On Computational and Combinatorial Properties of the Total Co-independent Domination Number of Graphs

On Computational and Combinatorial Properties of the Total Co-independent Domination Number of... Abstract A subset D of vertices of a graph G is a total dominating set if every vertex of G is adjacent to at least one vertex of D. The total dominating set D is called a total co-independent dominating set if the subgraph induced by V−D is edgeless and has at least one vertex. The minimum cardinality of any total co-independent dominating set is the total co-independent domination number of G and is denoted by γt,coi(G). In this work we study some complexity and combinatorial properties of γt,coi(G). Specifically, we prove that deciding whether γt,coi(G)≤k for a given integer k is an NP-complete problem and give several bounds on γt,coi(G). Moreover, since any total co-independent dominating set is a total dominating set, we characterize all the trees having equal total co-independent domination number and total domination number. 1. INTRODUCTION Problems concerning domination in graphs are one of the most popular and highly investigated issues in the area of graph theory, and a rich literature on the topic is nowadays known in the research community. Such problems run from the theoretical point of view till several practical applications in real-life situations. The most interesting cases of such applications are probably regarding problems in computer science. One of the most interesting features of domination in graphs involves the existence of a very high number of variants of domination parameters. A very common combination of domination and vertex independence of graphs is known, and the way of combining both concepts groups a considerable number of possibilities. In this work, we center our attention into precisely study one of these combinations between domination and independence in graphs, namely the total co-independence domination parameter. We focus the investigation on some computational complexity aspects of this parameter as well as on combinatorial properties of it. Given a graph G with vertex set V(G) and edge set E(G), a set D⊂V(G) is a total dominating set of G if every vertex in V(G) is adjacent to at least one vertex in D. The total domination number of G is the minimum cardinality of any total dominating set in G and is denoted by γt(G). A γt(G)-set is a total dominating set of cardinality γt(G). For more information on total domination we suggest the relatively recent and fairly complete survey [1] and the book [2]. A set S of vertices is independent if S induced an edgeless graph. An independent set of maximum cardinality is a maximum independent set of G. The independence number of G is the cardinality of a maximum independent set of G and is denoted by β(G). An independent set of cardinality β(G) is called a β(G)-set. Relationships between (total) domination and independence in graphs have attracted the attention of several researchers in the last years. Several interesting connections among these parameters include independent dominating sets [3, 4], partitions into a dominating set and an independent set [5], (total) dominating sets which intersect every maximal independent set [6–8], and some other ones more, which we prefer to not mention here, since it is not the goal of this work. A total dominating set D of a graph G is called a total co-independent dominating set (or TC-ID) if the set of vertices of the subgraph induced by V−D is independent and not empty.1 The minimum cardinality of any TC-ID set is the total co-independent domination number of G and is denoted by γt,coi(G). A TC-ID set of cardinality γt,coi(G) is a γt,coi(G)-set. These concepts were previously introduced and barely studied in [9]. Moreover, in [10], the same parameter was introduced under the name of total outer-independent domination number. Since this article ([10]) is not published in any journal and the other one ([9]) is already published, we precisely follow the terminology and notation of the latter. Since total domination is not defined for graphs having isolated vertices, all the graphs considered herein have not isolated vertices. Moreover, in order to satisfy the total domination property and that the complement of a TC-ID set is not empty, it is required that 2≤γt,coi(G)≤n−1, if n is the order of G. Such trivial bounds were already noted in the work [9]. Throughout our exposition, we consider G=(V,E) as a simple graph of order n and size m. That is, graphs that are finite, undirected, and without loops or multiple edges. Given a vertex v of G, NG(v) represents the open neighborhood of v, i.e. the set of all neighbors of v in G and the degree of v is δ(v)=∣NG(v)∣. The minimum and maximum degrees of G are denoted by δ(G) and Δ(G), respectively (or δ and Δ, respectively, for short). If X and Y are two subsets of V(G), then we denote the set of all edges of G joining a vertex of X with a vertex of Y by E(X,Y). For a set S⊂V(G), the complement of S is S¯=V(G)⧹S. In this work, we represent an edgeless graph G of order n as Nn. For any other graph theory terminology and notation we follow the book [2]. Let T be a tree (a connected graph without cycles). A leaf or a pendant vertex of T is a vertex of degree one (it is similarly defined for non-tree graphs). A support vertex of T is a vertex adjacent to a leaf and a semi-support vertex is a vertex adjacent to a support vertex that is neither a leaf nor a support. By an isolated support vertex of T we mean an isolated vertex of the subgraph induced by the support vertices of T. The set of leaves of T is denoted by L(T), the set of support vertices by S(T), and the set of semi-support vertices by SS(T). Moreover, S*(T) is the set of isolated support vertices of T. We first notice that if H1, H2,…, Hr with r≥2, are the connected components of a graph H, then any TC-ID set of minimum cardinality in H is formed by a minimum total dominating set in the subgraphs Hj where ∣V(Hj)∣=2 and a minimum TC-ID set in the remaining subgraphs Hi with ∣V(Hi)∣≥3. That is stated in the following straightforward result. Remark 1 Let H1, H2,…, Hr with r≥2, be the connected components of a graph H different from the union of r copies of the path P2. Then γt,coi(H)=∑i∈{1,…,r}∣V(Hi)∣=2γt(Hi)+∑j∈{1,…,r}∣V(Hj)∣≥3γt,coi(Hj). In concordance with the result above, from now on, we only consider the study of the TC-ID sets of connected graphs and omit to refer to that fact throughout all our exposition. 2. COMPLEXITY OF THE DECISION PROBLEM We begin our exposition by considering the problem of deciding whether the total co-independent domination number of a graph is less than a given integer. That is stated in the following decision problem. TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r In order to deal with the complexity of the TOTAL CO-INDEPENDENT DOMINATION PROBLEM (TC-ID PROBLEM), we make a reduction from a very well-known decision problem concerning the independence number of graphs. MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r The problem above is one of the classical NP-complete problems appearing in the book [11]. Moreover, it remains NP-complete even when restricted to planar graphs. Theorem 2 MAXIMAL INDEPENDENT SET PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3 ([11]). Now on, in order to present our complexity results we need to introduce a family of graphs which is next defined. Let T6 be a tree with six vertices having two adjacent vertices u,v of degree three and the other four u1,u2,v1,v2 vertices are leaves. Clearly, each vertex of degree three has two adjacent leaves, say u1,u2∈N(u) and v1,v2∈N(v) (see Fig. 1(I)). Given a graph G of order n and n trees T6(1),…,T6(n) isomorphic to the tree T6, the graph GT is constructed by adding edges between the ith-vertex of G and the vertex u of the ith-tree T6(i). See Fig. 1 (II) for an example. Figure 1. View largeDownload slide The graph T6 (a) and a graph GT (b) where G is a complete graph minus one edge. Figure 1. View largeDownload slide The graph T6 (a) and a graph GT (b) where G is a complete graph minus one edge. We are now able to prove the NP-completeness of the TC-ID PROBLEM. Theorem 3 TOTAL CO-INDEPENDENT DOMINATION PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3. Proof The problem is clearly in NP since verifying that a given set is indeed a TC-ID set can be done in polynomial time. Let us now make a reduction from the MAXIMAL INDEPENDENT SET PROBLEM. Let G be a not edgeless graph of order n and construct the graph GT as described above. Let us denote by u(i),v(i) the vertices of degree three in the ith copy T6(i) of the tree T6 used to generate GT. We shall prove that γt,coi(GT)=3n−β(G). Let A be a β(G)-set and let D be the set of vertices of GT obtained from the complement of A in G together with the vertices u,v belonging to all the copies of the tree T6 used to generate GT, that is D=(V(G)⧹A)∪{⋃i=1n{u(i),v(i)}}. This set is clearly a total dominating set and its complement is an independent set. Thus, D is a TC-ID set in GT and, as a consequence, γt,coi(GT)≤n−∣A∣+⋃i=1n{u(i),v(i)}=3n−β(G). On the other hand, let D′ be a γt,coi(GT)-set. In order to totally dominate the leaves of every copy of T6 in GT, it must happen that ∣D′∩V(T6(i))∣≥2 for every i∈{1,…,n}. Also, V(G)∩D′≠∅, since otherwise the complement of D′ would not be independent. Moreover, the complement of V(G)∩D′ in G is an independent set in G. Thus, β(G)≥n−∣V(G)∩D′∣ and we obtain the following. γt,coi(GT)=∣D′∣=∣D′∩V(G)∣+⋃i=1n(D′∩V(T6(i)))≥n−β(G)+2n=3n−β(G). As a consequence, it follows that γt,coi(GT)=3n−β(G). Now, for j=3n−k, it is readily seen that γt,coi(GT)≤j if and only if β(G)≥k, which completes the reduction. We also observe that, if G is a planar graph, then GT is also planar. Therefore, since the MAXIMAL INDEPENDENT SET PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3, we also deduce that the TC-ID PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3 and the proof is completed.□ As a consequence of the result above, we deduce the following consequence. Corollary 4 The problem of computing the total co-independent domination number of graphs is NP-hard even when restricted to planar graphs of maximum degree at most 3. 3. BOUNDING THE TOTAL CO-INDEPENDENT DOMINATION NUMBER In order to present the first bounds for γt,coi(G) of any graph G, we need the next concepts. A set S of vertices of G is a vertex cover of G if every edge of G is incident with at least one vertex of S. The vertex cover number of G, denoted by α(G), is the smallest cardinality of a vertex cover of G. We refer to an α(G)-set in G as a vertex cover of cardinality α(G). The following well-known result, due to Gallai [12], states the relationship between the independence number and the vertex cover number of a graph. Theorem 5 For any graph Gof order n, α(G)+β(G)=n (Gallai [12]). On the other hand, it was shown in [9] the following relationship between γt,coi(G) and β(G). Theorem 6 For any graph Gof order n, γt,coi(G)≥n−β(G) ([9]). By using the two theorems above, we can easily deduce the lower bound of our next result. However, an upper bound for γt,coi(G) in terms of the vertex cover number can also be deduced. We first consider the case whether G is a star graph Sn for which is known that γt,coi(Sn)=2 and α(Sn)=1. For our purposes, an star graph Sn is a graph of order n+1 in which all but one of its vertices are leaves, unless n=1 where both vertices of S1 are leaves. Remark 7 For any star graph Sn, γt,coi(Sn)=2=2α(Sn). In concordance with the remark above, for our next result we exclude the case of star graphs and see that they behave in a different manner. Theorem 8 For any graph Gof order ndifferent from a star graph, α(G)≤γt,coi(G)≤2α(G)−1. Proof The lower bound follows from Theorems 5 and 6. If α(G)≥n/2, then γt,coi(G)≤n−1≤2(n/2)−1≤2α(G)−1. Thus, from now on in this proof we consider α(G)<n/2. Now, let C be an α(G)-set. We choose two vertices u,v∈C with the minimum possible distance between them and let P be a shortest u−v path. Clearly, V(P)∩C={u,v} and the distance between u and v is one or two (notice this also means 2≤∣V(P)∣≤3). For each vertex x∈C−{u,v}, choose a neighbor x′ of x. Then C∪V(P)∪{x′:x∈C−{u,v}} is a TC-ID set of cardinality 2∣C∣+∣V(P)∣−4≤2α(G)−1, which completes the proof of the upper bound.□ The bounds above are tight. For instance, a characterization of that trees achieving the equality in the lower bound was given in [13] (note that in [13] the trees T of order n satisfying equality in the bound γt,coi(T)=n−β(T) were characterized, which equals the lower bound of Theorem 8, in concordance with Theorem 5). The upper bound is attained for an infinite family of graphs, as we next show. To this end, we need the following operations for edges or induced paths P3 of a graph G. Subdivision: Given an edge uv, remove the edge, add a vertex w and the edges uw, wv. Inflation of size k: Given an induced path P3=uvw of G, in which v has degree two, remove the vertex v and the two incident edges, and replace them with k vertices v1,v2,…,vk and edges uvi,viw for every i∈{1,…,k}. Addition of tpendant vertices: Given a vertex x add t new vertices y1,…,yt and the edges xyi for every i∈{1,…,t}. Now, a graph Hn,a,b∈F1 is a graph obtained from a star graph Sn by making the following sequence of operations, which we will call as Sequence I. Apply the operation ‘Subdivision’ to a ( 1≤a≤n) edges of Sn. Apply the operation ‘Inflation of size ki’ with ki≥2 to b ( 0≤b≤a) paths P3(i) obtained from (i). Apply the operation ‘Addition of qipendant vertices’, qi≥0, to the b vertices corresponding to leaves of Sn obtained in the step (ii). Apply the operation ‘Addition of tipendant vertices’, ti≥1, to the leaves belonging to the remaining a−b paths obtained from (i), which were not ‘inflated’ in (ii). If a=n and b=0 (notice that in this case Hn,a,b is a tree such that the central vertex of the original star graph Sn has no adjacent leaves), then apply the operation ‘Addition of tpendant vertices’, t≥1, to the vertex corresponding to the central vertex of Sn. If a=n and b>0, then apply the operation ‘Addition of tpendant vertices’, t≥0, to the vertex corresponding to the central vertex of Sn. As an example, to obtain the cycle C4 (which belongs to F1) we begin with the star S1 (a path P2), next we apply the operation ‘Subdivision’ to the unique edge of S1 and then we apply the operation ‘Inflation of size 2’ to the path P3 obtained in the previous step. Note that different sequences of operations would lead to the same graph. For instance, the graph P5 can be obtained from the star S1 by subdividing its unique edge and then adding a pendant vertex to the leaf corresponding to the subdivision, as well as another pendant vertex to the center of S1 (coincidentally such center is also a leaf). Moreover, the graph P5 is obtained from the star P3 by subdividing one of its edges and then adding a pendant vertex to the leaf corresponding to such subdivision. On the other hand, we remark that three integers n,a,b would produce different graphs Hn,a,b depending on the addition of pendant vertices that would be done. However, since it is not significant for our work to denote them, we skip to use the notations for the addition of pendant vertices. A fairly representative graph of the family F1 is given in Fig. 2. Remark 9 For any graph Hn,a,b∈F1, α(Hn,a,b)=a+1 and γt,coi(Hn,a,b)=2a+1. Proof For any edge of Sn which was subdivided in step (i), it appears either a path P4 or a cycle C4 and all these subgraphs have in common only one vertex (the corresponding one to the center of Sn). Thus, in order to cover all the edges of Hn,a,b, at least a+1 vertices are required. Moreover, a set given by those a leaves corresponding to the a edges of the star Sn, which were subdivided, together with the central vertex form a vertex cover of cardinality a+1. Thus, the equality α(Hn,a,b)=a+1 follows. Now, let D be a γt,coi(Hn,a,b)-set. We analyze the following situations for every edge wu (assume w is the center of Sn) of the star which is initially subdivided. Case 1: There is only one path between w and u in Hn,a,b. Hence, the edge wu was subdivided with a vertex, say v, and not inflated, which made a required addition of at least one pendant vertex, say u′, to the leaf u. Thus, in order to totally dominate u′, ∣D∩{v,u,u′}∣≥2. Case 2: There are at least two paths between w and u in Hn,a,b. Clearly, this means wu was subdivided and then inflated with at least two vertices, say v1,…,vr, r≥2. Moreover, if some pendant vertices were added to u, then in order to totally dominate v1,…,vr,u and the extra leaves adjacent to u, at least two vertices of v1,…,vr,u are required. We next consider the vertex w separately. If a<n, then the vertex w has a least one adjacent leaf which needs to be totally dominated. Thus, w must belong to D. On the contrary, if a=n, then we must consider the value b. If b=0, then no path P3 was inflated ( Hn,a,b is a tree) and so, by step (v), w has at least one adjacent leaf which needs to be totally dominated, which means w must belong to D again. Finally, we assume b>0. Thus, at least one path P3 was inflated and this creates a cycle of order four, say C4*, to which w belongs. Also, it may happen w has no adjacent leaves. Now, note that if w∉D, then the two vertices of C4* adjacent to w must belong to D, since D¯ is an independent set. Moreover, the fourth vertex of C4* must belong to D too, in order to get the vertices of D totally dominated. As a consequence, at least three vertices of the cycle are in D, which is equivalent to have in D the vertex w, one of its neighbors in C4* and the vertex of C4* which is not adjacent to w. Consequently, we can deduce that for any set of vertices of a subgraph of Hn,a,b, induced by the vertices obtained in a subdivision of one of the a leaves of Sn, and the corresponding addition of some pendant vertices (if it is the case), at least two of these vertices are in D. Moreover, note that at least one extra vertex is required, and to satisfy this we simply consider the central vertex w of Sn. Thus, γt,coi(Hn,a,b)=∣D∣≥2a+1. On the other hand, by using Theorem 8, we obtain that γt,coi(Hn,a,b)≤2α(Hn,a,b)−1=2a+1 and the equality follows for γt,coi(Hn,a,b).□ Figure 2. View largeDownload slide A graph H5,5,3∈F1 where the six bolded vertices form an α(H) set and gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H) set, which has cardinality 11. Figure 2. View largeDownload slide A graph H5,5,3∈F1 where the six bolded vertices form an α(H) set and gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H) set, which has cardinality 11. Now, a graph H∈F2 is a graph obtained from the cycle C6 by making the following sequence of operations, which we will call as Sequence II. Apply the operation ‘Addition of tipendant vertices’, ti≥0 and i∈{1,2,3}, to the three vertices, say v1,v2,v3, of a β(C6)-set, respectively. If there is a ti=0 from the above operation and the degree of vi is two, then apply the operation ‘Inflation of size k’ with k≥2 to one of the two possible paths of order three between vi and the other two vertices in {v1,v2,v3}−{vi}. Apply the operation ‘Inflation of size ki’ with ki≥1 and i∈{1,2,3} to the three possible paths of order three between v1,v2,v3. An example of a graph of the family F2 appears in Fig. 3. Figure 3. View largeDownload slide A graph H∈F2 where the three bolded vertices form an α(H)-set and the two gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H)-set, which has cardinality five. Figure 3. View largeDownload slide A graph H∈F2 where the three bolded vertices form an α(H)-set and the two gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H)-set, which has cardinality five. The following result concerning the values of α(H) and γt,coi(H) for graphs H∈F2 is straightforward to observe. Remark 10 For any graph H∈F2, α(H)=3 and γt,coi(H)=5. According to the Remarks above, we can easily check that the upper bound of Theorem 8 is achieved for any graph G∈F1∪F2. Moreover, we next prove that precisely the graphs of these families are the only ones achieving the upper bound of Theorem 8. To this end, we need the following two lemmas whose proofs can be made by using some similar techniques as in the proof of Theorem 8. Lemma 11 If a graph Gcontains an α(G)-set which is not independent, then γt,coi(G)≤2α(G)−2. Proof Let C be an α(G)-set which is not independent. We choose two adjacent vertices u,v∈C. For each vertex x∈C−{u,v}, choose a neighbor x′ of x. Then C∪{x′:x∈C−{u,v}} is a TC-ID set of cardinality 2(∣C∣−2)+2≤2α(G)−2.□ Lemma 12 If an α(G)-set of a graph Gcontains four different vertices u1,u2,v1,v2such that a shortest u1−u2path and a shortest v1−v2path have length two and are vertex disjoint, then γt,coi(G)≤2α(G)−2. Proof Let C be an α(G)-set such that u1,u2,v1,v2∈C. For each vertex x∈C−{u1,u2,v1,v2}, choose a neighbor x′ of x and let w1,w2∉C be two vertices such that w1∈N(u1)∩N(u2) and w2∈N(v1)∩N(v2), which exist by assumption. Then C∪{w1,w2}∪{x′:x∈C−{u1,u2,v1,v2}} is a TC-ID set of cardinality 2(∣C∣−4)+6≤2α(G)−2.□ Theorem 13 Let Gbe a graph of order nsuch that 2α(G)≤n. Then γt,coi(G)=2α(G)−1if and only if G∈F1∪F2. Proof In one hand, if G∈F1∪F2, then it clearly happens that γt,coi(G)=2α(G)−1 according to Remarks 9 and 10. On the second hand, assume γt,coi(G)=2α(G)−1 and let D be any α(G)-set. We first notice that D must induce an independent set according to Lemma 11. We shall now proceed by proving some partial claims that will further give our required conclusion. Claim 1 G has no triangles (cycles of order three). Proof of Claim 1 If there is a triangle, then, in order to cover all its edges, at least two of its vertices must belong to D. So, this cover is not an independent set. Thus, we get a contradiction by using Lemma 11.□ Claim 2 G has no induced cycles of order five or larger than six. Proof of Claim 2 Suppose G contains a cycle Cr with r=5 or r≥7. In order to cover all the edges of Cr, and since r≠6, there must be two vertices in D∩V(Cr) at distance one (which means D is not independent), or there are four different vertices u1,u2,v1,v2∈D∩V(Cr) such that a shortest u1−u2 path and a shortest v1−v2 path have length two and are vertex disjoint. Thus, we obtain contradictions by using Lemmas 11 and 12.□ As a consequence of the Claims above, we have that G can only contain cycles of order four or six. We first analyze the case in which G contains a cycle of order six. Let V(C6)={v1,…,v6} where v1∼v2∼…∼v6∼v1 ( u∼v means u,v are adjacent). According to Lemma 11, it must happen D∩V(C6) is independent. Thus, without loss of generality we assume D∩V(C6)={v1,v3,v5}. We consider now several situations. Suppose v2 has degree larger than two. If v2∼vj with j∈{4,6}, then G has a triangle, which is not possible. If v2∼v5 and D={v1,v3,v5}, then D∪{v2} is a TC-ID set of cardinality four and α(G)=3, a contradiction. If v2∼v5 and {v1,v3,v5}⊊D, then for each vertex x∈D−{v1,v3,v5}, choose a neighbor x′ of x and we observe that the set D∪{v2}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. Thus, v2 has a neighbor z∉V(C6). Since v2∉D, it must happen z∈D. Since z≁v1 and z≁v3 (otherwise there would be a triangle), we obtain a contradiction with Lemma 12 by using the vertices z,v1,v3,v5. As a consequence, v2 must have degree two, and by symmetry also v4,v6 are of degree two too. Suppose v1 has degree two. If D={v1,v3,v5}, then {v2,v3,v5,v6} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, for each vertex x∈D−{v1,v3,v5}, we choose a neighbor x′ of x and observe that the set (D−{v1})∪{v2,v6}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. So, v1 must have degree at least three and, by symmetry also v3,v5 are of degree at least three too. We consider now a vertex x∈N(v1)−{v2,v6}. Notice that x≠v3,v5 (otherwise there would be a triangle). Also, x≠v4 because v4 has degree two as stated before. Suppose x has degree larger than one and let x′∈N(x)−{v1}. Since D is independent and the edge xx′ must be covered by D, x′∈D. If x′≠v3 and x′≠v5, then we obtain a contradiction with Lemma 12 by using the vertices x′,v1,v3,v5 (notice that x′≁v1 since D is independent). As a consequence, we obtain that any neighbor x of v1 is either of degree one or has a neighbor in V(C6)−{v1}. We next consider the latter situation whether x′∈V(C6)−{v1}. Clearly, x′≠v2,v4,v6 because v2,v4,v6 have degree two. Suppose x is neighbor of v3 and of v5. If D={v1,v3,v5}, then {x,v1,v3,v5} is a TC-ID set of cardinality four and α(G)=3, a contradiction. We may now choose a neighbor y′ of y for every y∈D−{v1,v3,v5} and hence, observe that the set D∪{x}∪{y′:y∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. Thus, x is a neighbor of either v3 or v5, in which case, it happens x has degree two. By symmetry, we obtain similar conclusions for v3 and v5. That is, for any vi with i∈{1,3,5}, N(vi) is given by leaves or vertices of degree two. In the latter case, if x∈N(vi)−V(C6), then N(x)={vi,y} where y∈D∩V(C6)−{vi}. As a consequence, we observe that G can be obtained from a cycle C6 by the Sequence II of operations described above, or equivalently G∈F2. We may now consider the case in which G contains a cycle C4, but G does not contain the cycle C6. Claim 3 G does not contain vertex disjoint cycles. Proof of Claim 3 We directly obtain a contradiction by Lemma 12, since in this case there are four different vertices u1,u2,v1,v2 (two of them in one cycle, the other two in the other cycle) such that a shortest u1−u2 path and a shortest v1−v2 path have length two and are vertex disjoint.□ Thus, if G contains more than one cycle C4, then they are not vertex disjoint. Moreover, we can next see that not two adjacent vertices of a cycle can be in any other cycle. Claim 4 If two cycles C4 of G have exactly two vertices in common, then these vertices are not adjacent. Proof of Claim 4 Suppose there are two cycles C4 having two adjacent vertices in common. Assume the cycles are C4(1)=v1v2v3v4v1 and C4(2)=v1v2v5v6v1. Hence, we note that exactly three vertices of {v1,…,v6} must belong to D, otherwise there are two adjacent vertices in D. Indeed, such vertices are either v1,v3,v5 or v2,v4,v6, say for instance v1,v3,v5. If precisely D={v1,v3,v5}, then {v1,v2,v3,v5} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, we may choose a neighbor x′ of x for every x∈D−{v1,v3,v5}, and observe that the set D∪{v2}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction.□ Now, according to the Claims above, if G contains more than one cycle C4, then only the following situations can occur. Any two cycles have exactly one vertex in common. Any two cycles have exactly two vertices in common which are not adjacent. Any two cycles have exactly three vertices in common. We note that the situation in which two cycles of G have exactly three vertices in common can be understood as G has three cycles with two vertices in common. We now turn our attention on the following. Claim 5 There is a vertex w∈D such that d(w,x)=2 for every x∈D−{w}. Proof of Claim 5 We first note that there are at least two vertices w,x∈D such that d(x,w)=2, otherwise there would be an edge not covered by D. Let h be a vertex adjacent to w and x. Suppose there is a vertex y∈D such that d(w,y)≠2 and d(x,y)≠2 (note that d(w,y)≠1 and d(x,y)≠1). Thus, since there are no cycles of order larger than four in G, there must happen one of the following situations. There is a shortest path joining yand hnot containing wnor x. Also, y is different from the neighbor of h, say h′, in such path. In such case, in order to cover the edge hh′, it must happen h′∈D. Thus, we obtain a contradiction by using Lemma 12 and the vertices h′,w,x,z where z is a vertex at distance two from x in the x−y path. Without loss of generality, there is a shortest path joining yand xcontaining w. Thus, there must be a vertex y′∈D belonging to this path such that d(y,y′)=2 (it cannot be d(y,y′)=1 since D is independent), otherwise there should be a not covered edge. Clearly w≠y′. Thus, we obtain a contradiction by using Lemma 12 and the vertices y,y′,w,x. As a consequence, the vertex y has distance two to x or to w. Moreover, if d(w,y)=2 and d(x,y)=2, then there is a cycle of order six, which is not possible, or there is a vertex y′∈N(y)∩N(x). In the latter case, if D={x,y,w}, then {x,y,w,y′} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, we may assume {x,y,w}⊊D. Now, we choose a neighbor x′ of x for every x∈D−{x,y,w}, and observe that the set D∪{y′}∪{x′:x∈D−{x,y,w}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction. Thus, y has distance two to exactly one vertex of x and w. From now on, we assume d(y,w)=2. We next prove that for any vertex z∈D−{x,y,w}, it follows d(z,w)=2 too. If D={x,y,w}, then we are done. So, me may suppose there is a vertex z′∈D−{x,y,w} such that d(z′,w)≠2 (clearly d(z′,w)>2). Consider now the shortest path between z′ and w, say z′z1′z2′…zq′w. Notice that q≥2. In order to cover the edge z1′z2′, it must be z2′∈D. So, we obtain a contradiction by using Lemma 12 and the vertices x,w,z′,z2′. Therefore, for any vertex z∈D−{w}, we obtain that d(z,w)=2 and the claim is proved.□ Next step gives some result on the distances between any two vertices x,y∈D−{w}. Claim 6 For any two vertices x,y∈D−{w}, any shortest path between x and y passes through w. Proof of Claim 6 From Claim 5, we know that d(x,w)=d(y,w)=2. Thus d(x,y)≤4. Clearly d(x,y)>1, since x,y cannot be adjacent. Let x′,y′∈N(w) such that x′∈N(x) and y′∈N(y). If x∼y′ or y∼x′ (say x∼y′) and D={x,y,w}, then the set D={x,y,w,y′} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Also, if x∼y′ and {x,y,w}⊊D, then we choose a neighbor z′ of z for every z∈D−{w,x,y} and observe that the set D∪{y′}∪{z′:z∈D−{w,x,y}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction. Thus, neither x∼y′ nor y∼x′. If there is a vertex z∈N(x)∩N(y), then wx′xzyy′w is a cycle C6 in G, which is not possible. Thus d(x,y)≠2. By using a similar reasoning, it can be deduced that d(x,y)≠3 and so, d(x,y)=4. If there is another path of length four between x and y not containing w, then we have one of the following situations. There is a vertex w′∈D such that x′,y′∈N(w′) (note that w′ must be in D in order to cover the edges w′y′, w′x′). In such case, we obtain a contradiction by using Lemma 12 and the vertices x,w,w′,y. There are three vertices x1,y1,w″≠x′,y′,w such that x1∈N(x), y1∈N(y) and x1,y1∈N(w″). In such situation, wx′xx1w″y1yy′w is an induced cycle of order eight in G, which is not possible. Similarly to the case above, if either x1=x′ or y1=y′, then we obtain an induced cycle of order six in G, which is also not possible. Therefore, any shortest path between x and y passes throughout w.□ We now give several facts which are consequences of the Claims above, in order to deduce the structure of the graph G. The set V(G)−D is independent (otherwise there is an edge not covered by D). If x,y∈D−{w}, then N(x)∩N(y)=∅. If z∈N(x) for some x∈D−{w}, then either z∈N(w) and z has degree two, or z is a vertex of degree one (otherwise we get a contradiction with Claim 6). If z′∈N(w) is not a vertex of degree one, then there is exactly one vertex x∈D such that N(z′)={w,x} (equivalently z′ has degree two). As a consequence of the items above, as well as from the Claims, and all the reasoning till this point, we observe that D is formed by w and a set of vertices v1,v2…,vr (satisfying the properties above). Clearly, for any vertex vi, the set of its neighbors are either leaves or vertices of degree two adjacent to w. Moreover, if vi has only one neighbor of degree two, then it must have at least one adjacent leaf (otherwise one can find a cover set of smaller cardinality). In this sense, such set of vertices can clearly be obtained from a leaf of a star by making a subdivision of the corresponding edge, an inflation of the path P3 obtained from the subdivision and a subsequent addition of some pendant vertices. On the other hand, if w has some adjacent leaves, then they could be obtained directly from a star, if subdivisions were not done to all the leaves of the star or, by a subsequent addition of leaves to the center of the original star, if all its leaves would have been subdivided. Therefore, it is then concluded that the graph G was obtained from a star by making the Sequence I of operations previously described, which means G∈F1 and the proof is completed.□ We close this section with two bounds for γt,coi(G) in terms of order, size and minimum and maximum degrees. Proposition 14 Let Gbe a graph of order n, minimum and maximum degrees δand Δ, respectively. Then γt,coi(G)≥nδΔ+δ−1. Proof Let D be a γt,coi(G)-set. Hence, the subgraph induced by V(G)−D is edgeless. So, (n−∣D∣)δ=(∣V(G)−D∣)δ≤E(V(G)−D,D)≤∣D∣(Δ−1). Furthermore, it follows that γt,coi(G)≥nδΔ+δ−1. □ Proposition 15. Let Gbe a graph of order n, size m, minimum and maximum degrees δ and Δ, respectively. Then γt,coi(G)≥2m+nδ3Δ+δ−2. Proof Let D be a γt,coi(G)-set. Hence, the subgraph induced by V(G)−D is edgeless. So, E(V(G)−D,D)+E(D,D)=m. Now, notice that E(V(G)−D,D)≤∣D∣(Δ−1) and E(D,D)≤∣D∣Δ−(n−∣D∣)δ2. Adding these inequations, we have m=E(V(G)−D,D)+E(D,D)≤∣D∣Δ−(n−∣D∣)δ2+∣D∣(Δ−1). Therefore, it follows that γt,coi(G)≥2m+nδ3Δ+δ−2. □ The two bounds above are attained for instance for the double stars Sk,k (a double star Sk,k is a tree with exactly two adjacent support vertices, each of degree k+1, and the remaining vertices are leaves, i.e. each support vertex is adjacent to k leaves), which has order 2(k+1), size m=2k+1, minimum degree δ=1, maximum degree Δ=k+1 and γt,coi(Sk,k)=2. 4. THE CASE OF TREES In order to easily proceed with our exposition, and based on the following known bound, from now on we say that a tree T belongs to the family Tγt, if γt,coi(T)=γt(T). Moreover, we assume in this section that ∣S(T)∣≥2, since the case ∣S(T)∣=0 ( T is a P2 and γt,coi(T) is not defined) and ∣S(T)∣=1 ( T is a star graph Sn and γt,coi(T)=2) are straightforward to study. Theorem 16 For any graph G, γt,coi(G)≥γt(G) ([9]). It is now our goal to characterize the family of trees achieving the equality in the bound above. To this end, we observe the following basic results, which can easily be obtained by using some known properties of minimum total dominating sets. Proposition 17 If Sis a minimal total dominating set of a connected graph G=(V,E), then each v∈Shas at least one of the following two properties ([14]). There exists a vertex w∈V−Ssuch that N(w)∩S={v}. The subgraph induced by S−{v}contains an isolated vertex. The next remark is one useful consequence of the proposition above. Remark 18 Let D be a γt,coi(T)-set of cardinality γt(T). Then, for every v∈D, at least one of the following conditions is satisfied. There exists a vertex u∈D such that N(u)∩D={v}. There exists a vertex w∈V−D such that N(w)∩D={v}. We may recall to notice that condition (ii) implies that vertex v is a support, because the set D¯ is independent. Lemma 19 Let T∈Tγtand let Dbe a γt,coi(T)-set containing no leaves. Then for every v∈V(T)−(D∪L(T))there exist a leaf hsuch that d(v,h)≤3. Proof Let v∈V(T)−(D∪L(T)). Since ∣N(v)∣≥2, we consider N(v)={v1,v2,…,vr} with r≥2. Clearly, N(v)⊂D since D¯ is independent. For every vi, with i∈{1,…,r}, by Remark 18, vi is adjacent to a leaf or there exist a vertex si∈D such that N(si)∩D={vi}. Hence, as si∈D, N(si)−{vi}⊂V(T)−D. We assume that for every i∈{1,…,r}, vi is not adjacent to a leaf h, otherwise d(v,h)=2. Now, we suppose that (N(si)−{vi})∩L(T)=∅. Also note that, by condition above, the vertices belonging to N(si) are totally dominated by other vertices of D. So, we observe that the set (D−{s1,s2,…,sr})∪{v} is a total dominating set of T of cardinality smaller than ∣D∣, a contradiction. Furthermore, there exist i∈{1,…,r} such that (N(si)−{vi})∩L(T)≠∅. Thus, for any h∈(N(si)−{vi})∩L(T), it follows d(v,h)=3, and this completes the proof.□ From this point, the set of leaves having distance three with respect to at least one other leaf is denoted by L3(T), and given a γt,coi(T)-set D, we denote by V2,3(T)⊂V(T)−D the set of vertices having distance two or three to some leaf and by V6(T)⊂V(T)−D the set of vertices having distance three to some vertex of V2,3(T). In order to provide a constructive characterization of the trees belonging to the family Tγt, we need the following five operations F1, F2, F3, F4 and F5 on a tree T (by attaching a path P to a vertex v of T we mean adding the path P and joining v to a vertex of P). Moreover, through all the next results we make use of the fact that any tree T always contains a γt,coi(T)-set which does not contain leaves. Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Let F be the family of trees defined as F={T∣T is obtained from P4 by a finite sequence of operations F1,F2,F3,F4 or F5}. Figure 4 contains a fairly representative example of a tree T∈F. We first show that every tree of the family F belongs to the family Tγt. Lemma 20 If T∈F, then T∈Tγt. Proof We proceed by induction on the number r(T) of operations required to construct the tree T. If r(T)=0, then T=P4 and T∈Tγt. This establishes the base case. Hence, we now assume that k≥1 is an integer and that each tree T′∈F with r(T′)<k satisfies that T′∈Tγt. Let T∈F be a tree for which r(T)=k. Since T can be obtained from a tree T′∈F with r(T′)=k−1 by one of the operations F1,F2,F3,F4 or F5, we shall prove that T∈Tγt, by considering a γt,coi(T′)-set D′ containing no leaves and through the following situations. Case 1. T is obtained from T′ by operation F1. Let u be the vertex added to T′ in order to obtain T. Since u is a leaf of T and is adjacent to a vertex of D′, the set D′ remains to be a total dominating set in T. Moreover, D′ is a γt(T)-set, since otherwise we would find a total dominating set in T′ of cardinality smaller than γt(T′). On the other hand, since (V(T′)−D′)∪{u} is independent, we deduce D′ is a TC-ID set in T. Thus, γt,coi(T)≤∣D′∣=γt,coi(T′)=γt(T′)=γt(T) (by also using the inductive hypothesis). Thus, by Theorem 16, we get the equality γt,coi(T)=γt(T), which means T∈Tγt. Case 2. T is obtained from T′ by operation F2. Assume T is obtained from T′ by adding the vertex u and the edge uv where v∈L3(T′). As v∈L3(T′), there exist a path vu1u2h in T′ where h is a leaf and u1,u2 are support vertices adjacent to v,h, respectively. Now, in T, the vertices u2,v are supports and belong to any TC-ID set in T. Hence, the set D=D′∪{v} is a TC-ID set in T, and so γt(T)≤γt,coi(T)≤γt,coi(T′)+1=γt(T′)+1 (1) (by also using Theorem 16 and the inductive hypothesis). Now, let A be a γt(T)-set containing no leaves. Notice that the vertex v is a support and so, it belong to A, also the vertex u1 belongs to A too, because v has degree two. Moreover, note that the set A−{v} is a total dominating set in T′, which leads to γt(T′)≤γt(T)−1. By using this, it follows that all the inequalities in (1) must be equalities. Thus γt,coi(T)=γt(T), and T∈Tγt. Case 3. T is obtained from T′ by operation F3. Assume T is obtained from T′ by adding the path P2=h1h2 to a vertex v∈L3(T′) through the edge vh1. By using some similar reasons as in the case above (now we must use D=D′∪{v,h1} instead of D=D′∪{v}), it is observed that T∈Tγt. Case 4. T is obtained from T′ by operation F4. Assume T is obtained from T′ by adding the path P3=h1u1h2 to a vertex v∈V2,3(T′) through the edge vh1. We notice that u1,h1 belong to any TC-ID set containing no leaves of T. Hence, the set D=D′∪{u1,h1} is a TC-ID set in T. Thus γt(T)≤γt,coi(T)≤γt,coi(T′)+2=γt(T′)+2 (by also using Theorem 16 and the inductive hypothesis). Now, let A be a γt(T)-set. Since the vertex u1 is a support, it belongs to A and so, ∣A∩{h1,u1,h2}∣≥2. Moreover, note that ∣A∩V(T′)∣≥γt(T′). Hence, γt(T)=∣A∣≥γt(T′)+2. Again, as in Case 2, we deduce γt,coi(T)=γt(T), which means T∈Tγt. Case 5. T is obtained from T′ by operation F5. Assume T is obtained from T′ by adding the path P3=h1u1h2 to a vertex v∈V6(T′) through the edge vh1. By using some similar reasons as in the case above, it can be deduced that γt,coi(T)=γt(T), which gives T∈Tγt.□ Figure 4. View largeDownload slide A tree T obtained from a path P4=v1v2v3v4, applying the five operations F1, F2, F3, F4 and F5. Firstly, operations F4 and F5 are applied by adding the path P3=v5v6v7 to the vertex v4 through the edge v4v5 and the path P3=v8v9v10 to the vertex v7 through the edge v7v8. Next, we apply the operation F1 twice by attaching the vertices w1 and w2 to the vertices v3 and v8, respectively. Moreover, we apply the operation F2 by adding the vertex u3 to w2. Finally, we apply the operation F3 by adding the path P2=u1u2 to the vertex w1 through the edge w1u1. Figure 4. View largeDownload slide A tree T obtained from a path P4=v1v2v3v4, applying the five operations F1, F2, F3, F4 and F5. Firstly, operations F4 and F5 are applied by adding the path P3=v5v6v7 to the vertex v4 through the edge v4v5 and the path P3=v8v9v10 to the vertex v7 through the edge v7v8. Next, we apply the operation F1 twice by attaching the vertices w1 and w2 to the vertices v3 and v8, respectively. Moreover, we apply the operation F2 by adding the vertex u3 to w2. Finally, we apply the operation F3 by adding the path P2=u1u2 to the vertex w1 through the edge w1u1. We now turn our attention to the opposite direction concerning the lemma above. In this sense, from now on we shall need the following terminology and notation in our results. Given a tree T and a set S⊂V(T), by T−S we denote a tree obtained from T by removing from T all the vertices in S and all its incident edges (if S={v} for some vertex v, then we simply write T−v). For an integer r≥2, by Qr we mean a graph which is obtained from a path Pr+2=vss1s2…sr by attaching a path P1 to every vertex of Pr+2−v. In Fig. 5 we show the example of Q5. Figure 5. View largeDownload slide The structure of the tree Q5. Figure 5. View largeDownload slide The structure of the tree Q5. We next show that every tree of the family Tγt belongs to the family F. Lemma 21 If T∈Tγt, then T∈F. Proof We proceed by induction on the order n≥4 of the trees T∈Tγt. If T is a double star, then T can be obtained from P4 by repeatedly applying operation F1. This establishes the base case. We assume next that k>4 is an integer and that each tree T′∈Tγt with ∣V(T′)∣<k satisfies T′∈F. Let T be a tree such that T∈Tγt and ∣V(T)∣=k. Let D be a γt,coi(T)-set containing no leaves and let B=V(T)−D. We analyze the following situations. Case 1: ∣S(T)∣<∣L(T)∣. We consider a support vertex v that is adjacent to at least two leaves. Let h∈N(v)∩L(T) and T′=T−h. Thus, the set D is a γt(T′)-set too, and by inductive hypothesis, T′∈F. Therefore, since T can be obtained from T′ by operation F1, it follows T∈F. Case 2: ∣S(T)∣=∣L(T)∣ and ∣SS(T)∣=0. In this case we note that V(T)=S(T)∪L(T) and clearly, S(T) is a γt,coi(T)-set (moreover ∣S(T)∣≥3 since otherwise T is a double star). Let s∈S(T) such that ∣N(s)∩S(T)∣=1 (note that such s always exists) and let h∈L(T) be the leaf adjacent to s. We first notice that there exists a leaf having distance three to the support s. Thus, we deduce that S′(T)=S(T)−{s} is a γt,coi(T′)-set, where T′=T−h. By induction hypothesis T′∈F and, since T can be obtained from T′ by operation F2, we get T∈F. Case 3: ∣S(T)∣=∣L(T)∣ and ∣SS(T)∣>0. Herein we denote by P(x,y) the set of vertices of one shortest path between x and y, including x and y. Let h,h′ be two leaves at the maximum possible distance in T such that there is v∈SS(T)∩P(h,h′) with d(v,h)=2 or d(v,h′)=2. Without loss of generality assume that d(v,h)=2 and let s be the support adjacent to h. Since ∣S(T)∣=∣L(T)∣ and by the maximality of the path between h and h′, we observe that N(s)⊂S(T)∪{h,v} and also, that every support vertex is adjacent to exactly one leaf. We have now some possible scenarios. Case 3.1 ∣N(s)∩S(T)∣=1. Hence, by the maximality of the path P(h,h′), it must happen that T has an induced subgraph isomorphic to a graph Qr, as previously described, obtained from the vertices v,s,h and some supports, say s1,s2,…sr∈S(T), with the leaves h1,h2,…,hr, adjacent to the supports s1,s2,…sr, respectively, and such that {s1,…,sr,h1,…,hr}∩P(h,h′)=∅. Assume r=1. Note that s,s1∈D and that h,h1∉D. Let T′=T−h. Notice that D is also a TC-ID set in T′, and so γt(T′)≤γt,coi(T′)≤γt,coi(T)=γt(T) (2) (by using Theorem 16 and hypothesis). On the other hand, let A be a γt(T′)-set containing no leaves. We observe that s1∈A because s1 is a support in T′, and s∈A because δ(s1)=2. Thus, A is clearly also a total dominating set in T. Hence γt(T)≤∣A∣=γt(T′). Thus, all the inequalities in the relation (2) must be equalities, from which follows γt,coi(T′)=γt(T′) and by the inductive hypothesis T′∈F. Since T can be obtained from T′ by operation F1, we obtain T∈F. Assume now r≥2. Note that s,s1,…,sr∈D and that there is a leaf at distance three from sr. Let T′=T−hr. Hence, D−{sr} is a TC-ID set in T′, and so γt(T′)≤γt,coi(T′)≤γt,coi(T)−1=γt(T)−1 (by using Theorem 16 and hypothesis). Moreover, the set D−{sr} is a γt(T′)-set, otherwise we would find a total dominating set of T of cardinality smaller than γt(T), which is not possible. So, γt(T′)=γt(T)−1 which leads to γt,coi(T′)=γt(T′), as in the previous case. Now, by the inductive hypothesis T′∈F, and since T can be obtained from T′ by operation F2, we deduce T∈F. Case 3.2 ∣N(s)∩S(T)∣>1. An analogous procedure to the one above (Case 3.1) leads to our desired conclusion, based on the fact that s must have at least two neighbors s1′,s1″∈S(T) and there are at least two induced subgraphs isomorphic to the graphs Qr′ and Qr″, which can be used instead of Qr of Case 3.1. Case 3.3 ∣N(s)∩S(T)∣=0. Clearly, s has degree two since it has one leaf neighbor, no support neighbors and cannot have more than one (it has exactly one) semi-support neighbor due to the maximality of P(h,h′). Also, it must happen v∈D, h∈B and s∈D. Assume the subgraph induced by P(h,h′) is hsvu1u2u3u4…s′h′, where h,h′∈L(T) and s,s′∈S(T). Note that, if u1 is not a semi-support, then N(v)⊂S(T)∪{u1}. We consider again some possible scenarios. Case 3.3.1 ∣N(v)∩S(T)∣>1. In this case, the vertex v is also totally dominated by another support sv different from s. Let hv be the leaf adjacent to the support sv. Notice that D′=D−{s} is a TC-ID set of T′=T−h. Moreover, we note that the vertex s is a leaf in T′ having distance three to the leaf hv. So, by using a similar procedure as above (Case 3.1 and r≥2) we obtain T′∈F. Therefore, due to that T can be obtained from T′ by operation F2, it follows T∈F. Case 3.3.2 ∣N(v)∩S(T)∣=1 and ∣N(u1)∣≥3. Clearly s,v have degree two and belong to D. We firstly consider the case whether u1∈D. By Remark 18 we note that N(u1)∩L(T)≠∅ (which is not possible since ∣N(v)∩S(T)∣=1), or there is a vertex w∈D with N(w)∩D={u1}. Hence, we assume that N(u1)∩L(T)=∅, and that there is a vertex w∈D such that N(w)∩D={u1}. We note that, by using Remark 18, (N(w)−{u1})∩L(T)≠∅. Let w′∈(N(w)−{u1})∩L(T). Hence, D′=D−{s,v} is a TC-ID set in T′=T−{h,s}. Thus, γt(T′)≤γt,coi(T′)≤γt,coi(T)−2=γt(T)−2 (by using Theorem 16 and hypothesis). Now, let D″ be a γt(T′)-set such that u1,w∈D″ and v,w′∉D″ (such set exists because v,w′ are leaves of T′). We easily notice that D″∪{v,s} is a total dominating set of T. Thus, γt(T)≤γt(T′)+2 which leads to the equality γt(T)=γt(T′)+2. Consequently, we deduce from the above, that γt,coi(T′)=γt(T′), and by inductive hypothesis, T′∈F. We also observe that v has distance three to the leaf r′. This means T can be obtained from T′ by operation F3, and so T∈F. Now, consider the case in which u1∈B. By the maximality of P(h,h′) and by the fact that ∣N(u1)∣≥3, there is a leaf distinct of h at distance two or three from u1. Hence, the set D′=D−{s,v} is a TC-ID set in T′=T−{h,s,v}. By using a similar procedure as in Case 3.1 where r≥2 we can prove that T′∈F and, due to that now T can be obtained from T′ by operation F4, we get T∈F. Case 3.3.3 ∣N(v)∩S(T)∣=1and ∣N(u1)∣=2. Clearly s,v,u1 have degree two and s,v belong to D. We only consider the case whether u1∈B, otherwise u1∈D implies that u2 is a leaf and u1 is a support, which is not possible since ∣N(v)∩S(T)∣=1. As u1∈B, we get u2∈D. Notice that, as u2 has to be totally dominated, there exist a vertex q∈D such that N(q)∩D={u2}. So, by Remark 18 and Lemma 19, it follows (N(q)−{u2})⊂L(T)∪V2,3(T). If (N(q)−{u2})∩L(T))≠∅, then this case is analogous to the Case 3.3.2 and u1∈B. If (N(q)−{u2})⊂V2,3(T), then we see that u1∈V6(T). So, the set D′=D−{s,v} is a TC-ID set in T′=T−{h,s,v} and also u1∈V6(T′). Again, by using a similar procedure as in Case 3.1 where r≥2, we prove that T′∈F. Finally, due to that T can be obtained from T′ by operation F5, we have T∈F, which completes the proof.□ As an immediate consequence of Lemmas 20 and 21, we have the following characterization. Theorem 22 Let Tbe a tree. Then T∈Tγtif and only if T∈F. We next see that all the operations F1 to F4 are required in the characterization above. First, we see that operation F1 is required to obtain a double star from the path P4. The operations F2,F3,F4 are required to obtain the paths P5,P6,P7, respectively, from the path P4, and the path P10 can only be obtained from P4 by a sequence of operations F4,F5. 5. CONCLUDING REMARKS We have studied several combinatorial and complexity properties of the total co-independent domination number of graphs. As a consequence of the study, a couple of questions could be remarked as possible future research lines. We have proved that computing the total co-independent domination number of graphs is NP-hard even when restricted to planar graphs of maximum degree at most 3. However, it would be interesting to find some non-trivial families of graphs in which the problem above can be solved in polynomial time. On the other hand, the bounds of Theorem 8 together with the fact that the problem of computing the vertex cover number can be approximated within a factor of 2, allow to claim that the problem of computing the total co-independent domination number can be approximated within a constant factor. In this sense, it would be interesting to give some other approximation (or inapproximation) results on this parameter. We have characterized the family of graphs achieving the upper bound of Theorem 8. According to the construction of such family, it seems one could also characterize the graphs G for which γt,coi(G)=2α(G)−k for some values of k like for instance k=2 or k=3. Moreover, it would be of interest to characterize the family of graphs attaining the lower bound of Theorem 8 (note that for instance the trees satisfying such bound were characterized in [13]). REFERENCES 1 Henning , M.A. ( 2009 ) A survey of selected recent results on total domination in graphs . Discrete Math. , 309 , 32 – 63 . Google Scholar CrossRef Search ADS 2 Henning , M.A. and Yeo , A. ( 2013 ) Total Domination in Graphs . Springer , New York, USA . Google Scholar CrossRef Search ADS 3 Berge , C. ( 1962 ) The Theory of Graphs and its Applications . Wiley, Methuen , London . 4 Ore , O. ( 1962 ) Theory of graphs . Amer. Math. Soc. Transl. Ser. 2 , 38 , 206 – 212 . 5 Löwenstein , C. ( 2010 ) In the complement of a dominating set. PhD Dissertation, Technische Universitat Ilmenau, Germany. 6 Abdollahzadeh Ahangar , H. , Samodivkin , V. and Yero , I.G. ( 2016 ) Independent transversal dominating sets in graphs: complexity and structural properties . FILOMAT , 30 , 293 – 303 . Google Scholar CrossRef Search ADS 7 Cabrera Martínez , A. , Sigarreta , J.M. and Yero , I.G. ( 2017 ) On the independence transversal total domination number of graphs . Discrete Appl. Math. , 219 , 65 – 73 . Google Scholar CrossRef Search ADS 8 Hamid , I.S. ( 2012 ) Independent transversal domination in graphs . Discuss. Math. Graph Theory , 32 , 5 – 17 . Google Scholar CrossRef Search ADS 9 Soner , N.D. , Dhananjaya Murthy , B.V. and Deepak , G. ( 2012 ) Total co-independent domination in graphs . Appl. Math. Sci. , 6 , 6545 – 6551 . 10 Krzywkowski , M. ( 2014 ) Total outer-independent domination in graphs. Unpublished manuscript. 11 Garey , M.R. and Johnson , D.S. ( 1979 ) Computers and Intractability: A Guide to the Theory of NP-Completeness . W. H. Freeman & Co , New York, USA . 12 Gallai , T. ( 1959 ) Über extreme Punkt- und Kantenmengen . Ann. Univ. Sci. Budapest. Eötvös Sect. Math. , 2 , 133 – 138 . 13 Cabrera-Martínez , A. , Hernández-Mira , F.A. , Sigarreta Almira , J.M. and Yero , I.G. ( 2017 ) A note on total co-independent domination in trees. Unpublished manuscript. 14 Cockayne , E.J. , Dawes , R.M. and Hedetniemi , S.T. ( 1980 ) Total domination in graphs . Networks , 10 , 211 – 219 . Google Scholar CrossRef Search ADS Footnotes 1 Notice that the condition of V−D to be not empty is not exactly necessary. However, if such condition is not required, then we readily seen that the only graphs containing a TC-ID set of minimum cardinality with empty complement are the union of paths P2. Author notes Handling editor: Daniel Paulusma © The British Computer Society 2018. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png The Computer Journal Oxford University Press

On Computational and Combinatorial Properties of the Total Co-independent Domination Number of Graphs

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Abstract

Abstract A subset D of vertices of a graph G is a total dominating set if every vertex of G is adjacent to at least one vertex of D. The total dominating set D is called a total co-independent dominating set if the subgraph induced by V−D is edgeless and has at least one vertex. The minimum cardinality of any total co-independent dominating set is the total co-independent domination number of G and is denoted by γt,coi(G). In this work we study some complexity and combinatorial properties of γt,coi(G). Specifically, we prove that deciding whether γt,coi(G)≤k for a given integer k is an NP-complete problem and give several bounds on γt,coi(G). Moreover, since any total co-independent dominating set is a total dominating set, we characterize all the trees having equal total co-independent domination number and total domination number. 1. INTRODUCTION Problems concerning domination in graphs are one of the most popular and highly investigated issues in the area of graph theory, and a rich literature on the topic is nowadays known in the research community. Such problems run from the theoretical point of view till several practical applications in real-life situations. The most interesting cases of such applications are probably regarding problems in computer science. One of the most interesting features of domination in graphs involves the existence of a very high number of variants of domination parameters. A very common combination of domination and vertex independence of graphs is known, and the way of combining both concepts groups a considerable number of possibilities. In this work, we center our attention into precisely study one of these combinations between domination and independence in graphs, namely the total co-independence domination parameter. We focus the investigation on some computational complexity aspects of this parameter as well as on combinatorial properties of it. Given a graph G with vertex set V(G) and edge set E(G), a set D⊂V(G) is a total dominating set of G if every vertex in V(G) is adjacent to at least one vertex in D. The total domination number of G is the minimum cardinality of any total dominating set in G and is denoted by γt(G). A γt(G)-set is a total dominating set of cardinality γt(G). For more information on total domination we suggest the relatively recent and fairly complete survey [1] and the book [2]. A set S of vertices is independent if S induced an edgeless graph. An independent set of maximum cardinality is a maximum independent set of G. The independence number of G is the cardinality of a maximum independent set of G and is denoted by β(G). An independent set of cardinality β(G) is called a β(G)-set. Relationships between (total) domination and independence in graphs have attracted the attention of several researchers in the last years. Several interesting connections among these parameters include independent dominating sets [3, 4], partitions into a dominating set and an independent set [5], (total) dominating sets which intersect every maximal independent set [6–8], and some other ones more, which we prefer to not mention here, since it is not the goal of this work. A total dominating set D of a graph G is called a total co-independent dominating set (or TC-ID) if the set of vertices of the subgraph induced by V−D is independent and not empty.1 The minimum cardinality of any TC-ID set is the total co-independent domination number of G and is denoted by γt,coi(G). A TC-ID set of cardinality γt,coi(G) is a γt,coi(G)-set. These concepts were previously introduced and barely studied in [9]. Moreover, in [10], the same parameter was introduced under the name of total outer-independent domination number. Since this article ([10]) is not published in any journal and the other one ([9]) is already published, we precisely follow the terminology and notation of the latter. Since total domination is not defined for graphs having isolated vertices, all the graphs considered herein have not isolated vertices. Moreover, in order to satisfy the total domination property and that the complement of a TC-ID set is not empty, it is required that 2≤γt,coi(G)≤n−1, if n is the order of G. Such trivial bounds were already noted in the work [9]. Throughout our exposition, we consider G=(V,E) as a simple graph of order n and size m. That is, graphs that are finite, undirected, and without loops or multiple edges. Given a vertex v of G, NG(v) represents the open neighborhood of v, i.e. the set of all neighbors of v in G and the degree of v is δ(v)=∣NG(v)∣. The minimum and maximum degrees of G are denoted by δ(G) and Δ(G), respectively (or δ and Δ, respectively, for short). If X and Y are two subsets of V(G), then we denote the set of all edges of G joining a vertex of X with a vertex of Y by E(X,Y). For a set S⊂V(G), the complement of S is S¯=V(G)⧹S. In this work, we represent an edgeless graph G of order n as Nn. For any other graph theory terminology and notation we follow the book [2]. Let T be a tree (a connected graph without cycles). A leaf or a pendant vertex of T is a vertex of degree one (it is similarly defined for non-tree graphs). A support vertex of T is a vertex adjacent to a leaf and a semi-support vertex is a vertex adjacent to a support vertex that is neither a leaf nor a support. By an isolated support vertex of T we mean an isolated vertex of the subgraph induced by the support vertices of T. The set of leaves of T is denoted by L(T), the set of support vertices by S(T), and the set of semi-support vertices by SS(T). Moreover, S*(T) is the set of isolated support vertices of T. We first notice that if H1, H2,…, Hr with r≥2, are the connected components of a graph H, then any TC-ID set of minimum cardinality in H is formed by a minimum total dominating set in the subgraphs Hj where ∣V(Hj)∣=2 and a minimum TC-ID set in the remaining subgraphs Hi with ∣V(Hi)∣≥3. That is stated in the following straightforward result. Remark 1 Let H1, H2,…, Hr with r≥2, be the connected components of a graph H different from the union of r copies of the path P2. Then γt,coi(H)=∑i∈{1,…,r}∣V(Hi)∣=2γt(Hi)+∑j∈{1,…,r}∣V(Hj)∣≥3γt,coi(Hj). In concordance with the result above, from now on, we only consider the study of the TC-ID sets of connected graphs and omit to refer to that fact throughout all our exposition. 2. COMPLEXITY OF THE DECISION PROBLEM We begin our exposition by considering the problem of deciding whether the total co-independent domination number of a graph is less than a given integer. That is stated in the following decision problem. TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r TOTAL CO-INDEPENDENT DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γt,coi(G) is less than r In order to deal with the complexity of the TOTAL CO-INDEPENDENT DOMINATION PROBLEM (TC-ID PROBLEM), we make a reduction from a very well-known decision problem concerning the independence number of graphs. MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r MAXIMAL INDEPENDENT SET PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether the independence number of G is larger than r The problem above is one of the classical NP-complete problems appearing in the book [11]. Moreover, it remains NP-complete even when restricted to planar graphs. Theorem 2 MAXIMAL INDEPENDENT SET PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3 ([11]). Now on, in order to present our complexity results we need to introduce a family of graphs which is next defined. Let T6 be a tree with six vertices having two adjacent vertices u,v of degree three and the other four u1,u2,v1,v2 vertices are leaves. Clearly, each vertex of degree three has two adjacent leaves, say u1,u2∈N(u) and v1,v2∈N(v) (see Fig. 1(I)). Given a graph G of order n and n trees T6(1),…,T6(n) isomorphic to the tree T6, the graph GT is constructed by adding edges between the ith-vertex of G and the vertex u of the ith-tree T6(i). See Fig. 1 (II) for an example. Figure 1. View largeDownload slide The graph T6 (a) and a graph GT (b) where G is a complete graph minus one edge. Figure 1. View largeDownload slide The graph T6 (a) and a graph GT (b) where G is a complete graph minus one edge. We are now able to prove the NP-completeness of the TC-ID PROBLEM. Theorem 3 TOTAL CO-INDEPENDENT DOMINATION PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3. Proof The problem is clearly in NP since verifying that a given set is indeed a TC-ID set can be done in polynomial time. Let us now make a reduction from the MAXIMAL INDEPENDENT SET PROBLEM. Let G be a not edgeless graph of order n and construct the graph GT as described above. Let us denote by u(i),v(i) the vertices of degree three in the ith copy T6(i) of the tree T6 used to generate GT. We shall prove that γt,coi(GT)=3n−β(G). Let A be a β(G)-set and let D be the set of vertices of GT obtained from the complement of A in G together with the vertices u,v belonging to all the copies of the tree T6 used to generate GT, that is D=(V(G)⧹A)∪{⋃i=1n{u(i),v(i)}}. This set is clearly a total dominating set and its complement is an independent set. Thus, D is a TC-ID set in GT and, as a consequence, γt,coi(GT)≤n−∣A∣+⋃i=1n{u(i),v(i)}=3n−β(G). On the other hand, let D′ be a γt,coi(GT)-set. In order to totally dominate the leaves of every copy of T6 in GT, it must happen that ∣D′∩V(T6(i))∣≥2 for every i∈{1,…,n}. Also, V(G)∩D′≠∅, since otherwise the complement of D′ would not be independent. Moreover, the complement of V(G)∩D′ in G is an independent set in G. Thus, β(G)≥n−∣V(G)∩D′∣ and we obtain the following. γt,coi(GT)=∣D′∣=∣D′∩V(G)∣+⋃i=1n(D′∩V(T6(i)))≥n−β(G)+2n=3n−β(G). As a consequence, it follows that γt,coi(GT)=3n−β(G). Now, for j=3n−k, it is readily seen that γt,coi(GT)≤j if and only if β(G)≥k, which completes the reduction. We also observe that, if G is a planar graph, then GT is also planar. Therefore, since the MAXIMAL INDEPENDENT SET PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3, we also deduce that the TC-ID PROBLEM is NP-complete even when restricted to planar graphs of maximum degree at most 3 and the proof is completed.□ As a consequence of the result above, we deduce the following consequence. Corollary 4 The problem of computing the total co-independent domination number of graphs is NP-hard even when restricted to planar graphs of maximum degree at most 3. 3. BOUNDING THE TOTAL CO-INDEPENDENT DOMINATION NUMBER In order to present the first bounds for γt,coi(G) of any graph G, we need the next concepts. A set S of vertices of G is a vertex cover of G if every edge of G is incident with at least one vertex of S. The vertex cover number of G, denoted by α(G), is the smallest cardinality of a vertex cover of G. We refer to an α(G)-set in G as a vertex cover of cardinality α(G). The following well-known result, due to Gallai [12], states the relationship between the independence number and the vertex cover number of a graph. Theorem 5 For any graph Gof order n, α(G)+β(G)=n (Gallai [12]). On the other hand, it was shown in [9] the following relationship between γt,coi(G) and β(G). Theorem 6 For any graph Gof order n, γt,coi(G)≥n−β(G) ([9]). By using the two theorems above, we can easily deduce the lower bound of our next result. However, an upper bound for γt,coi(G) in terms of the vertex cover number can also be deduced. We first consider the case whether G is a star graph Sn for which is known that γt,coi(Sn)=2 and α(Sn)=1. For our purposes, an star graph Sn is a graph of order n+1 in which all but one of its vertices are leaves, unless n=1 where both vertices of S1 are leaves. Remark 7 For any star graph Sn, γt,coi(Sn)=2=2α(Sn). In concordance with the remark above, for our next result we exclude the case of star graphs and see that they behave in a different manner. Theorem 8 For any graph Gof order ndifferent from a star graph, α(G)≤γt,coi(G)≤2α(G)−1. Proof The lower bound follows from Theorems 5 and 6. If α(G)≥n/2, then γt,coi(G)≤n−1≤2(n/2)−1≤2α(G)−1. Thus, from now on in this proof we consider α(G)<n/2. Now, let C be an α(G)-set. We choose two vertices u,v∈C with the minimum possible distance between them and let P be a shortest u−v path. Clearly, V(P)∩C={u,v} and the distance between u and v is one or two (notice this also means 2≤∣V(P)∣≤3). For each vertex x∈C−{u,v}, choose a neighbor x′ of x. Then C∪V(P)∪{x′:x∈C−{u,v}} is a TC-ID set of cardinality 2∣C∣+∣V(P)∣−4≤2α(G)−1, which completes the proof of the upper bound.□ The bounds above are tight. For instance, a characterization of that trees achieving the equality in the lower bound was given in [13] (note that in [13] the trees T of order n satisfying equality in the bound γt,coi(T)=n−β(T) were characterized, which equals the lower bound of Theorem 8, in concordance with Theorem 5). The upper bound is attained for an infinite family of graphs, as we next show. To this end, we need the following operations for edges or induced paths P3 of a graph G. Subdivision: Given an edge uv, remove the edge, add a vertex w and the edges uw, wv. Inflation of size k: Given an induced path P3=uvw of G, in which v has degree two, remove the vertex v and the two incident edges, and replace them with k vertices v1,v2,…,vk and edges uvi,viw for every i∈{1,…,k}. Addition of tpendant vertices: Given a vertex x add t new vertices y1,…,yt and the edges xyi for every i∈{1,…,t}. Now, a graph Hn,a,b∈F1 is a graph obtained from a star graph Sn by making the following sequence of operations, which we will call as Sequence I. Apply the operation ‘Subdivision’ to a ( 1≤a≤n) edges of Sn. Apply the operation ‘Inflation of size ki’ with ki≥2 to b ( 0≤b≤a) paths P3(i) obtained from (i). Apply the operation ‘Addition of qipendant vertices’, qi≥0, to the b vertices corresponding to leaves of Sn obtained in the step (ii). Apply the operation ‘Addition of tipendant vertices’, ti≥1, to the leaves belonging to the remaining a−b paths obtained from (i), which were not ‘inflated’ in (ii). If a=n and b=0 (notice that in this case Hn,a,b is a tree such that the central vertex of the original star graph Sn has no adjacent leaves), then apply the operation ‘Addition of tpendant vertices’, t≥1, to the vertex corresponding to the central vertex of Sn. If a=n and b>0, then apply the operation ‘Addition of tpendant vertices’, t≥0, to the vertex corresponding to the central vertex of Sn. As an example, to obtain the cycle C4 (which belongs to F1) we begin with the star S1 (a path P2), next we apply the operation ‘Subdivision’ to the unique edge of S1 and then we apply the operation ‘Inflation of size 2’ to the path P3 obtained in the previous step. Note that different sequences of operations would lead to the same graph. For instance, the graph P5 can be obtained from the star S1 by subdividing its unique edge and then adding a pendant vertex to the leaf corresponding to the subdivision, as well as another pendant vertex to the center of S1 (coincidentally such center is also a leaf). Moreover, the graph P5 is obtained from the star P3 by subdividing one of its edges and then adding a pendant vertex to the leaf corresponding to such subdivision. On the other hand, we remark that three integers n,a,b would produce different graphs Hn,a,b depending on the addition of pendant vertices that would be done. However, since it is not significant for our work to denote them, we skip to use the notations for the addition of pendant vertices. A fairly representative graph of the family F1 is given in Fig. 2. Remark 9 For any graph Hn,a,b∈F1, α(Hn,a,b)=a+1 and γt,coi(Hn,a,b)=2a+1. Proof For any edge of Sn which was subdivided in step (i), it appears either a path P4 or a cycle C4 and all these subgraphs have in common only one vertex (the corresponding one to the center of Sn). Thus, in order to cover all the edges of Hn,a,b, at least a+1 vertices are required. Moreover, a set given by those a leaves corresponding to the a edges of the star Sn, which were subdivided, together with the central vertex form a vertex cover of cardinality a+1. Thus, the equality α(Hn,a,b)=a+1 follows. Now, let D be a γt,coi(Hn,a,b)-set. We analyze the following situations for every edge wu (assume w is the center of Sn) of the star which is initially subdivided. Case 1: There is only one path between w and u in Hn,a,b. Hence, the edge wu was subdivided with a vertex, say v, and not inflated, which made a required addition of at least one pendant vertex, say u′, to the leaf u. Thus, in order to totally dominate u′, ∣D∩{v,u,u′}∣≥2. Case 2: There are at least two paths between w and u in Hn,a,b. Clearly, this means wu was subdivided and then inflated with at least two vertices, say v1,…,vr, r≥2. Moreover, if some pendant vertices were added to u, then in order to totally dominate v1,…,vr,u and the extra leaves adjacent to u, at least two vertices of v1,…,vr,u are required. We next consider the vertex w separately. If a<n, then the vertex w has a least one adjacent leaf which needs to be totally dominated. Thus, w must belong to D. On the contrary, if a=n, then we must consider the value b. If b=0, then no path P3 was inflated ( Hn,a,b is a tree) and so, by step (v), w has at least one adjacent leaf which needs to be totally dominated, which means w must belong to D again. Finally, we assume b>0. Thus, at least one path P3 was inflated and this creates a cycle of order four, say C4*, to which w belongs. Also, it may happen w has no adjacent leaves. Now, note that if w∉D, then the two vertices of C4* adjacent to w must belong to D, since D¯ is an independent set. Moreover, the fourth vertex of C4* must belong to D too, in order to get the vertices of D totally dominated. As a consequence, at least three vertices of the cycle are in D, which is equivalent to have in D the vertex w, one of its neighbors in C4* and the vertex of C4* which is not adjacent to w. Consequently, we can deduce that for any set of vertices of a subgraph of Hn,a,b, induced by the vertices obtained in a subdivision of one of the a leaves of Sn, and the corresponding addition of some pendant vertices (if it is the case), at least two of these vertices are in D. Moreover, note that at least one extra vertex is required, and to satisfy this we simply consider the central vertex w of Sn. Thus, γt,coi(Hn,a,b)=∣D∣≥2a+1. On the other hand, by using Theorem 8, we obtain that γt,coi(Hn,a,b)≤2α(Hn,a,b)−1=2a+1 and the equality follows for γt,coi(Hn,a,b).□ Figure 2. View largeDownload slide A graph H5,5,3∈F1 where the six bolded vertices form an α(H) set and gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H) set, which has cardinality 11. Figure 2. View largeDownload slide A graph H5,5,3∈F1 where the six bolded vertices form an α(H) set and gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H) set, which has cardinality 11. Now, a graph H∈F2 is a graph obtained from the cycle C6 by making the following sequence of operations, which we will call as Sequence II. Apply the operation ‘Addition of tipendant vertices’, ti≥0 and i∈{1,2,3}, to the three vertices, say v1,v2,v3, of a β(C6)-set, respectively. If there is a ti=0 from the above operation and the degree of vi is two, then apply the operation ‘Inflation of size k’ with k≥2 to one of the two possible paths of order three between vi and the other two vertices in {v1,v2,v3}−{vi}. Apply the operation ‘Inflation of size ki’ with ki≥1 and i∈{1,2,3} to the three possible paths of order three between v1,v2,v3. An example of a graph of the family F2 appears in Fig. 3. Figure 3. View largeDownload slide A graph H∈F2 where the three bolded vertices form an α(H)-set and the two gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H)-set, which has cardinality five. Figure 3. View largeDownload slide A graph H∈F2 where the three bolded vertices form an α(H)-set and the two gray vertices form a possible set to be added to the bolded vertices to get a γt,coi(H)-set, which has cardinality five. The following result concerning the values of α(H) and γt,coi(H) for graphs H∈F2 is straightforward to observe. Remark 10 For any graph H∈F2, α(H)=3 and γt,coi(H)=5. According to the Remarks above, we can easily check that the upper bound of Theorem 8 is achieved for any graph G∈F1∪F2. Moreover, we next prove that precisely the graphs of these families are the only ones achieving the upper bound of Theorem 8. To this end, we need the following two lemmas whose proofs can be made by using some similar techniques as in the proof of Theorem 8. Lemma 11 If a graph Gcontains an α(G)-set which is not independent, then γt,coi(G)≤2α(G)−2. Proof Let C be an α(G)-set which is not independent. We choose two adjacent vertices u,v∈C. For each vertex x∈C−{u,v}, choose a neighbor x′ of x. Then C∪{x′:x∈C−{u,v}} is a TC-ID set of cardinality 2(∣C∣−2)+2≤2α(G)−2.□ Lemma 12 If an α(G)-set of a graph Gcontains four different vertices u1,u2,v1,v2such that a shortest u1−u2path and a shortest v1−v2path have length two and are vertex disjoint, then γt,coi(G)≤2α(G)−2. Proof Let C be an α(G)-set such that u1,u2,v1,v2∈C. For each vertex x∈C−{u1,u2,v1,v2}, choose a neighbor x′ of x and let w1,w2∉C be two vertices such that w1∈N(u1)∩N(u2) and w2∈N(v1)∩N(v2), which exist by assumption. Then C∪{w1,w2}∪{x′:x∈C−{u1,u2,v1,v2}} is a TC-ID set of cardinality 2(∣C∣−4)+6≤2α(G)−2.□ Theorem 13 Let Gbe a graph of order nsuch that 2α(G)≤n. Then γt,coi(G)=2α(G)−1if and only if G∈F1∪F2. Proof In one hand, if G∈F1∪F2, then it clearly happens that γt,coi(G)=2α(G)−1 according to Remarks 9 and 10. On the second hand, assume γt,coi(G)=2α(G)−1 and let D be any α(G)-set. We first notice that D must induce an independent set according to Lemma 11. We shall now proceed by proving some partial claims that will further give our required conclusion. Claim 1 G has no triangles (cycles of order three). Proof of Claim 1 If there is a triangle, then, in order to cover all its edges, at least two of its vertices must belong to D. So, this cover is not an independent set. Thus, we get a contradiction by using Lemma 11.□ Claim 2 G has no induced cycles of order five or larger than six. Proof of Claim 2 Suppose G contains a cycle Cr with r=5 or r≥7. In order to cover all the edges of Cr, and since r≠6, there must be two vertices in D∩V(Cr) at distance one (which means D is not independent), or there are four different vertices u1,u2,v1,v2∈D∩V(Cr) such that a shortest u1−u2 path and a shortest v1−v2 path have length two and are vertex disjoint. Thus, we obtain contradictions by using Lemmas 11 and 12.□ As a consequence of the Claims above, we have that G can only contain cycles of order four or six. We first analyze the case in which G contains a cycle of order six. Let V(C6)={v1,…,v6} where v1∼v2∼…∼v6∼v1 ( u∼v means u,v are adjacent). According to Lemma 11, it must happen D∩V(C6) is independent. Thus, without loss of generality we assume D∩V(C6)={v1,v3,v5}. We consider now several situations. Suppose v2 has degree larger than two. If v2∼vj with j∈{4,6}, then G has a triangle, which is not possible. If v2∼v5 and D={v1,v3,v5}, then D∪{v2} is a TC-ID set of cardinality four and α(G)=3, a contradiction. If v2∼v5 and {v1,v3,v5}⊊D, then for each vertex x∈D−{v1,v3,v5}, choose a neighbor x′ of x and we observe that the set D∪{v2}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. Thus, v2 has a neighbor z∉V(C6). Since v2∉D, it must happen z∈D. Since z≁v1 and z≁v3 (otherwise there would be a triangle), we obtain a contradiction with Lemma 12 by using the vertices z,v1,v3,v5. As a consequence, v2 must have degree two, and by symmetry also v4,v6 are of degree two too. Suppose v1 has degree two. If D={v1,v3,v5}, then {v2,v3,v5,v6} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, for each vertex x∈D−{v1,v3,v5}, we choose a neighbor x′ of x and observe that the set (D−{v1})∪{v2,v6}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. So, v1 must have degree at least three and, by symmetry also v3,v5 are of degree at least three too. We consider now a vertex x∈N(v1)−{v2,v6}. Notice that x≠v3,v5 (otherwise there would be a triangle). Also, x≠v4 because v4 has degree two as stated before. Suppose x has degree larger than one and let x′∈N(x)−{v1}. Since D is independent and the edge xx′ must be covered by D, x′∈D. If x′≠v3 and x′≠v5, then we obtain a contradiction with Lemma 12 by using the vertices x′,v1,v3,v5 (notice that x′≁v1 since D is independent). As a consequence, we obtain that any neighbor x of v1 is either of degree one or has a neighbor in V(C6)−{v1}. We next consider the latter situation whether x′∈V(C6)−{v1}. Clearly, x′≠v2,v4,v6 because v2,v4,v6 have degree two. Suppose x is neighbor of v3 and of v5. If D={v1,v3,v5}, then {x,v1,v3,v5} is a TC-ID set of cardinality four and α(G)=3, a contradiction. We may now choose a neighbor y′ of y for every y∈D−{v1,v3,v5} and hence, observe that the set D∪{x}∪{y′:y∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, a contradiction. Thus, x is a neighbor of either v3 or v5, in which case, it happens x has degree two. By symmetry, we obtain similar conclusions for v3 and v5. That is, for any vi with i∈{1,3,5}, N(vi) is given by leaves or vertices of degree two. In the latter case, if x∈N(vi)−V(C6), then N(x)={vi,y} where y∈D∩V(C6)−{vi}. As a consequence, we observe that G can be obtained from a cycle C6 by the Sequence II of operations described above, or equivalently G∈F2. We may now consider the case in which G contains a cycle C4, but G does not contain the cycle C6. Claim 3 G does not contain vertex disjoint cycles. Proof of Claim 3 We directly obtain a contradiction by Lemma 12, since in this case there are four different vertices u1,u2,v1,v2 (two of them in one cycle, the other two in the other cycle) such that a shortest u1−u2 path and a shortest v1−v2 path have length two and are vertex disjoint.□ Thus, if G contains more than one cycle C4, then they are not vertex disjoint. Moreover, we can next see that not two adjacent vertices of a cycle can be in any other cycle. Claim 4 If two cycles C4 of G have exactly two vertices in common, then these vertices are not adjacent. Proof of Claim 4 Suppose there are two cycles C4 having two adjacent vertices in common. Assume the cycles are C4(1)=v1v2v3v4v1 and C4(2)=v1v2v5v6v1. Hence, we note that exactly three vertices of {v1,…,v6} must belong to D, otherwise there are two adjacent vertices in D. Indeed, such vertices are either v1,v3,v5 or v2,v4,v6, say for instance v1,v3,v5. If precisely D={v1,v3,v5}, then {v1,v2,v3,v5} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, we may choose a neighbor x′ of x for every x∈D−{v1,v3,v5}, and observe that the set D∪{v2}∪{x′:x∈D−{v1,v3,v5}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction.□ Now, according to the Claims above, if G contains more than one cycle C4, then only the following situations can occur. Any two cycles have exactly one vertex in common. Any two cycles have exactly two vertices in common which are not adjacent. Any two cycles have exactly three vertices in common. We note that the situation in which two cycles of G have exactly three vertices in common can be understood as G has three cycles with two vertices in common. We now turn our attention on the following. Claim 5 There is a vertex w∈D such that d(w,x)=2 for every x∈D−{w}. Proof of Claim 5 We first note that there are at least two vertices w,x∈D such that d(x,w)=2, otherwise there would be an edge not covered by D. Let h be a vertex adjacent to w and x. Suppose there is a vertex y∈D such that d(w,y)≠2 and d(x,y)≠2 (note that d(w,y)≠1 and d(x,y)≠1). Thus, since there are no cycles of order larger than four in G, there must happen one of the following situations. There is a shortest path joining yand hnot containing wnor x. Also, y is different from the neighbor of h, say h′, in such path. In such case, in order to cover the edge hh′, it must happen h′∈D. Thus, we obtain a contradiction by using Lemma 12 and the vertices h′,w,x,z where z is a vertex at distance two from x in the x−y path. Without loss of generality, there is a shortest path joining yand xcontaining w. Thus, there must be a vertex y′∈D belonging to this path such that d(y,y′)=2 (it cannot be d(y,y′)=1 since D is independent), otherwise there should be a not covered edge. Clearly w≠y′. Thus, we obtain a contradiction by using Lemma 12 and the vertices y,y′,w,x. As a consequence, the vertex y has distance two to x or to w. Moreover, if d(w,y)=2 and d(x,y)=2, then there is a cycle of order six, which is not possible, or there is a vertex y′∈N(y)∩N(x). In the latter case, if D={x,y,w}, then {x,y,w,y′} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Hence, we may assume {x,y,w}⊊D. Now, we choose a neighbor x′ of x for every x∈D−{x,y,w}, and observe that the set D∪{y′}∪{x′:x∈D−{x,y,w}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction. Thus, y has distance two to exactly one vertex of x and w. From now on, we assume d(y,w)=2. We next prove that for any vertex z∈D−{x,y,w}, it follows d(z,w)=2 too. If D={x,y,w}, then we are done. So, me may suppose there is a vertex z′∈D−{x,y,w} such that d(z′,w)≠2 (clearly d(z′,w)>2). Consider now the shortest path between z′ and w, say z′z1′z2′…zq′w. Notice that q≥2. In order to cover the edge z1′z2′, it must be z2′∈D. So, we obtain a contradiction by using Lemma 12 and the vertices x,w,z′,z2′. Therefore, for any vertex z∈D−{w}, we obtain that d(z,w)=2 and the claim is proved.□ Next step gives some result on the distances between any two vertices x,y∈D−{w}. Claim 6 For any two vertices x,y∈D−{w}, any shortest path between x and y passes through w. Proof of Claim 6 From Claim 5, we know that d(x,w)=d(y,w)=2. Thus d(x,y)≤4. Clearly d(x,y)>1, since x,y cannot be adjacent. Let x′,y′∈N(w) such that x′∈N(x) and y′∈N(y). If x∼y′ or y∼x′ (say x∼y′) and D={x,y,w}, then the set D={x,y,w,y′} is a TC-ID set of cardinality four and α(G)=3, a contradiction. Also, if x∼y′ and {x,y,w}⊊D, then we choose a neighbor z′ of z for every z∈D−{w,x,y} and observe that the set D∪{y′}∪{z′:z∈D−{w,x,y}} is a TC-ID set of cardinality 2(∣D∣−3)+4≤2α(G)−2, which is a contradiction. Thus, neither x∼y′ nor y∼x′. If there is a vertex z∈N(x)∩N(y), then wx′xzyy′w is a cycle C6 in G, which is not possible. Thus d(x,y)≠2. By using a similar reasoning, it can be deduced that d(x,y)≠3 and so, d(x,y)=4. If there is another path of length four between x and y not containing w, then we have one of the following situations. There is a vertex w′∈D such that x′,y′∈N(w′) (note that w′ must be in D in order to cover the edges w′y′, w′x′). In such case, we obtain a contradiction by using Lemma 12 and the vertices x,w,w′,y. There are three vertices x1,y1,w″≠x′,y′,w such that x1∈N(x), y1∈N(y) and x1,y1∈N(w″). In such situation, wx′xx1w″y1yy′w is an induced cycle of order eight in G, which is not possible. Similarly to the case above, if either x1=x′ or y1=y′, then we obtain an induced cycle of order six in G, which is also not possible. Therefore, any shortest path between x and y passes throughout w.□ We now give several facts which are consequences of the Claims above, in order to deduce the structure of the graph G. The set V(G)−D is independent (otherwise there is an edge not covered by D). If x,y∈D−{w}, then N(x)∩N(y)=∅. If z∈N(x) for some x∈D−{w}, then either z∈N(w) and z has degree two, or z is a vertex of degree one (otherwise we get a contradiction with Claim 6). If z′∈N(w) is not a vertex of degree one, then there is exactly one vertex x∈D such that N(z′)={w,x} (equivalently z′ has degree two). As a consequence of the items above, as well as from the Claims, and all the reasoning till this point, we observe that D is formed by w and a set of vertices v1,v2…,vr (satisfying the properties above). Clearly, for any vertex vi, the set of its neighbors are either leaves or vertices of degree two adjacent to w. Moreover, if vi has only one neighbor of degree two, then it must have at least one adjacent leaf (otherwise one can find a cover set of smaller cardinality). In this sense, such set of vertices can clearly be obtained from a leaf of a star by making a subdivision of the corresponding edge, an inflation of the path P3 obtained from the subdivision and a subsequent addition of some pendant vertices. On the other hand, if w has some adjacent leaves, then they could be obtained directly from a star, if subdivisions were not done to all the leaves of the star or, by a subsequent addition of leaves to the center of the original star, if all its leaves would have been subdivided. Therefore, it is then concluded that the graph G was obtained from a star by making the Sequence I of operations previously described, which means G∈F1 and the proof is completed.□ We close this section with two bounds for γt,coi(G) in terms of order, size and minimum and maximum degrees. Proposition 14 Let Gbe a graph of order n, minimum and maximum degrees δand Δ, respectively. Then γt,coi(G)≥nδΔ+δ−1. Proof Let D be a γt,coi(G)-set. Hence, the subgraph induced by V(G)−D is edgeless. So, (n−∣D∣)δ=(∣V(G)−D∣)δ≤E(V(G)−D,D)≤∣D∣(Δ−1). Furthermore, it follows that γt,coi(G)≥nδΔ+δ−1. □ Proposition 15. Let Gbe a graph of order n, size m, minimum and maximum degrees δ and Δ, respectively. Then γt,coi(G)≥2m+nδ3Δ+δ−2. Proof Let D be a γt,coi(G)-set. Hence, the subgraph induced by V(G)−D is edgeless. So, E(V(G)−D,D)+E(D,D)=m. Now, notice that E(V(G)−D,D)≤∣D∣(Δ−1) and E(D,D)≤∣D∣Δ−(n−∣D∣)δ2. Adding these inequations, we have m=E(V(G)−D,D)+E(D,D)≤∣D∣Δ−(n−∣D∣)δ2+∣D∣(Δ−1). Therefore, it follows that γt,coi(G)≥2m+nδ3Δ+δ−2. □ The two bounds above are attained for instance for the double stars Sk,k (a double star Sk,k is a tree with exactly two adjacent support vertices, each of degree k+1, and the remaining vertices are leaves, i.e. each support vertex is adjacent to k leaves), which has order 2(k+1), size m=2k+1, minimum degree δ=1, maximum degree Δ=k+1 and γt,coi(Sk,k)=2. 4. THE CASE OF TREES In order to easily proceed with our exposition, and based on the following known bound, from now on we say that a tree T belongs to the family Tγt, if γt,coi(T)=γt(T). Moreover, we assume in this section that ∣S(T)∣≥2, since the case ∣S(T)∣=0 ( T is a P2 and γt,coi(T) is not defined) and ∣S(T)∣=1 ( T is a star graph Sn and γt,coi(T)=2) are straightforward to study. Theorem 16 For any graph G, γt,coi(G)≥γt(G) ([9]). It is now our goal to characterize the family of trees achieving the equality in the bound above. To this end, we observe the following basic results, which can easily be obtained by using some known properties of minimum total dominating sets. Proposition 17 If Sis a minimal total dominating set of a connected graph G=(V,E), then each v∈Shas at least one of the following two properties ([14]). There exists a vertex w∈V−Ssuch that N(w)∩S={v}. The subgraph induced by S−{v}contains an isolated vertex. The next remark is one useful consequence of the proposition above. Remark 18 Let D be a γt,coi(T)-set of cardinality γt(T). Then, for every v∈D, at least one of the following conditions is satisfied. There exists a vertex u∈D such that N(u)∩D={v}. There exists a vertex w∈V−D such that N(w)∩D={v}. We may recall to notice that condition (ii) implies that vertex v is a support, because the set D¯ is independent. Lemma 19 Let T∈Tγtand let Dbe a γt,coi(T)-set containing no leaves. Then for every v∈V(T)−(D∪L(T))there exist a leaf hsuch that d(v,h)≤3. Proof Let v∈V(T)−(D∪L(T)). Since ∣N(v)∣≥2, we consider N(v)={v1,v2,…,vr} with r≥2. Clearly, N(v)⊂D since D¯ is independent. For every vi, with i∈{1,…,r}, by Remark 18, vi is adjacent to a leaf or there exist a vertex si∈D such that N(si)∩D={vi}. Hence, as si∈D, N(si)−{vi}⊂V(T)−D. We assume that for every i∈{1,…,r}, vi is not adjacent to a leaf h, otherwise d(v,h)=2. Now, we suppose that (N(si)−{vi})∩L(T)=∅. Also note that, by condition above, the vertices belonging to N(si) are totally dominated by other vertices of D. So, we observe that the set (D−{s1,s2,…,sr})∪{v} is a total dominating set of T of cardinality smaller than ∣D∣, a contradiction. Furthermore, there exist i∈{1,…,r} such that (N(si)−{vi})∩L(T)≠∅. Thus, for any h∈(N(si)−{vi})∩L(T), it follows d(v,h)=3, and this completes the proof.□ From this point, the set of leaves having distance three with respect to at least one other leaf is denoted by L3(T), and given a γt,coi(T)-set D, we denote by V2,3(T)⊂V(T)−D the set of vertices having distance two or three to some leaf and by V6(T)⊂V(T)−D the set of vertices having distance three to some vertex of V2,3(T). In order to provide a constructive characterization of the trees belonging to the family Tγt, we need the following five operations F1, F2, F3, F4 and F5 on a tree T (by attaching a path P to a vertex v of T we mean adding the path P and joining v to a vertex of P). Moreover, through all the next results we make use of the fact that any tree T always contains a γt,coi(T)-set which does not contain leaves. Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Operation F1: Attach a path P1 to a vertex of T, which is in some γt,coi(T)-set. Operation F2: Attach a path P1 to a vertex of T, which is in L3(T). Operation F3: Attach a path P2 to a vertex of T, which is in L3(T). Operation F4: Attach a path P3 to a vertex of T, which is in V2,3(T). Operation F5: Attach a path P3 to a vertex of T, which is in V6(T). Let F be the family of trees defined as F={T∣T is obtained from P4 by a finite sequence of operations F1,F2,F3,F4 or F5}. Figure 4 contains a fairly representative example of a tree T∈F. We first show that every tree of the family F belongs to the family Tγt. Lemma 20 If T∈F, then T∈Tγt. Proof We proceed by induction on the number r(T) of operations required to construct the tree T. If r(T)=0, then T=P4 and T∈Tγt. This establishes the base case. Hence, we now assume that k≥1 is an integer and that each tree T′∈F with r(T′)<k satisfies that T′∈Tγt. Let T∈F be a tree for which r(T)=k. Since T can be obtained from a tree T′∈F with r(T′)=k−1 by one of the operations F1,F2,F3,F4 or F5, we shall prove that T∈Tγt, by considering a γt,coi(T′)-set D′ containing no leaves and through the following situations. Case 1. T is obtained from T′ by operation F1. Let u be the vertex added to T′ in order to obtain T. Since u is a leaf of T and is adjacent to a vertex of D′, the set D′ remains to be a total dominating set in T. Moreover, D′ is a γt(T)-set, since otherwise we would find a total dominating set in T′ of cardinality smaller than γt(T′). On the other hand, since (V(T′)−D′)∪{u} is independent, we deduce D′ is a TC-ID set in T. Thus, γt,coi(T)≤∣D′∣=γt,coi(T′)=γt(T′)=γt(T) (by also using the inductive hypothesis). Thus, by Theorem 16, we get the equality γt,coi(T)=γt(T), which means T∈Tγt. Case 2. T is obtained from T′ by operation F2. Assume T is obtained from T′ by adding the vertex u and the edge uv where v∈L3(T′). As v∈L3(T′), there exist a path vu1u2h in T′ where h is a leaf and u1,u2 are support vertices adjacent to v,h, respectively. Now, in T, the vertices u2,v are supports and belong to any TC-ID set in T. Hence, the set D=D′∪{v} is a TC-ID set in T, and so γt(T)≤γt,coi(T)≤γt,coi(T′)+1=γt(T′)+1 (1) (by also using Theorem 16 and the inductive hypothesis). Now, let A be a γt(T)-set containing no leaves. Notice that the vertex v is a support and so, it belong to A, also the vertex u1 belongs to A too, because v has degree two. Moreover, note that the set A−{v} is a total dominating set in T′, which leads to γt(T′)≤γt(T)−1. By using this, it follows that all the inequalities in (1) must be equalities. Thus γt,coi(T)=γt(T), and T∈Tγt. Case 3. T is obtained from T′ by operation F3. Assume T is obtained from T′ by adding the path P2=h1h2 to a vertex v∈L3(T′) through the edge vh1. By using some similar reasons as in the case above (now we must use D=D′∪{v,h1} instead of D=D′∪{v}), it is observed that T∈Tγt. Case 4. T is obtained from T′ by operation F4. Assume T is obtained from T′ by adding the path P3=h1u1h2 to a vertex v∈V2,3(T′) through the edge vh1. We notice that u1,h1 belong to any TC-ID set containing no leaves of T. Hence, the set D=D′∪{u1,h1} is a TC-ID set in T. Thus γt(T)≤γt,coi(T)≤γt,coi(T′)+2=γt(T′)+2 (by also using Theorem 16 and the inductive hypothesis). Now, let A be a γt(T)-set. Since the vertex u1 is a support, it belongs to A and so, ∣A∩{h1,u1,h2}∣≥2. Moreover, note that ∣A∩V(T′)∣≥γt(T′). Hence, γt(T)=∣A∣≥γt(T′)+2. Again, as in Case 2, we deduce γt,coi(T)=γt(T), which means T∈Tγt. Case 5. T is obtained from T′ by operation F5. Assume T is obtained from T′ by adding the path P3=h1u1h2 to a vertex v∈V6(T′) through the edge vh1. By using some similar reasons as in the case above, it can be deduced that γt,coi(T)=γt(T), which gives T∈Tγt.□ Figure 4. View largeDownload slide A tree T obtained from a path P4=v1v2v3v4, applying the five operations F1, F2, F3, F4 and F5. Firstly, operations F4 and F5 are applied by adding the path P3=v5v6v7 to the vertex v4 through the edge v4v5 and the path P3=v8v9v10 to the vertex v7 through the edge v7v8. Next, we apply the operation F1 twice by attaching the vertices w1 and w2 to the vertices v3 and v8, respectively. Moreover, we apply the operation F2 by adding the vertex u3 to w2. Finally, we apply the operation F3 by adding the path P2=u1u2 to the vertex w1 through the edge w1u1. Figure 4. View largeDownload slide A tree T obtained from a path P4=v1v2v3v4, applying the five operations F1, F2, F3, F4 and F5. Firstly, operations F4 and F5 are applied by adding the path P3=v5v6v7 to the vertex v4 through the edge v4v5 and the path P3=v8v9v10 to the vertex v7 through the edge v7v8. Next, we apply the operation F1 twice by attaching the vertices w1 and w2 to the vertices v3 and v8, respectively. Moreover, we apply the operation F2 by adding the vertex u3 to w2. Finally, we apply the operation F3 by adding the path P2=u1u2 to the vertex w1 through the edge w1u1. We now turn our attention to the opposite direction concerning the lemma above. In this sense, from now on we shall need the following terminology and notation in our results. Given a tree T and a set S⊂V(T), by T−S we denote a tree obtained from T by removing from T all the vertices in S and all its incident edges (if S={v} for some vertex v, then we simply write T−v). For an integer r≥2, by Qr we mean a graph which is obtained from a path Pr+2=vss1s2…sr by attaching a path P1 to every vertex of Pr+2−v. In Fig. 5 we show the example of Q5. Figure 5. View largeDownload slide The structure of the tree Q5. Figure 5. View largeDownload slide The structure of the tree Q5. We next show that every tree of the family Tγt belongs to the family F. Lemma 21 If T∈Tγt, then T∈F. Proof We proceed by induction on the order n≥4 of the trees T∈Tγt. If T is a double star, then T can be obtained from P4 by repeatedly applying operation F1. This establishes the base case. We assume next that k>4 is an integer and that each tree T′∈Tγt with ∣V(T′)∣<k satisfies T′∈F. Let T be a tree such that T∈Tγt and ∣V(T)∣=k. Let D be a γt,coi(T)-set containing no leaves and let B=V(T)−D. We analyze the following situations. Case 1: ∣S(T)∣<∣L(T)∣. We consider a support vertex v that is adjacent to at least two leaves. Let h∈N(v)∩L(T) and T′=T−h. Thus, the set D is a γt(T′)-set too, and by inductive hypothesis, T′∈F. Therefore, since T can be obtained from T′ by operation F1, it follows T∈F. Case 2: ∣S(T)∣=∣L(T)∣ and ∣SS(T)∣=0. In this case we note that V(T)=S(T)∪L(T) and clearly, S(T) is a γt,coi(T)-set (moreover ∣S(T)∣≥3 since otherwise T is a double star). Let s∈S(T) such that ∣N(s)∩S(T)∣=1 (note that such s always exists) and let h∈L(T) be the leaf adjacent to s. We first notice that there exists a leaf having distance three to the support s. Thus, we deduce that S′(T)=S(T)−{s} is a γt,coi(T′)-set, where T′=T−h. By induction hypothesis T′∈F and, since T can be obtained from T′ by operation F2, we get T∈F. Case 3: ∣S(T)∣=∣L(T)∣ and ∣SS(T)∣>0. Herein we denote by P(x,y) the set of vertices of one shortest path between x and y, including x and y. Let h,h′ be two leaves at the maximum possible distance in T such that there is v∈SS(T)∩P(h,h′) with d(v,h)=2 or d(v,h′)=2. Without loss of generality assume that d(v,h)=2 and let s be the support adjacent to h. Since ∣S(T)∣=∣L(T)∣ and by the maximality of the path between h and h′, we observe that N(s)⊂S(T)∪{h,v} and also, that every support vertex is adjacent to exactly one leaf. We have now some possible scenarios. Case 3.1 ∣N(s)∩S(T)∣=1. Hence, by the maximality of the path P(h,h′), it must happen that T has an induced subgraph isomorphic to a graph Qr, as previously described, obtained from the vertices v,s,h and some supports, say s1,s2,…sr∈S(T), with the leaves h1,h2,…,hr, adjacent to the supports s1,s2,…sr, respectively, and such that {s1,…,sr,h1,…,hr}∩P(h,h′)=∅. Assume r=1. Note that s,s1∈D and that h,h1∉D. Let T′=T−h. Notice that D is also a TC-ID set in T′, and so γt(T′)≤γt,coi(T′)≤γt,coi(T)=γt(T) (2) (by using Theorem 16 and hypothesis). On the other hand, let A be a γt(T′)-set containing no leaves. We observe that s1∈A because s1 is a support in T′, and s∈A because δ(s1)=2. Thus, A is clearly also a total dominating set in T. Hence γt(T)≤∣A∣=γt(T′). Thus, all the inequalities in the relation (2) must be equalities, from which follows γt,coi(T′)=γt(T′) and by the inductive hypothesis T′∈F. Since T can be obtained from T′ by operation F1, we obtain T∈F. Assume now r≥2. Note that s,s1,…,sr∈D and that there is a leaf at distance three from sr. Let T′=T−hr. Hence, D−{sr} is a TC-ID set in T′, and so γt(T′)≤γt,coi(T′)≤γt,coi(T)−1=γt(T)−1 (by using Theorem 16 and hypothesis). Moreover, the set D−{sr} is a γt(T′)-set, otherwise we would find a total dominating set of T of cardinality smaller than γt(T), which is not possible. So, γt(T′)=γt(T)−1 which leads to γt,coi(T′)=γt(T′), as in the previous case. Now, by the inductive hypothesis T′∈F, and since T can be obtained from T′ by operation F2, we deduce T∈F. Case 3.2 ∣N(s)∩S(T)∣>1. An analogous procedure to the one above (Case 3.1) leads to our desired conclusion, based on the fact that s must have at least two neighbors s1′,s1″∈S(T) and there are at least two induced subgraphs isomorphic to the graphs Qr′ and Qr″, which can be used instead of Qr of Case 3.1. Case 3.3 ∣N(s)∩S(T)∣=0. Clearly, s has degree two since it has one leaf neighbor, no support neighbors and cannot have more than one (it has exactly one) semi-support neighbor due to the maximality of P(h,h′). Also, it must happen v∈D, h∈B and s∈D. Assume the subgraph induced by P(h,h′) is hsvu1u2u3u4…s′h′, where h,h′∈L(T) and s,s′∈S(T). Note that, if u1 is not a semi-support, then N(v)⊂S(T)∪{u1}. We consider again some possible scenarios. Case 3.3.1 ∣N(v)∩S(T)∣>1. In this case, the vertex v is also totally dominated by another support sv different from s. Let hv be the leaf adjacent to the support sv. Notice that D′=D−{s} is a TC-ID set of T′=T−h. Moreover, we note that the vertex s is a leaf in T′ having distance three to the leaf hv. So, by using a similar procedure as above (Case 3.1 and r≥2) we obtain T′∈F. Therefore, due to that T can be obtained from T′ by operation F2, it follows T∈F. Case 3.3.2 ∣N(v)∩S(T)∣=1 and ∣N(u1)∣≥3. Clearly s,v have degree two and belong to D. We firstly consider the case whether u1∈D. By Remark 18 we note that N(u1)∩L(T)≠∅ (which is not possible since ∣N(v)∩S(T)∣=1), or there is a vertex w∈D with N(w)∩D={u1}. Hence, we assume that N(u1)∩L(T)=∅, and that there is a vertex w∈D such that N(w)∩D={u1}. We note that, by using Remark 18, (N(w)−{u1})∩L(T)≠∅. Let w′∈(N(w)−{u1})∩L(T). Hence, D′=D−{s,v} is a TC-ID set in T′=T−{h,s}. Thus, γt(T′)≤γt,coi(T′)≤γt,coi(T)−2=γt(T)−2 (by using Theorem 16 and hypothesis). Now, let D″ be a γt(T′)-set such that u1,w∈D″ and v,w′∉D″ (such set exists because v,w′ are leaves of T′). We easily notice that D″∪{v,s} is a total dominating set of T. Thus, γt(T)≤γt(T′)+2 which leads to the equality γt(T)=γt(T′)+2. Consequently, we deduce from the above, that γt,coi(T′)=γt(T′), and by inductive hypothesis, T′∈F. We also observe that v has distance three to the leaf r′. This means T can be obtained from T′ by operation F3, and so T∈F. Now, consider the case in which u1∈B. By the maximality of P(h,h′) and by the fact that ∣N(u1)∣≥3, there is a leaf distinct of h at distance two or three from u1. Hence, the set D′=D−{s,v} is a TC-ID set in T′=T−{h,s,v}. By using a similar procedure as in Case 3.1 where r≥2 we can prove that T′∈F and, due to that now T can be obtained from T′ by operation F4, we get T∈F. Case 3.3.3 ∣N(v)∩S(T)∣=1and ∣N(u1)∣=2. Clearly s,v,u1 have degree two and s,v belong to D. We only consider the case whether u1∈B, otherwise u1∈D implies that u2 is a leaf and u1 is a support, which is not possible since ∣N(v)∩S(T)∣=1. As u1∈B, we get u2∈D. Notice that, as u2 has to be totally dominated, there exist a vertex q∈D such that N(q)∩D={u2}. So, by Remark 18 and Lemma 19, it follows (N(q)−{u2})⊂L(T)∪V2,3(T). If (N(q)−{u2})∩L(T))≠∅, then this case is analogous to the Case 3.3.2 and u1∈B. If (N(q)−{u2})⊂V2,3(T), then we see that u1∈V6(T). So, the set D′=D−{s,v} is a TC-ID set in T′=T−{h,s,v} and also u1∈V6(T′). Again, by using a similar procedure as in Case 3.1 where r≥2, we prove that T′∈F. Finally, due to that T can be obtained from T′ by operation F5, we have T∈F, which completes the proof.□ As an immediate consequence of Lemmas 20 and 21, we have the following characterization. Theorem 22 Let Tbe a tree. Then T∈Tγtif and only if T∈F. We next see that all the operations F1 to F4 are required in the characterization above. First, we see that operation F1 is required to obtain a double star from the path P4. The operations F2,F3,F4 are required to obtain the paths P5,P6,P7, respectively, from the path P4, and the path P10 can only be obtained from P4 by a sequence of operations F4,F5. 5. CONCLUDING REMARKS We have studied several combinatorial and complexity properties of the total co-independent domination number of graphs. As a consequence of the study, a couple of questions could be remarked as possible future research lines. We have proved that computing the total co-independent domination number of graphs is NP-hard even when restricted to planar graphs of maximum degree at most 3. However, it would be interesting to find some non-trivial families of graphs in which the problem above can be solved in polynomial time. On the other hand, the bounds of Theorem 8 together with the fact that the problem of computing the vertex cover number can be approximated within a factor of 2, allow to claim that the problem of computing the total co-independent domination number can be approximated within a constant factor. In this sense, it would be interesting to give some other approximation (or inapproximation) results on this parameter. We have characterized the family of graphs achieving the upper bound of Theorem 8. According to the construction of such family, it seems one could also characterize the graphs G for which γt,coi(G)=2α(G)−k for some values of k like for instance k=2 or k=3. Moreover, it would be of interest to characterize the family of graphs attaining the lower bound of Theorem 8 (note that for instance the trees satisfying such bound were characterized in [13]). REFERENCES 1 Henning , M.A. ( 2009 ) A survey of selected recent results on total domination in graphs . Discrete Math. , 309 , 32 – 63 . Google Scholar CrossRef Search ADS 2 Henning , M.A. and Yeo , A. ( 2013 ) Total Domination in Graphs . Springer , New York, USA . Google Scholar CrossRef Search ADS 3 Berge , C. ( 1962 ) The Theory of Graphs and its Applications . Wiley, Methuen , London . 4 Ore , O. ( 1962 ) Theory of graphs . Amer. Math. Soc. Transl. Ser. 2 , 38 , 206 – 212 . 5 Löwenstein , C. ( 2010 ) In the complement of a dominating set. PhD Dissertation, Technische Universitat Ilmenau, Germany. 6 Abdollahzadeh Ahangar , H. , Samodivkin , V. and Yero , I.G. ( 2016 ) Independent transversal dominating sets in graphs: complexity and structural properties . FILOMAT , 30 , 293 – 303 . Google Scholar CrossRef Search ADS 7 Cabrera Martínez , A. , Sigarreta , J.M. and Yero , I.G. ( 2017 ) On the independence transversal total domination number of graphs . Discrete Appl. Math. , 219 , 65 – 73 . Google Scholar CrossRef Search ADS 8 Hamid , I.S. ( 2012 ) Independent transversal domination in graphs . Discuss. Math. Graph Theory , 32 , 5 – 17 . Google Scholar CrossRef Search ADS 9 Soner , N.D. , Dhananjaya Murthy , B.V. and Deepak , G. ( 2012 ) Total co-independent domination in graphs . Appl. Math. Sci. , 6 , 6545 – 6551 . 10 Krzywkowski , M. ( 2014 ) Total outer-independent domination in graphs. Unpublished manuscript. 11 Garey , M.R. and Johnson , D.S. ( 1979 ) Computers and Intractability: A Guide to the Theory of NP-Completeness . W. H. Freeman & Co , New York, USA . 12 Gallai , T. ( 1959 ) Über extreme Punkt- und Kantenmengen . Ann. Univ. Sci. Budapest. Eötvös Sect. Math. , 2 , 133 – 138 . 13 Cabrera-Martínez , A. , Hernández-Mira , F.A. , Sigarreta Almira , J.M. and Yero , I.G. ( 2017 ) A note on total co-independent domination in trees. Unpublished manuscript. 14 Cockayne , E.J. , Dawes , R.M. and Hedetniemi , S.T. ( 1980 ) Total domination in graphs . Networks , 10 , 211 – 219 . Google Scholar CrossRef Search ADS Footnotes 1 Notice that the condition of V−D to be not empty is not exactly necessary. However, if such condition is not required, then we readily seen that the only graphs containing a TC-ID set of minimum cardinality with empty complement are the union of paths P2. Author notes Handling editor: Daniel Paulusma © The British Computer Society 2018. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices)

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The Computer JournalOxford University Press

Published: Apr 13, 2018

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