1 23
Bulletin of the Malaysian Mathematical Sciences Society ISSN 0126-6705
Bull. Malays. Math. Sci. Soc.
DOI 10.1007/s40840-019-00758-0
According to Their Total Domination Number
Majid Hajian, Michael A. Henning &
Nader Jafari Rad
1 23
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https://doi.org/10.1007/s40840-019-00758-0
A Classification of Cactus Graphs According to Their Total Domination Number
Majid Hajian1·Michael A. Henning2·Nader Jafari Rad3
Received: 21 November 2018 / Revised: 13 March 2019
© Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2019
Abstract
A setSof vertices in a graphGis a total dominating set ofGif every vertex inGis adjacent to some vertex inS. The total domination number,γt(G), is the minimum cardinality of a total dominating set ofG. A cactus is a connected graph in which every edge belongs to at most one cycle. Equivalently, a cactus is a connected graph in which every block is an edge or a cycle. LetGbe a connected graph of ordern ≥2 withk≥0 cycles andℓleaves. Recently, the authors have proved thatγt(G)≥ 12(n−ℓ+2)−k.
As a consequence of this bound,γt(G)= 12(n−ℓ+2+m)−k for some integer m≥0. In this paper, we characterize the class of cactus graphs achieving equality in this bound, thereby providing a classification of all cactus graphs according to their total domination number.
Keywords Total dominating sets·Total domination number·Cactus graphs Mathematics Subject Classification 05C69
Communicated by Xueliang Li.
Michael A. Henning: Research supported in part by the University of Johannesburg.
B
Michael A. Henning [email protected] Majid Hajian[email protected] Nader Jafari Rad
1 Department of Mathematics, Shahrood University of Technology, Shahrood, Iran 2 Department of Mathematics and Applied Mathematics, University of Johannesburg,
Johannesburg, South Africa
3 Department of Mathematics, Shahed University, Tehran, Iran
1 Introduction
Atotal dominating set, abbreviated TD-set, of a graphG with no isolated vertex is a setSof vertices such that every vertex inG is adjacent to a vertex inS. Thetotal domination number, denoted byγt(G), ofGis the minimum cardinality of a TD-set ofG. We call a TD-set of cardinalityγt(G)aγt-set ofG. For a recent book on total domination in graphs we refer the reader to [6]. Acactus is a connected graph in which every edge belongs to at most one cycle. Equivalently, a (non-trivial) cactus is a connected graph in which every block is an edge or a cycle. Our aim in this paper is to provide a characterization of all cactus graphs according to their total domination number.
For notation and graph theory terminology we generally follow [6]. Theorderof a graphG=(V(G),E(G))with vertex setV(G)and edge set E(G)is denoted by n(G)= |V(G)|and itssizebym(G)= |E(G)|. Two verticesvandwareneighborsin Gif they are adjacent; that is, ifvw∈E(G). Theopen neighborhoodof a vertexvin Gis the set of neighbors ofv, denotedNG(v), and theclosed neighborhoodofvis the setNG[v] =NG(v)∪ {v}. Thedegreeof a vertexvinGis denoteddG(v)= |NG(v)|.
For a setSof vertices in a graphG, the subgraph induced bySis denoted byG[S].
Further, the subgraph obtained from G by deleting all vertices in S and all edges incident with vertices inSis denoted byG−S. IfS = {v}, we simply denoteG− {v}
byG−v. Two verticesuandvin a graphGareconnectedif there exists a(u, v)-path inG. If every two vertices in Gare connected, then the graphGisconnected. The distancebetween two verticesuandvin a connected graphGis the minimum length of a(u, v)-path inG. Thediameter, diam(G), ofGis the maximum distance among pairs of vertices inG. AblockofGis a maximal connected subgraph ofGwhich has no cut-vertex of its own. Acycle edgeof a graphGis an edge that belongs to a cycle ofG.
Aleaf of a graphGis a vertex of degree 1 inG, while asupport vertex ofGis a vertex adjacent to a leaf. The set of all leaves ofGis denoted byL(G), and we let ℓ(G)= |L(G)|be the number of leaves inG. The set of all support vertices ofGby S(G). A treeT of ordern ≥2 is astarifn=2 orn≥3 andT contains exactly one vertex that is not leaf. Adouble staris a tree with exactly two (adjacent) vertices that are not leaves. Further, if one of these vertices is adjacent torleaves and the other to sleaves, then we denote the double star byS(r,s). We denote the path and cycle on nvertices byPnandCn, respectively.
Letvbe a vertex of a treeT. We call the vertexvabad leaf ofT ifvis a leaf and noγt-set ofT containsv. The vertexvis astable vertexof T ifvis not a support vertex andγt(T −v)≥γt(T). We remark that the total domination of a tree can be computed in linear time. In particular, to determine if a leaf of a tree is a bad leaf can be determined in linear time. The bad leaves of a tree can also be efficiently computed using results of Cockayne et al. [2].
Arooted tree T distinguishes one vertexrcalled theroot. For each vertexv =r ofT, theparentofvis the neighbor ofvon the unique(r, v)-path, while achildofv is any other neighbor ofv. The set of children ofvis denoted byC(v). Adescendant ofvis a vertexu =vsuch that the unique(r,u)-path containsv. In particular, every child ofv is a descendant ofv. Agrandchild and agreat grandchild ofv inT are
descendants ofvat distance 2 and 3 fromv, respectively. We letD(v)denote the set of descendants ofv, and we define D[v] = D(v)∪ {v}. Themaximal subtreeatvis the subtree ofT induced byD[v]and is denoted byTv.
We use the standard notation[k] = {1, . . . ,k}.
2 Main Results
Our aim in this paper is to provide a characterization of all cactus graphs according to their total domination number. More precisely, we shall prove the following result whereGkmis a family of graphs defined in Sect.3for each integerk≥0 andm≥0.
Theorem 1 Let m≥0be an integer. If G is a cactus graph of order n≥2with k≥0 cycles andℓleaves, thenγt(G)= 12(n−ℓ+2+m)−k if and only if G∈Gkm.
We proceed as follows. In Sect.3we define the families Gkm of graphs for each integerk≥0 andm≥0. Known results on the total domination number are given in Sect.4. In Sect.5we present a proof of our main result.
3 The FamiliesGm
k
In this section, we define the familiesGm
k of graphs for each integerk≥0 andm≥0.
3.1 The FamilyG0
k
The familyG0
0of trees was defined by Chellali and Haynes [1] as follows. LetG0
0be the class of treesT that can be obtained from a sequenceT1, . . . ,Tℓ of trees, where ℓ≥1 and where the treeT1is the pathP4with support verticesxandy, and where the treeT =Tℓ. Further ifℓ≥2, then for eachi ∈ [ℓ], the treeTi can be obtained from the treeTi−1by applying one of the following three operationsO1,O2,O3defined below. For the initial treeT1(recall thatT1is a path P4with support verticesx and y), we define A(T1)= {x,y}, and fori ≥2, we define the setA(Ti)of vertices inTi
recursively according to the rules given below. Further, we defineH to be a pathP4 with support verticesuandv.
• OperationO1. Add a new vertex toTi−1and join it to a vertex ofA(Ti−1). Let A(Ti)=A(Ti−1).
• OperationO2. Add a vertex disjoint copy ofH toTi−1, and add an edge from a leaf ofHto a leaf ofTi−1. LetA(Ti)=A(Ti−1)∪ {u, v}.
• OperationO3. Add a vertex disjoint copy ofHtoTi−1, and add a new vertexwand an edge fromwto a support ofHand a leaf ofTi−1. LetA(Ti)=A(Ti−1)∪ {u, v}.
Fork ≥ 1, Hajian et al. [4] recursively define the familyG0
i of graphs for each i ∈ [k]by the following procedure.
• Procedure A: Fori ∈ [k], a graphGi belongs to the familyG0
i if it contains an edgee=x ysuch that the graphGi−ebelongs to the familyG0
i−1and the vertices xandyare leaves inGi−ethat are connected by a unique path inGi−e.
3.2 The FamilyG1
k
Hajian et al. [4] defined the family of treesG01as follows. LetT01,T02andT03be the class of trees defined as follows.
• LetT1
0 be the class of all treesT that can be obtained from a treeT′ ∈ G0
0 by addinga≥1 new vertices and joining all of them to the same leaf ofT′.
• LetT2
0 be the class of all treesT that can be obtained from a treeT′ ∈ G0
0 that contains a support vertexxall of whose neighbors, except for exactly one neighbor, are leaves inT′by removing all leaf-neighbors ofx.
• LetT3
0 be the class of all treesT that can be obtained from a treeT′ ∈ G0
0 by adding a vertex disjoint copy of a double starQand adding an edge from a leaf ofQto a vertex of degree at least 2 inT′.
LetG01be the class of treesT that can be obtained from a sequenceT1, . . . ,Tℓof trees, where ℓ ≥ 1 and where the tree T1 ∈ T01∪T02∪T03and the tree T = Tℓ. Further, ifℓ≥2, then for eachi ∈ [ℓ], the treeTi can be obtained from the treeTi−1 by applying operationO∗defined below.
• OperationO∗. Add a vertex disjoint copy of a double starQtoTi−1by adding an edge joining a leaf ofQand a leaf ofTi−1.
Fork≥1, Hajian et al. [4] defined the familyG1
i of graphs for eachi∈ [k]by the following two procedures.
• Procedure B: Fori ∈ [k], a graphGi belongs to the familyGi1if it contains an edgee=x ysuch that the graphGi−ebelongs to the familyG1
i−1and the vertices xandyare leaves inGi−ethat are connected by a unique path inGi−e.
• Procedure C: Fori ∈ [k], a graphGi belongs to the familyG1
i if it contains an edgee=x ysuch that the graphGi−ebelongs to the familyG0
i−1and the vertices xandyare connected by a unique path inGi−e. Further, exactly one ofxandy is a leaf inGi−e.
3.3 FamilyGm
0,m≥2 Recall that the familiesG0
0 andG1
0are defined in Sects.3.1and3.2, respectively. For m ≥2, we recursively define the familyGm
0 of graphs constructed from the families Gm−1
0 andGm−2
0 as follows.
• LetT0m,1be the class of all treesT that can be obtained from a treeT′∈G0m−1by addinga≥1 new vertices and joining all of them to precisely one bad leaf ofT′. Form=2, add the pathP2toT02,1, and for simplicity and uniformity, denote the resulting class byT02,1. (That is,T02,1∪ {P2}is denoted byT02,1.)
• LetTm,2
0 be the class of all treesT that can be obtained from a treeT′∈Gm−2
0 by
adding a vertex disjoint copy of a double starQand identifying a leaf ofQwith a stable vertex ofT′.
• LetTm,3
0 be the class of all treesT that can be obtained from a treeT′∈Gm−1
0 by
adding a vertex disjoint copy of a double starQand adding an edge from a leaf ofQto a vertex ofT′of degree at least 2.
LetGm
0 be the class of treesT that can be obtained from a sequenceT1, . . . ,Tℓof trees, whereℓ≥1 and where the treeT1∈Tm,1
0 ∪Tm,2
0 ∪Tm,3
0 and the treeT =Tℓ. Further, ifℓ≥2, then for eachi ∈ [ℓ]\{1}, the treeTi can be obtained from the tree Ti−1by applying operationO∗defined in Sect.3.2.
3.4 FamilyGm
k,m≥2 andk≥1
Form≥2 andk≥1, construct a familyGm
k fromGm−2
k−1,Gm−1
k−1 andGm
k−1, recursively, as follows.
• Procedure D: Fori ∈ [k], a graph Gi belongs to the familyGm
i if it contains an edgee=x ysuch that the graphGi −ebelongs to the familyGm−2
i−1 and the verticesxandyare non-leaves inGi−ethat are connected by a unique path in Gi −eandγt(Gi)=γt(Gi −e).
• Procedure E: Fori ∈ [k], a graphGi belongs to the family Gm
i if it contains an edgee=x ysuch that the graphGi −ebelongs to the familyGim−1−1 and the verticesxandyare connected by a unique path inGi−eandγt(Gi)=γt(Gi−e).
Further, exactly one ofxandyis a leaf inGi−e.
• Procedure F: Fori ∈ [k], a graphGi belongs to the familyGm
i if it contains an edgee=x ysuch that the graphGi−ebelongs to the familyGm
i−1and the vertices x and y are connected by a unique path inGi −e andγt(Gi) = γt(Gi −e).
Further, bothxandyare leaves inGi−e.
4 Known Results
In this section, we present some known results and observations. We begin with the following elementary properties of a total dominating set in a graphG.
Observation 1 The following hold in a graph G with no isolated vertex.
(a) Every TD-set in G contains the set of support vertices of G.
(b) If G is connected anddiam(G)≥3, then there exists aγt-set of G that contains no leaf of G.
Lower and upper bounds on the total domination number of a graph are well studied in the literature. A detailed discussion of such bounds can be found in the 2013 book [6]
on total domination in graphs, and in particular in [8]. For subsequent recent papers on bounds on the total domination number we refer the reader to [3–5,9].
A discussion of lower and upper bounds on the total domination number of a tree can be found in [7]. Chellali and Haynes [1] were the first to establish a lower bound on the total domination number of a tree in terms of the order, number of leaves, and number of support vertices in the tree.
Theorem 2 [1]If T is a tree of order n≥2withℓleaves, thenγt(T)≥(n−ℓ+2)/2, with equality if and only if T ∈G0
0.
The authors [4] have recently generalized the Chellali–Haynes result to connected graphs.
Theorem 3 [4]If G is a connected graph of order n ≥ 2with k ≥ 0cycles andℓ leaves, then the following holds.
(a) γt(G)≥ 21(n−ℓ+2)−k, with equality if and only if G∈G0
k. (b) γt(G)= 12(n−ℓ+3)−k if and only if G∈G1
k.
As an immediate consequence of Theorem3, we have the following result.
Corollary 4 [4]If G is a connected graph of order n ≥ 2with k ≥ 0cycles andℓ leaves, thenγt(G)= 12(n−ℓ+2+m)−k for some integer m≥0.
5 Proof of Main Result
In this section, we present a proof of our main result, namely Theorem 1. For this purpose, we first prove Theorem1in the special case whenk =0, that is, when the cactus is a tree.
Theorem 5 Let m≥0be an integer. If T is a tree of order n≥2withℓleaves, then γt(T)=12(n−ℓ+2+m)if and only if T ∈Gm
0.
Proof LetTbe a tree of ordern ≥2 withℓleaves. We proceed by induction onm≥0, namelyfirst induction, to show thatγt(T)= 12(n−ℓ+2+m)if and only ifT ∈G0m. Ifm = 0 andm = 1, then the result follows by Theorem3(a) and Theorem3(b), respectively. This establishes the base step of the induction. Letm ≥ 2 and assume that ifm′is an integer where 0≤ m′ < mandT′is a tree of ordern′ ≥ 2 withℓ′ leaves, thenγt(T′)= 12(n−ℓ′+2+m′)if and only ifT′∈Gm′
0 . LetT be a tree of ordern ≥2 withℓleaves. We show thatγt(T)= 12(n−ℓ+2+m)if and only if T ∈Gm0.
( ⇒) Assume thatγt(T)=12(n−ℓ+2+m). We show thatT ∈G0m. IfT =P2, then by definition of the familyT2,1
0 , we haveT ∈ T2,1
0 ⊆ G2
0. In this case when T = P2, we note thatγt(T)=2= 12(n−ℓ+2+2), and som=2 andT ∈Gm
0. If T is a star andn≥3, then by the definition of the familyT02we haveT ∈T02⊆G01. Thus, by Theorem3(b),γt(T)= 21(n−ℓ+2+1), and som =1 andT ∈ Gm
0. If T is a double star, then by the definition of the familyG0
0, we haveT ∈G0
0. Thus, by Theorem3(a),γt(T)= 12(n−ℓ+2), and som =0 andT ∈ Gm
0. Hence, we may assume that diam(T)≥4, for otherwiseT ∈Gm
0, as desired. In particular,n≥5.
We now root the treeT at a vertexrat the end of a longest pathPinT. Letube a vertex at maximum distance fromr, and sodT(u,r)=diam(T). Necessarily,r and uare leaves. Letvbe the parent ofu, letwbe the parent ofv, letxbe the parent ofw, and letybe the parent ofx. Possibly,y=r. Sinceuis a vertex at maximum distance from the rootr, every child ofvis a leaf. By Observation1(b), there exists aγt-set, Dsay, ofT that contains no leaf ofT, implying that{v, w} ⊆D. LetdT(v)=t1. ⊓⊔ Claim 1 If the vertexwhas at least two neighbors in D, then T ∈Gm
0.
Proof Suppose that|NT(w)∩D| ≥2. As observed earlier,v∈ D. Letv′be a neighbor ofw, different fromv, that belong to the set D. We now consider the treeT′obtained
fromT by deleting all leaf-neighbors ofv. LetT′have ordern′, and letT′haveℓ′ leaves. We note thatn′ = n−(t1−1)andℓ′ = ℓ−(t1−1)+1 = ℓ−t1+2.
By Observation1(b), there exists aγt-set, D′ say, ofT′that contains no leaf ofT′. Since the vertexv is a leaf-neighbor ofwinT′, we note thatw ∈ D′andv /∈ D′. The set D′ can be extended to a TD-set of T by adding to it the vertex v, and so γt(T)≤ |D′| +1=γt(T′)+1. Conversely, since the setD\{v}is a TD-set ofT′, we note thatγt(T′)≤ |D| −1=γt(T)−1. Consequently,γt(T′)=γt(T)−1. Thus,
γt(T′)=γt(T)−1
= 12(n−ℓ+2+m)−1
= 12(n−ℓ+m)
= 12((n′+t1−1)−(ℓ′+t1−2)+m)
= 12(n′−ℓ′+2+(m−1)).
Applying the inductive hypothesis to the treeT′, we haveT′∈G0m−1. If there is a γt-set ofT′that contains the leafv, then such a set is a TD-set ofT′, implying that γt(T)≤γt(T′), contradicting our earlier observation thatγt(T)=γt(T′)+1. Hence, the vertexvis a bad leaf ofT, implying thatT ∈Tm,1
0 ⊆Gm
0. ⊓⊔
By Claim1, we may assume that the vertexwhas exactly one neighbor inD, for otherwiseT ∈Gm
0 as desired. As observed earlier, the vertexvbelongs to D. Thus, NT(w)∩D = {v}. In particular,x ∈/ D. Recall that theγt-setDof T contains no leaf ofT. By Observation 1(a), the set Dcontains the set S(T)of support vertices ofT. Thus, ifdT(w)≥3, then every child ofwdifferent fromvis a leaf inT. Let dT(w)=t2. We note thatt2≥2. Recall thatxis the parent ofwin the rooted treeT. Claim 2 If dT(x)≥3, then T ∈Gm
0.
Proof Suppose that dT(x) ≥ 3. We now consider the tree T′ obtained fromT by deleting all vertices in the maximal subtree of T induced by D[w]; that is, T′ = T−V(Tw). Ift2=2, then the subtreeTwis a star withvas its central vertex. Ift2≥3, then by our earlier observations, the subtreeTwis a double star withvandwas the two vertices that are not leaves in the double star. LetT′have ordern′, and letT′haveℓ′ leaves. We note thatn′=n−(t1+t2−1)andℓ′=ℓ−(t1−1)−(t2−2)=ℓ−t1−t2+3.
We show thatγt(T′)=γt(T)−2. Everyγt-set ofT′can be extended to a TD-set ofT by adding to it the verticesvandw, implying thatγt(T)≤γt(T′)+2. We prove next the reverse inequality. By supposition,dT(x)≥ 3. By our earlier observations, x∈/ D. Thus, since the set Dcontains the setS(T)of support vertices ofT, we note thatxis not a support vertex ofT. Thus, no child ofxis a leaf. Letw′be an arbitrary child ofx different fromw. If every child ofw′is a leaf, then since Dcontains no leaf ofT this implies that bothw′andxbelong toD, a contradiction. Therefore, at least one child ofw′is not a leaf. Letv′be an arbitrary child ofw′that is not a leaf.
By maximality of the pathP, every child ofv′is a leaf. Thus, by Observation1, both v′andw′belong to the setD, implying that the setD\{v, w}is a TD-set ofT′. Thus, γt(T′)≤ |D| −2=γt(T)−2. Consequently,γt(T′)=γt(T)−2. Thus,
γt(T′)=γt(T)−2
=21(n−ℓ+2+m)−2
=12(n−ℓ+m−2)
=21((n′+t1+t2−1)−(ℓ′+t1+t2−3)+m−2)
=21(n′−ℓ′+2+(m−2)).
Applying the inductive hypothesis to the treeT′, we haveT′∈Gm−2
0 . Everyγt-set ofT′−xcan be extended to a TD-set ofTby adding to it the verticesvandw, implying thatγt(T′)+2=γt(T)≤ γt(T′−x)+ |{v, w}| =γt(T′−x)+2. Consequently, we must have equality throughout this inequality chain. Hence,γt(T′)=γt(T′−x).
Thus, the vertexxis a stable vertex ofT′. LetQbe obtained from the double starTw by adding to it a new vertexx′and the edgex′w. We note thatQis a double star with vandwas the two vertices that are not leaves inQ. SinceT can be obtained from the treeT′∈Gm−2
0 by adding a vertex disjoint copy of the double starQand identifying the leafx′ofQwith the stable vertexxofT′, the treeT belongs to the familyTm,2
0 .
Thus,T ∈Tm,2
0 ⊆Gm
0. ⊓⊔
By Claim2, we may assume thatdT(x)= 2, for otherwiseT ∈ Gm
0 as desired.
By our earlier observations, the subtreeTx is a double star withvandwas the two vertices ofTxthat are not leaves. We note thatTx ∈G0
0. Claim 3 Ifdiam(T)=4, then T ∈Gm
0.
Proof Suppose that diam(T)=4. Thus, the vertexyis the rootr of the treeT, and soT −r=Tx. Thus,T is obtained from the treeTx∈G00by adding the new vertexz and joining it with an edge to the leafxofTx. Hence,T ∈T1
0 ⊆G1
0. Thus,T ∈Gm
0
wherem=1. ⊓⊔
By Claim3, we may assume that diam(T)≥5, for otherwiseT ∈G0m as desired.
We now consider the treeT′obtained fromT by deleting all vertices in the maximal subtree of T induced by D[x]; that is,T′ = T −V(Tx). As observed earlier, the subtreeTxis a double star withvandwas the two vertices that are not leaves in the double star. Further,Tx ∈G0
0. LetT′have ordern′, and letT′haveℓ′leaves. We note thatn′=n−t1−t2.
We show thatγt(T′)=γt(T)−2. Everyγt-set ofT′can be extended to a TD-set ofT by adding to it the verticesvandw, implying thatγt(T)≤γt(T′)+2. We prove next the reverse inequality. By our earlier observations,x ∈/ D. Thus, the restriction of the setDto the treeT′ is a TD-set ofT′, implying thatγt(T′) ≤ |D\{v, w}| =
|D| −2=γt(T)−2. Consequently,γt(T′)=γt(T)−2.
Claim 4 If dT(y)=2, then T ∈Gm
0.
Proof Suppose thatdT(y) = 2. In this case,ℓ′ = ℓ−(t1−1)−(t2−2)+1 = ℓ−t1−t2+4. Thus,
γt(T′)=γt(T)−2
= 12(n−ℓ+2+m)−2
= 12(n−ℓ+m−2)
= 12((n′+t1+t2)−(ℓ′+t1+t2−4)+m−2)
= 12(n′−ℓ′+2+m).
Applying the inductive hypothesis to the treeT′, we haveT′∈ Gm
0. As observed earlier,Tx is a double star withvandwas the two vertices that are not leaves in the double star. Thus, the tree T can be obtained from the tree T′ by adding toT′ the double starTx and adding the edgex yjoining the leafxofTx and the leafyofT′. Hence,T can be obtained from the treeT′∈Gm
0 by applying operationO∗, implying
thatT ∈G0m. ⊓⊔
By Claim4, we may assume thatdT(y)≥ 3, for otherwiseT ∈ G0m as desired.
Recall thatT′=T−V(Tx). Thus,dT′(y)=dT(y)−1≥2, and soyis not a leaf of T′. In this case,ℓ′=ℓ−(t1−1)−(t2−2)=ℓ−t1−t2+3. Thus,
γt(T′)=γt(T)−2
= 12(n−ℓ+m−2)
= 12((n′+t1+t2)−(ℓ′+t1+t2−3)+m−2)
= 12(n′−ℓ′+2+(m−1)).
Applying the inductive hypothesis to the treeT′, we have T′ ∈ Gm−1
0 . The tree T can be obtained from the treeT′ ∈ G0m−1by adding toT′the double starTx and adding the edgex yjoining the leafxofTx and the vertexyof degree at least 2 inT′. Thus, the treeTbelongs to the familyT0m,3. Hence,T ∈T0m,3⊆G0m. This completes the necessity part of the proof of Theorem5.
(⇐ ) Conversely, assume thatT ∈Gm
0, wherem≥ 2 and where recall thatT is a tree of ordern ≥2 withℓleaves. As shown earlier, ifT is a star andn ≥ 3, then T ∈ G1
0, while ifT is a double star, thenT ∈ G0
0. In both cases we contradict the assumption thatT ∈Gm
0 wherem≥2. Hence,T is neither a star of ordern≥3 nor a double star. Thus, ifT = P2, then diam(T)≥4. By definition of the familyGm
0, the treeT is obtained from a sequenceT1, . . . ,Tq of trees, whereq ≥1 and where the treeT1∈T0m,1∪T0m,2∪T0m,3and the treeT =Tq. Further, ifq ≥2, then for each i ∈ [q]\{1}, the treeTi can be obtained from the treeTi−1by applying operationO∗ defined in Sect.3.2. We proceed by induction onq ≥1, namelysecond induction, to show thatγt(T)=12(n−ℓ+2+m).
Claim 5 If q=1, thenγt(T)= 12(n−ℓ+2+m).
Proof Suppose thatq =1. Thus,T =T1∈Tm,1
0 ∪Tm,2
0 ∪Tm,3
0 . We consider each
of the three possibilities in turn. ⊓⊔
Claim 5.1 If T ∈T0m,1, thenγt(T)=12(n−ℓ+2+m).
Proof Suppose thatT ∈ Tm,1
0 . IfT = P2, then as observed earlierT ∈ G2
0. In this case,m=2 andγt(T)=2= 12(n−ℓ+2+2)=21(n−ℓ+2+m). Hence, we may assume thatT =P2, for otherwise the desired result follows. Thus,n≥3 and the tree T is neither a star nor a double star; we note that diam(T)≥4. The treeT ∈T0m,1can be obtained from a treeT′ ∈Gm−1
0 by addinga ≥1 new vertices and joining all of them to precisely one bad leaf,xsay, ofT′. Thus, the vertexxdoes not belong to any γt-set ofT′. LetT′have ordern′, and letT′haveℓ′leaves. We note thatn′=n−a andℓ′=ℓ−a+1. Applying the first induction hypothesis to the treeT′∈Gm−10 , we haveγt(T′)= 12(n′−ℓ′+2+(m−1)).
We show next thatγt(T)=γt(T′)+1. Everyγt-set ofT′can be extended to a TD- set ofT by adding to it the vertexx, and soγt(T)≤γt(T′)+1. By Observation1(b), there exists aγt-set,Dsay, ofTthat contains no leaf ofT, implying thatx∈ Dand that no leaf ofxbelongs toD. Thus, the setDis a TD-set ofT′that contains the vertexx.
Since noγt-set ofT′contains the vertexx, we note thatγt(T)= |D| ≥γt(T′)+1.
Consequently,γt(T)=γt(T′)+1. Thus,
γt(T)=γt(T′)+1
= 12(n′−ℓ′+1+m)+1
= 12((n−a)−(ℓ−a+1)+1+m)+1
= 12(n−ℓ+2+m).
This completes the proof of Claim5.1. ⊓⊔
Claim 5.2 If T ∈Tm,2
0 , thenγt(T)=21(n−ℓ+2+m).
Proof Suppose thatT ∈T0m,2. Thus, the treeTcan be obtained from a treeT′∈G0m−2 by adding a vertex disjoint copy of a double starQand identifying a leaf ofQwith a stable vertex,xsay, ofT′. Letuandvbe the two vertices inQthat are not leaves, and soS(Q)= {u, v}. Sincexis a stable vertex ofT′, we note thatxis not a support vertex andγt(T′−x)≥γt(T′). LetT′have ordern′, and letT′haveℓ′leaves. Further, let Qhavet leaves, and so|L(Q)| =tandn(Q)=t+2. We note thatn′ =n−t−1 andℓ′=ℓ−t+1. Applying the first induction hypothesis to the treeT′∈Gm−2
0 , we haveγt(T′)= 12(n′−ℓ′+2+(m−2))= 12(n′−ℓ′+m).
We show next thatγt(T)= γt(T′)+2. Everyγt-set ofT′ can be extended to a TD-set ofT by adding to it the verticesuandv, implying thatγt(T)≤ γt(T′)+2.
Recall that diam(T)≥4. By Observation1(b), there exists aγt-set, Dsay, ofT that contains no leaf ofT, implying that{u, v} ⊆ Dand that neither leaf-neighbor ofu norvinT belongs toD. Thus, the setD\{u, v}is a TD-set ofT′−x, implying that γt(T′)≤ γt(T′−x) ≤ |D| −2 =γt(T)−2. Consequently,γt(T)=γt(T′)+2.
Thus,
γt(T)=γt(T′)+2
= 12(n′−ℓ′+m)+2
= 12((n−t−1)−(ℓ−t+1)+m)+2
= 12(n−ℓ+2+m).
This completes the proof of Claim5.2. ⊓⊔ Claim 5.3 If T ∈T0m,3, thenγt(T)=12(n−ℓ+2+m).
Proof Suppose thatT ∈T0m,3. Thus, the treeTcan be obtained from a treeT′∈G0m−1 by adding a vertex disjoint copy of a double starQand adding an edge from a leaf, sayy, ofQto a vertex,zsay, ofT′of degree at least 2. We note thatdT(y)=2. Letuandvbe the two vertices inQthat are not leaves, and soS(Q)= {u, v}. LetT′have ordern′, and letT′haveℓ′leaves. Further, letQhavetleaves, and so|L(Q)| =tandn(Q)=t+2.
We note thatn′=n−t−2 andℓ′=ℓ−t+1. Applying the first induction hypothesis to the treeT′∈Gm−1
0 , we haveγt(T′)= 12(n′−ℓ′+2+(m−1))=12(n′−ℓ′+1+m).
We show next thatγt(T)= γt(T′)+2. Everyγt-set ofT′ can be extended to a TD-set ofT by adding to it the verticesuandv, implying thatγt(T)≤ γt(T′)+2.
By Observation 1(b), there exists a γt-set, Dsay, of T that contains no leaf of T, implying that{u, v} ⊆ Dand that no leaf-neighbor ofunorvinT belongs to D. If y∈ D, then we can replaceyinDwith an arbitrary neighbor ofzinT′to produce a newγt-set ofT. Hence, we may assume thaty∈/ D. With this assumption, the set D\{u, v}is a TD-set ofT′, implying thatγt(T′)≤ |D|−2=γt(T)−2. Consequently, γt(T)=γt(T′)+2. Thus,
γt(T)=γt(T′)+2
=21(n′−ℓ′+1+m)+2
=21((n−t−2)−(ℓ−t+1)+1+m)+2
=21(n−ℓ+2+m).
This completes the proof of Claim5.3. ⊓⊔
By Claims5.1,5.2and5.3, ifT ∈T0m,1∪T0m,2∪T0m,3, thenγt(T)= 12(n−ℓ+ 2+m). This completes the proof of Claim5. ⊓⊔ By Claim5, ifq =1, thenγt(T)= 12(n−ℓ+2+m). This establishes the base step of the second induction. Letq ≥ 2 and assume that ifq′ is an integer where 1 ≤q′ <qand ifT′ ∈Gm
0 is a tree obtained from a sequenceT1, . . . ,Tq′ of trees, whereT1∈Tm,1
0 ∪Tm,2
0 ∪Tm,3
0 and where the treeT′=Tq′ and ifq′≥2, then for eachi ∈ [q′] \ {1}, the treeTican be obtained from the treeTi−1by applying operation O∗.
Recall that the tree T is obtained from a sequence T1, . . . ,Tq of trees, where T1∈T0m,1∪T0m,2∪T0m,3and the treeT =Tq, and where for eachi ∈ [q]\{1}, the treeTican be obtained from the treeTi−1by applying operationO∗. In particular, the treeT is obtained from the treeTq−1by adding to it a vertex disjoint copy of a double starQand adding an edge joining a leaf ofQand a leaf ofTq−1. Letuandvbe the two vertices inQthat are not leaves, and soS(Q)= {u, v}. LetT′=Tq−1, and letT′have ordern′andℓ′leaves. Further, letQhavetleaves, and so|L(Q)| =tandn(Q)=t+2.
We note thatn′=n−t−2 andℓ′=ℓ−(t−1)+1=ℓ−t+2. Applying the second induction hypothesis to the treeTq−1∈ Gm
0, we haveγt(T′)= 12(n′−ℓ′+2+m).
Proceeding analogously as in the proof of the last paragraph of Claim5.3, we have thatγt(T)=γt(T′)+2. Thus,
γt(T)=γt(T′)+2
= 12(n′−ℓ′+2+m)+2
= 12((n−t−2)−(ℓ−t+2)+2+m)+2
= 12(n−ℓ+2+m).
This completes the proof of Theorem5. ⊓⊔
We are now in a position to prove Theorem1. Recall its statement.
Theorem 1 Let m≥0be an integer. If G is a cactus graph of order n≥2with k≥0 cycles andℓleaves, thenγt(G)= 12(n−ℓ+2+m)−k if and only if G∈Gkm. Proof Letm ≥ 0 be an integer, and letG be a cactus graph of order n ≥ 2 with k ≥ 0 cycles and ℓleaves. We proceed by induction onk to show thatγt(G) =
1
2(n −ℓ+2+m)−kif and only if G ∈ Gm
k . Ifk = 0, then the result follows from Theorem5. This establishes the base case. Letk ≥1 and assume that ifG′is a cactus graph of ordern′ ≥ 2 withk′ cycles andℓ′leaves where 0≤ k′ <k, then γt(G′)=12(n′−ℓ′+2+m′)−k′if and only ifG∈Gm′
k′ . LetGbe a cactus graph of ordern≥2 withk≥0 cycles andℓleaves. We show thatγt(G)=12(n−ℓ+2+m)−k if and only ifG ∈ Gm
k. Ifm =0, then the result follows by Theorem3(a), while if m=1, then the result follows by Theorem3(b). Hence, we may assume thatm≥2.
( ⇒) Assume thatγt(G)= 12(n−ℓ+2+m)−k. We show thatG∈ Gm
k. For
this purpose, we first prove the following claim. ⊓⊔
Claim 6 The graph G contains a cycle edge e such thatγt(G−e)=γt(G).
Proof LetC:v1v2. . . vℓv1be a cycle inG, and letSbe aγt-set ofG. IfV(C)⊆S, then lete=v1v2. In this case, the setSis a TD-set ofG−e, and soγt(G−e)≤ |S| =γt(G).
Since removing a cycle edge from a graph cannot decrease the total domination of the graph, we note thatγt(G) ≤ γt(G−e). Consequently, γt(G) = γt(G−e).
Hence, we may assume that at least one vertex of the cycleCdoes not belong to the setS. Renaming vertices ofC if necessary, we may assume thatv2 ∈/ S. Ifv1 ∈/ S, then lettinge =v1v2we note that the setS is a TD-set ofG−e, and so as before γt(G)=γt(G−e). Hence, we may assume thatv1∈ S. Analogously, we may assume thatv3∈S. But then as before, lettinge=v1v3the setSis a TD-set ofG−e, implying
thatγt(G)=γt(G−e). ⊓⊔
By Claim6, the graphGcontains a cycle edgeesuch thatγt(G−e)=γt(G). Let e=uv, and consider the graphG′=G−e. LetG′have ordern′withk′≥0 cycles and ℓ′leaves. We note thatn′=n. Further, sinceGis a cactus graph,k′=k−1. Removing the cycle edgeefrom Gproduces at most two new leaves, namely the ends of the edgee, implying thatℓ′−2≤ℓ≤ℓ′. By Corollary4,γt(G′)= 12(n′−ℓ′+2+m′)−k′ for some integerm′ ≥ 0. Applying the inductive hypothesis to the cactus graphG′, we have that G′ ∈ Gm′
k′ = Gm′
k−1. We note that 12(n−ℓ+2+m)−k = γt(G)= γt(G′)=12(n′−ℓ′+2+m′)−k′, implying thatm−ℓ=m′−ℓ′+2. SinceGis a cactus, the verticesuandvare connected inG′=G−eby a unique path.
Suppose thatℓ = ℓ′. In this case, neitheru norv is a leaf ofG′, implying that bothu andvhave degree at least 2 inG′. Further, the equationm−ℓ=m′−ℓ′+2
