Blocking total dominating sets via edge contractions
aa r X i v : . [ c s . D M ] S e p Blocking total dominating sets via edge contractions
E. Galby, F. Mann, B. Ries
University of Fribourg, Department of Informatics,Fribourg, Switzerland
Abstract
In this paper, we study the problem of deciding whether the total domination numberof a given graph G can be reduced using exactly one edge contraction (called γ t ) ). We focus on several graph classes and determine the computationalcomplexity of this problem. By putting together these results, we manage to obtain acomplete dichotomy for H -free graphs (see Theorem 3). In this paper, we consider the problem of reducing the total domination number of a graphby contracting a single edge. More precisely, given a graph G = ( V, E ), we want to knowwhether there exists an edge e ∈ E such that the total domination number of the graph G ′ ,obtained from G by contracting the edge e , is strictly less than the total domination numberof G . This problem fits into the general framework of so-called blocker problems which havebeen studied intensively in the literature (see for instance [1, 2, 3, 4, 5, 7, 8, 9, 14, 16, 17, 18]).In this framework, we ask for a specific graph parameter π to decrease: given a graph G ,a set O of one or more graph operations and an integer k ≥
1, the question is whether G can be transformed into a graph G ′ by using at most k operations from O such that π ( G ′ ) ≤ π ( G ) − d for some threshold d ≥
0. Such problems are called blocker problems as theset of vertices or edges involved can be viewed as “blocking” the parameter π . Identifyingsuch sets may provide important information about the structure of the graph G . Blockerproblems can be seen as a kind of graph modification problems. Indeed, in such problemswe are usually interested in modifying a given graph G , via a small number of operations,into some other graph G ′ that has a certain desired property which often describes a certaingraph class to which G ′ must belong. Here we consider graph parameters instead of graphclasses.Blocker problems are also related to other well-known graph problems as shown for in-stance in [7, 16]. So far, the literature mainly focused on the following graph parameters: thechromatic number, the independence number, the clique number, the matching number, thevertex cover number and the domination number. Furthermore, the set O usually consistedof a single graph operation, namely either vertex deletion, edge contraction, edge deletion oredge addition. Since these blocker problems are usually NP -hard in general graphs, a par-ticular attention has been paid to their computational complexity when restricted to specialgraph classes.Recently, the authors in [9] studied the blocker problem with respect to the dominationnumber and edge contractions. More precisely, they consider the problem of deciding if for agiven connected graph G it is possible to obtain a graph G ′ by contracting at most k edges,where k ≥ γ ( G ′ ) ≤ γ ( G ) −
1, where γ represents the domination number?1or k = 1, they provided an almost dichotomy (only one family of graphs remained open)for this problem when restricted to H -free graphs. Very recently, this last open problem wassolved as well (see [10]). In this paper, we continue this line of research by considering thetotal domination number.More specifically, let G = ( V, E ) be a graph. The contraction of an edge uv ∈ E removesvertices u and v from G and replaces them by a new vertex that is made adjacent to preciselythose vertices that were adjacent to u or v in G (without introducing self-loops nor multipleedges). A graph obtained from G by contracting an edge e will be denoted by G/e . A set D ⊆ V is called a dominating set , if every vertex in V \ D has at least one neighbor in D .The domination number of a graph G , denoted by γ ( G ), is the smallest size of a dominatingset in G . A set D ⊆ V is called a total dominating set , if every vertex in V has at least oneneighbor in D . The total domination number of a graph G , denoted by γ t ( G ), is the smallestsize of a total dominating set in G . We consider the following problem in this paper: γ t ) Instance:
A connected graph G = ( V, E ). Question:
Does there exist an edge e ∈ V such that γ t ( G/e ) ≤ γ t ( G ) − ct γ t ( G ) the minimum number of edge contractions requiredto transform a given graph G into a graph G ′ such that γ t ( G ′ ) ≤ γ t ( G ) −
1. They provethat for a connected graph G , we have ct γ t ( G ) ≤
3. In other words, one can always reduceby at least 1 the total domination number of a connected graph G by using at most 3 edgecontractions ([13, Theorem 4.3]). They also prove the following theorem, which is a crucialresult for our work. Theorem 1 ([13]) . For a connected graph G , ct γ t ( G ) = 1 if and only if there exists aminimum total dominating D set in G such that the graph induced by D contains a P . As mentioned above, the authors in [9] considered the domination number, i.e. they con-sidered the problem above but with γ ( G ) instead of γ t ( G ) denoted by γ ) .In particular they showed that if H is not an induced subgraph of P + pP + tK , for p ≥ t ≥ γ ) is polynomial-time solvable on H -free graphs ifand only if H is an induced subgraph of P + tK , for t ≥
0. Recently, it was shown that theproblem can be solved in polynomial time in H -free graphs when H is an induced subgraphof P + pP + tK , for p, t ≥ Theorem 2. γ ) is polynomial-time solvable for H -free graphs ifand only if H is an induced subgraph of P + tK with t ≥ , or H is an induced subgraph of P + pK + tK with p, t ≥ . In this paper, we provide a complete dichotomy for γ t ) in H -freegraphs. Our main result is as follows. Theorem 3. γ t ) is polynomial-time solvable for H -free graphs ifand only if H is an induced subgraph of P + tK with t ≥ , or H is an induced subgraph of P + qP + pK + tK with q, p, t ≥ . It has been shown in [11] that the complexities of the problems
Dominating set (i.e.,given a graph G and an integer k ≥
0, does there exist a dominating set of size at most k ?) and Total dominating set (i.e., given a graph G and an integer k ≥
0, does thereexist a total dominating set of size at most k ?) agree in H -free graphs for any graph H .The results above show that there are not only hereditary but even monogenic graph classes2i.e. H -free graphs for some graph H ) for which the complexities of the problems γ t ) and γ ) differ.This paper is organised as follows. In Section 2, we present definitions and notationsthat are used throughout the paper. Section 3 is devoted to the hardness results of γ t ) while Section 4 presents cases when γ t ) is polynomial-time solvable. In Section 5 we put these results together to proof our mainresult, Theorem 2. We conclude the paper by presenting final remarks and future researchdirections in Section 6. Throughout this paper, we only consider graphs which are finite, simple and connected. Werefer the reader to [6] for any terminology and notation not defined here.For n ≥
1, the path and cycle on n vertices are denoted by P n and C n respectively. Apath on n vertices may also be called an n -path . The claw is the complete bipartite graphwith one partition of size one and the other of size three.Given a graph G , we denote by V ( G ) its vertex set and by E ( G ) its edge set. For any ver-tex v ∈ V ( G ), the neighborhood of v in G , denoted by N G ( v ) or simply N ( v ) if it is clear fromthe context, is the set of vertices adjacent to v , that is, N G ( v ) = { w ∈ V ( G ) : vw ∈ E ( G ) } ;the closed neighborhood of v in G , denoted by N G [ v ] or simply N [ v ] if it is clear from thecontext, is the set of vertices adjacent to v together with v , that is, N G [ v ] = N G ( v ) ∪ { v } .For any subset S ⊆ V ( G ), the neighborhood of S in G , denoted by N G ( S ) or simply N ( S ) ifit is clear from context, is the set ∪ v ∈ S N ( v ), and the closed neighborhood of S in G , denotedby N G [ S ] or simply N [ S ] if it is clear from the context, is the set N G ( S ) ∪ S . For an edge xy ∈ E ( G ), we may write N G ( xy ) (resp. N G [ xy ]) in place of N G ( { x, y } ) (resp. N G [ { x, y } ])for simplicity. Similarly, for a family F ⊆ E ( G ) of edges, we may write N G ( F ) (resp. N G [ F ])in place of ∪ e ∈F N G ( e ) (resp. ∪ e ∈F N G [ e ]) for simplicity. Let A, B ⊆ V ( G ). We say that A is complete (resp. anticomplete) to B , if every vertex in A is adjacent (resp. non adjacent)to every vertex in B . For a subset S ⊆ V ( G ), we let G [ S ] denote the subgraph induced by S , which has vertex set S and edge set { xy ∈ E ( G ) : x, y ∈ S } . Given a subset S ⊆ V ( G )and a graph H , we say that S contains an (induced) H if G [ S ] contains H as an (induced)subgraph. The length of a path in G is its number of edges. For any two vertices u, v ∈ V ( G ),the distance from u to v in G , denoted by d G ( u, v ) or simply d ( u, v ) if it is clear from thecontext, is the length of a shortest path from u to v in G . Similarly, for any two subset S, S ′ ⊆ V ( G ), the distance from S to S ′ in G , denoted by d G ( S, S ′ ) or simply d ( S, S ′ ) ifit is clear from the context, is the minimum length of a shortest from a vertex in S to avertex in S ′ , that is, d G ( S, S ′ ) = min x ∈ S,y ∈ S ′ d G ( x, y ). If S consists of a single vertex, say S = { x } , we may write d G ( x, S ′ ) in place of d G ( { x } , S ′ ) for simplicity. The k -subdivisionof an edge uw ∈ E ( G ) consists in replacing it with a path uv . . . , v k w , where v , . . . , v k arenew vertices.A subset K ⊆ V ( G ) is a clique of G if any two vertices of K are adjacent in G . A subset S ⊆ V ( G ) is an independent set of G if any two vertices of S are nonadjacent in G . Giventwo subsets S, S ′ ⊆ V ( G ), we say that S dominates S ′ if N ( v ) ∩ S = ∅ for every v ∈ S ′ .Given a vertex v ∈ V ( G ) and a subset S ⊆ V ( G ), we say that the vertex v dominates S ifthe set { v } dominates S . If D is a total dominating set of G and v ∈ D , we say that a vertex w ∈ V ( G ) is a private neighbour of v with respect to D , or simply a private neighbor of v ifit is clear from the context, if N ( w ) ∩ D = { v } . For a subset S ⊆ V ( G ), we say that a vertex w ∈ V ( G ) \ S is a private neighbour of S with respect to D , or simply a private neighbor of S if it is clear from the context, if N ( w ) ∩ D ⊆ S and | N ( w ) ∩ D | = 1. A subset D ⊆ V ( G )is a dominating set of G if every vertex in V ( G ) \ D is adjacent to at least one vertex in D ;3he domination number of G , denoted by γ ( G ), is the size of a minimum dominating set of G . The (Even) Dominating Set problem takes as input a graph G and an (even) integer k , and asks whether G has a dominating set of size at most k . The NP -hardness of these twoproblems follows from [12].For a family { H , . . . , H p } of graphs, we say that G is { H , . . . , H p } -free if G contains noinduced subgraph isomorphic to a graph in { H , . . . , H p } . If p = 1, we may write H -freein place of { H } -free for simplicity. The union of two simple graphs G and H is the graph G + H with vertex set V ( G ) ∪ V ( H ) and edge set E ( G ) ∪ E ( H ). The union of k disjointcopies of G is denoted by kG .For n ∈ N we denote by [ n ] the set { , . . . , n } . In this section, we will present several hardness results regarding γ t ) with respect to H -free graphs, where H is a family of at most two graphs. Theorem 4. γ t ) is NP -hard when restricted to { P , P + P } -freegraphs.Proof. We reduce from
Even Dominating Set with domination number at least 4. Givenan instance ( G, ℓ ) (with γ ( G ) ≥
4) of this problem, we construct an equivalent instance G ′ of γ t ) as follows. Let V ( G ) = { v , v , . . . , v n } the vertex set of G .The vertex set of G ′ consists of 2 ℓ vertices x , x , . . . , x ℓ and 2 ℓ + 1 copies of V ( G ), denotedby V , V , . . . V ℓ . For any 0 ≤ i ≤ ℓ , we denote the vertices of V i by v i , v i , . . . , v in . Theadjacencies in G ′ are then defined as follows (see Fig. 1): · V is a cliqueand for any 1 ≤ i ≤ ℓ , · V i is an independent set; · for any 1 ≤ j ≤ n , v ij is adjacent to { v k , v k ∈ N G [ v j ] } ; · x i is adjacent to every vertex in V ∪ V i ; · if i (mod 2) = 1, then x i is adjacent to x i +1 .Note that the fact that for any 1 ≤ i ≤ ℓ and 1 ≤ j ≤ n , v ij is adjacent to { v k , v k ∈ N G [ v j ] } is not made explicit in Fig. 1 for the sake of readability. We now claim the following. Claim 1. γ t ( G ′ ) = min { γ ( G ) , ℓ } .Proof. It is clear that { x , x , . . . , x ℓ − , x ℓ } is a total dominating set of G ′ ; thus, γ t ( G ′ ) ≤ ℓ .If γ ( G ) ≤ ℓ and { v i , v i , . . . , v i k } is a minimum dominating set of G , it follows from theconstruction above that { v i , v i , . . . , v i k } is a total dominating set of G ′ (recall that V is a clique). Thus, γ t ( G ′ ) ≤ γ ( G ) and so, γ t ( G ′ ) ≤ min { γ ( G ) , ℓ } . Now suppose that γ t ( G ′ ) < min { γ ( G ) , ℓ } and consider a minimum total dominating set D of G ′ . Then theremust exist i ∈ { , . . . , ℓ } such that x i D ; indeed, if for all i ∈ { , . . . , ℓ } , x i ∈ D then wewould have that | D | ≥ ℓ , thereby contradicting the fact that γ t ( G ′ ) < min { γ ( G ) , ℓ } . Butthen, D ′ = D ∩ ( V ∪ V i ) must dominate every vertex in V i and so, | D ′ | ≥ γ ( G ). But | D ′ | ≤ | D | which implies that γ ( G ) ≤ | D | , a contradiction. Therefore, γ t ( G ′ ) = min { γ ( G ) , ℓ } . y We now show that ( G, ℓ ) (with γ ( G ) ≥
4) is a
Yes -instance for
Even DominatingSet if and only if G ′ is a Yes -instance for γ t ) .4 x V x V x V x V · · · x ℓ − V ℓ − x ℓ V ℓ Figure 1:
The graph G ′ (thick lines indicate that the vertex x i is adjacent to every vertex of V ∪ V i for any 1 ≤ i ≤ ℓ ). First assume that γ ( G ) ≤ ℓ . Then by Claim 1, γ t ( G ′ ) = γ ( G ) and if { v i , v i , . . . , v i k } isa minimum dominating set of G then { v i , v i , . . . , v i k } is a minimum total dominating set of G ′ containing a P (recall that V is a clique and γ ( G ) ≥ G ′ is a Yes -instance for γ t ) .Conversely, assume that G ′ is a Yes -instance for γ t ) , that is,there exists a minimum total dominating set D of G ′ containing a P (see Theorem 1). Thenthere must exist i ∈ { , . . . , ℓ } such that x i D ; indeed, if for all i ∈ { , . . . , ℓ } , x i ∈ D then | D | ≥ ℓ and we conclude by Claim 1 that in fact equality holds. It follows that D consists of x , x , . . . , x ℓ − , x ℓ ; in particular, D contains no P , a contradiction. Thus, thereexists 1 ≤ i ≤ ℓ such that x i / ∈ D and so, D ′ = D ∩ ( V ∪ V i ) must dominate every vertexin V i . It follows that | D ′ | ≥ γ ( G ) and since | D ′ | ≤ | D | , we conclude that γ ( G ) ≤ | D | ≤ ℓ by Claim 1, that is, ( G, ℓ ) is a Yes -instance for
Even Dominating Set .Finally, it is easy to see that G ′ is P -free as well as ( P + P )-free which concludes theproof. Theorem 5. γ t ) is coNP -hard when restricted to P -free graphs.Proof. We reduce from as follows. Given an instance Φ of this problem, with variableset X and clause set C , we construct a graph G Φ such that Φ is satisfiable if and only if G Φ is a No -instance for γ t ) , as follows. For any variable x ∈ X , weintroduce the gadget G x depicted in Figure 2 with one distinguished positive literal vertex x and one distinguished negative literal vertex ¯ x . For any clause c ∈ C , we introduce a clause vertex c which is made adjacent to the (positive or negative) literal vertices whosecorresponding literal occurs in c . Finally, we add an edge between any two clause verticesso that the set of clause vertices induces a clique denoted by K in the following. x ¯ xu x v x Figure 2:
The variable gadget G x . bservation 1. For any total dominating set D of G Φ and any variable x ∈ X , | D ∩ V ( G x ) | ≥ and u x ∈ D . In particular, γ t ( G Φ ) ≥ | X | . Indeed, since v x should be dominated, necessarily u x ∈ D and since u x should be domi-nated, D ∩ { v x , x, ¯ x } 6 = ∅ . Claim 2. Φ is satisfiable if and only if γ t ( G Φ ) = 2 | X | .Proof. Assume that Φ is satisfiable and consider a truth assignment satisfying Φ. We con-struct a total dominating set of G Φ as follows. For any variable x ∈ X , if x is true then weadd x and u x to D ; otherwise, we add ¯ x and u x to D . Clearly, D is a total dominating setas every clause is satisfied and we conclude by Observation 1 that D is minimum.Conversely, assume that γ t ( G Φ ) = 2 | X | and consider a minimum total dominating set D of G Φ . First observe that by Observation 1, | D ∩ V ( G x ) | = 2 and | D ∩ { x, ¯ x }| ≤ x ∈ X , which implies in particular that D ∩ K = ∅ . It follows that for any clausevertex c , there must exist x ∈ X such that the (positive or negative) literal vertex whosecorresponding literal occurs in c belongs to D . We may thus construct a truth assignmentsatisfying Φ as follows. For any variable x ∈ X , if the positive literal vertex x belongs to D then we set x to true; if the negative literal vertex ¯ x belongs to D then we set x to false;otherwise, we set x to true. y Claim 3. γ t ( G Φ ) = 2 | X | if and only if G Φ is a No -instance for γ t ) .Proof. Assume that γ t ( G Φ ) = 2 | X | and let D be a minimum total dominating set of G Φ .Then by Observation 1, | D ∩ V ( G x ) | = 2 for any x ∈ X which implies in particular that D ∩ K = ∅ . But then, it is clear that D contains no P , and hence G Φ is a No -instance for γ t ) according to Theorem 1.Conversely, assume that G Φ is a No -instance for γ t ) and considera minimum total dominating set D of G Φ . First observe that since D contains no P (seeTheorem 1), necessarily | D ∩ V ( G x ) | ≤ x ∈ X ; we then conclude by Observation 1that in fact equality holds for any x ∈ X . We now claim that D ∩ K = ∅ . Indeed, supposeto the contrary that there exists a clause vertex c such that c ∈ D and consider a variable x occuring in c , say x occurs positive in c without loss of generality. Then ( D \ { v x , ¯ x } ) ∪ { x } is a minimum total dominating set containing a P , a contradiction. Thus, D ∩ K = ∅ andso, | D | = 2 | X | . y Now by combining Claims 2 and 3, we obtain that Φ is satisfiable if and only if G Φ is a No -instance for γ t ) . Since G Φ is obviously 2 P -free, this concludesthe proof. Theorem 6. γ t ) is coNP -hard when restricted to claw-free graphs.Proof. We reduce from
Positive Cubic 1-In-3 3-Sat which was shown to be NP -hardin [15]. It is a variant of the problem where each variable occurs only nonnegatedand in exactly three clauses, and the formula is satisfiable if and only if there exists a truthassignment to the variables such that each clause has exactly one true literal. Given aninstance Φ of this problem, with variable set X and clause set C , we contruct a graph G Φ such that Φ is satisfiable if and only if G Φ is a No -instance for γ t ) ,as follows. For each variable x ∈ X contained in clauses c, c ′ and c ′′ , we introduce the gadget G x depicted in Figure 3. For each clause c ∈ C containing variables x, y and z , we introducethe gadget G c which is the disjoint union of the graphs G Tc and G Fc depicted in Figure 4;then for all ℓ ∈ { x, y, z } , we add an edge between t ℓc and t cℓ , and f ℓc and f cℓ .We begin with the following easy observations.6 x v x T x F x a c ′ x a cx a c ′′ x b cx d cx c cx t cx b c ′ x d c ′ x c c ′ x t c ′ x b c ′′ x d c ′′ x c c ′′ x t c ′′ x g c ′ x g cx g c ′′ x h cx j cx i cx f cx h c ′ x j c ′ x i c ′ x f c ′ x h c ′′ x j c ′′ x i c ′′ x f c ′′ x Figure 3:
The variable gadget G x for a variable x contained in clauses c, c ′ and c ′′ (a rectangleindicates that the corresponding set of vertices induces a clique). u c a yc a xc a zc c xc b xc d xc t xc c yc b yc d yc t yc c zc b zc d zc t zc (a) The graph G Tc (the rectangle indicatesthat the corresponding set of vertices inducesa clique). v c w c g yc g xc g zc f xc f yc f zc (b) The graph G Fc (the rectangle indicatesthat the corresponding set of vertices inducesa clique). Figure 4:
The clause gadget G c is the disjoint union of G Tc and G Fc for a clause c containing variables x, y and z . Observation 2.
Let D be a total dominating set of G Φ . Then for each clause c ∈ C withvariables x, y and z , the following holds.(i) | D ∩ { g xc , g yc , g zc , v c , w c }| ≥ and v c ∈ D .(ii) For any ℓ ∈ { x, y, z } , D ∩ { a ℓc , c ℓc } 6 = ∅ and | D ∩ { a ℓc , b ℓc , c ℓc , d ℓc , t ℓc }| ≥ .In particular, | D ∩ V ( G c ) | ≥ . (i) Indeed, since w c must be dominated, necessarily v c ∈ D and since v c must be domi-nated, D ∩ { g xc , g yc , g zc , w c } 6 = ∅ .(ii) Indeed, since c ℓc must be dominated, either a ℓc ∈ D or c ℓc ∈ D . If a ℓc ∈ D then D ∩ { c ℓc , t ℓc } 6 = ∅ as d ℓc should be dominated; and if c ℓc ∈ D then D ∩ { a ℓc , b ℓc , d ℓc } 6 = ∅ as c ℓc should be dominated. ⋄ Observation 3.
Let D be a total dominating set of G Φ . Then for each variable x ∈ X contained in clauses c, c ′ and c ′′ , the following holds.(i) | D ∩ { u x , v x , T x , F x }| ≥ and u x ∈ D .(ii) For any ℓ ∈ { c, c ′ , c ′′ } , D ∩ { b ℓx , d ℓx } 6 = ∅ and | D ∩ { a ℓx , b ℓx , c ℓx , d ℓx , t ℓx }| ≥ .(iii) for any ℓ ∈ { c, c ′ , c ′′ } , D ∩ { h ℓx , j ℓx } 6 = ∅ and | D ∩ { g ℓx , h ℓx , i ℓx , j ℓx , f ℓx }| ≥ .In particular, | D ∩ V ( G x ) | ≥ . (i) Indeed, since v x must be dominated, necessarily u x ∈ D and since u x must be domi-nated, D ∩ { v x , T x , F x } 6 = ∅ .(ii) Indeed, since c ℓx must be dominated, either b ℓx ∈ D or d ℓx ∈ D . If b ℓx ∈ D then D ∩ { a ℓx , c ℓx , d ℓx } 6 = ∅ as b ℓx should be dominated; and if d ℓx ∈ D then D ∩ { b ℓx , c ℓx , t ℓx } 6 = ∅ as7 ℓx should be dominated. The proof for (iii) is symmetric. ⋄ We now prove the following two claims.
Claim 4. γ t ( G Φ ) = 14 | X | + 8 | C | if and only if Φ is satisfiable.Proof. Assume first that Φ is satisfiable and consider a truth assignment satisfying Φ. Weconstruct a total dominating set D for G Φ as follows. For any variable x ∈ X contained inclauses c, c ′ and c ′′ , if x is true then add { u x , T x , d cx , d c ′ x , d c ′′ x , t cx , t c ′ x , t c ′′ x , h cx , h c ′ x , h c ′′ x , j cx , j c ′ x , j c ′′ x } to D ; otherwise add { u x , F x , j cx , j c ′ x , j c ′′ x , f cx , f c ′ x , f c ′′ x , b cx , b c ′ x , b c ′′ x , d cx , d c ′ x , d c ′′ x } to D . For anyclause c ∈ C containing variables x, y and z , we proceed as follows. Assume without loss ofgenerality that x is true (and thus y and z are false). Then add { c xc , a xc , d yc , c yc , d zc , c zc , g xc , v c } to D . Clearly, D is a total dominating set and we conclude by Observations 2 and 3 that D is minimum. Thus, γ t ( G Φ ) = 14 | X | + 8 | C | .Conversely, assume that γ t ( G Φ ) = 14 | X | + 8 | C | . Let us first make several observations.The following is a straightforward consequence of Observation 2. Observation 4.
Let D be a total dominating set of G Φ . Then for any clause c ∈ C con-taining variables x, y and z , if | D ∩ V ( G c ) | = 8 then D ∩ { f xc , f yc , f zc , u c } = ∅ . Observation 5.
Let D be a total dominating set of G Φ . Then for any variable x ∈ X contained in c, c ′ , and c ′′ , if | D ∩ V ( G x ) | = 14 , the following holds.(i) For any ℓ ∈ { c, c ′ , c ′′ } , if t ℓx ∈ D then D ∩ { a ℓx , b ℓx , c ℓx , d ℓx , t ℓx } = { d ℓx , t ℓx } .(ii) For any ℓ ∈ { c, c ′ , c ′′ } , if a ℓx ∈ D then D ∩ { a ℓx , b ℓx , c ℓx , d ℓx , t ℓx } = { a ℓx , b ℓx } .(iii) For any ℓ ∈ { c, c ′ , c ′′ } , if f ℓx ∈ D then D ∩ { g ℓx , h ℓx , i ℓx , j ℓx , f ℓx } = { j ℓx , f ℓx } .(iv) For any ℓ ∈ { c, c ′ , c ′′ } , if g ℓx ∈ D then D ∩ { g ℓx , h ℓx , i ℓx , j ℓx , f ℓx } = { g ℓx , h ℓx } . (i) Indeed, note first that since | D ∩ V ( G x ) | = 14, we have by Observation 3 that | D ∩{ a ℓx , b ℓx , c ℓx , d ℓx , t ℓx }| = 2 for all ℓ ∈ { c, c ′ , c ′′ } . Thus, if t ℓx ∈ D then by Observation 3(ii), | D ∩ { b ℓx , d ℓx }| = 1 (note that in particular, a ℓx / ∈ D ); but if b ℓx ∈ D then b ℓx is not dominatedas a ℓx / ∈ D . Therefore, if t ℓx ∈ D then d ℓx ∈ D . The proof for (iii) is symmetric.(ii) Similarly, if a ℓx ∈ D then by Observation 3(ii), | D ∩ { b ℓx , d ℓx }| = 1 (note that in partic-ular, t ℓx / ∈ D ); but if d ℓx ∈ D then d ℓx is not dominated as t ℓx / ∈ D . Therefore, if a ℓx ∈ D then b ℓx ∈ D . The proof for (iv) if symmetric. ⋄ Observation 6.
Let D be a total dominating set of G Φ . For any variable x ∈ X containedin clauses c, c ′ and c ′′ , if | D ∩ V ( G x ) | = 14 and | D ∩ V ( G ℓ ) | = 8 for all ℓ ∈ { c, c ′ , c ′′ } , thenthe following holds.(i) If there exists ℓ ∈ { c, c ′ , c ′′ } such that t ℓx ∈ D then T x ∈ D .(ii) If there exists ℓ ∈ { c, c ′ , c ′′ } such that f ℓx ∈ D then F x ∈ D . (i) Indeed, note first that since | D ∩ V ( G x ) | = 14, we have by Observation 3 that | D ∩{ a ℓx , b ℓx , c ℓx , d ℓx , t ℓx }| = 2 for all ℓ ∈ { c, c ′ , c ′′ } . Similarly, since | D ∩ V ( G ℓ ) | = 8 for all ℓ ∈{ c, c ′ , c ′′ } , it follows from Observation 2 that | D ∩ { a xℓ , b xℓ , c xℓ , d xℓ , t xℓ }| = 2 for all ℓ ∈ { c, c ′ , c ′′ } .Now assume that there exists ℓ ∈ { c, c ′ , c ′′ } such that t ℓx ∈ D , say c without loss of generality,and suppose to the contrary that T x / ∈ D . Then by Observation 5(i), D ∩ { a cx , b cx , c cx , d cx , t cx } = { d cx , t cx } . Thus since a cx should be dominated and T x / ∈ D , there must exist p ∈ { c ′ , c ′′ } such that a px ∈ D , say c ′ without loss of generality. But then by Observation 5(ii), D ∩{ a c ′ x , b c ′ x , c c ′ x , d c ′ x , t c ′ x } = { a c ′ x , b c ′ x } and so, t xc ′ ∈ D for otherwise t c ′ x would not be dominated.But | D ∩ { a xc ′ , b xc ′ , c xc ′ , d xc ′ , t xc ′ }| = 2 and D ∩ { a xc ′ , c xc ′ } 6 = ∅ by Observation 2, which impliesthat d xc ′ / ∈ D and so, t xc ′ is not dominated, a contradiction. Thus, T x ∈ D .8ii) Assume that there exists ℓ ∈ { c, c ′ , c ′′ } such that f ℓx ∈ D , say c without lossof generality, and suppose to the contrary that F x / ∈ D . Then by Observation 5(iii), D ∩ { g cx , h cx , i cx , j cx , f cx } = { j cx , f cx } . Thus since g cx should be dominated and F x / ∈ D , theremust exist p ∈ { c ′ , c ′′ } such that g px ∈ D , say c ′ without loss of generality. But then by Ob-servation 5(iv), D ∩ { g c ′ x , h c ′ x , i c ′ x , j c ′ x , f c ′ x } = { g c ′ x , h c ′ x } and so, f xc ′ must belong to D ( f c ′ x wouldotherwise not be dominated) which contradicts Observation 4 (recall that | D ∩ V ( G c ′ ) | = 8).Thus, F x ∈ D . ⋄ Remark 1. If γ t ( G Φ ) = 14 | X | + 8 | C | and D is a minimum total dominating set of G Φ ,then the following hold. For any clause c ∈ C containing variable x, y and z , we have byObservation 2 that(i) | D ∩ { g xc , g yc , g zc , v c , w c }| = 2 and v c ∈ D ; and(ii) for any ℓ ∈ { x, y, z } , D ∩ { a ℓc , c ℓc } 6 = ∅ and | D ∩ { a ℓc , b ℓc , c ℓc , d ℓc , t ℓc }| = 2.Similarly by Observation 3, we have that for any variable x ∈ X contained in clauses c, c ′ and c ′′ ,(i) | D ∩ { u x , v x , T x , F x }| = 2 and u x ∈ D ;(ii) for any ℓ ∈ { c, c ′ , c ′′ } , D ∩ { b ℓx , d ℓx } 6 = ∅ and | D ∩ { a ℓx , b ℓx , c ℓx , d ℓx , t ℓx }| = 2; and(iii) for any ℓ ∈ { c, c ′ , c ′′ } , D ∩ { h ℓx , j ℓx } 6 = ∅ and | D ∩ { g ℓx , h ℓx , i ℓx , j ℓx , f ℓx }| = 2.Turning back to the proof of Claim 4, let D be a minimum total dominating set of G Φ .We claim the following. Observation 7. If γ t ( G Φ ) = 14 | X | + 8 | C | then for any minimum total dominating set D ′ and any clause c ∈ C containing variables x, y and z , there exists ℓ ∈ { x, y, z } such that t cℓ ∈ D ′ and for any p ∈ { x, y, z } \ { ℓ } , f cp ∈ D ′ . Indeed, suppose to the contrary that for all ℓ ∈ { x, y, z } , t cℓ / ∈ D ′ . Then { d xc , d yc , d zc } ⊂ D as t xc , t yc and t zc should be dominated, and so D ∩ { t xc , t yc , t zc } = ∅ as D ∩ { a ℓc , c ℓc } 6 = ∅ and | D ∩ { a ℓc , b ℓc , c ℓc , d ℓc , t ℓc }| = 2 for any ℓ ∈ { x, y, z } . But then by Observation 2, | D ′ ∩ { a pc , c pc }| = 1for any p ∈ { x, y, z } which implies that c pc ∈ D ′ for any p ∈ { x, y, z } for otherwise at leastone of d xc , d yc and d zc would not be dominated. It follows that D ′ ∩ { a xc , a yc , a zc } = ∅ and so u c is not dominated, a contradiction. Thus, there exists ℓ ∈ { x, y, z } such that t cℓ ∈ D ′ , say x without loss of generality. Then by Observation 6(i), T x ∈ D ′ and so necessarily F x / ∈ D ′ by Remark 1. But then, f cx / ∈ D ′ for otherwise by Observation 6(ii), F x would belong to D ′ ,and so g xc ∈ D ′ ( f xc would otherwise not be dominated). It then follows from Observations 2and 4 that D ′ ∩ { f xc , f yc , f zc , g xc , g yc , g zc , v c , w c } = { g xc , v c } which implies that for p ∈ { y, z } , f cp ∈ D ′ for otherwise f pc would not be dominated; in particular, F p ∈ D ′ for p ∈ { y, z } byObservation 6(ii). ⋄ Combining Remark 1 and Observations 6 and 7, we conclude that for any variable x ∈ X , | D ∩ { T x , F x }| = 1 and for any clause c containing variables x, y and z , there exists exactlyone variable ℓ ∈ { x, y, z } such that T ℓ ∈ D . Therefore, we may construct a truth assignmentsatisfying Φ as follows: for any variable x ∈ X , if T x ∈ D we set x to true, otherwise we set x to false. This concludes the proof of Claim 4. y Claim 5. γ t ( G Φ ) = 14 | X | +8 | C | if and only if G Φ is a No -instance for γ t ) .Proof. Assume first that γ t ( G Φ ) = 14 | X | + 8 | C | and let D be a minimum total dominatingset of G Φ (note that Remark 1 holds). Let us show that D contains no P .9irst, consider a clause c ∈ C containing variables x, y and z . Note that by Obser-vation 4 and Remark 1, | D ∩ V ( G Fc ) | = 2 and thus D ∩ V ( G Fc ) cannot contain any P nor can it be part of a P . Now by Observation 7, there exists ℓ ∈ { x, y, z } such that t cℓ ∈ D and for p ∈ { x, y, z } \ { ℓ } , f cp ∈ D . Assume without loss of generality that ℓ = x and denote by c ′ and c ′′ the two other clauses in which x occurs. It follows fromObservation 6 and Remark 1 that T x ∈ D and F x D . Then, necessarily t px ∈ D for p ∈ { c ′ , c ′′ } ; indeed, since by Observation 7, there exists a variable t contained in c ′ suchthat t c ′ t ∈ D and f c ′ r ∈ D for the other variables r = t in c ′ , necessarily t = x for otherwisewe would conclude by Observation 6 that F x ∈ D , a contradiction (the same reasoning ap-plies for c ′′ ). It then follows from Observation 5(i) that D ∩ ( { t px , p ∈ { c, c ′ , c ′′ }} ∪ { d px , p ∈{ c, c ′ , c ′′ }} ∪ { c px , p ∈ { c, c ′ , c ′′ }} ∪ { b px , p ∈ { c, c ′ , c ′′ }} ∪ { a px , p ∈ { c, c ′ , c ′′ }} ∪ { T x , F x , u x , v x } ) = { t px , p ∈ { c, c ′ , c ′′ }}∪{ d px , p ∈ { c, c ′ , c ′′ }}∪{ T x , u x } . On the other hand, since by Observation 4, f xp / ∈ D for any p ∈ { c, c ′ , c ′′ } , necessarily j px ∈ D ( f px would otherwise not be dominated).But then, h px ∈ D for any p ∈ { c, c ′ , c ′′ } as j px and g px should be dominated (recall that F x / ∈ D ) and so, D ∩ ( { f px , p ∈ { c, c ′ , c ′′ }} ∪ { j px , p ∈ { c, c ′ , c ′′ }} ∪ { i px , p ∈ { c, c ′ , c ′′ }} ∪ { h px , p ∈{ c, c ′ , c ′′ }} ∪ { g px , p ∈ { c, c ′ , c ′′ }} ) = { j px , p ∈ { c, c ′ , c ′′ }} ∪ { h px , p ∈ { c, c ′ , c ′′ }} . Thus, D ∩ V ( G x )does not contain any P . Now denote by k and k ′ the two others clauses in which y occurs. Then, a reasoning similar to the above shows that f py ∈ D for p ∈ { k, k ′ } (re-call that by assumption, f cy ∈ D ) and so by Observations 3 and 5(iii), we conclude that D ∩ ( { f py , p ∈ { c, k, k ′ }} ∪ { j py , p ∈ { c, k, k ′ }} ∪ { i py , p ∈ { c, k, k ′ }} ∪ { h py , p ∈ { c, k, k ′ }} ∪ { g py , p ∈{ c, k, k ′ }} ∪ { T y , F y , u y , v y } ) = { f py , p ∈ { c, k, k ′ }} ∪ { j py , p ∈ { c, k, k ′ }} ∪ { F y , u y } . On theother hand, since T y / ∈ D necessarily t py / ∈ D for any p ∈ { c, k, k ′ } (we would otherwiseconclude by Observation 6(i) that T y ∈ D ). We claim that then d py ∈ D for all p ∈ { c, k, k ′ } .Indeed, if d py / ∈ D for some p ∈ { c, k, k ′ } then necessarily t yp ∈ D as t py should be dominated.But then since t py / ∈ D , it must be that d yp ∈ D for otherwise t yp would not be dominated. But | D ∩ { t yp , d yp , c yp , b yp , a yp }| = 2 and so D ∩ { a yp , c yp } = ∅ thereby contradicting Observation 2(ii).Thus d cy , d ky , d k ′ y ∈ D which implies that b py ∈ D for any p ∈ { c, k, k ′ } as d py and a py shouldbe dominated (recall that T y / ∈ D ). In particular, D ∩ V ( G y ) does not contain any P (thesame reasoning shows that D ∩ V ( G z ) does not contain any P either). Now since t cy / ∈ D necessarily d yc ∈ D as t yc should be dominated. We then conclude by Remark 1 that c yc ∈ D ;indeed, | D ∩ { c yc , a yc }| = 1 and if a yc ∈ D then d yc is not dominated. Similarly, we concludethat D ∩ { t zc , d zc , c zc , b zc , a zc } = { d zc , c zc } . Thus, since u c should be dominated as well as a xc and d xc , we obtain by Observation 4 that D ∩ { u c , a xc , b xc , c xc , d xc , t xc } = { a xc , c xc } . Thus, D ∩ V ( G Tc )contains no P nor can it be part of a P and so, D contains no P .Conversely, assume that G Φ is a No -instance for γ t ) and let D be a minimum total dominating set of G Φ . Consider a variable x ∈ X contained in clauses c, c ′ and c ′′ . First observe that since D ∩ { F x , t x , u x , v x } does not contain a P , it followsfrom Observation 3(i) that | D ∩ { F x , T x , u x , v x }| = 2, and we conclude by Observation 3(i)that in fact equality holds (recall that u x ∈ D ). Similarly for any p ∈ { c, c ′ , c ′′ } , | D ∩{ a px , b px , c px , d px , t px }| ≤
3. Now suppose to the contrary that there exists p ∈ { c, c ′ , c ′′ } such that | D ∩ { a px , b px , c px , d px , t px }| = 3. Then | D ∩ { b px , c px , d px }| ≤
1; indeed, clearly | D ∩ { b px , c px , d px }| < | D ∩ { b px , c px , d px }| = 2 then we may assume without loss of generality that D ∩{ b px , c px , d px } = { b px , d px } . But then D ∩ { t px , a px } 6 = ∅ and so, D ∩ { a px , b px , c px , d px , t px } contains a P , a contradiction. Thus | D ∩ { b px , c px , d px }| ≤ | D ∩ { b px , c px , d px }| = 1 (in fact, either b px ∈ D or d px ∈ D ). It follows that t px , a px ∈ D and so,necessarily T x / ∈ D for otherwise a px , T x , u x would induce a P (recall that by Observation 3(i), u x ∈ D ). But then, ( D \ { a px , b cx } ) ∪ { T x , d cx } is a minimum total dominating set of G Φ containing a P , a contradiction. Thus, we conclude that for any p ∈ { c, c ′ , c ′′ } , | D ∩{ a px , b px , c px , d px , t px }| ≤
2; and by symmetry, we also conclude that | D ∩ { g px , h px , i px , j px , f px }| ≤ p ∈ { c, c ′ , c ′′ } . It then follows from Observation 3 that for any variable x ∈ X , | D ∩ V ( G x ) | = 14.Consider now a clause c ∈ C containing variables x, y and z . First observe that since D ∩ { g xc , g yc , g zc , v c , w c } does not contain a P , it follows from Observation 2(i) that | D ∩{ g xc , g yc , g zc , v c , w c }| = 2 and v c ∈ D . Now if there exists p ∈ { x, y, z } such that f pc ∈ D then g pc / ∈ D and so, f cp ∈ D for otherwise f pc would not be dominated. It follows that j cp / ∈ D ( D would otherwise contain a P ) and i cp / ∈ D (( D \ { i cp } ∪ { j cp } ) would otherwisecontain a P ) which implies that h cp ∈ D as i cp would otherwise not be dominated. But then,( D \ { f pc } ) ∪ { j cp } is a minimum total dominating set of G Φ containing a P , a contradiction.Thus, | D ∩ V ( G Fc ) | ≤
2. Now for any p ∈ { x, y, z } , | D ∩ { a pc , b pc , c pc , d pc , t pc }| ≤ D ∩{ a pc , b pc , c pc , d pc , t pc } would otherwise contain a P . Suppose to the contrary that there exists p ∈ { x, y, z } such that | D ∩ { a pc , b pc , c pc , d pc , t pc }| = 3. If | D ∩ { a pc , b pc , c pc }| = 1 then d pc , t pc ∈ D and so necessarily c pc / ∈ D . Then, since b pc should be dominated, it must be that a pc ∈ D . Butthen, t cp / ∈ D ( t cp , t pc and d pc would otherwise induce a P ) and so D ′ = ( D \ { d pc } ) ∪ { t cp } is aminimum total dominating set of G Φ with | D ′ ∩ { t cp , d cp , c cp , b cp , a cp }| ≥
3, thereby contradictingthe above. Thus, | D ∩ { a pc , b pc , c pc }| = 2 (indeed, clearly | D ∩ { a pc , b pc , c pc }| <
3) and we mayassume without loss of generality that D ∩ { a pc , b pc , c pc } = { a pc , c pc } . It follows that d pc / ∈ D ( d pc , a pc and c pc would otherwise induce a P ) and so, t pc ∈ D . But then, it must be that t cp ∈ D ( t pc would otherwise not be dominated) and so d cp / ∈ D ( d cp , t cp and t pc would otherwise inducea P ). It follows that D ′ = ( D \ { t pc } ) ∪ { d cp } ) is a minimum total dominating set of G Φ with | D ′ ∩ { t cp , d cp , c cp , b cp , a cp }| ≥ p ∈ { x, y, z } , | D ∩ { a px , b pc , c pc , d pc , t pc }| ≤ u c / ∈ D ; indeed, if u c ∈ D then | D ∩ { a xc , a yc , a zc }| ≤ D ∩ { u c , a xc , a yc , a zc } would otherwise contain a P ) which implies that D ′ = ( D \ { u c } ) ∪ { a pc } with p ∈ { x, y, z } such that a pc / ∈ D , is a minimum total dominating set of G Φ with | D ′ ∩ { a pc , b pc , c pc , d pc , t pc }| ≥ | D ∩ V ( G c ) | = 8 for any clause c ∈ C and so, γ t ( G Φ ) = | D | = 14 | X | + 8 | C | . y Now by combining Claims 4 and 5, we obtain that Φ is satisfiable if and only if G Φ is a No -instance for γ t ) thus concluding the proof. Lemma 7.
Let G be a graph on at least three vertices, and let G ′ be the graph obtained by4-subdividing every edge of G . Then ct γ t ( G ) = 1 if and only if ct γ t ( G ′ ) = 1 .Proof. Let G = ( V, E ) be a graph with | V | ≥
3. In the following, given an edge e = uv of G ,we denote by e , e e and e the four new vertices resulting from the 4-subdivision of theedge e (where e is adjacent to u and e is adjacent to v ). We first prove the following. Claim 6. If H is the graph obtained from G by 4-subdividing one edge, then γ t ( H ) = γ t ( G ) + 2 .Proof. Assume that H is obtained by 4-subdividing the edge e = uv and consider a minimumtotal dominating set D of G . We construct a total dominating set of H as follows (see Fig.5). If D ∩ { u, v } = ∅ , then D ∪ { e , e } is a total dominating set of H . If | D ∩ { u, v }| = 1, say u ∈ D without loss of generality, then D ∪ { e , e } is a total dominating set of H . Finally,if { u, v } ⊂ D then D ∪ { e , e } is a total dominating set of H . We thus conclude that γ t ( H ) ≤ γ t ( G ) + 2.Conversely, let D be a minimum total dominating set of H . First note that if e ∈ D and u D , necessarily e ∈ D for otherwise e would not be dominated. Similarly, if e ∈ D and v D then e ∈ D . Thus, if e , e ∈ D then ( D \{ e , e , e , e } ) ∪ { u, v } is a total dominatingset of G of size at most γ t ( H ) −
2. Now suppose that e D . Then, necessarily e ∈ D forotherwise e would not be dominated, and if v / ∈ D then e ∈ D for otherwise e would not11 ve u e e e e vu ve u e e e e vu ve u e e e e vG H Figure 5:
Constructing a total dominating set of H from a total dominating set of G (vertices inred belong to the corresponding total dominating set). be dominated. Thus, if e ∈ D and e D , either v ∈ D in which case D \ { e , e , e , e } is atotal dominating set of G of size at most γ t ( H ) −
2; or v D and ( D \ { e , e , e , e } ) ∪ { v } is a total dominating set of size at most γ ( H ) −
2. By symmetry, we conclude similarlyif e ∈ D and e D . Now if both e and e do not belong to D then e , e ∈ D andso, D \ { e , e , e , e } is a total dominating set of G of size at most γ t ( H ) −
2. Therefore, γ t ( G ) ≤ γ t ( H ) − y Remark 2.
Note that the minimum total dominating set D of G constructed from a mini-mum total dominating D ′ of H according to the proof of Claim 6 has the following property:if e ∈ D ′ (resp. e ∈ D ′ ) then v ∈ D (resp. u ∈ D ).We now prove the statement of the lemma. Let G ′ be the graph obtained by 4-subdividingevery edge of G . Then, γ t ( G ′ ) = γ t ( G ) + 2 | E | by Claim 6.First assume that ct γ t ( G ) = 1. Then by Theorem 1, there exists a minimum totaldominating set D of G containing a P , say u, v, w . Let D ′ be the minimum total dominatingset of G ′ constructed from D according to the proof of Claim 6. Then D ′ contains a P ,namely e , v, f where e = uv , f = vw and f is the vertex resulting from the 4-subdivisionof f adjacent to v .Conversely, assume that ct γ t ( G ′ ) = 1. Then by Theorem 1, there exists a minimum totaldominating set D ′ of G ′ containing a P which we denote by P in the following. Now let D be the minimum total dominating set of G constructed from D ′ according to the proofof Claim 6. If P is made up of the vertices e , v, f , where e = uv , f = vw and f is thevertex resulting from the 4-subdivision of f adjacent to v , then u, v, w ∈ D by construction(see Remark 2). If P is made up of the vertices u, e , e , where e = uv , then we may assumethat u has no other neighbor in D than e (we would otherwise fall back into the previouscase). Suppose first that v ∈ D ′ . Then, e D ′ for otherwise D ′ \ { e } would be a totaldominating set of G ′ of size strictly less than that of D ′ , a contradiction. It follows that v has a neighbor f belonging to D , with f = vw ( v would otherwise not be dominated); butthen w ∈ D by construction (see Remark 2) and so, D contains u, v, w . Thus, suppose that v D ′ . Then e D ′ ; indeed, if e ∈ D ′ then e ∈ D ′ ( e would otherwise not be dominated)but then, D ′ \ { e } is a total dominating set of G ′ of size strictly less than that of D ′ , acontradiction. It follows that v has a neighbor f belonging to D ′ , with f = vw ( v wouldotherwise not be dominated). But then, v, w ∈ D by construction (see Remark 2) and so, D contains u, v, w . Suppose finally that P is made up of the vertices e , e , e with e = uv andassume that u D ′ (we would otherwise fall back into the previous case). Then v D ′ forotherwise D ′ \ { e } would be a total dominating set of G ′ of size stricly less than that of D ′ , acontradiction. Suppose first that e ∈ D ′ . If v has another neighbor in D ′ , say f ∈ D ′ with f = vw , then by construction D contains u, v, w (see Remark 2). Thus, we may assume that v has no other neighbor in D ′ than e . Now since | V | ≥ G is connected, one of u and v has a neighbor in V \ { u, v } , say f = vw ∈ E without loss of generality. Note that we may12ssume that w D ′ for otherwise D would contain u, v, w . Now since f D ′ by assumption,necessarily f ∈ D ′ ( f would otherwise not be dominated) and f ∈ D ′ ( f would otherwisenot be dominated) and so, by considering D ′′ = ( D ′ \ { e , e } ) ∪ { v, f } , we fall back intothe previous case (indeed, D ′′ contains v, f , f ). Second, suppose that e D . Clearly, v has a neighbor f ∈ D , with f = vw ( v would otherwise not be dominated), and f ∈ D ( f would otherwise not be dominated). But then, by considering D ′′ = ( D ′ \ { e } ) ∪ { v } ,we fall back into the previous case (indeed, D ′′ contains v, f , f ). Thus, G has a minimumtotal dominating set containing a P and we conclude by Theorem 1 that ct γ t ( G ) = 1.By applying a 4-subdivision to an instance of γ t ) sufficientlymany times, we deduce the following from Lemma 7. Theorem 8.
For any l ≥ , γ t ) is NP -hard when restricted to { C , . . . , C l } -free graphs. In this section, we deal with the cases in which γ t ) is tractable. Afirst simple approach to this problem, from which we obtain Proposition 9, is based on bruteforce. Proposition 9. γ t ) can be solved in polynomial-time solvable on agraph class C , if one of the following holds:(a) C is closed under edge contraction and Total Dominating Set is solvable in poly-nomial time on C ; or(b) for every G ∈ C , γ t ( G ) ≤ q where q is a fixed constant; or(c) C is the class of ( H + K ) -free graphs where | V ( H ) | = q is a fixed constant and γ t ) is polynomial-time solvable on H -free graphs.Proof. In order to prove (a), it suffices to note that if we can compute γ t ( G ) and γ t ( G/e )for any edge e of G in polynomial time, then we can determine in polynomial time whether G is a Yes -instance for γ t ) .For (b), we proceed as follows. Given a graph G of C , we first check whether G has adominating edge. If it is the case, then G is a No -instance for γ t ) .Otherwise, we may consider any subset S ⊆ V ( G ) with | S | ≤ q and check whether it is a totaldominating set of G . Since there are at most O ( n q ) possible such subsets, we can determinethe total domination number of G and check whether the conditions given in Theorem 1 aresatisfied in polynomial time.So as to prove (c), we provide the following algorithm. Let H and q and let G be aninstance of γ t ) on ( H + K )-free graphs. We first test whether G is H -free (note that this can be done in time O ( n q )). If this is the case, we use the polynomial-time algorithm for γ t ) on H -free graphs. Otherwise, there is a set S ⊆ V ( G ) such that G [ S ] is isomorphic to H ; but since G is a ( H + K )-free graph, S mustthen be a dominating set of G and so, γ t ( G ) ≤ q . We then conclude by Proposition 9(b)that γ t ) is also polynomial-time solvable in this case. Theorem 10. γ t ) is polynomial-time solvable on P -free graphs.Proof. Let G be a P -free graph. If γ t ( G ) = 2, then G is clearly a No -instance for γ t ) . Now, assume that γ t ( G ) ≥ D of G . Let us now show that then G is a Yes -instance for γ t ) .If γ t ( G ) = 3 then it is clear that D contains a P as every vertex in S has a neighbor in D ;13hus, by Theorem 1, G is a Yes -instance for γ t ) . Next, suppose that γ t ( G ) ≥ u, v ∈ D such that d G ( u, v ) = max x,y ∈ D d G ( x, y ). If d G ( u, v ) = 1,then G [ D ] is a clique and G is therefore a Yes -instance for γ t ) byTheorem 1. Thus, we may assume that d G ( u, v ) ≥
2. Furthermore, we may assume that u and v have no common neighbor in D for otherwise we are done by Theorem 1. Denote by x (resp. y ) a neighbor of u (resp. v ) in D and assume that x and y are not adjacent (if xy ∈ E ( G ) then we are done by Theorem 1). Since G is P -free, d G ( x, y ) ≤ d G ( x, y ) = 3 and let a (resp. b ) be the neighbor of x (resp. y ) ona shortest path from x to y . Then u is adjacent to either a or b but not both; indeed, u is adjacent to either a or b as u, x, a, b, y would otherwise induce a P . By symmetry, thesame holds for v . But if u is adjacent to both, since v is adjacent to either a or b , wewould have d G ( u, v ) = 2 < d G ( x, y ) thereby contradicting the choice of u and v . A similarreasoning shows that if u is adjacent to a (resp. b ) then v is adjacent to b (resp. a ). Assumewithout loss of generality that u is adjacent to a (and thus, v is adjacent to b ). Then, N G ( u ) ∪ N G ( v ) ⊆ N G ( a ) ∪ N G ( b ); indeed, if t is a neighbor of u then t is nonadjacent to v (recall that d G ( u, v ) ≥ d G ( x, y ) = 3) and thus, t is adjacent to either a or b for otherwise t, u, a, b, v would induce a P . We conclude similarly if t is a neighbor of v . But then, D ′ = ( D \{ u, v } ) ∪ { a, b } is a minimum total dominating set of G containing a P ; indeed, D ′ is clearly dominating and if a vertex w ∈ D were dominated by either u or v , then w is dominated a or b in D ′ . We then conclude by Theorem 1 that G is a Yes -instance for γ t ) .Now, suppose that d G ( x, y ) = 2 and denote by a the vertex on a shortest path from x to y . Then, a is adjacent to either u or v for otherwise u, x, a, y, v induce a P . Suppose firstthat a is adjacent to both u and v . We may assume that both x and y have at least oneprivate neighbor with respect to D ; if it weren’t the case for x , then ( D \{ x } ) ∪ { a } wouldbe a minimum total dominating set of G containing a P (the same argument holds for y ).Let t (resp. s ) be a private neighbor of x (resp. y ). Clearly, t and s must be nonadjacentsince otherwise, x, t, s, y, v induce a P . Also, at least t or s is adjacent to a , otherwise t, x, a, y, s induce a P . Without loss of generality, we may assume that s is adjacent to a . If t is nonadjacent to a then every private neighbor r of y must be adjacent to a for otherwise t, x, a, y, r would induce a P ; thus, ( D \{ y } ) ∪ { a } is a minimum total dominating set of G containing a P and so by Theorem 1, we have that ct γ t ( G ) = 1. Thus, we may assumenow that t is also adjacent to a , and hence every private neighbor of x and y is adjacentto a (if there exists a private neighbor of x or y which is nonadjacent to a , we conclude aspreviously), and therefore ( D \{ y } ) ∪ { a } is a minimum total dominating set of G containinga P . The result then follows from Theorem 1. Now, if a is nonadjacent to one of u and v ,say v without loss of generality, then any neighbor t of u is adjacent to either a , y or v forotherwise tuayv would induce a P ; but then, ( D \{ u } ) ∪ { a } is a minimum total dominatingset of G containing a P and thus by Theorem 1, ct γ t ( G ) = 1. Theorem 11.
For any fixed k ≥ , γ t ) is polynomial-time solvableon ( P + kP ) -free graphs.Proof. First observe that since
Total Dominating Set is polynomial-time solvable on P -free graphs [11, Theorem 25], γ t ) is polynomial-time solvable on P -free graphs by Proposition 9(a) (note indeed that the class of P -free graphs is closed underedge contraction). Now assume that k ≥ G be a ( P + kP )-free graph containingan induced P + ( k − P . Let A ⊆ V ( G ) be such that G [ A ] is isomorphic to P + ( k − P ,set B to be the set of vertices at distance one from A and let C = V ( G ) \ ( A ∪ B ). Notethat since G is ( P + kP )-free, G [ C ] is a disjoint union of cliques.Let K be the set of maximal cliques in G [ C ]. Let K ′ ⊆ K be the subset of cliques such14hat the closed neighborhood of each clique in K ′ does not contain an induced P and noclique in K ′ is complete to a vertex in B . We call a clique K ∈ K ′ a regular clique if thereexist k other cliques K , . . . , K k ∈ K ′ such that K, K , . . . , K k have pairwise distance at leastfour from one another. We denote by R the set of regular cliques. Note that we can identifythis set in polynomial time. Claim 7.
Let K ∈ K be a clique of at least two vertices such that N [ K ] is P -free. Thenthere exist x, y ∈ K such that N [ xy ] = N [ K ] and for any v ∈ N [ K ] , either N [ K ] ⊆ N [ vx ] or N [ K ] ⊆ N [ vy ] .Proof. Let x, y ∈ K be such that | N [ xy ] | is maximum amongst all pairs of vertices in K .Suppose for a contradiction that there exists a vertex b ∈ N [ K ] nonadjacent to both x and y , and let c ∈ K be a neighbor of b . Suppose that x has a neighbor p x which is adjacentto neither c nor y , and that y has a neighbor p y which is adjacent to neither c nor x . Then p x and p y must be adjacent for otherwise p x , x, y, p y would induce a P ; but then, p x , p y , y, c induce a P , a contradiction. It follows that N [ xy ] is dominated by either cx or cy , say N [ xy ] ⊆ N [ cx ] without loss of generality. Now since b ∈ N [ cx ] \ N [ xy ], we conclude that | N [ xy ] | < | N [ cx ] | thereby contradicting the maximality of | N [ xy ] | . Hence, xy dominates N [ K ] and thus, N [ xy ] = N [ K ]. Now consider v ∈ N [ K ] \ { x, y } and assume without loss ofgenerality that v is adjacent to x . Suppose that N [ xy ] N [ vx ], that is, y has a neighbor p y which is adjacent to neither v nor x . Then y must be adjacent to v for otherwise v, x, y, p y would induce a P . Now if there exists a vertex p x ∈ N ( x ) which is adjacent to neither y nor v , then either p x is not adjacent to p y in which case p x , x, y, p y induce a P , or p x is adjacentto p y and v, x, p x , p y induce a P , a contradiction in both cases. Thus, we conclude that if N [ xy ] N [ vx ] then N [ xy ] ⊆ N [ vy ]. y Claim 8.
Let K , . . . , K k +1 ∈ R be regular cliques which are pairwise at distance at leastfour from one another. For any i ∈ [ k + 1] , if v ∈ V ( G ) \ N [ K i ] is adjacent to a vertex in N ( K i ) ∩ B then there exists j ∈ [ k + 1] , j = i , such that v is complete to N ( K j ) ∩ B .Proof. Assume that there exists a vertex v ∈ V ( G ) \ N [ K i ] where i ∈ [ k + 1], which isadjacent to a vertex b i ∈ N ( K i ) ∩ B . Let c i ∈ K i ∩ N ( b i ) and c ′ i ∈ K i \ N ( b i ) (recall thatsince K i is a regular clique, b i is not complete to K i ). Suppose for a contradiction that thereexists no j ∈ [ k + 1] such that v is complete to N ( K j ) ∩ B . Then for every j ∈ [ k + 1] \ { i } ,there exists a vertex b j ∈ N ( K j ) ∩ B which is nonadjacent to v . For every j ∈ [ k + 1] \ { i } ,let c j ∈ K j ∩ N ( b j ) and c ′ j ∈ K j \ N ( b j ). Then S k +1 i =1 { b i , c i , c ′ i } ∪ { v } induces a P + kP , acontradiction. y Claim 9.
Let D be a minimum total dominating set of G and let K ∈ R be a regular clique.Then | D ∩ N [ K ] | = 2 .Proof. It is clear from the definition that | D ∩ N [ K ] | ≥ K . Nowsuppose for a contradiction that there exists a clique K ∈ R such that | D ∩ N [ K ] | ≥ K , . . . , K k +1 be k regular cliques such that K , K , . . . , K k +1 are pairwise at distance atleast four from one another. It follows from the above that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) D ∩ [ ≤ i ≤ k +1 N [ K i ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ k. Now by Claim 7, we have that for any i ∈ [ k +1], there exist b i ∈ B ∩ N [ K i ] and c i ∈ N ( b i ) ∩ K i such that b i c i dominates N [ K i ]. But then D ′ = ( D \ S i ∈ [ k +1] N [ K i ]) ∪ S i ∈ [ k +1] { b i , c i } is atotal dominating set of G; indeed, if v ∈ V ( G ) is adjacent to a vertex in D ∩ N [ K i ], for some15 ∈ [ k + 1], then either v ∈ N [ K i ] in which case v ∈ N [ b i c i ], or v ∈ V ( G ) \ N [ K i ] and weconclude by Claim 8 that v is complete to N ( K j ) ∩ B for some j ∈ [ k + 1] (in particular, v is adjacent to b j ). But | D ′ | < | D | , a contradiction to the minimality of D . y Remark 3.
Note that by the proof of Claim 9, we have that for any minimum total domi-nating set D of G , any k + 1 regular cliques K , . . . , K k +1 ∈ R which are pairwise at distanceat least four from one another, any b i ∈ B ∩ N ( K i ) and c i ∈ V ( K i ) ∩ N ( b i ) such that N [ K i ] ⊆ N [ b i c i ] with i ∈ [ k + 1], k +1 [ i =1 { b i , c i } ∪ D \ [ ≤ i ≤ k +1 N [ K i ] is a minimum total dominating set of G . Claim 10.
If there are two regular cliques at distance at most three from one another then G is a Yes -instance for γ t ) .Proof. Assume that such two regular cliques exist and let K and K ′ be two regular cliquessuch that d ( K , K ′ ) = min K,K ′ ∈R d ( K, K ′ ) (note that by assumption, d ( K , K ′ ) ≤ G is a No -instance for γ t ) .Let K , . . . , K k +1 (resp. K ′ , . . . , K ′ k +1 ) be k regular cliques such that K , . . . , K k +1 (resp. K ′ , . . . , K ′ k +1 ) have pairwise distance at least four from one another, and denote by S = K ∪ . . . ∪ K k +1 and S ′ = K ′ ∪ . . . ∪ K ′ k +1 . By Claim 7, we have that for every i ∈ [ k + 1],there exist b i ∈ N ( K i ) ∩ B and c i ∈ K i ∩ N ( b i ) (resp. b ′ i ∈ N ( K ′ i ) ∩ B and c ′ i ∈ K ′ i ∩ N ( b ′ i ))such that N [ K i ] ⊆ N [ b i c i ] (resp. N [ K ′ i ] ⊆ N [ b ′ i c ′ i ]). In the following, let D be a minimumtotal dominating set of G .Suppose first that K and K ′ have a common neighbor v , that is, d ( K , K ′ ) = 2. Then byClaim 7, there exists c ∈ K ∩ N ( v ) such that N [ K ] ⊂ N [ cv ]. As c has a neighbor in N ( K ′ ) ∩ B (namely v ), it follows from Claim 8 that there exists j ∈ [ k + 1] such that c is complete to N [ K ′ j ] ∩ B . By Remark 3, we then have that D ′ = ( D \ N [ S ]) ∪ { c, v, b , c , . . . , b k +1 , c k +1 } is a minimum total dominating set of G . Similarly by Remark 3, we conclude that D ′′ =( D ′ \ N [ S ′ ]) ∪ (cid:8) b ′ , c ′ , . . . , b ′ k +1 , c ′ k +1 (cid:9) is a minimum total dominating set of G . But as c belongs to D ′′ and is adjacent to b ′ j , it follows that D ′′ contains a P , a contradiction byTheorem 1. Thus, d ( K , K ′ ) ≥
3; in particular, no two regular cliques in G have a commonneighbor by minimality of d ( K , K ′ ).Now since d ( K , K ′ ) ≤ w ∈ N ( K ) ∩ B and w ′ ∈ N ( K ′ ) ∩ B such that w and w ′ are adjacent. By Claim 7, there exist v ∈ K ∩ N ( w ) and v ′ ∈ K ′ ∩ N ( w ′ ) such that N [ K ] ⊆ N [ w v ] and N [ K ′ ] ⊆ N [ w ′ v ′ ]. But then by Claim 8,we have that D ′ = ( D \ N [ S ∪ S ′ ]) ∪ { w , v , w ′ , v ′ } ∪ S k +1 i =2 { b i , b ′ i , c i , c ′ i } is a total dominatingset of G of size at most that of D ; indeed, since no two regular cliques have a commonneighbor, it follows from Claim 9 that | D ∩ N [ S ∪ S ′ ] | = 2 (cid:12)(cid:12)(cid:8) K , . . . , K k +1 , K ′ , . . . , K ′ k +1 (cid:9)(cid:12)(cid:12) (Note that this number is not necessarily 2( k + 1)). But D ′ contains a P , a contradictionby Theorem 1. y In the following, given a total dominating set D of G , we call an edge xy , with x, y ∈ D ,a B -edge (resp. C -edge ; B − C -edge ) if x, y ∈ B (resp. x, y ∈ C ; x ∈ B and y ∈ C ). Given a B − C -edge xy , we call its endvertices the B -vertex and the C -vertex, according to the setin which they are contained. Recall that by Theorem 1, if G is a No -instance for γ t ) then every minimum total dominating set is an induced matching. Claim 11.
Let D be a minimum total dominating set of G . If G is a No -instance for γ t ) then there are at most ( k + 1) ( k + | A | ) − B -edges in D whichhave no private neighbors in C . roof. Assume that G is a No -instance for γ t ) and let ℓ = k + | A | .Suppose for a contradiction that there exist ℓ ( k + 1) B -edges x y , . . . , x ℓ ( k +1) y ℓ ( k +1) withno private neighbor in C . For every i ∈ [ k + 1], let X i = (cid:8) x ( i − ℓ +1 , . . . , x ( i − ℓ + k + | A | (cid:9) . Iffor some i ∈ [ k + 1] there were no vertex v ∈ C such that N ( v ) ∩ D ⊆ X i , then ( D \ X i ) ∪ A would be a total dominating set of G containing fewer vertices than D , a contradiction. Nowconsider v ∈ C such that N ( v ) ∩ D ⊆ X and let v , . . . , v q ∈ C be a longest sequence ofvertices defined as follows: for every i ∈ [ q ] \ { } , N ( v i ) ∩ D ⊆ X i and v i is anti-complete to { v , . . . , v i − } . We claim that q = k + 1. Indeed, if q < k + 1 then every vertex v ∈ C with N ( v ) ∩ D ⊆ X q +1 is adjacent to some vertex in { v , . . . , v q } by maximality of the sequence.But then ( D \ X q +1 ) ∪{ v , . . . , v q }∪ A is a total dominating set of G of cardinality at most | D | containing a P , which cannot be by Theorem 1. Thus, q = k + 1. Now for every i ∈ [ k + 1], v i has to be adjacent to at least two vertices in X i as no vertex in X i has a private neighborin C , say v i is adjacent to x ( i − ℓ +1 and x ( i − ℓ +2 without loss of generality. But now the setof vertices { y , x , v , x } ∪ k +1 [ i =2 (cid:8) x ( i − ℓ +1 , x ( i − ℓ +2 , v i (cid:9) induces a P + kP , a contradiction. y Let D be a total dominating set of G . We say that a B -edge xy can be turned into a B − C -edge if there exists a vertex z ∈ C such that ( D \ { x } ) ∪ { z } or ( D \ { y } ) ∪ { z } is atotal dominating set. Analogously, if xy is a B − C -edge with x ∈ B and y ∈ C , we say that xy can be turned into a C -edge if there is a vertex z ∈ C such that ( D \ { x } ) ∪ { z } is a totaldominating set of G . Claim 12.
Let D be a minimum total dominating set of G . If G is a No -instance for γ t ) and there exists a B -edge which has a private neighbor in C andcannot be turned into a B − C -edge, then there are at most | A | + 2 k B -edges.Proof. Assume that G is a No -instance for γ t ) and there existsa B -edge x y such that x has a private neighbor p ∈ C and x y cannot be turnedinto a B − C -edge. Suppose to the contrary that there exist | A | + 2 k additional B -edges x y , . . . , x | A | +2 k +1 y | A | +2 k +1 . If every private neighbor of y is adjacent to p , then ( D \{ y } ) ∪{ p } is a minimum total dominating set of G , that is, we can turn the B -edge x y into a B − C -edge which is contrary to our assumption. Thus, y has a private neighbor p ′ whichis not adjacent to p . Set V = { p, x , y , p ′ } and let V , . . . , V q be a longest sequence of setsof vertices defined as follows. For every i ∈ [ q ] \ { } , there is a j i ∈ [2 , . . . , | A | + 2 k + 1]and a vertex v i ∈ C such that V i = V i − ∪ { x j i , y j i , v i } , x j i V i − , v i is adjacent to x j i and v i is anti-complete to V i − ∪ y j i . We now claim that q ≥ k + 1. Indeed, observefirst that for any j ∈ [2 , . . . , | A | + 2 k + 1] such that x j V q , every neighbor of x j in C must be adjacent to some vertex in V q ∪ { y j } by maximality of the sequence. Now if q < k + 1, there are at least | A | + k + 1 indices j ∈ [ | A | + 2 k + 1] such that x j / ∈ V q and so, D ′ = ( D \ { x j : j ∈ [ | A | + 2 k + 1] , x j V q } ) ∪ V q ∪ A is a total dominating set of G of size atmost | D | . But D ′ contains a P , a contradiction by Theorem 1. Thus, q ≥ k + 1; but now V k +1 induces a P + kP , a contradiction. y Claim 13.
Let D be a minimum total dominating set. If G is a No -instance for γ t ) then all but at most k + | A | B − C -edges can be turned into C -edges.Proof. Assume that G is a No -instance for γ t ) . First, supposefor a contradiction that there exist | A | B − C -edges b c , . . . , b | A | c | A | , where b i ∈ B and c i ∈ C for all i ∈ [ | A | ], such that c i has no private neighbor in C for every i ∈ [ | A | ].17hen ( D \ (cid:8) c , . . . , c | A | (cid:9) ) ∪ A is a minimum total dominating set of G containing a P , acontradiction by Theorem 1. Thus, there exist at most | A | − B − C -edges such that the C -vertex has no private neighbor in C .Now suppose to the contrary that there exist k + | A | +1 B − C -edges b c , . . . , b k + | A | +1 c k + | A | +1 ,where b i ∈ B and c i ∈ C for all i ∈ [ k + | A | + 1], which cannot be turned into C -edges. Bythe above, we can assume without loss of generality that c has a private neighbor p ′ in C .If every private neighbor of b is adjacent to p ′ then ( D \ { b } ) ∪ { p ′ } is a minimum totaldominating set of G thereby contradicting the fact that b c cannot be turned into a C -edge.Thus, b must have a private neighbor p which is not adjacent to p ′ . We now claim thatthere are at most k − i ∈ { , . . . , | A | + k + 1 } such that there exists a vertex v i ∈ C which is adjacent to c i but not to b i and p . Indeed, if there were k such indices, say, withoutloss of generality, for every i ∈ [ k + 1] \ { } , there exists v i ∈ C such that v i is adjacent to c i but not to b i and p , then p ′ , c , b , p, v , c , b , . . . , v k +1 , c k +1 , b k +1 would induce a P + kP ;indeed, since for every i, j ∈ [ k + 1], c i and c j are non-adjacent and as v i is contained in C it follows that if v i is adjacent to c i then it is non-adjacent to c j and v j . Thus, we obtain acontradiction. It follows that there are at least | A | + 1 indices i ∈ { , . . . , | A | + k + 1 } suchthat every neighbor of c i in C is adjacent to b i or p , say without loss of generality indices2 to | A | + 2. But then ( D \ (cid:8) c , . . . , c | A | +2 (cid:9) ) ∪ A ∪ { p } is a minimum total dominating setcontaining a P , a contradiction by Theorem 1. y Claim 14. If G is a No -instance for γ t ) then every minimum totaldominating set of G is an induced matching and there exists a minimum total dominatingset of G in which all but at most ( k + 2) ( k + | A | ) + | A | − edges are contained in C .Proof. First note that since G is a No -instance for γ t ) , every min-imum total dominating set of G is an induced matching by Theorem 1. Now let D be aminimum total dominating set of G containing as few B -edges as possible amongst all min-imum total dominating sets of G . Then either no B -edge has a private neighbor in C inwhich case D contains at most ( k + 1) ( k + | A | ) − B -edges by Claim 11; or there exists a B -edge with a private neighbor in C in which case D contains at most | A | + 2 k B -edges byClaim 12 (indeed, note that by the choice of D , no B -edge can be turned into a B − C -edge).It then follows from Claim 13 that we can modify D in order to obtain a minimum totaldominating set of G which has at most k + | A | B − C -edges, and as the number of edgesintersecting A is trivially not more than | A | , the claim follows. y Claim 15.
Assume that G is a No -instance for γ t ) and let D be aminimum total dominating set of G . Let x y and x y be two C -edges contained in K and K respectively, where K , K ∈ K . If N [ K ] and N [ K ] are P -free then K and K havedistance at least four from one another.Proof. Assume that N [ K ] and N [ K ] are P -free. First, suppose for a contradiction that d ( K , K ) = 2, that is, there exists a vertex b ∈ N ( K ) ∩ N ( K ). By Claim 7, there exist c ∈ V ( K ) ∩ N ( b ) and c ∈ V ( K ) ∩ N ( b ) such that N [ K ] ⊆ N [ bc ] and N [ K ] ⊆ N [ bc ].But then, ( D \ { x , x , y , y } ) ∪ { c , c , b } is a total dominating set of G containing fewervertices than D , a contradiction. Thus, d ( K , K ) > d ( K , K ) = 3, that is, there exist b ∈ N ( K ) ∩ B and b ∈ N ( K ) ∩ B such that b and b are adjacent. By Claim 7, there exist c ∈ V ( K ) ∩ N ( b ) and c ∈ V ( K ) ∩ N ( b ) such that N [ K ] ⊆ N [ c b ] and N [ K ] ⊆ N [ c b ].But then, ( D \ { x , y , x , y } ) ∪ { b , b , c , c } is a minimum total dominating set containinga P , a contradiction by Theorem 1. Thus, d ( K , K ) > y The following claim is a straightforward corollary of Claim 15.18 laim 16.
Assume that G is a No -instance of γ t ) and let D be aminimum total dominating set of G . If there exist k + 1 cliques K , . . . , K k +1 ∈ K containing C -edges such that for any i ∈ [ k + 1] , N [ K i ] is P -free and there exists no vertex b ∈ B complete to K i , then K , . . . , K k +1 are regular cliques. Claim 17.
Assume that G is a No -instance to γ t ) and let D be aminimum total dominating set of G . Then there are at most | A |− cliques K ∈ K containinga C -edge, for which there exists a vertex b ∈ N ( K ) ∩ B such that N [ K ] ⊆ N [ b ] .Proof. Suppose for a contradiction that there are | A | cliques K , . . . , K | A | ∈ K containing C -edges x y , . . . , x | A | y | A | respectively, where for any i ∈ [ | A | ], there exists a vertex b i ∈ N ( K i ) ∩ B such that N [ K i ] ⊆ N [ b i ]. Then, ( D \ (cid:8) x , y , . . . , x | A | , y | A | (cid:9) ) ∪ (cid:8) b , . . . , b | A | (cid:9) ∪ A is a minimum total dominating set containing a P , a contradiction by Theorem 1. y Claim 18.
Let D be a minimum total dominating set of G and let C be a set of C-edgespairwise at distance at least three from one another. Then for any subset E ⊆ C of cardinality k + 1 for which there exists an induced path G E of length three containing exactly two verticesfrom two different edges of E , there exist a set T ⊆ N ( C ) of cardinality at most k − anda set N ⊆ { xy ∈ C : N ( xy ) ∩ T = ∅ } of cardinality at most k − k + 2 , such that every privateneighbor of an edge in C \ N is adjacent to V ( G E ) ∪ T , and { xy ∈ C : { x, y } ∩ V ( G E ) = ∅ } ⊆ N .Proof. Let E ⊆ C be a set of cardinality k + 1 for which there exists an induced P , denotedby G E , containing exactly two vertices from two different edges of E , and denote by xy and x ′ y ′ the two edges of E such that { x, y } ∩ V ( G E ) = ∅ and { x ′ , y ′ } ∩ V ( G E ) = ∅ . Let S ⊆ N ( C \ { xy, x ′ y ′ } ) ∩ B be a maximum independent set such that every vertex in S isadjacent to exactly one endvertex from an edge in C and not adjacent to G E , and every edgein C has at most one neighbor in S . Note that V ( G E ) ∪ S ∪{ v, w : vw ∈ C , N ( vw ) ∩ S = ∅ } induces a P + | S | P since the vertices in S are not adjacent to G E or to each other byconstruction, and any edge vw ∈ C such that N ( vw ) ∩ S = ∅ has exactly one neighbor u ∈ S where either u ∈ N ( v ) \ N ( w ) or u ∈ N ( w ) \ N ( v ) by construction. It follows that | S | < k . We construct a sequence of sets of vertices according to the following procedure.1. Initialize i = 1 and set C = E = { e ∈ C : N ( e ) ∩ S = ∅ } ∪ { xy, x ′ y ′ } .2. Increase i by one.3. Let S i ⊆ N ( C ) ∩ B \ ( N ( G E ) ∪ N ( C i − )) be a maximum stable set such that everyvertex in S i is adjacent to exactly one endvertex of an edge in C and every edge in C has at most one neighbor in S i .Set E i = { e ∈ C : N ( e ) ∩ S i = ∅ } and C i = C i − ∪ E i .4. If | S i | = | S i − | , stop the procedure. Otherwise, return to 2.Consider the value of i at the end of the above procedure (note that i ≥ vw in C \ C i − , every private neighbor b of vw isadjacent to G E or S i − for otherwise the procedure would have output S i − ∪ { b } in placeof S i − . Furthermore, any private neighbor b of an edge in E i − must be adjacent to G E or S i for otherwise the procedure would have output S i ∪ { b } in place of S i − (recall that byconstruction, | S i − | = | S i | ). Thus, it suffices to set T = S i ∪ S i − and N = C i − if i > N = { xy, x ′ y ′ } ). Observe that for any 1 ≤ p < q ≤ i − | S q | < | S p | which impliesthat i ≤ k + 1 as | S | ≤ k −
1; in particular, | S j | ≤ k − j for any j ∈ [ i −
1] and so | N | = | C i − | = 2 + i − X j =1 | E j | = 2 + i − X j =1 | S j | ≤ k − X j =1 ( k − j ) = k − k . Claim 19.
Assume that G is a No -instance of γ t ) and let D be aminimum total dominating set of G . Let C be a set of C -edges pairwise at distance at leastthree from one another, such that for every subset E ⊆ C of cardinality k + 1 , there exists aninduced P containing exactly two vertices from two different edges in E . Then |C| < (cid:18) k k (cid:19) (cid:18) k k (cid:19) + k + 1 . Proof.
Suppose for a contradiction that |C| ≥ (cid:18) k k (cid:19) (cid:18) k k (cid:19) + k + 1 . For a set E ⊆ C of cardinality k + 1, denote by G E an induced P which contains exactlytwo vertices from two different edges in E . Note that as any two edges in C have distance atleast three from one another, no other edge in C can be adjacent to G E .Now let E ⊆ C be a set of k +1 edges and let G E be as defined above. By Claim 18, thereexist a set T ⊆ N ( C ) of cardinality at most 2( k −
1) and a set N ⊆ { xy ∈ C : N ( xy ) ∩ T = ∅ } of cardinality at most k − k + 2, such that every private neighbor of an edge in C \ N is adjacent to V ( G E ) ∪ T , and { vw ∈ C : { v, w } ∩ V ( G E ) = ∅ } ⊆ N . We construct asequence of sets of vertices according to the following procedure.1. Initialize i = 1 and set F = N ∪ { e ∈ C : N ( e ) ∩ T = ∅ } .2. If i > k + k , stop the procedure. Otherwise, increase i by one.3. Let E i ⊆ C \ F i − be a set of k + 1 edges. By Claim 18, there exist a set T i ⊆ B ∩ N ( C \ F i − ) of cardinality at most 2( k −
1) and a set N i ⊆ { xy ∈ C : N ( xy ) ∩ T i = ∅ } of cardinality at most k − k + 2, such that every private neighbor of an edge in C \ ( F i − ∪ N i ) is adjacent to V ( G E i ) ∪ T i , and { vw ∈ C : { v, w } ∩ V ( G E i ) = ∅ } ⊆ N i .Set F i = F i − ∪ N i ∪ { e ∈ C : N ( e ) ∩ T i = ∅ } and return to 2.Note that the procedure ends with i = k + k + 1. Indeed, at each iteration, the set F j increases by at most k − k + 2 + 2( k −
1) and thus at the j th iteration, there are at least |C| − j ( k − k k − ≥ |C| − ( k k k − k k − ≥ k + 1edges left in C to form the set E j +1 . Finally, we increase once more i by one (that is, set i = k + k + 2) and let E i ⊆ C \ F i − be a subset of cardinality k + 1. Applying Claim 18 with C (rather than C \ F i − as above) and E i , we obtain that there exist a set T i ⊆ B ∩ N ( C ) ofcardinality at most 2( k −
1) and a set N i ⊆ { xy ∈ C : N ( xy ) ∩ T i = ∅ } of cardinality at most k − k + 2, such that every private neighbor of an edge in C \ N i is adjacent to V ( G E i ) ∪ T i ,and { vw ∈ C : { v, w } ∩ V ( G E i ) = ∅ } ⊆ N i . Observe that T i could intersect the sets V ( G E j )or T j for j < i , but every private neighbor of an edge in C \ N i is adjacent to G E i or T i .Note that by construction for any j, j ′ ∈ [ i −
1] the sets V ( G E j ) ∪ T j and V ( G E j ′ ) ∪ T j ′ aredisjoint and there exists no edge in C which is adjacent to both V ( G E j ) ∪ T j and V ( G E j ′ ) ∪ T j ′ .Since T i contains at most 2 ( k −
1) vertices and every vertex in T i is adjacent to exactly oneedge in C , there are at most 2( k −
1) edges in C which are adjacent to T i . As for each of theseedges there is at most one index j ∈ [ i −
1] such that the edge is adjacent to V ( G E j ) ∪ T j , itfollows that there are at least k − k +2+1 indices j ∈ [ i −
1] such that the sets V ( G E j ) ∪ T j and V ( G E i ) ∪ T i are disjoint and there exists no edge in C which is adjacent to both V ( G E j ) ∪ T j V ( G E i ) ∪ T i . Thus, there are at least k − k + 2 + 1 − | N i | ≥ j ∈ [ i −
1] suchthat the sets V ( G E j ) ∪ T j and V ( G E i ) ∪ T i are disjoint, there exists no edge in C which isadjacent to both V ( G E j ) ∪ T j and V ( G E i ) ∪ T i , and every private neighbor of an edge in (cid:8) vw ∈ C : { v, w } ∩ N ( V ( G E j ) ∪ T j ) = ∅ (cid:9) is adjacent to V ( G E i ) ∪ T i . Let j be one of theseindices. By construction, it holds that every private neighbor of an edge in E i is also adjacentto V ( G E j ) ∪ T j . Let x j y j and x ′ j y ′ j be the two edges in C which are adjacent to G E j andassume, without loss of generality, that x j and x ′ j are contained in V ( G E j ). Let x i y i and x ′ i y ′ i be the two edges in C which are adjacent to G E i and assume, without loss of generality,that x i and x ′ i are contained V ( G E i ). Let Y = { y ∈ V ( G ) : ∃ xy ∈ C , x ∈ N ( T i ∪ T j ) } be theset of all vertices which are adjacent to neither T i nor T j but belong to an edge in C that isadjacent to T i or T j . Then, D \ ( { y i , y ′ i , y j , y ′ j } ∪ Y ) ∪ V ( G E i ) ∪ V ( G E j ) ∪ T i ∪ T j is a minimumtotal dominating set containing a P , a contradiction by Theorem 1. y Claim 20. If G is a No -instance for γ t ) then there exists a mini-mum total dominating set of G in which all but at most | A | −
1) + (cid:18) k k (cid:19) (cid:18) k k (cid:19) + k + 1 edges are contained in regular cliques.Proof. Assume that G is a No -instance for γ t ) and let D be the setof all minimum total dominating sets of G in which all but at most ( k + 2) ( k + | A | ) + | A | − C -edges (note that D is nonempty by Claim 14). For every D ∈ D and S ⊆ B , let N e ( D, S ) be the number of C -edges which are adjacent to S . For any D ∈ D , let S ( D ) = arg max S ⊆ B N e ( D, S ) − | S | . In the following, let D ∈ D be a minimum total dominating set of G such that max S ⊆ B N e ( D, S ) −| S | is maximum amongst all sets in D .First, suppose for a contradiction that there exists a set S ⊆ B which is adjacent to s = N e ( D, S ) C -edges x y , . . . , x s y s , such that s − | S | ≥ | A | . Assume without loss ofgenerality that S is adjacent to x i for every i ∈ [ s ]. Then ( D \ { y , . . . , y s } ) ∪ S ∪ A is aminimum total dominating set of G containing a P , a contradiction by Theorem 1.Now consider a set S ∈ S ( D ) of minimum cardinality amongst all sets in S ( D ). By theabove, we have that N e ( D, S ) ≤ | A | − | S | . We claim that every v ∈ S must be adjacentto at least two C -edges to which no other vertex in S is adjacent. Indeed, if v were adjacentto no C -edge to which no other vertex in S was adjacent then S ′ = S \ { v } would be suchthat N e ( D, S ′ ) = N e ( D, S ) and | S ′ | < | S | . But then, N e ( D, S ′ ) − | S ′ | > N e ( D, S ) − | S | thereby contradicting the fact that S belongs to S ( D ). If v were adjacent to only one C -edgeto which no other vertex in S is adjacent, then removing v from S would decrease both | S | and N e ( D, S ) by one, leaving the difference unchanged and thus contradicting minimality of | S | . It follows that 2 | S | ≤ N e ( D, S ) which combined with the inequality above implies that | S | < | A | and so, N e ( D, S ) ≤ | A | − C be the set of C -edges which are not adjacent to S . We may assume that D satisfies the following property: for any edge xy ∈ C contained in a clique K ∈ K , if there exist x ′ , y ′ ∈ V ( K ) such that N [ xy ] ⊆ N [ x ′ y ′ ] then N [ xy ] = N [ x ′ y ′ ]. Indeed, suppose that thereexists an edge xy ∈ C contained in a clique K ∈ K for which there exist x ′ , y ′ ∈ K such that N [ xy ] ( N [ x ′ y ′ ]. Then, it suffices to consider D ′ = ( D \{ x, y } ) ∪{ x ′ , y ′ } in place of D . Clearly, D ′ is still a minimum total dominating set of G and the value of max S ′ ⊂ B N e ( D ′ , S ′ ) − | S ′ | does not increase by choice of D . Furthermore, the value of max S ′ ⊂ B N e ( D ′ , S ′ ) − | S ′ | doesnot decrease: since N [ xy ] ( N [ x ′ y ′ ], we have that N e ( D, S ) ≤ N e ( D ′ , S ) and so by choice21f D and because S ∈ S ( D ), we conclude that max S ′ ⊆ B N e ( D, S ′ ) − | S ′ | = N e ( D, S ) − | S | = N e ( D ′ , S ) − | S | = max S ′ ⊆ B N e ( D ′ , S ′ ) − | S ′ | . In particular, it follows that x ′ y ′ cannot beadjacent to S and thus, the set of C -edges in D ′ not adjacent to S has the same cardinalityas C . By replacing each such edge of C , the resulting total dominating set satisfies the aboveproperty. In the following, we denote by D the resulting dominating set and by C the set of C -edges not adjacent to S for simplicity.Now if there were two edges in C with a common neighbor v , then S ′ = S ∪ { v } wouldbe such that | S ′ | = | S | + 1 and N e ( D, S ′ ) = N e ( D, S ) + 2. But then, N e ( D, S ′ ) − | S ′ | >N e ( D, S ) − | S | thereby contradicting the fact that S belongs to S ( D ). Thus, any two edgesin C are at distance at least three from one another.Now denote by C ⊆ C the set of all C -edges xy for which there exists a vertex in b ∈ B such that b is complete to the clique in K containing xy . Then by Claim 17, |C | ≤ | A | − C ⊆ C \ C the set of edges which contain an induced P in their closedneighborhood and suppose that C has cardinality at least k + 1. Now consider a set E ⊆ C of cardinality k + 1 and denote by P e an induced P contained in N [ e ], for any e ∈ E . Ifany two edges in E were pairwise at distance at least four from one another then S e ∈ E P e would be isomorphic to ( k + 1) P , a contradiction. Thus, there exist xy, x ′ y ′ ∈ E such that d ( xy, x ′ y ′ ) = 3, say d ( xy, x ′ y ′ ) = d ( x, x ′ ) without loss of generality; in particular, there existsan induced path P from x to x ′ of length three containing exactly two vertices (namely x and x ′ ) from two different edges of E (namely xy and x ′ y ′ ). Since this holds for any subset E ⊆ C of cardinality k + 1, we conclude by Claim 19 that |C | < (cid:16) k + k (cid:17) (cid:16) k + k + 1 (cid:17) + k + 1.Now consider an edge xy ∈ C \ ( C ∪ C ) which is contained in a clique K ⊆ K and let usshow that N [ xy ] = N [ K ]. Suppose to the contrary that there is a vertex b ∈ N [ K ] \ N [ xy ]and let c ∈ K be a neighbor of b . By assumption on D , N [ xy ] cannot be a strict subsetof N [ xc ] or N [ yc ], so there exists a vertex p x ∈ N ( x ) which is adjacent to neither y nor c and there exists a vertex p y ∈ N ( y ) which is adjacent to neither x nor c . As xy / ∈ C itfollows that p x , x, y, p y cannot induce a P and thus p x and p y must be adjacent. But then p x , p y , y, c induce a P , a contradiction. Hence, every edge xy ∈ C \ ( C ∪ C ) is contained ina clique in K ∈ K whose closed neighborhood is P -free (recall that N [ xy ] = N [ K ] by theabove property) and for which there exists no b ∈ B such that N [ K ] ⊆ N [ b ]. Furthermoreby Claim 16, either C \ ( C ∪ C ) has cardinality at most k or contains only regular cliques.Thus, all but at most2 ( | A | −
1) + |C | + (cid:18) k k (cid:19) (cid:18) k k (cid:19) + k + 1 .C -edges are contained in regular cliques, which proves the claim since |C | ≤ | A | − y We now present an algorithm which can determine in polynomial-time whether G is a Yes -instance for γ t ) or not. In the following, we denote by f ( k ) =3 ( | A | −
1) + (cid:16) k + k (cid:17) (cid:16) k + k + 1 (cid:17) + k + 1.1. Determine A , B , C and R .1.1 If R = ∅ , then check if there exists a minimum total dominating set of size atmost f ( k ).1.1.1 If the answer is no, then output Yes .1.1.2 Else apply Proposition 9(b).1.2 Else go to 2.2. Check whether there exist two regular cliques in R which have distance at most threefrom one another. If so, output Yes . 22. Let V be the set of vertices at distance one from N [ R ] and let V = V ( G ) \ ( N [ R ] ∪ V ).If V = ∅ , output No .4. Determine S = { S ⊆ V ∪ V : | S | ≤ f ( k ) , ∀ x ∈ V , N ( x ) ∩ S = ∅ } . If S = ∅ , output Yes .5. Let S ′ be the family of all sets in S which have minimum size.(i) If there exists a set S ∈ S ′ containing a P , output Yes .(ii) If there exists a set S ∈ S ′ such that S ∩ V = ∅ , output Yes .6. Output No .Finally, let us show that this algorithm outputs the correct answer. In case R = ∅ , thenClaim 20 tells us that G is a Yes -instance for γ t ) if there existsno minimum total dominating set of size at most f ( k ) (see step 1.1.1). If such a minimumtotal dominating set exists, then we conclude using Proposition 9(b) (see step 1.1.2). If instep 2, two regular cliques at distance at most three from one another are found then byClaim 10, G is a Yes -instance for γ t ) . Otherwise, any two regularcliques have distance at least four to one another and by Remark 3, there exists a minimumtotal dominating set D of G such that for any regular clique K ∈ R , D ∩ N [ K ] = { b K , c K } where b K ∈ N ( K ) ∩ B and c K ∈ K . In the following, we denote by D ′ = S K ∈R { b K , c K } .Note that by Claim 8, for any x ∈ N [ R ] ∪ V , N ( x ) ∩ D ′ = ∅ . Now if V = ∅ then weconclude by Claim 9 and the fact that any two regular cliques are at distance at least fourfrom one another, that any minimum total dominating set of G is an induced matching, thatis, G is a No -instance for γ t ) (see step 3). Otherwise V = ∅ andif G is a No -instance for γ t ) then by Claim 20, there must exist aset S ⊆ V ∪ V of cardinality at most 2 f ( k ) such that for any x ∈ V , N ( x ) ∩ S = ∅ . Thus,if S = ∅ then G is a Yes -instance for γ t ) (see step 4). Otherwise S 6 = ∅ , and for any S ∈ S ′ , S ∪ D ′ is a minimum total dominating set of G . It then followsfrom Theorem 1 that if there exists S ∈ S ′ such that S contains a P then G is a Yes -instancefor γ t ) (see step 5(i)); otherwise, any S ∈ S ′ is an induced matchingand if there exists a set S ∈ S ′ such that S ∩ V = ∅ then by Claim 8, S ∪ D ′ contains a P and so, G is a Yes -instance by Theorem 1 (see step 5(ii)). Otherwise, for any S ∈ S ′ , S isan induced matching and S ∩ V = ∅ which implies that S ∪ D ′ is an induced matching andthus, G is a No -instance for γ t ) . As every step can clearly be donein polynomial time, this concludes the proof. We here give a proof of our main result, Theorem 3.Let H be a graph. If H contains a cycle then γ t ) is NP -hardon H -free graphs by Theorem 8. Thus, we may assume that H is a forest. Now if H contains a vertex of degree at least three, then H contains an induced claw and so, γ t ) is coNP -hard on H -free graphs by Theorem 6. Assume hence-forth that H is a linear forest. If H contains a path on at least six vertices, then γ t ) is NP -hard on H -free graphs by Theorem 4. Thus, we may assume thatevery connected component of H induces a path on at most five vertices. Now suppose that H contains a P . If H has another connected component on more than one vertex then γ t ) is NP -hard by Theorem 4. Otherwise, every other connected com-ponent of H (if any) contains exactly one vertex in which case γ t ) is polynomial-time solvable on H -free graphs by Theorem 10 and Proposition 9. Assumenow that H contains a P . Then if H has another connected component isomorphic to23 , γ t ) is coNP -hard by Theorem 5. Otherwise, every other con-nected component has at most three vertices and we conclude by Theorem 11 that γ t ) is polynomial-time solvable on H -free graphs. Finally, if the longestpath in H has length at most two then we also conclude by Theorem 11 that γ t ) is polynomial-solvable on H -free graphs, which concludes the proof. In this paper, we considered the problem of deciding whether the total domination num-ber of a given graph G can be reduced using exactly one edge contraction (called γ t ) ). We recall that if we were allowed to use 3 edge contractions, the answerto this problem would always be yes due to a result of Huang et al. ([13], Theorem 4.3).We focused on several graph classes and determined the computational complexity of thisproblem. By putting together these results, we managed to obtain a complete dichotomy for H -free graphs (see Theorem 3).In [9], the same problem was considered with respect to the domination number (called γ ) ). Here the authors provided an almost dichotomy for H -freegraphs (see Theorem 2). As mentioned in the introduction, the remaining cases left openhave recently been solved. Interestingly, the two problems do not behave the same way on H -free graphs from a complexity point of view. This is even more interesting since it hasbeen shown in [11] that the complexities of Dominating set and
Total dominating set agree on H -free graphs for any graph H .In fact, it was even shown in [11] that the complexities of Dominating set , Semi-total dominating set (given a graph G and an integer k , does there exist a dominatingset S ⊆ V ( G ) with | S | ≤ k such that every vertex in S is at distance at most two to anothervertex in S ) and Total dominating set agree on H -free graphs for any graph H . Thus,it would be of interest to look at our problem with respect to the semi-total dominationnumber and find out how it behaves on H -free graphs. A few of our results can be adaptedto the semi-total case, but in order to obtain a complete dichotomy for H -free graphs, newapproaches are needed. References [1]
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