Domination above r-independence: does sparseness help?

06/21/2019
by   Carl Einarson, et al.
Birkbeck, University of London
0

Inspired by the potential of improving tractability via gap- or above-guarantee parametrisations, we investigate the complexity of Dominating Set when given a suitable lower-bound witness. Concretely, we consider being provided with a maximal r-independent set X (a set in which all vertices have pairwise distance at least r + 1) along the input graph G which, for r >= 2, lower-bounds the minimum size of any dominating set of G. In the spirit of gap-parameters, we consider a parametrisation by the size of the 'residual' set R := V (G) N [X]. Our work aims to answer two questions: How does the constant r affect the tractability of the problem and does the restriction to sparse graph classes help here? For the base case r = 2, we find that the problem is paraNP -complete even in apex- and bounded-degree graphs. For r = 3, the problem is W[2]-hard for general graphs but in FPT for nowhere dense classes and it admits a linear kernel for bounded expansion classes. For r >= 4, the parametrisation becomes essentially equivalent to the natural parameter, the size of the dominating set.

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Introduction

The research of above/below guarantee parameters as first used by Mahajan and Raman [22] was an important step towards studying problems whose natural parameters provided only trivial and unsatisfactory answers. Case in point, the motivation for Mahajan and Raman was the observation that every CNF-SAT formula with  clauses trivially has an assignment that satisfies clauses, thus question for the maximum number of satisfied clauses is only interesting if , which of course renders the parametrised approach unnecessary. They therefore proposed to study parametrisations ‘above guarantee’: going with the previous example, we would ask to satisfy  clauses or ‘ above guarantee’. After some isolated results in that direction ( [15, 17]) the programme took up steam after Mahajan et alpresented several results and pointers in new directions [23] ( [3, 16, 18, 17]). In particular, Cygan et albroke new ground for Multiway Cut and Vertex Cover with algorithms that run in  time, where  is the gap parameter between an appropriate LP-relaxation and the integral optimum [5]. Lokshtanov et alimproved the Vertex Cover case to  using a specialized branching algorithm [21].

The latter result highlights an important realization: these alternative, smaller parameters might not only provide the means to investigate problems without ‘good’ natural parameters, it might also provide us with faster algorithms in practise! Gap- and above-guarantee parameters are attractive because there is a reasonable chance that they are small in real-world scenarios, something we often cannot expect from natural parameters.

To the best of our knowledge, so far no gap-parameter results are known for domination problems and an above/below-guarantee result is only known in bounded-degree graphs [23]

. This is probably due to the fact that there are no simple ‘natural’ upper/lower bounds and in the case of gap-parameters the LP-dualities do not provide much purchase. We therefore explore this topic under the most basic assumptions: we are provided with a witness for a lower bound on the domination number

as input and consider parametrisations that arise from this additional information. In the case of prob]Dominating Set@Dominating SetDominating Set, the witness takes the form of a -independent set, that is, a set in which all vertices have pairwise distance at least three. Note that this approach also captures a form of duality: the LP-dual of dominating set describes a -independent set, however, in general the two optima are arbitrarily far apart. Recently, Dvořák highlighted this connection [7] and proved that in certain sparse classes the gap between the dual optima is bounded by a constant.

Thus, assume we are given a maximal -independent set  alongside the input graph . A parametrisation by  would go against the spirit of gap-parameters, instead we parametrise by the size of residual set , that is, all vertices that lie at distance two from  (since is maximal, no vertex can have distance three or more). We choose this particular parameter for two reasons:

  1. For  the problem is decidable in polynomial time since the domination number of the graph is precisely .

  2. The set  is a dominating set of .

The first property is of course an important pre-requisite for the problem to be in  under this parametrisation, while the second property guarantees us that the dominating set size lies in-between  and .

Our first investigatory dimension is the constant  in the -independent set: intuitively, increasing the minimum distance between vertices in  increases the size of the parameter  and imposes more structure on the input instance. Our second dimension encompasses an approach that has been highly successful in improving tractability of domination problems: restricting the inputs to sparse graphs. While prob]Dominating Set@Dominating SetDominating Set is -complete in general graphs, Alber et alshowed that it is fpt in planar graphs [1]; Alon and Gutner later proved that assuming degeneracy is sufficient [4]. Philip, Raman, and Sikdar extended this result yet further to graphs excluding a fixed bi-clique and also proved that it admits a polynomial kernel [25]. A related line of research was the hunt for linear kernels in sparse classes. Beginning with such a kernel on planar graphs by Alber, Fellows, and Niedermeier [2], results on apex-minor free graphs [10], graphs excluding a minor [11] and classes excluding a topological minor [12] were soon proven. Recently, a linear kernel for graphs of bounded expansion [6] (and an almost-linear kernel for nowhere dense graphs [9]) has subsumed all previous results.

Our investigation of prob]Dominating Set@Dominating SetDominating Set parametrised above an -independent set, for , led us to the following results. For , the problem is -complete already for , squashing all hope for an or even  algorithm. This also holds true if the inputs are restricted to sparse graph classes (apex-graphs/graphs of maximum degree six).

For , the problem is -hard in general graphs but admits an -algorithm. In nowhere dense and bounded expansion classes, it is fixed-parameter tractable. We further show, in the probably most technical part of this paper, that it admits a linear kernel in bounded expansion classes.

Finally, for  the problem remains -hard in general graphs and essentially degenerates to prob]Dominating Set@Dominating SetDominating Set (hence, all the above mentioned results in sparse classes translate in the parametrisation above -independence).

1 Preliminaries

A set  is -independent if each pair of distinct vertices in  have distance at least , thus an independent set is -independent. We write  and , respectively, for the neighbourhood and the closed neighbourhood of a vertex . We extend this notation to sets as follows: for we let  be all vertices not in that have a neighbour in  and . We let  be all vertices not in that are at most distance from any vertex in and we let . A vertex set  is dominated by a set  if for every vertex  we have , is then called a -dominator. We let  denote the size of a minimum dominating set of .

Dominating Set above -independence parametrised by 

Note that  is trivially a dominating set and that, for , it holds that , thus we will tacitly assume in the following that  since all other instances are trivial.

We will frequently invoke the terms bounded expansion and nowhere dense to describe graph classes. The definitions of these terms requires the introduction of several concepts which will not be useful for the remainder of the paper, we refer the reader to the book by Nešetřil and Ossona de Mendez [24]. In this context, it is important to know that bounded expansion classes generalize most structurally sparse classes (planar, bounded genus, bounded degree, -minor free, -topological minor free) and nowhere dense classes contain bounded expansion classes in turn. The following lemma and propositions for those two sparse graph classes will be needed in the remainder of this paper:

[Twin class lemma [13, 26]] For every bipartite nowhere dense class there exists a constant  and a function such that for every member of the class it holds that

  1. , and

  2. .

where . If  is from a bounded expansion class, can be assumed a constant as well.

We will also frequently invoke the following result regarding first-order (FO) model checking in bounded expansion and nowhere dense classes:

Proposition (Dvořák, Král, and Thomas [8]).

For every bounded expansion class, the first-order model checking problem is solvable in linear fpt-time parametrised by the size of the input formula.

This result has since been extended to nowhere dense classes as well. Here, almost linear fpt-time means running time of the form  for some function .

Proposition (Grohe, Kreutzer, and Siebertz [14]).

For every nowhere dense class, he first-order model checking problem is solvable in almost linear fpt-time parametrised by the size of the input formula.

2 Above -independence: hard as nails

In this section we will show that when we let , we find that the problem is -complete for , hence this parametrisation does not even admit an -algorithm. In the following we first present a reduction from prob]3SAT@3SAT3SAT and then discuss how to modify it to reduce into sparse graph classes.

Since is a (maximal) -independent set, we know that each vertex in is a neighbour of some vertex in , otherwise we could add this vertex to . Let us now describe the reduction. Let be -instance with variables  and clauses . We construct  as follows (Figure 1):

  1. For each variable , add a triangle with vertices , , .

  2. For each clause  add a vertex . If the variable  occurs positively in , add the edge ; if it occurs negatively, add the edge .

  3. Add a single vertex  to the graph and connect it to each clause variable . Add two further vertices  and add the edges  and .

We further set  as our -independent set; notice that the only vertex not contained in  is . Hence, .

Figure 1: Sketch of reduction from prob]3SAT@3SAT3SAT to prob]Dominating Set@Dominating SetDominating Set above -independent set. The left side shows the basic reduction, the right side shows the bounded-degree replacement gadget for the clause part, with the tree-gadget  highlighted on the bottom right. The -independent set  is coloured blue and is shaded in grey. In both constructions the set  consists only of .

is satisfiable iff has a dominating set of size .

Proof.

Assume  is satisfiable and fix one satisfying assignment . We construct a dominating set  as follows: if  is true under , add  to ; otherwise add . Since  satisfies every clause of  the dominating set so far dominates every clause vertex and, of course, every variable gadget. The remaining undominated vertices are , thus adding  to  yields a dominating set of  of size .

In the other direction, assume that  is a dominating set of  of size . Since  is a pendant vertex we can assume that  (if  would be in  we could exchange it for ). That leaves  vertices in , precisely the number of variable-gadgets. Since every variable gadget must include at least one vertex of , we conclude the every such gadget contains precisely one dominating vertex. Since that depletes our budget, no other vertex is contained in .

By the usual exchange argument we may assume that the dominating vertex in each variable gadget is either  or  and not  for ; hence the dominating vertices inside the variable gadgets encode a variable assignment  of . Finally, note that the clause vertices are not dominated by  and  is not contained in . Hence, they must be completely dominated by vertices contained in the variable gadgets. Then, by construction, the assignment  satisfies  and the claim follows. ∎

We conclude that many-one reduces to prob]Dominating Set@Dominating SetDominating Set above -independence already with . We obtain the following two corollaries that demonstrate that sparseness cannot help tractability here:

prob]Dominating Set@Dominating SetDominating Set above -independence is -complete in apex-graphs.

Proof.

We use the above construction but reduce from a planar variant of prob]3SAT@3SAT3SAT. To ensure that we can construct variable-gadgets without edge crossings, we choose to reduce from Lichtenstein’s prob]Planar 3SAT@Planar 3SATPlanar 3SAT variant [20] which ensures that the following graph  derived from the prob]Planar 3SAT@Planar 3SATPlanar 3SAT instance  is planar:

  1. Every variable  of  is represented by two literal vertices with the edge 

  2. Each clause  is represented by a vertex . If the variable  occurs positively in , the edge exits; if it occurs negatively, the edge  exists.

To complete  to  we have to add the vertices  and connect them to . This is clearly possible without breaking planarity (picture placing  on the middle of the line segment  and moving it perpendicular by a small amount, then the edges  and  can be embedded without crossing other edges). The vertices  can be placed anywhere; finally the vertex  will break planarity (the embedding does not guarantee that the clause vertices lie on the outer face of the graph) and we conclude that  is indeed an apex-graph. ∎

prob]Dominating Set@Dominating SetDominating Set above -independence is -complete in graphs of maximum degree six.

Proof.

We reduce from prob](3,4)SAT@(3,4)SAT(3,4)SAT (-hardness shown in [27]) in which every clause has size three and every variable occurs in at most four clauses. We use the above construction with one modification. Instead of connecting all clause vertices to one vertex we create a bounded-degree tree with the clause-vertices as its leaves.

We begin by partitioning the clause vertices in pairs 

; if there is an odd number of vertices the last group will have three vertices. Then, for each group

, we add two vertices , connect the clause-vertices  and  to , and add the edge . We further add each to our -independent set and then create a set consisting of each . Now we iteratively construct the next level of the tree, starting with :

  1. If , create a single vertex , connect it to  and finish, otherwise proceed with the next step.

  2. Partition the vertices in into  groups  of pairs. If is odd, the last group will be a triple  instead.

  3. For each group, create a tree-gadget , with vertices (and if the group contains a third vertex), and edges , (), , , (), and .

  4. Add each to , let now be the set of all (for ) and continue with Step 1.

Figure 1, on the right side, shows an example of this construction. We note that, in the last tree-gadget, and are the same vertices as and respectively in the figure. We conclude the construction by adding each from the variable-gadgets to and setting . Notice that the only vertex not contained in  is  and thus .

Since each variable in prob](3,4)SAT@(3,4)SAT(3,4)SAT can be in up to four clauses, the maximum degree for and is six. All clause-vertices have degree at most four and all other types of vertices have a degree not higher than that, hence the claimed degree-bound holds. It is left to show that is satisfiable iff there is a dominating set of size in the graph.

Let us assume that is satisfiable and fix one satisfying assignment . We construct a dominating set as follows, beginning in the same way as in Lemma 2: if is true under , add to , otherwise add . Since satisfies every clause of the dominating set so far dominates every clause vertex and every variable gadget. Now, the remaining undominated vertices are the tree-vertices, and our remaining budget is which is equal to the amount of tree-gadgets. Since every clause vertex is already dominated we can, for each tree-gadget , add to the dominating set. This will dominate and the corresponding or in the tree-gadget below it, hence we can dominate all vertices of the graph within the budget .

In the other direction, assume that is a dominating set of of size . Since is a pendant vertex we can assume that . Thus, in the last tree-gadget () is in the dominating set. This means that in order for and to be dominated, in both tree-gadgets above has to be in the dominating set. This holds for all tree-gadgets, all the way up to the clause vertices. Since we now have one vertex per tree-gadget in the dominating set this leaves vertices. Just as in Lemma 2, we note that, for each variable gadget, either or is in the dominating set. We know that the clause variables are not dominated by anything in the tree gadgets and thus must be dominated by the variable vertices. As stated in Lemma 2, the dominating vertices inside the variable gadgets encode a variable assignment that satisfies and the claim follows. ∎

3 Above -independence: sparseness matters

3.1 -hardness in general graphs

In the following we present a result for prob]Dominating Set@Dominating SetDominating Set above -independence, namely that it is -hard in general graphs. We show this by reduction from prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set parametrised by the number of colours :

Colourful Dominating Set parametrised by 

It is easy to verify that prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set is -hard by reducing from prob]Red-Blue Dominating Set@Red-Blue Dominating SetRed-Blue Dominating Set: we copy the blue set times and make each copy a colour set , , and let be the red set.

Figure 2: Sketch of reduction from prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set to prob]Dominating Set@Dominating SetDominating Set above -independent set for . The set  contains only the vertex , the remaining vertices  are all contained in the residual .

prob]Dominating Set@Dominating SetDominating Set above -independence is -hard for .

Proof.

Let be an instance of prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set. We construct an instance  of prob]Dominating Set@Dominating SetDominating Set above -independence as follows (Figure 2):

  1. Begin with  equal to ; then

  2. for each block , , add edges to make a complete graph and add an additional vertex  with neighbourhood ; then

  3. add a vertex and connect it to all vertices in .

Let be the -independent set (which it clearly is for any ) and thus . Note that the graph  trivially has a dominating set of size ; the set . We now claim that has a colourful dominating set of size iff has a dominating set of size .

Assume that has a solution of size  and fix one such colourful dominating set . Note that  in  a) dominates all of  (because it dominates  in ) and b) dominates each set , (because  intersects each such ), and of course the vertex . Hence, is a dominating set of  as well.

In the other direction, assume that is a dominating set of  of size . Since each , , is only connected to vertices in the corresponding set , at least one vertex from each set  needs to be in . Since there are  such sets we conclude that  intersects each set  in precisely one vertex. We can further modify any such solution to not take the -vertices by taking an arbitrary vertex from  instead, thus assume that  has this form in the following. But then  is of course also a dominating set of size  for , as claimed.

3.2 Tractability in sparse graphs

In the following we present two positive results, namely that prob]Dominating Set@Dominating SetDominating Set above -independence is fpt in nowhere dense classes and that it admits a polynomial kernel in bounded expansion classes. The algorithm further implies an algorithm in general graphs. The following annotated domination problem will occur as a subproblem:

Annotated Dominating Set parametrised by 

In the following algorithm we will group vertices of  according to their neighbourhood in  (or a subset of ). We will call those groups -neighbourhood classes. We will write  to denote the size of an optimal solution of prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set.

prob]Dominating set@Dominating setDominating set above -independence can be solved in linear fpt-time in any graph class of bounded expansion.

Proof.

First we guess the intersection  of an optimal solution (should it exist) with  in  time. Let  be those vertices of  that are not dominated by . Define

as the neighbourhoods induced in  by vertices in . By Lemma 1, we have that  since  is from a class with bounded expansion (note that the partition into such neighbourhoods is computable in linear time using the partition-refinement data structure [19]). Accordingly, in time , we can guess a subset  such that an optimal solution covers exactly the neighbourhoods  (if  does not cover  we abort this branch of the computation). We are now left with the task of choosing vertices from  to a) cover the neighbourhoods  and b) dominate the vertices in  which are not dominated by .

Let us introduce the following notation to ease our task: for a collection of -neighbourhoods and a set , let . That is, contains those -neighbourhood classes in  whose neighbourhood is contained in . Let  be an ordering of  and let ; we will describe a dynamic-programming algorithm over the ordering . Let  be the minimum size of a partial solution in that covers the neighbourhoods  and together with  dominates all of . We initialize  for all  and , then compute the following entries with the recurrence111A proof for the correctness of the recurrence can be found in the Appendix

Note that  is the minimum size of a set that dominates  while choosing at least one vertex from each member of  (if we assume that ). The latter constraint corresponds to dominating the neighbourhoods of  in  by using vertices from . Once the DP table  has been computed, the size of an optimal solution is the value in .

It remains to be noted that every neighbourhood graph admits a dominating set of size one, hence the annotated dominating set has size at most . Thus, the problem of finding an annotated dominating set is FO-expressible by a formula of size  and we can solve the subproblem of computing  in time  for some function  using Proposition 1. As a result, we obtain a linear dependence on the input size (note that ) in the running time and thus the problem is solvable in linear fpt-time.∎

The same proof works for nowhere dense classes by applying Proposition 1 instead of Proposition 1:

prob]Dominating set@Dominating setDominating set above -independence can be solved in almost linear fpt-time in any nowhere dense class.

We finally note that the algorithm described in the proof of Lemma 3.2 only needs a black-box fpt-algorithm for prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set to run in fpt time, thus it is very likely that prob]Dominating set@Dominating setDominating set above -independence is in for other highly structured but not necessarily sparse graph classes. We can run the same algorithm on general graphs to obtain an -algorithm using the bound  and a simple brute-force in -time on the prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set subinstance during the DP.

prob]Dominating set@Dominating setDominating set above -independence is in .

3.3 Kernelization in sparse graphs

Let us now set up the necessary machinery for the kernelization. A boundaried graph  is a tuple  where  is a graph and  is the boundary. We also write  to denote the boundary. For a graph  and an induced subgraph , the boundary  are those vertices of  that have neighbours in . Thus for every subgraph  of  there is a naturally associated boundaried graph .

For a boundaried graph  and subsets  a set  is an -dominator of  if  and  dominates the set . We let  denote the size of a minimum -dominator of . A replacement for  is a boundaried graph with . The operation of replacing  by  in , written as , consist of removing the vertices  from , then adding  to  with  in  identified with  in  (we assume that the vertices  do not occur in ).

Let  be an instance of prob]Dominating Set@Dominating SetDominating Set above -independence where  is from a bounded expansion class. Let , and . Then, in fpt-time with parameter , we can compute a replacement  for  of size  such that . Moreover, the replacement  is a subgraph of  and contains .

Proof.

For every pair of subsets  we compute a minimal -dominator  for . Since this problem is expressible by an FO-formula of size , we can employ Proposition 1 to compute the set  in linear fpt-time with parameter . Note that will, besides the vertices in , contain at most additional vertices, since vertices suffice to dominate and the vertex dominates all of .

Let  be the union of all such computed solutions and the boundary. By construction, . Since all -dominators live inside , we can safely remove the edges from  that do not have any endpoint in , call the resulting graph . Let  and let  be a partition of  into -twin classes (thus all vertices in  have the same neighbourhood in  and no other class has this neighbourhood). Note that  is an independent set in .

Let  be obtained from  by removing all but two representatives from every twin-class, denote those by . Clearly, every -dominator of  is still an -dominator of , we need to proof the other direction. Let  be an -dominator of . If  then  is still an -dominator for . The same holds if  intersects every twin-class in at most one vertex: each such class either has size one, in which case it has size one in  as well, or it has size two and the vertex not contained in  is dominated by a vertex in . In either case, dominates all of  and thus is an -dominator of  and therefore of . Thus, assume that  fully contains some class . Clearly, only one vertex of  is enough to dominate  (and potentially vertices in ), thus we can modify  by picking the central vertex  instead to dominate the other vertex of  (which of course also dominates all of  in ). This can, of course, only happen once, otherwise we would reduce the size of the supposedly minimal set . This leads us back to the previous case and we conclude that there exists an -dominator of  and thus  of equal size.

We conclude that  for every choice of ; which implies that . Finally, by the twin-class lemma, since  is a constant in bounded expansion classes. ∎

Let  be an instance of prob]Dominating Set@Dominating SetDominating Set above -independence where  is from a bounded expansion class. Let  be a subset and let  be those vertices with . Let  be the induced subgraph on . Assuming is not empty we can, in fpt-time with parameter , compute a replacement  for  of size  alongside an offset  such that . Moreover, the replacement  is a subgraph of .

Proof.

Let  and define the graphs  for . We first we apply Lemma 3.3 to replace every subgraph  by a subgraph  of size  in linear fpt-time with parameter . For simplicity, let us call the resulting graph  and relabel the graphs  to  (note that Lemma 3.3 ensures that dominating set size does not change and that  is still a -independent set of the resulting graph).

For every  we compute a

characteristic vector

  indexed by pairs of subsets of  with the following semantic: for  we set . If  is larger than  we simply set .

Note that the constraint subproblem to compute  is FO-expressible, by a formula of size , thus we can compute the vectors  for  in linear time fpt-time with parameter  using Proposition 1. Let  be the equivalence relation over the graphs  defined as and let  be the corresponding partition into equivalence-classes under . Note that  since that is the number of possible characteristic vectors.

The construction of  is now simple: in every equivalence-class  we select  subgraphs and remove the rest; clearly  is a subgraph of  of the claimed size. We let the offset  to be equal to the number of subgraphs removed in this way.

We are left to show that . Consider any minimal dominating set  for . We call a graph  interesting under  if  intersects  in any vertex besides .

Claim.

There exists a solution  of size equal to  under which at most  graphs per equivalence class  are interesting.

Proof.

Consider any such class . If  we are done, so assume otherwise. Let  be the set of vertices that are dominated through vertices in graphs contained in  and select up to  many graphs that together already dominate ; we let  to be equivalent to  on these graphs. Any graph  not selected in this way only needs to dominate itself and we add its centre vertex  to . Since every graph needs to intersect any dominating set in at least one vertex, and since  is minimal we must have . Finally, only the  selected graphs are interesting under , as claimed. ∎

Thus, let us assume in the following that  is such a minimal solution under which at most  graphs per class  are interesting. For such a solution  of , we construct a solution  of  of size as follows. For a class , let  be those graphs that are contained in  and let be the graphs that are interesting under . Since , we can pair every graph  with a graph . Fix such a pair , let  and let  be those vertices of  that are exclusively dominated vertices in . Since , there must exist a set of  vertices in  that dominate  and . If we repeat this construction for every graph  with their respective pair in  and then pick the centre vertex for all , then the resulting set  has size  and dominates all of .

The same proof works in reverse if we start with a dominating set  of  to construct a dominating set  of  with ; thus we conclude that  and the claim follows. ∎

prob]Dominating Set@Dominating SetDominating Set above -independence has a linear kernel in bounded expansion graphs.

Proof.

Let  be the input instance and let  be the members of the -independent set . We define the graphs  and their respective -neighbours  for . Let  be as in Lemma 1. We partition the graphs  into two sets , where iff ; and  contains all remaining graphs.

Let us first reduce . Let be the -neighbourhoods of the graphs collected in . By the twin class lemma, , however, for each member  we might have many graphs in  that intersect  in precisely this set .

Fix  for now and let  be those graphs of  that intersect  in . Let  be the joint graph of the subgraphs in . Since , and  is a constant depending only on the graph class, we can apply Lemma 3.3 to compute a replacement  of size  with offset  in polynomial time. We apply the replacement  and decrease  by . Repeating this procedure for all  yields a graph  (a subgraph of ) in which the small graphs  in total contain at most vertices, a -independent set  of , and a new input . This concludes our reduction for .

Let us now deal with  in . By the twin class lemma, . But then has size bounded in and we conclude that the size of a minimal dominating set for is bounded by , hence we assume that is bounded by (otherwise the instance is positive and we can output a trivial instance). We now apply the existing linear kernel [6] for Dominating Set to . The output of the kernelization is a subgraph of the original graph, hence we collect the remaining vertices of the -independent set in order to output a well-formed instance . This concludes the proof. ∎

4 Above -independence: simple domination

In the case where the lower-bound set  is -independent we now have that for distinct  it holds that , thus for each  the set  can only be dominated from  and . Let us call an instance  reduced if for every  the intersection is non-empty. We can easily pre-process our input instance to enforce this property: if such an  would exist we can simply remove  from  to obtain an equivalent instance. In a reduced instance the parameter  is necessarily big compared to :

Observation .

For every reduced instance  of prob]Dominating Set@Dominating SetDominating Set above -independence it holds that .

Let  be a graph class for which prob]Dominating Set@Dominating SetDominating Set is in . Then prob]Dominating Set@Dominating SetDominating Set above -independence is in  for  as well.

Let  be a graph class for which prob]Dominating Set@Dominating SetDominating Set admits a polynomial kernel. Then prob]Dominating Set@Dominating SetDominating Set above -independence admits a polynomial kernel for  as well.

On the other hand, we showed in Section 3.1 that the problem remains -hard for any  in general graphs.

5 Conclusion

We considered prob]Dominating Set@Dominating SetDominating Set parametrised by the residual of a given -independent set and investigated how the value of  and the choice of input graph classes affect its tractability. We observed that the tractability does improve from  to  as it goes from being -complete to ‘merely‘ -hard and at least admits an -algorithm. Larger values of , however, do not increase the tractability as the problem becomes essentially equivalent to prob]Dominating Set@Dominating SetDominating Set.

If we consider sparse classes (bounded expansion and nowhere dense), the improvement in tractability from  to  is much more pronounced; changing from -complete to and even admitting a linear kernel in bounded expansion classes. We very much believe that the kernel can be extended to nowhere dense classes, but leave that quite technical task as an open question.

Acknowledgments: We thank our anonymous reviewer for helpfully pointing out how to achieve a linear kernel in Theorem 3.3.

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Appendix

We claim that that  is the size of a partial solution in that covers the neighbourhoods  and together with  dominates all of .

As the inductive base, we set  for all  and . Clearly, a non-empty cannot possibly be dominated (note that since we guessed , no vertex in is contained in the dominating set). The cost of dominating nothing in is of course , hence the choice of . We conclude that the tables of the base cases describe the claimed quantity.

Since the base cases hold, let us assume that the table for all describe the claimed partial solutions. Recall that the recurrence is given by

Let be a minimal set which covers and dominates all of . We argue in the following that .

For the first direction, let be the dominators of that lie outside of . Let then contain those sets of that are covered by vertices in and let be those that are covered by . Since covers all of , we know that and hence this pair appears in the above recurrence. By induction, we know that