Relating the Annihilation Number and the 2-Domination Number of a Tree 1Wyatt J. Desormeaux∗, 1Michael A. Henning∗, 2Douglas F. Rall and 1,3Anders Yeo† 1Department of Mathematics University of Johannesburg Auckland Park, 2006, South Africa Email: [email protected] Email: [email protected] 2Department of Mathematics Furman University Greenville, SC 29613 USA Email: [email protected] 3Engineering Systems and Design Singapore University of Technology and Design 20 Dover Drive Singapore, 138682, Singapore Email: [email protected] Abstract Aset S ofvertices inagraphGisa2-dominating setifevery vertex ofGnotinS is adjacentto atleast twovertices inS. The 2-domination number γ2(G)isthe minimum cardinality of a 2-dominating set in G. The annihilation number a(G) is the largest integer k such that the sumofthe first k terms ofthe nondecreasing degree sequence of G is at most the number of edges in G. The conjecture-generating computer program, Graffiti.pc, conjectured that γ2(G) ≤ a(G)+1 holds for every connected graph G. It is known that this conjecture is true when the minimum degree is at least 3. The conjecture remains unresolved for minimumdegree 1or 2. In this paper, we prove that theconjecture isindeedtruewhenGisatree,andwecharacterizethetreesthatachieve equality in the bound. It is known that if T is a tree on n vertices with n1 vertices of degree 1, then γ2(T) ≤ (n+n1)/2. As a consequence of our characterization, we also characterize trees T that achieve equality in this bound. ∗Researchsupported inpartbytheUniversityofJohannesburg andtheSouthAfricanNationalResearch Foundation. †Research supported inpart by the University of Johannesburg. 1 Keywords: 2-domination,2-dominationnumber, annihilationnumber AMS subject classification (2010): 05C69 1 Introduction In this paper, we study upper bounds on the 2-domination numbers of trees in terms of their annihilation numbers. For k ≥ 1, a k-dominating set of a graph G is a set S of vertices in G such that every vertex outside S is adjacent to at least k vertices in S. Every graph G has a k-dominating set, since V(G) is such a set. The k-domination number of G, denoted by γ (G), is the minimum cardinality of a k-dominating set of G. In particular, a k 1-dominatingsetisadominatingset,andthe1-dominationnumberγ (G)isthedomination 1 number γ(G). A k-dominating set of G of cardinality γ (G) is called a γ -set of G. The k k concept of a k-dominating set was first introduced by Fink and Jacobson in 1985 [6] and is now well-studied in the literature. We refer the reader to the two books on domination by Haynes, Hedetniemi, and Slater [9, 10], as well as to the excellent survey on k-domination in graphs by Chellali, Favaron, Hansberg, and Volkmann [2]. As explained in [11], the annihilation number of a graph was first introduced by Pepper in [13]. Originally it was defined in terms of a reduction process on the degree sequence similar to the Havel-Hakimi process (see [7, 14]). In [13], Pepper showed an equivalent way to define the annihilation number, this is the version we will use in this work. The annihilation number of a graph G, denoted a(G), is the largest integer k such that the sum of the first k terms of the degree sequence of G arranged in nondecreasing order is at most the number ofedges. That isifd ,...,d isthe degree sequence ofa graphGwithm edges, 1 n where d ≤ ··· ≤ d , then the annihilation number of G is the largest integer k such that 1 n k d ≤ m or, equivalently, the largest integer k such that k d ≤ n d . i=1 i i=1 i i=k+1 i PThe conjecture-generating computer program, Graffiti.pc, mPade the fPollowing conjecture relating the 2-domination number of a graph and its annihilation number. Conjecture1. ([5])IfGisaconnectedgraphwithatleast2vertices,thenγ (G)≤ a(G)+1. 2 ItisknownthatConjecture1istruewhentheminimumdegreeisatleast3. Conjecture1 is stillunresolved when the minimumdegree of G is 1 or 2. Provingthe conjecture for trees may be the most interesting case. Our aim in this paper is threefold: First to prove that Conjecture 1 is indeed true for trees. Secondly to characterize the extremal trees achieving equality in the upper bound of Conjecture 1. Thirdly to characterize trees with the largest possible 2-domination number. 1.1 Notation In this paper, the word “graph” is used to denote a “simple graph” with no loops or multipleedges. Fornotationandgraphtheoryterminologynotdefined herein,weingeneral 2 follow [9]. We write V(G) and E(G) for the vertex set and edge set of a graph G. Usually, we use n for the number of vertices and m for the number of edges. We write N (v) and G d (v) for the neighborhood and degree of a vertex v ∈ V(G). We extend the notion of G neighborhood to sets by letting N (S) = N(v) for any S ⊆ V(G). If the graph G is G v∈S clear from the context, we simply write N(v), N(S), and d(v) rather than N (v), N (S), S G G and d (v), respectively. The minimum degree among the vertices of G is denoted by δ(G). G The matching number is the maximum size of a matching in G and is denoted by α0(G). A vertex of degree 1 is called a leaf, its neighbor is a support vertex, and its incident edge is a pendant edge. We denote the set of leaves of a tree T by L(T). A star is a tree with at most one non-leaf vertex. The corona of a graph G, denoted G◦K , is formed from G by 1 adding for each v ∈ V(G), a new vertex v0 and the pendant edge vv0. For a set S ⊆ V(G), we let G[S] denote the subgraph induced by S. The graph obtained from G by deleting the vertices in S and all edges incident with vertices in S is denoted by G−S. In the special case when S = {v}, we also denote G−S by G−v for simplicity. For a set S ⊆ V(G) and v ∈ V(G), we denote by d (v) the number of all vertices in S S that are adjacent to v. In particular, when S = V(G), we note d (v) = d(v). For a subset S S ⊆ V(G), we define Σ(S,G)= d (v). G v∈S X For a graph G with m edges, we define an a-set of G to be a (not necessarily unique) set S of vertices in G such that |S| = a(G) and d (v) ≤ m. We define an a -set of G v∈S G min to be an a-set S of G, such that Σ(S,G)is a minimum. Hence if S is an a -set of G, then P min S is a set of (not necessarily unique) vertices corresponding to the first a(G) vertices in the nondecreasing degree sequence of G. In ordertoproveConjecture1fortrees,weintroduceaslightvariationoftheannihilation number of a graph. We define the upper annihilation number of a graph G, denoted a∗(G), to be the largest integer k such that the sum of the first k terms of the degree sequence of G arranged in nondecreasing order is at most |E(G)|+1. That is if d ,...,d is the degree 1 n sequence of a graph G with m edges, where d ≤ ··· ≤ d , then the upper annihilation 1 n number of G is the largest integer k such that k d ≤ m+1. We define an a∗ -set of i=1 i min G to be a (not necessarily unique) set S∗ of vertices in G such that |S∗| = a∗(G) and S∗ P corresponds to the first a∗(G) vertices in the nondecreasing degree sequence of G. 1.2 Known Results and Observations Intheirintroductorypaperonk-domination,FinkandJacobson[6]establishedthefollowing lower bound on the k-domination number of a tree. Theorem 1. ([6]) For k ≥ 1, if T is a tree with n vertices, then γ (T)≥ ((k−1)n+1)/k. k As a special case of Theorem 1, if T is a tree with n vertices, then γ (T) ≥ (n+1)/2. 2 The following upper bound on the 2-domination number of a tree was observed in several papers. 3 Theorem2. ([4,8,12])IfT is atreewith n verticesandn leaves,thenγ (T)≤ (n+n )/2. 1 2 1 Caro and Roditty [1] and Stracke and Volkmann [15] established the following upper bound on the k-domination number of a graph. Theorem 3. ([1, 15]) For every graph G with n vertices and every integer k ≥ 1, if δ(G)≥ 2k−1, then γ (G)≤ bn/2c. k In the special case when k = 2, the result of Theorem 3 states that if G is a graph with n vertices and δ(G)≥ 3, then γ (G)≤ bn/2c. Since α0(G)≤ bn/2c for any graph G with n 2 vertices, this result was improved in the following theorem. Theorem 4. ([3]) Let k be a positive integer. If G is any graph with δ(G) ≥ 2k−1, then γ (G)≤ α0(G). k We remark that both Theorems 2 and 3 follow from a more general result in Hansberg et al. [8]. Theorem 5. ([8]) If G is an r-partite graph with n vertices and k is a positive integer, then 1 γ (G)≤ ((r−1)n+|x∈ V(G):d (x)≤ k−1|). k G r Before we continue, a definition is in order. Definition 1. ([12]) A set of vertices is j-independent if each vertex of the set has at most j − 1 neighbors in the set. The j-independence number of G, denoted α (G), is j the cardinality of a largest j-independent set in G (when j = 1, this is just the standard independence number and a 1-independent set is a standard independent set). As can be seen in[12], Theorems2,3 and 5allfollowascorollariesof thefollowingresult. Theorem 6. ([12]) Given positive integers j, k, m, and n, let G be a graph with n vertices, and let H be the subgraph of G induced by the vertices having degree at least m. If m m = k+j −1, then γ (G)≤ n−α (H ). k j m Pepper was the first to observe that if G is an n-vertex graph with n ≥ 2, then a(G) ≥ bn/2c. This observationof Pepper’s as well as Theorem 3 when k = 2, lead to the following observation due to West. Observation 7. ([16]) Conjecture 1 is true if δ(G)≥ 3. We remark that Pepper constructed an infinite family F of graphs G for which δ(G)= 2 and γ (G) = a(G)+ 1, as follows. Let F be the family of graphs formed from r disjoint 2 copies of C in the followingmanner. Add to this graph a matchingwithr−1 edges, where 5 each edge ofthe matchingjoins two5-cycles and on each 5-cycle the vertices incidentto the matching are nonadjacent. When r = 3, an example of a graph in the familyF is shown in Figure 1. As observed by Pepper, if G ∈ F, then G has 6r−1 edges, with a(G) = 3r−1 and γ (G)= 3r. 2 4 Figure 1: A graph G∈ F. 2 The Family T For integers r and s such that r,s ≥ 1, a double star S(r,s) is a tree with exactly two verticesthatarenot leaves,oneof whichis adjacenttor leavesand the othertos leaves. In order to state our main result, we construct a family T of trees as follows. For any integer j greater than 1, let T denote the double star S(1,j−1). 2,j Definition 2. (The Family T) For integers i and j, where 2 ≤ i ≤ j, we construct the family T of trees defined recursively as follows. Initially we let T ∈ T . For every tree i,j 2,j 2,j T ∈T , do the following. i,j O : If v ∈ V(T) is a leaf in T, then add the set {t,s ,s ,...,s } of `+1 new vertices to 1 1 2 ` V(T), where ` ≥ i− 1 is arbitrary, and add the edge ts and the edges vs for all 1 i i= 1,2,...,` to E(T). Add the resulting tree to the family T . i,min{j,`+1} O : If v ∈ V(T) has d (v)≤ min{i,j−1}, then add the set {t,s ,s ,...,s } of `+1 new 2 T 1 2 ` vertices to V(T), where ` ≥ max{d (v)+1,i}−1 is arbitrary, and add the edge tv T and the edges s t for all i = 1,2,...,` to E(T). Add the resulting tree to the family i T . max{dT(v)+1,i},min{j,`+1} For an integer i≥ 2, let T = T and let T = {K }∪ T . i i,j 2 i   j≥i i≥2 [ [   Figure 2 illustrates the operations O and O on the tree T . 1 2 2,5 We remark that for 2 ≤ i ≤ j, the family T is defined as a set of trees. Further, since i,j it is the union of a set of trees, T is also a set of trees, as is the family T. In [16] it is i mentioned that Pepper found an infinite familyoftrees forwhich the 2-dominationnumber equals the upper bound of Conjecture 1. This is the family of trees formed by taking any tree, forming the corona of that tree and then forming the corona of the resulting tree. We remark that Pepper’s family of trees is contained in our family T constructed above. 3 Main Results Observation 7 states that Conjecture 1 is true when the minimum degree is at least 3. However, as remarked earlier Conjecture 1 remains unresolved when δ(G) ∈ {1,2}. We 5 s 3 t v s 2 s 1 s 3 v s 2 t s 1 (a) T (b) Operation O (c) Operation O 2,5 2 1 Figure 2: Operations O and O applied to the tree T . 1 2 2,5 prove that Conjecture 1 is true for trees and we characterize the trees achieving equality in Conjecture 1. We shall prove the following results. A proof of Theorem 8 is given in Section 5. Theorem 8. For a tree T, the following hold. (a) γ (T)≤ a∗(T). 2 (b) γ (T)≤ a(T)+1. 2 (c) γ (T)= a(T)+1 if and only if T ∈ T. 2 Recall that by Theorem 2, if T is a tree with n vertices and n leaves, then γ (T) ≤ 1 2 (n+n )/2. As a consequence of our main result, namely Theorem 8, we characterize the 1 trees achieving equality in this bound. Recall that T = T , where the union is taken 2 2,j over all integers j with j ≥ 2. A proof of Theorem 9 is given in Section 6. S Theorem 9. If T is a tree with n vertices and n leaves, then γ (T) ≤ (n+n )/2, with 1 2 1 equality if and only if T ∈ T ∪{K }. 2 2 4 Preliminary Results In this section, we establish some preliminary results that we will need when proving our main results. Recall that for a subset S ⊆ V(G), we define Σ(S,G) = d (v). We v∈S G begin with the following trivial observation. P Observation 10. For an a∗ -set S∗ in a tree T with m edges, the following hold. min (a) If Σ(S∗,T)≤ m, then a∗(T)= a(T). (b) If Σ(S∗,T)= m+1, then a∗(T)= a(T)+1. Proof. For an a∗ -set S∗ in a tree T with m edges, |S∗| = a∗(T) and Σ(S∗,T) ≤ m+1. min By definition, a∗(T) ≥ a(T). If Σ(S∗,T) ≤ m, then a(T) ≥ |S∗| = a∗(T), implying that a∗(T)= a(T),which establishes Part (a). If Σ(S∗,T)= m+1, then let v ∈ S∗ be arbitrary and note that Σ(S∗ \ {v},T) ≤ m, and so a(T) ≥ |S∗ \ {v}| = |S∗| − 1 = a∗(T) − 1. Since S∗ is a set of vertices corresponding to the first a∗(T) vertices in the nondecreasing 6 degree sequence of T and Σ(S∗,T) = m+ 1, we have a(T) ≤ a∗(T)− 1. Consequently, a∗(T)= a(T)+1. 2 As aconsequence of Observation10,if T isatree then a∗(T)= a(T)ora∗(T)= a(T)+1. We next establish some useful properties of trees that belong to the family T \{K }. We 2 shall need the following notation. For a set S ⊆ V(G), we define ∆ (G)= max{d (v)|v ∈ S} S G and δ (G)= min{d (v)|v ∈V(G)\S} S G Thus, ∆ (G) is the maximumdegree in G amongall the vertices in S, while δ (G) is the S S minimum degree in G among all the vertices that do not belong to S. Lemma 11. Let T0 ∈ T \{K }, and so T0 ∈ T for some integers j ≥ i ≥ 2. If S0 is an 2 i,j a -set in T0 and S∗0 is an a∗ -set in T0, then the following hold. min min (a) δ (T0) = i and δ (T0)= j. S0 S∗0 (b) ∆ (T0)≤ i and ∆ (T0)= i. S0 S∗0 (c) Σ(S∗0,T0) = m(T0)+1. (d) a∗(T0) = a(T0)+1. (e) γ (T0)= a∗(T0). 2 Proof. We proceed by induction on the minimum number k of operations needed to construct a tree T0 in the family T0 ∈ T \{K }. When k = 0, we have T0 = T ∈ T 2 2,j 2,j for some j ≥ 2. Let u and v denote the two vertices of T0 that are not leaves, where u has one leaf neighbor and v has j −1 leaf neighbors. Now γ (T0) = j +1, a(T0) = j and 2 a∗(T0) = j +1. In particular, a∗(T0) = a(T0)+1 and γ (T0) = a∗(T0). The set S0 = L(T0) 2 of leaves of T0 is the unique a -set in T0, implying that d (u) = δ (T0) = 2 = i and min T0 S0 ∆ (T0) = 1 < i. Furthermore, the set L(T0) ∪ {u} is the unique a∗ -set in T0, unless S0 min j = 2, in which case L(T0)∪{v} is also an a∗ -set in T0. Hence if S∗0 is an a∗ -set in T0, min min then d (v)= δ (T0) = j, d (u)= ∆ (T0)= 2 = i, and Σ(S∗0,T0)= m(T0)+1= j+2. T0 S∗0 T0 S∗0 Hence the tree T0 satisfies properties (a)–(e). This establishes the base case when k = 0. Suppose that k ≥ 0, and let T ∈ T \ {K } be constructed using k operations. Thus, 2 T ∈ T for some integers j ≥ i ≥ 2. For the inductive hypothesis, assume that the tree i,j T satisfies properties (a)–(e). Let T0 be obtained from T by using Operation O or O 1 2 in Definition 2, and so T0 is constructed using k+1 operations. We show that properties (a)–(e) also hold for the tree T0, which will complete the proof by induction. We consider two possibilities. Case 1. T0 is obtained from T by applying Operation O . We adopt the notation used 1 in Definition 2. Let L = {s ,...,s }∪{t}. Let S∗ be a a∗ -set in T. Let z be a vertex T0 2 ` min in S∗ of highest possible degree in T, and so dT(z) = ∆S∗(T). By Property (b), we have d (z)= i ≥ 2,andbyProperty(c)wededuce thatthesetS∗\{z}isan a -set inT. Since T min 7 v is a leaf in T, z ∈ S∗, and d (z) > 1, we note v ∈ S∗. Let S∗0 = (S∗∪L ∪{s })\{v}. T T0 1 Applying Properties (a)–(e) for the tree T, we have Σ(S∗0,T0) = Σ(S∗\{v},T)+`+2 = Σ(S∗,T)+`+1 = (m(T)+1)+`+1 = m(T0)+1. By Property (b), all vertices in S∗ have degree at most i in T. Since the degrees of vertices in S∗\{v} remain unchanged in T0 and since i≥ 2, all vertices in S∗0 have degree at most i in T0. Recall that z ∈ S∗ and d (z) = i. Since v 6= z, we have z ∈ S∗0 and T d (z)= i, implying ∆ (T0)= i. T0 S∗0 By Property (a), δ (T)= j. Hence since V(T0)\S∗0 = (V(T)\S∗)∪{v}, we have S∗ δS∗0(T0)= min{δS∗(T),dT0(v)}= min{j,`+1}. Since T0 ∈ T , it follows that the second statement of property (a) is satisfied. i,min{j,`+1} Inparticular,since`+1 ≥ iandj ≥ i,everyvertexinT0thatdoesnotbelongtothesetS∗0 has degree at least i in T0. As observed earlier,Σ(S∗0,T0)= m(T0)+1. Furthermore, every vertex in S∗0 has degree at most i in T0, and there exists a vertex in S∗0 of degree exactly i inT0. Therefore,thesetS∗0 isana∗ -setinT0,and soa∗(T0)= |S∗0|= |S∗|+` = a∗(T)+`. min As shown earlier, ∆ (T0)= i and δ (T0)= min{j,`+1}. Since S∗0 is defined as a set of S∗0 S∗0 verticescorrespondingtothefirsta∗(T0)verticesinthenondecreasingdegreesequenceofT0, and Σ(S∗0,T0)= m(T0)+1, we have a(T0)≤ a∗(T0)−1. Consequently, by Observation 10, a∗(T0) = a(T0)+1. By our properties of the set S∗0, it follows that if S0 is an a -set in min T0, then δ (T0)= i, so ∆ (T0)≤ i. S0 S0 It remainstoshowthatγ (T0) = a∗(T0). LetD be aγ -setofT0. SincethesetD contains 2 2 allleavesofT0,wehaveL ⊂ D. Ifs ∈ D,thenwecanreplaces inDbyv. Hencewemay T0 1 1 assume that s ∈/ D. In order to 2-dominate s , we have v ∈ D. Therefore the set D\L 1 1 T0 is a 2-dominating set in T, so γ (T)≤ |D|−|L |= γ (T0)−`. Conversely, noting that v, 2 T0 2 whichisa leafin T,belongs toevery2-dominatingsetof T,wehavethataddingthe setL T0 to an arbitrary γ -set of T produces a 2-dominating set of T0, and so γ (T0) ≤ γ (T)+`. 2 2 2 Consequently,γ (T0) = γ (T)+`. SinceT satisfiesProperty(e),wehaveγ (T)= a∗(T). As 2 2 2 established earlier, a∗(T0)= a∗(T)+`. Therefore, γ (T0)= γ (T)+` = a∗(T)+` = a∗(T0). 2 2 Hence in Case 1, Properties (a)–(e) hold for the tree T0. Case 2. T0 is obtained from T by applying Operation O . We once again adopt the 2 notationusedinDefinition2. InthiscaseletL = {s ,s ,...,s }. Bydefinition,T0 ∈ T , T0 1 2 ` i0,j0 where i0 = max{d (v)+1,i} and j0 = min{j,`+1}. The definition of ` yields ` ≥ i0−1. T Let S∗ be a a∗ -set in T. By the inductive hypothesis applied to the tree T, we have min Σ(S∗,T) = m(T)+1 and ∆ (T) = i. This implies that S∗ contains at least one vertex S∗ of degree i. By the choice of v, we have d (v) ≤ i. If v ∈/ S∗, then d (v) = i and we can T T 8 simply replace a vertex of S∗ of degree i in T with v. Hence we may choose S∗ so that v ∈ S∗. Let S∗0 = S∗∪L . Since v ∈ S∗, we have T0 Σ(S∗0,T0) = (Σ(S∗,T)+1)+` = (m(T)+1)+`+1 = m(T0)+1. We next consider the degrees of the vertices of S∗0 in the tree T0. If x ∈ L , then T0 d (x) = 1. If x ∈ S∗, then either x = v, in which case d (v) = d (v)+1, or x 6= v, in T0 T0 T which case d (x) = d (x) ≤ ∆ (T) = i. Hence, ∆ (T0) = max{d (v)+1,∆ (T)} = T0 T S∗ S∗0 T S∗ max{d (v)+ 1,i}; that is, ∆ (T0) = i0. Thus all vertices in S∗0 have degree at most i0 T S∗0 in T0. By Property (a), δ (T)= j. Hence since V(T0)\S∗0 = (V(T)\S∗)∪{t}, we have S∗ δS∗0(T0)= min{δS∗(T),dT0(t)} = min{j,`+1} = j0. We also note that by the choice of v, we have j ≥ d (v)+ 1. Further, j ≥ i, and so T j ≥ max{d (v)+1,i}= i0. Moreover, `+1≥ i0. Therefore, T j0 = min{j,`+1}≥ i0. Hence all vertices in T0 that do not belong to the set S∗0 have degree at least i0 in T0. As observed earlier, all vertices in S∗0 have degree at most i0 in T0, and there exists a vertex in S∗0 of degree exactly i0 in T0. Thus since Σ(S∗0,T0) = m(T0)+1, the set S∗0 is an a∗ -set min in T0, and so a∗(T0) = |S∗0|= |S∗|+` = a∗(T)+`. As shown earlier, ∆ (T0) = i0 and δ (T0)= j0. Since S∗0 is defined as a set of vertices S∗0 S∗0 corresponding to the first a∗(T0) vertices in the nondecreasing degree sequence of T0 and Σ(S∗0,T0) = m(T0) + 1, we have a(T0) ≤ a∗(T0) − 1. Consequently, by Observation 10, a∗(T0) = a(T0)+1, thus proving property (d). Let S0 be an a -set in T0. Since a∗(T0) = min a(T0)+ 1 and ∆ (T0) = i0, it follows that S∗0 has exactly one more vertex of degree i0 S∗0 than S0 does. Hence δ (T0) = i0, implying that ∆ (T0)≤ i0. S0 S0 Itremainstoshowthatγ (T0)= a∗(T0). LetD beaγ -setinT0. SincethesetD contains 2 2 all leaves of T0, we have L ⊂ D. Note that t is not a leaf in T0 and N (t) = L ∪{v}. T0 T0 T0 If t ∈ D, then we can replace t in D by v. Hence we may assume t ∈/ D. Therefore the set D\L is a 2-dominating set in T, so γ (T) ≤ |D|−|L | = γ (T0)− `. Conversely, T0 2 T0 2 adding the set L to an arbitrary γ -set of T produces a 2-dominating set of T0, and so T0 2 γ (T0) ≤ γ (T) + `. Consequently, γ (T0) = γ (T) + `. Since T satisfies Property (e), 2 2 2 2 we have γ (T) = a∗(T). As established earlier, a∗(T0) = a∗(T)+ `. Therefore γ (T0) = 2 2 γ (T)+` = a∗(T)+` = a∗(T0). Hence in Case 2, Properties (a)–(e) hold for the tree T0. 2 This completes the proof of Lemma 11. 2 9 5 Proof of Theorem 8 In this section, we present a proof of Theorem 8. Recall the statement of the theorem. Since our proof by induction involves constructing a tree T0 from a tree T ∈ T we state Theorem 8 in terms of T0. Theorem 8. For a tree T0, the following hold. (a) γ (T0) ≤ a∗(T0). 2 (b) γ (T0)≤ a(T0)+1. 2 (c) γ (T0)= a(T0)+1 if and only if T0 ∈ T. 2 Proof. We proceed by induction on n, the number of vertices of the tree T0. When n = 1, we have γ (T0) = 1 = a∗(T0) = a(T0) and T0 6∈ T. When n = 2, we have 2 γ (T0) = 2 = a∗(T0) = a(T0)+ 1 and T0 ∈ T. This establishes the base cases. For the 2 induction hypothesis, consider n ≥ 3and assumethat every tree T with lessthan n vertices satisfies Properties (a), (b) and (c) in the statement of the theorem. Let T0 be a tree with n vertices. If T0 is a star K with s ≥ 2, then γ (T0) = s = a∗(T0) = a(T0) and T0 6∈ T. Hence we 1,s 2 mayassumethatT0 isnotastar,forotherwisethetreeT0satisfiesthedesiredproperties(a), (b)and (c) and we aredone. Hence no vertex dominatesallotherverticesin the graph. Let r be any vertex in T0 and call r the root of T0. We now consider the tree rooted at r. (We note that r is not necessarily a leaf in T0.) Let x be a vertex in T0 at maximum distance fromr. SinceT0 isnotastar,xisnotaneighborofr. Thus,x 6∈ N (r). Byourchoiceofx, T0 the vertex x is a leaf in T0. Let y be the unique neighbor of x. Since no vertex dominates all other vertices in T0, we have y 6= r and d (y)≥ 2. Let z be the vertex adjacent to y on T0 the unique path from r to y in T0; z is the parent of y in T0 when T0 is rooted at r. Since x is a vertex at maximum distance from r, every neighbor of y other than z must be a leaf. We now consider the following two cases, which exhaust all possibilities. Case1: d (y)≥ 3. LetQdenotethesetofallleavesadjacenttoy,andsoQ = N(y)\{z}. T0 Let T = T0−(Q∪{y}) and note that |E(T0)|= |E(T)|+|Q|+1. Let S be a a∗ -set in T, min and so Σ(S,T)≤ |E(T)|+1. Letting S = S∪Q, we have 1 Σ(S ,T0) ≤ (Σ(S,T)+1)+|Q| 1 ≤ |E(T)|+1+|Q|+1 = |E(T0)|+1, so a∗(T0)≥ |S |= |S|+|Q|= a∗(T)+|Q|. Every 2-dominatingset in T can be extended to 1 a 2-dominating set in T0 by combining it with Q, so γ (T0)≤ γ (T)+|Q|. Since T satisfies 2 2 Properties (a), (b) and (c), we now have γ (T0) ≤ γ (T)+|Q| 2 2 ≤ a∗(T)+|Q| ≤ (a∗(T0)−|Q|)+|Q| = a∗(T0). 10