ertex. For a connected graph G = (V,E), an edge e E is called a bridge if G e is disconnected. We subdiide an edge e = {u,w} E by remoing e and adding a new ertex e such that N( e ) = {u,w}. A graph G = (V,E) is outerplanar if it has a crossing-free embedding in the plane with eery ertex on the boundary of the unbounded face. A cut-ertex of a connected graph G = (V,E) is a ertex V such that G is disconnected. A graph is nonseparable if it is connected and does not hae a cut-ertex. A block is a maximal nonseparable induced

3 subgraph. A graph G is a cactus graph if eery block of G is either a cycle or a single edge, and G is a block-clique graph if eery block is a clique. Cactus graphs are a well studied family of outerplanar graphs. Throughout this paper, gien a graph G = (V,E) and a tree coer T of G, we use T T to denote the tree containing V. 1. Preliminaries In this section, we gie some preliminary results that will be used throughout the remainder of the paper. It is shown in Propostion. of [] that deleting a leaf of a graph does not affect the tree coer number of the graph and that subdiiding an edge does not affect the tree coer number of the graph. These two facts will be used repeatedly in the proofs throughout this paper. Theorem 1. [9] Suppose G i,i = 1,...,h are graphs, there is a ertex for all i j, G i G j = {} and G = h i=1 G i. Then, ( h ) M + (G) = M + (G i ) h+1. i=1 This is known as the cut-ertex reduction formula. The authors of [7] gie an analogous cut-ertex reduction formula for computing the tree coer number and we use this technique multiple times throughout this paper. Proposition. [7] Suppose G i,i = 1,...,h, are graphs, there is a ertex for all i j, G i G j = {} and G = h i=1 G i. Then ( h ) T(G) = T(G i ) h+1. i=1 In the case that h from Proposition is two, we say G is the ertex-sum of G 1 and G and write G = G 1 G. Bozeman et al. [5] gie a bound on the tree coer number of a graph in terms of the independence number of the graph. Proposition. [5] Let G = (V,E) be a connected graph, and let S V(G) be an independent set. Then, T(G) G S. In particular, T(G) G α(g), where α(g) is the independence number of G. Furthermore, this bound is tight. It is also shown in Proposition 6 of [5] that for a graph G = (V,E) and a bridge e E, e belongs to some tree in eery minimum tree coer. Embedded in the proof of this proposition is the following lemma, and we include the proof of the lemma for completeness. Lemma 4. Let G = (V,E) be a connected graph and e = {u,} a bridge in E. Let G 1 and G be the connected components of G e. Then T(G) = T(G 1 )+T(G ) 1 = T(G e) 1.

4 For a graph G = (V,E) and an edge e E, M + (G) 1 M + (G e) M + (G) +1 [7] and T(G) 1 T(G e) T(G) +1 [5]. It is also known that for V, M + (G) 1 M + (G ) M + (G) + deg() 1 (see Fact 11 of page of [8]). We show that an analogous bound holds for T(G). Proposition 5. For a graph G = (V,E) and ertex V, T(G) 1 T(G ) T(G)+deg() 1. Proof. Since any tree coer of G together with the tree consisting of the single ertex is a tree coer for G, then T(G) T(G ) + 1, which gies the lower bound. To see the upper bound, let E denote the set of edges incident to, and let G E denote the graph resulting from deleting the edges in E. Note that E = deg(), and that T(G E ) = T(G ) + 1. Since the deletion of an edge can raise the tree coer number by at most 1, then T(G ) + 1 = T(G E ) T(G) + deg(), and the upper bound follows. Graphs with T(G) M + (G) In this section, we proe that T(G) M + (G) for certain line graphs, for G (defined below) where G is any graph, for graphs whose complements hae sufficiently small tree-width, and for graphs with a sufficiently large number of edges. We first show that for any connected graph G on n ertices, T(G) n. Lemma 6. Let G be a connected graph on n ertices. Then there exists an induced subgraph H of G such that H = K 1,p for some p 1 and G H is connected. (See Figure 1). Proof. We proe the lemma by induction. For n = the claim holds. Let G be a graph on n 4 ertices and suppose the lemma holds for all graphs on k n 1 ertices. It is known that eery connected graph has at most n cut ertices (since a spanning tree of the graph has at least two leaes and the remoal of these leaes will not disconnect the graph). Let be a ertex in V(G) that is not a cut ertex. By hypothesis, there exists an induced subgraph H = K 1,p for some p 1 in G whose deletion does not disconnect G. First we consider the case with p = 1, and then we consider the case with p. Case 1: Suppose p = 1 (i.e., H = K ), and let a,b be the ertices of H. If has a neighbor in G[V(G) \{a,b}], then G[V(G) \{a,b}] is connected, and the claim holds with H = H. Otherwise has a neighbor in {a,b}. Assume first that is adjacent to exactly one of a and b. Without loss of generality, suppose is adjacent to a and not adjacent to b. Then, H = G[a,b,] = K 1, and G H is connected. Now suppose that is adjacent to both a and b. Since G is connected, then either a or b has a neighbor in G[V(G)\{,a,b}]. Without loss of generality, let a hae a neighbor in G[V(G) \{,a,b}]. Then H = G[{,b}] = K 1,1 and G H is connected. Case : Suppose p. If has a neighbor in G[V(G) \V(H )], then set H = H and the claim holds. Otherwise has neighbors only in V(H ). Recall that H is a star. First 4

5 suppose that is adjacent to a leaf w V(H ). If w is not a cut ertex of G, then for H = G[{,w}] = K 1,1, G H is connected. If w is a cut ertex of G, then w has a neighbor in G[V(G) \{V(H ) }]. Then H = G[V(H )\{w}] = K 1,q for some q 1 and G H is connected. Next suppose that is not adacent to a leaf in H. Then it must be adjacent to the center ertex. Then H = G[V(H ) {}] is a star, and G H is connected. This completes the proof. Figure 1: Two examples of Lemma 6, where induced subgraphs H are black. Theorem 7. For any simple connected graph G = (V,E) on n ertices, T(G) n. Proof. The theorem holds for n =. Let G beagraph on n such that the claim holds on all graphs with fewer than n ertices. By Lemma 6, there exists an induced tree H = K 1,p, for some p 1, of G such that G = G H is connected. By the induction hypothesis, T(G ) G n. Then T(G) T(G )+1 n. It is shown in [6] that for a triangle-free graph G, M + (G) n. The next corollary is a result of Theorem 7 and the fact that M + (G) = T(G) [] for outerplanar graphs. Corollary 8. If G is a connected outerplanar graph on n ertices, then M + (G) n. Some examples of graphs with T(G) = n are the complete graphs Kn and the well known Friendship graphs (graphs on n = k + 1 ertices, k 1, consisting of exactly k triangles all joined at a single ertex.) Connected outerplanar graphs haing T(G) = n are characterized in Section 4. For a graph G = (V,E), the line graph of G, denoted L(G), is the graph whose ertex set is the edge set of G, and two ertices are adjacent in L(G) if and only if the corresponding edges share an endpoint in G. Theorem 9. Let G be a graph on n ertices and m n edges. Then T(L(G)) M + (L(G)). Proof. The adjacency matrix of L(G) is A(L(G)) = B T B I m, where B is the ertex-edge incidence matrix of G and I m is the m midentity matrix. Note that A(L(G))+I m = B T B is in S + (L(G)) and that rank(b) n. So, mr + (L(G)) rank(b T B) = rank(b) n. It follows that M + (L(G)) m n m T(L(G)), where the second inequality follows from the fact that m n and the last inequality follows from Theorem 7. 5

6 The next theorem shows that the conjecture T(G) M + (G) holds true for graphs with a large number of edges. Theorem 10. Let G be a graph on n ertices and m n 4 edges. Then T(G) M +(G). Proof. Let T be a spanning tree of G. Then M + (T) n (see [1, Theorem.16 and Corollary.17]). Note that G can be obtained from T by adding at most m (n 1) edges, so G can be obtained from T by deleting at most m (n 1) edges. Since edge deletion decreases the positie semidefinite maximum nullity by at most 1, then M + (G) M + (T) (m (n 1)) (n ) (m (n 1)) n, where the last inequality follows from the fact that m n 4. Since M +(G) is an integer, by Theorem 7, we hae that M + (G) T(G). Definition 11. For a graph G = (V,E), let G be the graph constructed from G such that for each edge e = {u,} E, add a new ertex w e where w e is adjacent to exactly u and. The ertices w e are called edge-ertices of G. Figure : K 4, where the edge-ertices are black. Theorem 1. For a connected graph G = (V,E) on n ertices and m edges, T(G ) M + (G ). Proof. We show that mr + (G ) = α(g ) and then apply Proposition. It is always the case that a connected graph H has α(h) mr + (H) (see Corollary.7 in [4]), so we show ( that ) mr + (G ) α(g I m ). Let B be the ertex-edge incidence matrix of G, and let X =, B ( ) where I m is the m m identity matrix. Then XX T = B BB T S + (G ), where the first m rows and columns are indexed by the edge-ertices and the last n rows and columns are indexed by the ertices in V. Note that the set of edge-ertices of G is an independent set of size m and that the rank of XX T is m. So mr + (G ) m α(g), and therefore mr + (G ) = α(g ). By Proposition, T(G ) m+n mr + (G ) = M + (G ). The tree-width of a graph G, denoted tw(g), is a widely studied parameter, and there are multiple ways in which it is defined. Here we define the tree-width in terms of chordal 6 I m B T

7 completions. A graph is chordal if it has no induced cycle on four or more ertices. If G is a subgraph of H such that V(G) = V(H) and H is chordal, then H is called a chordal completion of G. The tree-width of G is defined as tw(g) = min{ω(h) 1 H is a chordal completion of G}. Proposition 1. Let G be a graph on n ertices with tw(g) n 4. Then T(G) M +(G). Proof. If tw(g) k, then mr + (G) k + [11], i.e., M + (G) n k. For k = n 4, it follows that M + (G) n. Since M +(G) is an integer, we hae that M + (G) n T(G), where the last inequality follows from Theorem 7. A k tree is constructed inductiely by starting with a complete graph K k+1 and at each step a new ertex is added and this ertex is adjacent to exactly k ertices in an existing K k. A graph G that is a tree is outerplanar by definition, so T(G) = M + (G). Obseration 14 and Theorem 15 show that T(G) M + (G) for -trees and 5-trees. Obseration 14. If G is a graph with T(G), then it holds that T(G) M + (G). This is because of the fact that M + (G) = 1 implies G is a tree [10] (so T(G) = 1), and the facts that T(G) Z + (G) [7] (where Z + (G) is the positie semidefinite zero forcing number of a graph, defined in []) and M + (G) = implies Z + (G) = [7] (so T(G) ). Theorem 15. [1] Let G be a k tree with k odd. If G is a k tree, then T(G) = k+1. Corollary 16. For k {,5}, T(G) M + (G). T(G) of graphs with girth at least 5 For many graphs, the tree coer number is much lower than n. Thenext theorem improes this bound for graphs with girth at least 5. Theorem 17. Let G be a connected graph on n 6 ertices with girth at least 5. Then T(G) n. Proof. The proof is by induction on n. A connected graph on 6 ertices with girth at least 5 is either a tree, C 6, or C 5 with a leaf adjacent to one of the ertices on the cycle. In each case, the tree coer number is at most, so the theorem holds. Let n 7. If G has a leaf, then T(G) = T(G ) n 1. Suppose G has no leaes. Let P = (x,y,z) be an induced path in G. We consider the connected components of G P (see Figure ). Note that since G has no leaes and no three or four cycles, G P cannot hae an isolated ertex as a connected component. We now show that if G P has a connected component H with H {,4,5}, then the theorem holds. Suppose G P has a connected component H of order (i.e., H is a path on three ertices). Note that G H is a connected graph (since the remaining components of G P are all connected to P), so if G H 6, by applying the hypothesis to G H and coering H with a path to get that T(G) 1 + n = n. Otherwise, G H = 5 since n 7 and G P does not hae an isolated ertex as a 7

8 P x y z... H H 1 H k Figure : The partition described in proof of Theorem 17, where H,H 1,...,H k are the connected components of G P. component, so G H P = K. By assumption G has no leaes and no three or four cycles, so G H = C 5, G is one of the two graphs shown in Figure 4 and the theorem holds. Suppose that G P has a connected component H of order 4. Then H is a tree. If G H 6, then T(G) 1+ n 4 = n 1. If G H = P, then T(G) =, n = 7, and the theorem holds. Otherwise G H = C 5, T(G) (since G H may be coered with trees and H is a tree, n = 9, and the theorem holds. Consider G P haing a connected component H of order 5. Then H is either a tree or H = C 5. Assume first that H is a tree. If G H 6, then T(G) 1 + n 5 = n. If G H = P, then T(G) =,n = 8, and the theorem holds. Otherwise, G H = C 5, T(G), n = 10, and the theorem holds. Suppose H = C 5 = (u 1,...,u 5 ), and without loss of generality, assume that u 1 has a neighbor on P = (x,y,z). If G H = P, then n = 8 and for T 1 = G[{u,u,u 4,u 5 }] and T = G[{x,y,z,u 1 }], T = {T 1,T } is a tree coer of size. Otherwise, for path P = (u,u,u 4,u 5 ), G P is a connected graph on at least 6 ertices, so T(G) 1+ n 4 = n 1. We may now assume that each component of G P is K or has at least 6 ertices. If all components of G P are of order at least 6, then by the induction hypothesis, T(G) 1+ n = n. Suppose G P has exactly one component that is K = (u,). Since G has no leaes then each of u and must be adjacent to a ertex of P, and since G has no three or four cycles, then u must be adjacent to x and must be adjacent to z. Furthermore, since n 7, then G P must hae a component H with at least 6 ertices. Note that H has a ertex that is adjacent to some r {x,y,z}. By adding r to H, we partition G into a tree (namely, the tree with ertex set {x,y,z,u,}\{r}) and connected components of order at least 6. Thus, T(G) 1+ n 4 = n 1. Suppose G P has s components that are K. We first show that the ertices of P = (x,y,z) and the ertices of each K can be coered with two trees: recall that since G has no leaes then each endpoint of a K must be adjacent to a ertex of P, and since G has no three or four cycles, then for each K, one endpoint must be adjacent to x and 8

9 the other end must be adjacent to z. Let X be the set of endpoints that are adjacent to x and let Z be the set of of endpoints that are adjacent to z. Then for T 1 = G[X {x}] and T = G[Z {z,y}], T = {T 1,T } is a tree coer of size two that coers the ertices of P and the ertices of G P belonging to a K. We apply the induction hypothesis to each component of G P with at least 6 ertices to get that T(G) + n s n 1. x y z x y z Figure 4: Graphs in Proof of Theorem 17 Corollary 18. If G is a connected outerplanar graph on n ertices with girth at least 5, then M + (G) n. Triangle-free graphs are a family of widely studied graphs, so an interesting question is whether or not the bound gien in Corollary 18 holds when girth is at least 4. The cycle on four ertices demonstrates that the bound no longer holds. Howeer, computations in Sage suggest the next conjecture. Conjecture 19. For all connected triangle-free graphs, T(G) n. 4 M + and T for connected outerplanar graphs We now turn our attention specifically to connected outerplanar graphs on n. We hae seen that M + (G) = T(G) n, and in this section we characterize graphs that achiee this upper bound. Let F denote the block-clique graphs such that each clique is K (see Figure 5), and obsere that eery graph in F has an odd number of ertices. We begin by stating the results that proide this characterization. Figure 5: A block-clique graph such that each clique is K. 9

10 Theorem 0. Let G be a connected outerplanar graph of odd order n. Then T(G) = n if and only if G F. Corollary 1. Let G be a connected outerplanar graph of odd order n. Then M + (G) = n if and only if G F. Obsere that the only connected graph of order n =, K, has tree coer number n = 1. Theorem. For a connected outerplanar graph G = (V,E) of een order n 4, T(G) = n if and only if one of the following holds: (1) G is obtained from some G F by adding one leaf. () G is obtained from some G 1,G F by connecting them with a bridge. () G is constructed from the following iteratie process: Start with G [0] {C 4,K 4 e,c r ( for some r )}. For i 1, pick a V(G [i 1] ) and let G [i] = G [i 1] K. Corollary. Let G = (V,E) be a connected outerplanar graph of een order n. Then M + (G) = n if and only if one of (1), (), () of Theorem holds. For a block-clique graph where each clique is a cycle, note that any two cycles share at most one common ertex. Two blocks are said to be adjacent if they hae one common ertex. A pendant block is a block that is adjacent to exactly one other block. Lemma 4. [1] Any block-clique graph where each clique is a cycle has at least two pendant blocks. To proe Theorem 0, we use the next two lemmas. Lemma 5. If G is a connected graph with n ertices and T(G) = n, then there exist adjacent ertices u, V(G) such that G = G[V(G) \ {u,}] remains connected. Furthermore, T(G ) = n. Proof. ByLemma6, wemay remoeaninducedsubgraphh = K 1,p suchthat G H remains connected. First note that if p, then T(G) 1+ n 4 < n, which is a contradiction, so p. If p = 1, then we are done. Suppose p = (i.e., H is a path (x,y,z)). If x and z are both leaes in G, then T(G) = T(G {x,z}) n, which is a contradiction to T(G) = n. So without loss of generality, x has a neighbor in G H, and the theorem holds with u = y and = z. It is easy to see that if T(G ) < n, then T(G) < n. So, T(G ) = n. Lemma 6. Let G = (V,E) be a connected graph and suppose u, V are adjacent ertices such that G = G[V \{u,}] is connected. Let T be a minimum tree coer of G, and suppose there exists w V(G ) such that (1.) V(T w ) = {w,x} (.) y N(w) N(x) such that N(x) V(T y ) = {y}. If u is adjacent to w and is not adjacent to w, then T(G) T(G ). 10

11 Proof. For T = (T \{T w T y }) G[{u,,w}] G[V(T y ) {x}],t is a tree coer of G of size T(G ). Proof of Theorem 0. Let G be a graph on n = k+1 ertices and first supposet(g) = n. We proe that G F by induction on k. If k = 1, then G = K. Let n = k + 1 where k and suppose that the claim holds for graphs with (k 1) +1 ertices. By Lemma 5, we can delete an edge H (including the endpoints) such that G H is connected. Note that since T(G) = n (see the next figure)., then T(G H) = n u. By the induction hypothesis, G H F G Furthermore, by usinglemma 5, G H has a minimum tree coer, T, such that one tree has exactly one ertex and the remainingtrees hae exactly two ertices. Let V(H) = {u,}. We show that G F by showing 1) if u is adjacent to a ertex w V(G H), then must alsobeadjacent towand)u(andtherefore) isadjacent toexactly oneertex inv(g H). To see 1), suppose u is adjacent to w V(G H) and is not adjacent to w. If V(T w ) = {w}, then T = (T \T w ) G[{w,,u}] is a tree coer of G of size n, which is a contradiction. Otherwise, T w = P = (w,x) for some x V(G H). Since each edge of G H belongs to a triangle, then there exists y V(G H) such that y N(w) N(x). Since any two triangles in G H share at most one ertex, it follows from Lemma 6 that T(G) n. (See the next figure.) So must be adjacent to w. u u w x w x y y G G 11

12 To see ), suppose that and u were adjacent to x,y V(G H). We show that G is not outerplanar. Since G H is connected, then there is a path, P, in G H with endpoints x to y. Then G[V(P) {,u}] has a K 4 minor, which contradicts G being outerplanar. We show theconerse by inductionon k. Let G F. For k = 1, G = K, so T(G) = n. For k, by Lemma 4, G has a pendant block so G = G K for some V, where G is a graph on n ertices and G F. It follows from the induction hypothesis and Proposition that T(G) = T(G )+T(K ) 1 = n +1 = n. To proe Theorem, we use an additional lemma. Lemma 7. Let G = (V,E) be a connected graph of een order n with T(G) = n that satisfies the following conditions: (a) G does not hae a bridge. (b) δ(g), and if z V is a ertex such that N(z) = {z,z }, then z and z are adjacent. Let u, V be adjacent ertices in G such that G = G[V \{u,}] remains connected and T(G ) = n. If G does not hae a leaf, then one of the following holds: 1. G satisfies (a) and (b).. G satisfies () of Theorem.. G satisfies () of Theorem. Proof. Assume G has no leaes. Suppose first that e = {g 1,g } is a bridge in G (i.e., G does not satisfy (a)) and let G 1,G betheconnected components of G e, whereg 1 V(G 1 ) and g V(G ). We show that G satisfies (). By hypothesis, T(G ) = n and by Lemma 4, T(G ) = T(G 1 ) +T(G ) 1. It follows that G i is odd and T(G i ) = Gi for i = 1,. Note that since G has no leaes, G i for i = 1,, and by Theorem 0, G i F. Let W be the set of ertices in V(G ) \{g 1,g } that are adjacent to either u or. We first show that for each w W, w is adjacent to both u and. Without loss of generality, let u hae a neighbor w in W, suppose is not adjacent to w, and suppose that w V(G 1 ). Let T be a tree coer of G such that each tree has exactly two ertices (T is guaranteed by Lemma 5) and let T w = G[{w,x}] be the tree containing w. Note that x V(G 1 ) since w g 1. Since G 1 F, w and x hae a common neighbor y. Let V(T y ) = {y,y }, and note that y / N(x) (if y V(G 1 ) then this follows from the fact that G 1 F, and if y V(G ), then this follows from the fact that x has no neighbor in V(G )). By Lemma 6, T(G) T(G ) = n, which contradicts T(G) = n. Thus, is adjacent to w, and this shows that u and hae the same set of neighbors in V(G )\{g 1,g }. Furthermore, since G is outerplanar, it follows that W 1. If W =, then G[{u,,g 1,g }] is K 4 e (since u and are not leaes, G is outerplanar, e is not a bridge in G, and the neighbors of a degree two ertex in G must be adjacent). Since G 1,G F, it follows that G satisfies () with G [0] = K 4 e. Consider W = 1, let W = {w}, and without loss of generality, suppose w V(G 1 ). Since e is not a bridge in G, then u or must be adjacent to a ertex in V(G ), and since 1

13 W = 1, this ertex must be g. Without loss of generality, suppose u is adjacent to g, and note that cannot also be adjacent to g since G is outerplanar. Suppose first that N(u) (V(T g )\{g }) =. If N() (V(T w )\{w}) =, then (T \(T w T g )) G[V(T w ) {}] G[V(T g ) {u}] is a tree coer of G of size n, which contradicts T(G) = n. Thus, N() (V(T w )\{w}), so it must be the case that V(T w ) = {w,g 1 } and is adjacent to g 1. But then G[{u,,w,g 1,g }] has a K 4 minor (see the next figure), which is a contradiction to G being outerplanar. u w g 1 g So, N(u) (V(T g )\{g }), and it must be the case that V(T g ) = {g 1,g } and u is adjacent to g 1. Note that since G is outerplanar, is not adjacent to g 1 nor g. It follows that G satisfies () with G [0] = C r, where C r = (u,w,x 1,...,x j,g 1 ) and (w,x 1,...,x j,g 1 ) is the path between w and g 1 in G 1 (see the next figure). g u w x x x x 1 j g 1 Suppose now that G does not satisfy (b). We show that G satisfies (). Since G does not satisfy (b), then there exists a ertex z V(G ) of degree whose neighbors z and z are not adjacent. By contracting the edge {z,z }, we obtain a graph H from G on n ertices with T(H) = n. So, H F. Then for the triangle (z,z,y) in H, it follows that (z,z,z,y) is a 4 cycle in G, and G satisfies () of Theorem with G [0] = C 4. Proof of Theorem. Let G be a graph on n = k ertices and first suppose T(G) = n. If G has a leaf, then T(G) = T(G ), and G is in F by Theorem 0. Thus (1) holds. If G has a bridge e and the connected components of G e are G 1 and G, then by Lemma 4, T(G) = T(G e) 1 = T(G 1 )+T(G ) 1. Note that G 1 and G must both be een or both are odd since n is een. If both are een then T(G) = T(G 1 )+T(G ) 1 G 1 + G 1 = n 1, which contradicts T(G) = n. Thus, G 1 and G are both odd, and G1 n = T(G) = T(G 1)+T(G ) 1 + G 1 = G G +1 1 = n. It follows that T(G i ) = for i = 1,, so () holds. Suppose G can be obtained from some graph Gi G by subdiiding an edge of G. Since subdiiding an edge does not change the tree coer number, then n = T(G) = T(G ) and by Theorem 0, G F. Note that subdiiding an edge of a graph in F results in a graph in () with G [0] = C 4, so G satisfies (). We may now assume that δ(g), G does not hae a bridge, and for each V with deg() =, the neighbors of are adjacent. For the remainder of the proof, u and 1

14 are the adjacent ertices from Lemma 5 such that G = G[V(G)\{u,}] is connected and T(G ) = n. We consider two cases, G has a leaf and G does not hae a leaf. Case 1. Suppose G has a leaf l and let l be its neighbor. We show G satisfies (). We first show that u and hae the same set of neighbors in V(G ) \ {l,l }. Note that T(G l) = T(G ) = n and by Theorem 0, G l is in F. Suppose u has a neighbor w in V(G )\{l,l } and is not adjacent to w. Let T be a tree coer of G such that each tree has exactly two ertices (T is gauranteed by Lemma 5) and let T w = {w,x} be the tree containing w. Since G l F, then w and x hae a common neighbor y such that V(T y ) = {y,z} T and z / N(x). By Lemma 6, T(G) n, contradicting T(G) = n. Thereforeuand hae thesameset ofneighborsinv(g )\{l,l }, andsincegisouterplanar, this set has cardinality at most one. Since G has no leaes, we may assume that u is adjacent to l. Suppose first that is also adjacent to l. Since {l,l } is not a bridge in G, then either u or must hae a neighbor in G l, and since G is outerplanar, u and cannot both hae a neighbor in G l (since if we contract each edge of G l to obtain a single ertex t, then the graph induced on {u,,l,t} would form a K 4 ). Without loss of generality, we let u hae a neighbor w V(G l). Since u and hae the same neighbors in V(G )\{l,l }, then w = l, G[{u,,l,l }] is K 4 e, and G satisfies () with G [0] = K 4 e. Assume is not adjacent to l. Since l has degree in G, by hypothesis u is adjacent to l, andsinceghasnoleaes, hasaneighborw V(G l). Ifw l, wehaealreadyseenthat u must also beadjacent to w and that N(u) (V(G)\{l,l }) = N() (V(G)\{l,l }) = {w}. Also note that if w l, then cannot also be adjacent to l since G is outerplanar, so N(u) = {,l,l,w} and N() = {u,w}. To see that G satisfies (), let (l,x 1,...,x j,w) be the shortest path from l to w in G (see the next figure). Since G l F, it follows that G satisfies () with G [0] = Cr and C r = (u,l,l,x 1,...,x j,w). u l x j x x x 1 w l G l F If is adjacent to l, then u and share the same set of neighbors in V(G )\{l}, and since G is outerplanar we must hae that N(u) = {,l,l },N() = {u,l } (see the next figure). Thus, G satisfies () with G [0] = K 4 e. 14

15 u l x j x x x 1 w l G l F Case : Suppose G does not hae a leaf. We proe this case by induction on n. Let n = 6. Then G is a graph on four ertices with tree coer number two. Since G does not hae a leaf, then G is K 4 e or C 4. Suppose first that G = C 4. If u has a neighbor w V(C 4 ) and is not adjacent to w, then for T 1 = G[{u,,w}] and T = G[V(C 4 )\{w}], {T 1,T } is a tree coer of G of size, contradiction T(G) =. So u and hae the same set of neighbors in V(C 4 ), and since G is outerplanar, u and hae exactly one neighbor in V(C 4 ) (see next figure), and () holds. u Consider G = K 4 e. It is well known that an outerplanar graph on n ertices has at most n edges (this can be proen by deleting a ertex of degree two and usinginduction on n). Thus G has at most nine edges. Since there are fie edges in K 4 e and one edge between u and, there are at most three edges between the sets {u,} and V(K 4 e), so either u or has degree two (since G has no leaes). Suppose N(u) = {,w} for some w V(K 4 e). By hypothesis, and w are adjacent. Note that since G has at most nine edges, can hae at most one additional neighbor. Suppose has an additional neighbor in V(K 4 e). Then G is one of the graphs gien in the next figure, and T(G) =, contradicting T(G) =. Thus, has no additional neighbors, and G satisfies () with G [0] = K 4 e. u u w w 15

16 Let n 8. Since G has no leaes, by Lemma 7 we either hae that G satisfies () (in which case the proof is complete), G has no bridge and is not a subdiision, or G satisfies (). Suppose that G has no bridge and is not a subdiision. We show that G satisfies (). Let G be the graph obtained from G after one more application of Lemma 5. If G has a leaf, G satisfies () by case 1. If G does not hae a leaf, then by the induction hypothesis G satisfies (). We now use the fact that G satisfies () to show that G satisfies () by showing that N(u) = {,w} and N() = {u,w}, for some w V(G ) (i.e., G = K w G ). Since u is not a leaf, let w V(G ) be a neighbor of u and suppose first that is not adjacent to w. We show that this contradicts T(G) = n. Let T be a minimum tree coer of G with each tree haing exactly two ertices and let V(T w ) = {w,x}. We consider two cases, there exists y N(w) N(x) and N(w) N(x) =. Let y N(w) N(x) and let V(T y ) = {y,z}. If x is not adjacent to z, by Lemma 6, T(G) n, so x is adjacent to z and G[{w,x,y,z}] is K 4 e. Since G is outerplanar and is not adjacent to w, then it can be seen by examination that G[{u,,w,x,y,z}] can be coered with two trees, contradicting T(G) = n. Consider N(w) N(x) =. Note that if G satisfies () with G [0] {K 4 e,cr }, then eery edge of G would belong to a triangle, so N(w) N(x) = implies that G satisfies () with G [0] = C 4. Furthermore, eery edge of G that is not an edge of C 4 belongs to a triangle, so {w,x} is an edge on C 4. Since is not adjacent to w, we may coer G[V(C 4 ) {u,}] with two trees, contradicting T(G) = n. So, must be adjacent to w, which shows that u and hae the same set of neighbors on G. Since G is outerplanar, u and must hae exactly one common neighbor in G, which shows that G satisfies (). We now show the conerse. The remoal of a leaf does not affect the tree coer number of a graph, so if G satisfies (1), then T(G) = n. If G satisfies (), then by Lemma 4, T(G) = T(G 1 ) + T(G ) 1 = G1 + G 1 = n. For a graph G satisfying (), n = T(G ) = T(G) since subdiiding does not affect tree coer number. Suppose G satisfies (). If G {C 4,K 4 e,cr }, then T(G) = n. Let G = G[k] for some k 1. By Proposition, and by induction, T(G) = T(G [k 1] )+T(K ) 1 = n. The next theorem gies a formula for computing T(G) (and therefore M + (G)) of cactus graphs. Theorem 8. Let G = (V,E) be a cactus graph on n ertices. Then M + (G) = T(G) = k+1, where k is the number of cycles in G. Proof. Let E E be the set of bridges of G. By Lemma 4, the deletion of a bridge increases the tree coer number by exactly one, so T(G) = T(G E ) E. Let H 1,...,H r be the connected components of G E. Obsere that H i does not hae a bridge for i = 1,...,r. We first show that for each i, T(H i ) = c i +1, where c i is the number of cycles in H i. Note that H i is either a single ertex, or it is a cactus graph where each block is a cycle (see Figure 6 for example). If H i is a single ertex, then T(H i ) = 1 = c i +1. Suppose H i is a cactus graph where each block is a cycle. Note that any two cycles in H i share at most one common 16

17 ertex. It follows from Proposition and induction on c i that T(H i ) = c i +1. Thus ( r ) ( r ( r T(G) = T(G E ) E = T(H i ) E = c i )+r E = c i )+1, i=1 where the last equality follows from the fact that r E = 1. Furthermore, since bridges ( are not edges of cycles, we hae that G and G E hae the same number of cycles (i.e., r ) c i = k), so T(G) = k +1. i=1 Example 9. For the graph G gien in Figure 6, T(G) = 6. i=1 i=1 y x y x z z Figure 6: Graph G and G E of Example 9 References [1] F. Barioli, W. Barrett, S. Butler, S. M. Cioabǎ, D. Cetkoi c, S. M. Fallat, C. Godsil, W. Heamers, L. Hogben, R. Mikkelson, S. Narayan, O. Pryporoa, I. Sciriha, W. So, D. Steanoi c, H. an der Holst, K. Vander Meulen, A. Wangsness Wehe. Zero forcing sets and the minimum rank of graphs. Lin. Alg. Appl., 48: , 008. [] F.Barioli, W.Barrett, S.Fallat, H.T.Hall, L.Hogben, B.Shader, P.an den Driessche, and H.an der Holst. Zero forcing parameters and minimum rank problems. Lin. Alg. Appl., 4: , 010. [] F. Barioli, S. Fallat, L. Mitchell, and S. Narayan. Minimum semidefinite rank of outerplanar graphs and the tree coer number, Elec. J. Lin. Alg., : 10 1, 011. [4] M. Booth, P. Hackney, B. Harris, C.R. Johnson, M. Lay, L.H. Mitchell, S.K. Narayan, A. Pascoe, K. Steinmetz, B.D. Sutton, and W. Wang, On the minimum rank among positie semidefinite matrices with a gien graph, SIAM J. Matrix Anal. Appl. 0: ,

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