On the Parameterized Complexity of [ 1 , j ]-Domination Problems

For a graph G, a set D ⊆ V (G) is called a [1, j]-dominating set if every vertex in V (G) \D has at least one and at most j neighbors in D. A set D ⊆ V (G) is called a [1, j]-total dominating set if every vertex in V (G) has at least one and at most j neighbors in D. In the [1, j]-(Total) Dominating Set problem we are given a graph G and a positive integer k. The objective is to test whether there exists a [1, j]-(total) dominating set of size at most k. The [1, j]-Dominating Set problem is known to be NP-complete, even for restricted classes of graphs such as chordal and planar graphs, but polynomial-time solvable on split graphs. The [1, 2]-Total Dominating Set problem is known to be NP-complete, even for bipartite graphs. As both problems generalize the Dominating Set problem, both are W[1]-hard when parameterized by solution size. In this work, we study [1, j]-Dominating Set on sparse graph classes from the perspective of parameterized complexity and prove the following results when the problem is parameterized by solution size: [1, j]-Dominating Set is W[1]-hard on d-degenerate graphs for d = j + 1; [1, j]-Dominating Set is FPT on nowhere dense graphs. We also prove that the known algorithm for [1, j]-Dominating Set on split graphs is optimal under the Strong Exponential Time Hypothesis (SETH). Finally, assuming SETH, we provide a lower bound for the running time of any algorithm solving the [1, 2]-Total Dominating Set problem parameterized by pathwidth. 2012 ACM Subject Classification Mathematics of computing → Graph algorithms, Theory of computation → Design and analysis of algorithms


Introduction
A dominating set of a graph G is a subset D of vertices such that each vertex in V (G) \ D is adjacent to at least one vertex in D. Various extensions of domination, such as independent, total, efficient, and perfect domination, have been introduced and widely studied both combinatorially and algorithmically.A discussion of these extensions can be found in [14].A

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We begin by some basic terminology and notation in Section 2. Then the next sections each address one of the results mentioned above.

Preliminaries
Graphs.We assume G is a simple graph with vertex set V (G) and edge set E(G).For brevity, we often denote these sets by V and E. We let n = |V (G)| denote the order of G.For a vertex v ∈ V , the open neighborhood of v, denoted by N (v), is defined as {u : {u, v} ∈ E} and the closed neighborhood N [v] is defined as N (v)∪{v}.For a set S ⊆ V , we use N (S) and N [S] to denote the open and closed neighborhood S, respectively.That is, and N (S) = N [S] \ S. For a set U ⊆ V , we use G[U ] to denote the subgraph of G induced on U .A tree decomposition of a graph G is a tree T in which each vertex x ∈ T has an assigned set of vertices B x ⊆ V (G) (called a bag) which satisfies the following properties: (i) x∈T B x = V (G); (ii) For any {u, v} ∈ E, there exists x ∈ V (T ) such that u, v ∈ B x ; (iii) For any v ∈ V (G), the subtree of T induced on {x ∈ V (T ) : v ∈ B x } is connected.The width of a tree decomposition T is max x∈V (T ) (|B x | − 1).The treewidth of G, denoted by tw(G), is the minimum width over all tree decompositions of G.The pathwidth of G, denoted by pw(G), is the minimum width over all tree decompositions T of G, where T is a path.Parameterized complexity.We now review some necessary concepts from parameterized complexity.For more details we refer the reader to [4,8].Given a finite alphabet Σ, a parameterization of Σ * is a function p : Σ * → N. A parameterized language L is a subset of {(x, k) | x ∈ Σ * ∧ k = p(x)}.Here k is called the parameter.A parameterized language L ⊆ Σ * × N is called fixed-parameter tractable (FPT) if there exist an algorithm A (called a FPT algorithm) and a computable function f : N → N such that given (x, k) ∈ Σ * × N, the algorithm A correctly decides whether (x, k) ∈ L in time f (k) • |(x, k)| O (1) .The class of all fixed-parameter tractable problems is denoted by FPT.Definition 1.Let L and L be two parameterized languages with parameterization functions p and p .An FPT reduction from L to L is a mapping ρ : Σ * → Σ * such that the following holds.
To classify problems that are not FPT, Downey and Fellows [8] introduced the Whierarchy.The hierarchy consists of complexity class W[t] for every integer t ∈ N such that For our purposes, it is sufficient to note that these classes are closed under FPT reduction.

ETH and SETH.
For q ≥ 3, let δ q be the infimum of the set of constants c for which there exists an algorithm solving q-SAT with n variables and m clauses in time 2 cn • m O (1) .The Exponential-Time Hypothesis (ETH) and Strong Exponential-Time Hypothesis (SETH) are then formally defined as follows.ETH conjectures that δ 3 > 0 and SETH that lim q→∞ δ q = 1.In other words, SETH conjectures that for all 0 < < 1, there exist a (large) q = q( ) such that q-SAT cannot be solved in time O(2 (1− )n ), where n is the number of variables in the input formula.

Split Graphs
] is an empty graph (i.e., V 2 is an independent set in G).Several domination-like problems such as domination, total domination, and k-tuple domination are known to be NP-complete even on split graphs.On the other hand, there is a n j • (lg n) O (1) time algorithm for [1, j]-Dominating Set on split graphs [2].In this section, we prove that this is optimal, in the sense that one cannot obtain an O(n j− ) algorithm for [1, j]-Dominating Set on split graphs unless SETH fails.

Reduction. Given an instance
of variables, we construct a graph G I as follows.
We partition the set X into j subsets X 1 , . . ., X j of size at most n/j .For each X i we add a set S i of 2 |Xi| vertices to the graph, each corresponding to one possible valuation of the variables in X i .We also add two distinguished vertices u i and v i and connect them to all of S i .We connect all the vertices in j i=1 S i .That is j i=1 S i forms a clique in G I .For every clause C i , we add a vertex c i and connect it to every vertex w of any S j where the valuation corresponding to w satisfies C i .This completes the construction of G I .See Figure 1 for an illustration.Our reduction algorithm will output an instance (G I , j) of [1, j]-Dominating Set.We now proceed to proving the correctness of the reduction.

Lemma 2. If I is satisfiable then
Proof.Take one satisfying valuation of I and let s 1 ∈ S 1 , s 2 ∈ S 2 , . . ., s j ∈ S j be the vertices of G I that correspond to this valuation.We claim that S = {s 1 , s 2 , . . ., s j } is a [1, j]-dominating set.Every vertex in any of the S i 's is dominated by all of S, every {u i , v i } pair is dominated only by the corresponding s i and every c i is dominated by a non-empty subset of S, i.e., the vertices whose corresponding valuation forces satisfaction of C i .Given that |S| = j, there can be at most j such vertices.

Lemma 3.
If G I has a [1, j]-dominating set of size j then I is satisfiable.
Proof.First note that any dominating set of G I of size j is also a [1, j]-dominating set.Let S be a dominating set of size at most j (and hence a [1, j]-dominating set) in G I .Since {N [u i ] : i ∈ {1, . . ., j}} are pairwise vertex disjoint we have that |S| = j and |S ∩ N [u i ]| = 1

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for all i ∈ {1, . . ., j}.Moreover, since N (u i ) = N (v i ), we conclude that S ⊆ ( j i=1 S i ).Consider the valuation of variables in X that corresponds to the vertices in S.This valuation satisfies every C i , because S dominates every c i .Theorem 4. For any ε < 1 and constant j, there is no O(n j−ε ) time algorithm for [1, j]-Dominating Set on split graphs unless SETH fails.
Proof.For the sake of contradiction assume that there is an O(n j−ε ) time algorithm A for [1, j]-Dominating Set on split graphs.Then we claim that SETH is false.Let = (1 − ε j ).Let q = q( ) be the constant defined in the SETH conjecture.Given a q-SAT instance I we construct an instance (G I , j) as mentioned in the reduction.By Lemmas 2 and 3 we know that I is satisfiable if and only if (G I , j) is a yes-instance.Therefore, we use algorithm A to solve q-SAT.Now consider the running time for solving q-SAT using A. The construction of (G I , j) takes time 2 n/j • n O (1) and the number of vertices in G I is at most 2j + m + j2 n/j = 2 n/j • n O (1) .Thus the algorithm A on instance (G I , j) takes time 1) .This refutes SETH and the proof of the theorem is complete.

Degenerate Graphs
A graph G is d-degenerate if every subgraph of G contains a vertex of degree at most d.
Equivalently, a graph G is d-degenerate if and only if there exists an elimination ordering on its vertices such that every vertex has at most d neighbors appearing later in the ordering.
In this section we prove the following result.
The following parameterized problem, proved to be W [1]-hard in [11], is used in our proof.In the Multicolored Independent Set problem, we are given a graph G and a proper vertex coloring of V (G) with k colors.The parameter k is equal to the number of colors and the goal is to find a k-sized independent set in G containing exactly one vertex from each color class (such independents sets are called k-colored independent sets).We shall reduce the Multicolored Independent Set problem (with parameter k) to the problem of finding a [1, j]-dominating set of size at most 2k + j − 1 in a graph of degeneracy j + 1.

The reduction.
Let k be an integer and G be a proper k-vertex colored graph such that its vertices are partitioned into k groups V 1 , V 2 , . . ., V k , where each group corresponds to an independent set of the same color.Now we construct an instance (G , 2k + j − 1) of [1, j]-Dominating Set as follows.For every edge e = {u, v} ∈ E(G), we replace it by a path uv e v, where v e is a new vertex corresponding to the edge e.Let S E = {v e : e ∈ E(G)}.For each group V i , 1 ≤ i ≤ k, we build a K 1,2k+j graph centered at a new vertex u i .We also add new vertices x i 1 and x i 2 and connect the vertices u i , x i 1 and x i 2 to the vertices of V i .Moreover, we build j − 1 star graphs K 1,2k+j centered at vertices r 1 , . . ., r j−1 and make r 1 , . . ., r j−1 adjacent to the vertices in S E .This concludes the construction of G .See Figure 2 Proof.The proof is by constructing a degeneracy ordering, which is an ordering on the vertices that we get from repeatedly removing a vertex of minimum degree in the remaining subgraph.First, we put all of the degree-one vertices in the degeneracy ordering and delete them.In the remaining subgraph, we select all the vertices in S E and put them in ordering, because every vertex in S E has degree j + 1 in G .After removing all vertices of S E from the graph, each vertex in any block V i , for 1 ≤ i ≤ k, has degree three because after removing S E such vertices are only connected to x i 1 , x i 2 , and u i .So, next we can put the vertices k i=1 V i in the ordering.Finally we add all the remaining vertices.Lemmas 6, 7, and 8 below, imply Theorem 5. Proof.Let D be a [1, j]-dominating set of size at most 2k + j − 1.First, note that since |D| ≤ 2k + j − 1, we have that {r 1 , . . ., r j−1 } ∪ {u i , . . ., u k } ⊆ D (because each u i and r i is connected to 2k +j degree one vertices).We claim that S = D \({r 1 , . . .,

Lemma 7. If there exists a k-colored independent set in G then there exists a [1, j]-
Then the vertex v e , where e = {u, v} is dominated j + 1 times by D (because v e is adjacent to {u, v} and {r 1 , . . ., r j−1 }).Therefore since D is a [1, j]-dominating set in G , S is a k-colored independent set in G This completes the proof of the lemma.Nowhere Dense Graphs The notion of nowhere denseness was introduced by Nešetřil and Ossona de Mendez [20,21] as a general model of uniform sparseness of graphs.Many familiar classes of sparse graphs, like planar graphs, graphs of bounded tree-width, graphs of bounded degree, and all classes that exclude a fixed (topological) minor, are nowhere dense.An important and related concept is the notion of a graph class of bounded expansion, which was also introduced by Nešetřil and Ossona de Mendez [17,18,19].

Definition 9.
Let H be a graph and let r ∈ N.An r-subdivision of H is obtained by replacing all edges of H by internally vertex disjoint paths of length at most r.Definition 10.A class C of graphs is nowhere dense if there exists a function t : N → N such that for all q ∈ N and for all G ∈ C we do not find an q-subdivision of the complete graph K t(q) as a subgraph of G. Otherwise, C is called somewhere dense.Definition 11.A class C of graphs has bounded expansion if there exists a function d : N → N such that for all r ∈ N and all graphs H, such that an r-subdivision of Every class of bounded expansion is nowhere dense, which in turn excludes some biclique as a subgraph and hence is biclique-free.For extensive background on bounded expansion and nowhere dense graphs we refer to the textbook of Nešetřil and Ossona de Mendez [22].
Before we state our result, we quickly recall the necessary definitions from logic.For our purpose, it suffices to consider first-order logic over the vocabulary of graphs.We refer to the textbook [9] for extensive background on logic.A (relational) vocabulary is a finite set of relation symbols, each with a prescribed arity.We let σ be a vocabulary.A σ-structure A consist of a (not necessarily finite) set V (A), the universe or vertex set of A, and for each k-ary relation symbol R ∈ σ a k-ary relation R(A) ⊆ V (A) k .A structure A is finite if its universe is.For example, graphs may be viewed as {E}-structures, where E is a binary relation symbol.First-order formulas of vocabulary σ are formed from atomic formulas x = y and R(x 1 , . . ., x k ), where R ∈ σ is a k-ary relation symbol and x, y, x 1 , . . ., x k are variables (we assume that we have an infinite supply of variables) by the usual Boolean connectives ¬ (negation), ∧ (conjunction), ∨ (disjunction), and existential and universal quantifications ∃x and ∀x, respectively.The set of all first-order formulas of vocabulary σ is denoted by FO[σ], and the set of all first-order formulas by FO.The free variables of a formula are those not in the scope of a quantifier, and we write φ(x 1 , . . ., x k ) to indicate that the free variables of the formula φ are among x 1 , . . ., x k .A sentence is a formula without free variables.To define the semantics, we inductively define a satisfaction relation |=, where for a σ-structure A, a formula φ(x 1 , . . ., x k ), and elements a 1 , . . ., a k ∈ V (A), A |= φ(a 1 , . . ., a k ) means that A satisfies φ if the free variables x 1 , . . ., x k are interpreted by a 1 , . . ., a k , respectively.If The meaning of the equality symbol, the Boolean connectives, and the quantifiers is the usual one.For example, consider the formula φ( in the vocabulary {E} of graphs.For every graph G and vertices Proof.Our proof is based on a result of Grohe, Kreutzer, and Siebertz [13], which states that for every first-order sentence ψ (or formula without free variables), every nowhere dense class C of graphs and every real > 0, there exists a constant f (|ψ|, ), such that given an n-vertex graph G ∈ C, one can decide in time f (|ψ|, ) • n 1+ whether ψ holds in G.
It is easy to verify that the [1, j]-Dominating Set problem is expressible in FO.We let ψ be the following sentence.
where the function φ i (u, v 1 , v 2 , . . ., v k ) is true where the vertex u is adjacent to exactly i vertices of v 1 , v 2 , . . ., v k , which can be represented using a formula of length bounded by a function of i and k.Note that the length of ψ is bounded by a function depending only on k (and on j, though only as a fixed constant).Then, by fixing any > 0 and using the result of [13], we conclude that [1, j]-Dominating Set is fixed-parameter tractable parameterized by solution size k on every nowhere dense class C.

Bounded Treewidth Graphs
It is well-known [4] that the Dominating Set problem can be solved in O * (3 t ) time on graphs of treewidth at most t.Lokshtanov et al. [15] showed that this is essentialy optimal, i.e., they showed that the problem cannot be solved in O * Proof.As mentioned before, [1, j]-Dominating Set is a special case of (σ, ρ)-Dominating Set, where σ = {0, 1, . ..} and ρ = {1, . . ., j}.The states needed in the representation of the solution in this special case are {ρ 0 , ρ 1 , . . ., ρ j } and * , where ρ i , 0 ≤ i ≤ j means the vertex in the bag is not in the dominating set and dominated i times and * means that the vertex is in the dominating set.Then by Proposition 13, [1, j]-Dominating Set is solvable in time O * ((j + 2) tw ).
[1, j]-Total Dominating Set is the special case of (σ, ρ)-Dominating Set, with σ = ρ = {1, . . ., j}.In this case the states needed in the representation of the solution are {σ 0 , ρ 0 , σ 1 , ρ 1 , . . ., σ j , ρ j }, where ρ i , 0 ≤ i ≤ j means the vertex in the bag is not in In what follows, we prove that the [1, 2]-Total Dominating Set problem cannot be solved in time O * (4 − ) pw (unless SETH fails) and which is also a lower bound in terms of the treewidth of the input graph.It remains open whether a similar result holds for [1,2]-Dominating Set.Our proof closely follow the work of Cygan et al. [5] and we use the notions of path decompositions, pathwidth, tree decompositions, treewidth, and mixed search games.We give a reduction from CNF-SAT to the [1,2]-Total Dominating Set problem and prove that the reduced graph has pathwidth at most n 2 + O( 1), where n is the number of variables in the input CNF-SAT formula.
The reduction.Given > 0 and an instance Φ of CNF-SAT with n variable and m clauses, we construct a graph G as follows.We assume that the number of variables n is even, otherwise we add a single dummy variable.We partition the variables of Φ into groups F 1 , F 2 , . . ., F n , each of size two, where n = n/2.We let a = m(n + 1).
For each 1 ≤ t ≤ n we create a path P t of length 4a = 4m(n + 1) consisting of vertices v α t,q and h α t,q , where 1 ≤ α ≤ 2 and 0 ≤ q < a.The vertices are arranged on the path in the following order: We let V and H denote the sets of all v α t,q vertices and h α t,q vertices, respectively.For each vertex v α t,q , we add a forcing gadget consisting of a 4-cycle with one additional pendant vertex connected to a vertex which we denote as the root vertex r α t,q .We add an edge between v α t,q and the root of the cycle r α t,q .Similarly, for each vertex h α t,q , we add a forcing gadget consisting of a 4-cycle rooted at vertex s α t,q (i.e. the pendant vertex is connected to s α t,q ).We add an edge between h α t,q and s α t,q (see Figure 3).Note that any dominating set in the graph will contain at least two vertex from a forcing gadget and if a [1, 2]-total dominating set contains only two vertex from a forcing gadget, then it include the root vertex and one of its neighbours on the 4-cycle (by the definition of [1,2]-total dominating set).
Next, we add three pairs of guard vertices p 1 t,q , p 2 t,q , and p 3 t,q , for each 1 ≤ t ≤ n and 0 ≤ q < a.Each of the vertices in these pairs are of degree two and are connected to other vertices as follows: (i) vertices in p 1 t,q are adjacent to v 1 t,q and v 2 t,q ; (ii) vertices in p 2 t,q are adjacent to v 2 t,q and v 2 t,q+1 ; and (iii) vertices in p 3 t,q are adjacent to h 1 t,q and h 2 t,q .This structure forces a dominating set to either contain the vertices from the guard sets or one of their two neighbors.For instance, to dominate the pair p 1 t,q , either both vertices in p 1 t,q must be in the dominating set or one of v 1 t,q or v 2 t,q .We use G to denote the set of all guard vertices.

On the Parameterized Complexity of [1, j]-Domination Problems
The intuition of the construction made so far is as follows.For each two-variable block F t we encode any assignment of the variables in F t as a choice of whether to take v 1 t,q or v 2 t,q and h 1 t,q or h 2 t,q into the dominating set.This concludes the construction of the "variable gadgets" which are required for encoding an assignment.The forcing gadgets attached to vertices in V and H will guarantee that those vertices are all dominated at least once.
We now add "clause gadgets" required for checking the satisfiability of Φ.For each clause C i , we build (n + 1) vertices c i,j , one for each 0 ≤ j ≤ n.Consider a clause C i and a group of variables to c i,j via a path of length five by adding four vertices w i,j , x i,j , y i,j and z i,j and edges {c i,j , w i,j }, {w i,j , x i,j }, {x i,j , y i,j }, {y i,j , z i,j }, and {z i,j , v 1 t,mj+i }.If x 1 t occurs negatively as the th literal in C i , we connect v 2 t,mj+i to c i,j again via a path of length five by adding four vertices w i,j , x i,j , y i,j and z i,j and edges {c i,j , w i,j }, {w i,j , x i,j }, {x i,j , y i,j }, {y i,j , z i,j }, and {z i,j , v 2 t,mj+i }.Similarly, if x 2 t occurs positively as the th literal in C i , we connect h 1 t,mj+i to c i,j via a path of length five and if it occurs negatively, we connect h 2 t,mj+i to c i,j via a path of length five.We let W, X , Y, and Z denote the sets of all w, x, y, and z vertices added between clause vertices and vertices in V ∪ H, respectively.This concludes the construction of the graph G.
Finally, since we have 2an pairs of guards with disjoint neighbors, exactly one vertex from {v 1 t,q , v 2 t,q } and one vertex from {h 1 t,q , h 2 t,q } must be in D, for each 1 ≤ t ≤ n and 0 ≤ q < a.This implies that each forcing gadget will contribute two vertices to D-the root vertex and one of its neighbour in the 4-cycle.Moreover |D ∩ (W ∪ X ∪ Y ∪ Z)| = 2(n + 1)Σ m i=1 |C i |.This implies that exactly two vertex from a path from c i,j to a vertex in V ∪ H is included in D.Moreover, this in turn implies that c i,j / ∈ D for any i, j.Now, for each 0 ≤ q < a, we construct an assignment φ q as follows.For each block F t = {x 1 t , x 2 t }, we define φ q (x 1 t ) as true if v 1 t,q ∈ D and we define φ q (x 1 t ) to be false if v 2 t,q ∈ D. Similarly, for x 2 t , we define φ q (x 2 t ) to be true if h 1 t,q ∈ D and we define φ q (x 2 t ) to be false if h 2 t,q ∈ D. Note that for each block F t = {x 1 t , x 2 t } and each 0 ≤ q < a, we have: If φ q (x 1 t ) is true then φ q+1 (x 1 t ) is true.Otherwise, both v 2 t,q and v 1 t,q+1 are not in the dominating set and vertices in p 2 t,q is not dominated by D. If φ q (x 2 t ) is false then φ q+1 (x 2 t ) is false.Otherwise, both h 2 t,q and h 1 t,q+1 are in the dominating set and vertex v 1 t,q+1 is dominated three times in D; by h 2 t,q , h 1 t,q+1 , and the root of the forcing gadget corresponding to v 1 t,q+1 .
For each variable x, we define a sequence φx = φ 0 (x)φ 1 (x) . . .φ a−1 (x).By the discussion above we infer that for each variable x, the sequence φx can change its value at most once.Hence, as a = m(n + 1), we conclude that there exists 0 ≤ j < n + 1 such that for all 0 ≤ i < m, the assignment φ mj+i (x) are equal.We claim that the assignment φ = φ mj (x) satisfies φ.Consider a clause C i and focus on the vertex c i,j .We know that c i,j / ∈ D. Thus one of its neighbors from W (say w i,j ) is contained in D. Therefore, by the definition of a [1, 2]-total dominating set, x i,j is in D. We have already argued that any path from c i,j to any vertex in V ∪ H can contain at most 2 vertices in D. This implies that z i,j is dominated by a vertex in V ∪ H and the literal corresponding to it will be set to true by φ.This completes the proof of the lemma.

Lemma 17. pw(G) ≤ n + O(1).
Proof.Let us first recall the definition of a mixed search game.In a mixed search game, the graph G represents a "system of tunnels".Initially, all edges are contaminated by a gas.An edge is cleared by placing searchers at both its endpoints simultaneously or by sliding a searcher along the edge.A cleared edge is re-contaminated if there is a path from an uncleared edge to the cleared edge without any searchers on its vertices or edges.A search is a sequence of operations that can be of the following types: (i) placement of a new searcher on a vertex; (ii) removal of a searcher from a vertex; (iii) sliding a searcher on a vertex along an incident edge and placing the searcher on the other end.A search strategy is winning if after its termination all edges are cleared.The mixed search number of a graph G, denoted by ms(G), is the minimum number of searchers required for a winning strategy of mixed searching on G. Takahashi et al. [25] obtained the following relationship between pw(G) and ms(G): pw(G) ≤ ms(G) ≤ pw(G) + 1.
We give a mixed strategy to clean the graph with n + O(1) searchers.The cleaning process will be done in a rounds.At each round 0 ≤ q < a, we put a searcher on clause variable c i,j and keep this searcher there until the cleaning round is completed.It is clear we could clean each forcing gadget by four searcher and keep searcher on the root vertex after cleaning the edges in the force gadget.For each 1 ≤ t ≤ n and 0 ≤ q < a, after cleaning four connected forcing gadgets we slide the searchers from the root of forcing gadgets to the vertices v 1 t,q , h 1 t,q , v 2 t,q , h 2 t,q .Then we put searchers on the the (at most two) z vertices connected to them and the guard vertices in sets p 1 t,q , p 2 t,q and p 3 t,q .As we have a searcher  on c i,j unil the end of this round we could clean the paths between vertex c i,j and those z vertices using one more searcher.The last step of the round is removing the searchers from vertices v 1 t,q , h 1 t,q , v 2 t,q , h 2 t,q and searchers on the gaurd sets p 1 t,q , p 2 t,q and p 3 t,q .We just keep the searcher on v 1 t,q+1 .Also, we remove searchers in path between c i,j and V ∪ H except for the one standing on the c i,j .To commence the next round, the searcher in c i,j is deleted and a new searcher is put on c i,j+1 .After the last round the whole graph G is cleaned.Since at any point in time we need at most 14 searchers (they are on c i,j , two z vertices, v 1 t,q , h 1 t,q , v 2 t,q , h 2 t,q , guard vertices in sets p 1 t,q , p 2 t,q and p 3 t,q , one searcher to clean the paths between two z vertices and c i,j ) and we reuse 14 searchers in the cleaning process, n + 14 searchers suffice to clean the graph.  (1.Given an instance of CNF-SAT, we can construct an instance of [1,2]-Total Dominating Set using the above construction and solve it with a (4 − ) pw |V (G)| O (1) time algorithm.The correctness of the algorithm follows from Lemmata 15 and 16.Lemma 17 implies that the running time of the algorithm is (4 − ) n 2 |V (G)| O (1) .However, we have (4 − )

Conclusion
We have shown that the [1, j]-Dominating Set problem parameterized by solution size is W [1]-hard on graphs of degeneracy (j + 1).It is thus natural to ask whether the problem becomes fixed-parameter tractable when the degeneracy of the graph is smaller or equal to j.In particular, is the [1, j]-Dominating Set problem fixed-parameter tractable on j-degenerate graphs?There is a very rich literature [6,1,10]

Lemma 6 .F
for an illustration.Now we output (G , 2k + j − 1) as an instance of [1, j]-Dominating Set.Clearly our reduction takes time polynomial in |V (G)| and k.The constructed graph G has degeneracy j + 1. Complexity of [1, j]-Domination Problems

2Lemma 8 .
vertices is dominated by the single vertex in V i ∩ S. All the vertices in S E are dominated by {r i : i ∈ {1, . . ., j − 1} and by at most one vertex from S (since S is an independent set).Moreover all the degree one vertices in G are dominated exactly once by D. If there exists a [1, j]-dominating set of size 2k + j − 1 in G then there exists a k-colored independent set in G.

F
and only if it has a (nonempty) dominating set of size at most 2.Theorem 12.The [1, j]-Dominating Set problem parameterized by solution size k is fixed-parameter tractable on nowhere dense classes of graphs.

Figure 3
Figure 3 Parts of the construction.

Figure 4
Figure 4 Parts of the construction.Dashed edges are connecting vertices with a forcing gadget.
on kernelization for the Dominating Set problem on sparse graphs and it would be interesteing to see where (if anywhere) the known techniques for the Dominating Set problem become applicable to the [1, j]-Dominating Set problem.Finally, we have shown a lower bound of O * ((4 − ) tw ) for [1, 2]-Total Dominating Set assuming SETH, while the known upper bound is O * (6 tw ).Closing this gap is an interesting open problem.On the other hand [1, 2]-Dominating Set can be solved in time O * (4 tw ) and getting a matching lower bound is another open problem.Moreover, can we get lower bounds for more general problems such as [1, j]-Dominating Set and [1, j]-Total Dominating Set?