Alternative parameterizations of Metric Dimension

04/27/2018 ∙ by Gregory Gutin, et al. ∙ 0

A set of vertices W in a graph G is called resolving if for any two distinct x,y∈ V(G), there is v∈ W such that dist_G(v,x)≠ dist_G(v,y), where dist_G(u,v) denotes the length of a shortest path between u and v in the graph G. The metric dimension md(G) of G is the minimum cardinality of a resolving set. The Metric Dimension problem, i.e. deciding whether md(G)< k, is NP-complete even for interval graphs (Foucaud et al., 2017). We study Metric Dimension (for arbitrary graphs) from the lens of parameterized complexity. The problem parameterized by k was proved to be W[2]-hard by Hartung and Nichterlein (2013) and we study the dual parameterization, i.e., the problem of whether md(G)< n- k, where n is the order of G. We prove that the dual parameterization admits (a) a kernel with at most 3k^4 vertices and (b) an algorithm of runtime O^*(4^k+o(k)). Hartung and Nichterlein (2013) also observed that Metric Dimension is fixed-parameter tractable when parameterized by the vertex cover number vc(G) of the input graph. We complement this observation by showing that it does not admit a polynomial kernel even when parameterized by vc(G) + k. Our reduction also gives evidence for non-existence of polynomial Turing kernels.

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1 Introduction

A set of vertices of a graph is a resolving set for if for any two distinct , there is such that , where denotes the length of a shortest path between and in the graph . The metric dimension of is the minimum cardinality of a resolving set for . The metric dimension of graphs was introduced independently by Slater [25] and Harary and Melter [16]. Metric Dimension as a computational problem was first mentioned in the literature by Garey and Johnson [14] and its decision version is defined as follows.

Metric Dimension parameterized by 

Garey and Johnson [14] proved this problem to be NP-complete in general. Their proof was never published, a reduction from 3SAT was provided by Khuller  [21]. Diaz  [6] showed that the problem is NP-complete even when restricted to planar graphs of bounded degree but that it is solvable in polynomial time on the class of outer-planar graphs.

Prior to this, not much was known about the computational complexity of this problem except that it is polynomial-time solvable on trees (see [25, 21]), although there are several results proving combinatorial bounds on the metric dimension of various graph classes [3]. Subsequently, Epstein et al. [10] showed that this problem is NP-complete on split graphs, bipartite and co-bipartite graphs. They also showed that the weighted version of Metric Dimension can be solved in polynomial time on paths, trees, cycles, co-graphs and trees augmented with edges for a fixed . Hoffmann and Wanke [19] extended the tractability results to a subclass of unit disk graphs, while Foucaud  [12] showed that this problem is NP-complete on interval graphs.

The parameterized complexity of Metric Dimension under the standard parameterization—the metric dimension of the input graph—was open until 2012, when Hartung and Nichterlein [17] proved that it is -hard. Foucaud  [12] showed the problem becomes fixed-parameter tractable when restricted to interval graphs. The parameterized complexity of Metric Dimension on graphs of bounded treewidth is currently unresolved (the question of whether it is polynomial-time solvable on graphs of treewidth 2 is still open), however, Belmonte  [2] proved that it is when parameterized by the treelength111The length of a tree decomposition is the maximum diameter of the bags in this tree-decomposition and the treelength of a graph is the minimum length over all tree decompositions. Note that this parameter is upper-bounded by treewidth. plus the solution size. In a different line of work, Eppstein [9] showed that Metric Dimension is when parameterized by the max-leaf number of the input graph alone.

In this paper we initiate the study of the parametric dual of Metric Dimension. To avoid confusion, we will use to denote the (standard) parameter and phrase the parameterized dual as follows:

Saving Landmarks parameterized by 

We call a set of vertices of a co-resolving set if is a resolving set of . Clearly, an instance of Saving Landmarks is positive if and only if there is a co-resolving set of size at least .

This choice of parameterization is informed by previous studies of the parametric dual (see e.g. [1, 4, 15, 24]): problems that are hard with respect to the standard parameter often admit an -algorithms or even polynomial kernels under the dual parameter. A classic example is the Independent Set problem which is -hard while its dual, the Vertex Cover problem is among the earliest problems shown to be in and even admits a linear vertex kernel.

We add yet another entry to the list of hard problems with tractable duals by showing that Saving Landmarks admits a polynomial kernel and a single-exponential algorithm. Concretely, we prove the following two results.

Theorem.

Saving Landmarks admits a kernel with at most  vertices.

Theorem.

Saving Landmarks can be solved in time .

We also study the Metric Dimension problem from the kernelization perspective when parameterized by the vertex cover number of the input graph. As Hartung and Nichterlein observed [17], parameterization of Metric Dimension by the vertex cover number of the input graph (denoted Metric Dimension[VC]) can be easily seen to be in . It is therefore natural to ask whether this structural parameterization allows a polynomial kernel in general graphs, a question we answer in the negative. In fact, we show that not only does the problem not admit a polynomial kernel with the vertex cover as the parameter, even adding the size of the solution (the metric dimension of the graph) to the parameter is unlikely to be helpful in this regard. Specifically, we prove the following result.

Theorem.

Metric Dimension[] does not admit a polynomial kernel unless the polynomial hierarchy collapses to its third level.

The reduction used in the proof of Theorem 1 also gives evidence for non-existence of polynomial Turing kernels, generalizations of (ordinary) kernels, informally introduced in the end of Section 4.

2 Preliminaries

For a graph  we denote by  the standard distance-metric where  is the length of a shortest path between vertices . We denote by  and  the open and closed neighbourhood of a vertex. We omit the subscript  if clear from the context in all these notations. As customary, the number of vertices of a graph under consideration will be denoted by

Two vertices  are true twins if  (implying that ) and they are false twins if . A twin class is a maximal vertex set in  in which all vertices are pairwise true twins or in which all vertices are pairwise false twins.

A vertex set  resolves a set  if for every pair of distinct vertices  there exists at least one vertex  such that . We will also say that a pair  is resolved by  if the above holds and further that sets  are distinguished by  if every pair , is resolved by . A vertex subset  is a resolving set of  if  resolves . We call the members of such a set  landmarks.

Parameterized complexity is a two dimensional framework for studying the computational complexity of a problem. One dimension is the input size and the other is a parameter . A problem is said to be fixed parameter tractable () or in the class , if it can be solved in time for some computable function . We refer to the books of Cygan  [5] and Downey and Fellows [8] for detailed introductions to parameterized complexity.

Kernelization offers a mathematically rigorous way of analysing and comparing preprocessing algorithms for -hard problems in general and for parameterized problems in particular. A kernel of size for a parameterized problem is a polynomial time algorithm that takes as input an instance of the problem (where is the parameter) and outputs another instance of the same problem such that is a yes- instance of the problem if and only if is a yes-instance of the problem and . The notion of “effective” preprocessing is captured by requiring the function to be polynomially bounded, in which case the kernel is called a polynomial kernel. The reader is referred to Cygan  [5], Downey and Fellows [8], Fomin [11] and the surveys [22, 23] for a comprehensive introduction to the topic of kernelization.

Definition (Pruned graph).

For a graph  we define the pruned graph  as the graph obtained (up to isomorphism) from  by iteratively removing vertices from twin-classes of size three or larger. We say that a graph is pruned if .

The following observation simply follows from the fact that among a twin class  in , all but one vertex of  must be contained in any resolving set.

Observation.

A graph  has a resolving set of size  if and only if the pruned graph has a resolving set of size .

Consequently, we call an instance  of Metric Dimension or Saving Landmarks reduced if is pruned.

3 Standard parameterization for Saving Landmarks

We present two positive results in this section, namely, that Saving Landmarks admits a polynomial kernel and a single-exponential algorithm.

We begin by describing the kernel. Assume in the following that the input instance  is pruned as per Observation 2. This will be the only reduction rule. In the following we will prove that the size of the instance is either bounded polynomially in  or it will be a trivial yes-instance. Let us collect some basic observations first.

Lemma.

If  contains either an independent set or a clique with  vertices, then  is a yes-instance.

Proof

Let  be a set of size  such that  is either a clique or an independent set. Since  is pruned there are at least  distinct twin-classes in , which must be distinguished by their neighbourhoods outside of . Hence selecting one vertex from each twin-class in  gives a co-resolving set of size at least , and  is a yes-instance.

Let us define the function . Note that if , then  and  are distinguished from each other by any set of  landmarks, simply by virtue of having a necessarily different set of landmarks as neighbours. Let us therefore construct an auxiliary graph  on  where

Observe that if  contains an independent set  of size  then  is a yes-instance: the set  has size  and as such will still resolve all of . This indicates that  must be rather dense, however, we can also argue that it cannot have arbitrarily high degree:

Lemma.

Let  have degree at least  in . Then  is a yes-instance.

Proof

Let . Note that for every pair  it holds that

Now turn our attention to . First consider the case in which every vertex in  has degree less than . Then greedily packing closed neighbourhoods gives an independent set in  of size at least , and by Lemma 3, is a yes-instance.

Thus consider the alternative that  contains a vertex  of degree at least . Define  and pick any vertex . Note that since  it follows that

Consequently, and  share at least  neighbours in  (removing one extra since ). We can repeat this procedure to construct a sequence of distinct vertices  and subsets  where  and  is chosen arbitrarily. The sequence terminates with , giving a clique in of size . Since for every , we get since . Thus and , …, induces a clique of size at least  in , and again by Lemma 3 we conclude that  is a yes-instance.

With these pieces in place, we can prove the first result of this section.

See 1

Proof

By Lemma 3 we either have that is a yes-instance or that the auxiliary graph  has a maximum degree less than . Assuming the latter, if  then  contains an independent set of size at least  and, as observed above, is a yes-instance.

The kernel for Saving Landmarks is therefore the following procedure: for a given instance , compute the reduced instance . If  contains more than  vertices, return a trivial yes-instance. Otherwise, return .

Let us now move on to the second result, the single-exponential algorithm. To better describe the algorithm, let us introduce a definition. For a set , we say that two vertices and are -equidistant if for every , i.e., if fails to resolve and . Note that this induces an equivalence relation over .

The main ingredient will be fact that a solution to Saving Landmarks is witnessed already by a small resolving set.

Lemma.

Let  be a co-resolving set of a graph . Then there exists a set  of size at most  that resolves .

Proof

We construct  iteratively as follows. Begin with  and pick a pair  of -equidistant vertices in . Since  resolves , there exists a vertex  that distinguishes  and . Add  to  and partition  into equivalence classes of -equidistant vertices. Pick a new pair of -equidistant vertices from one of the classes and repeat. Observe that the number of equivalence classes increase with every addition to , hence after at most  steps the set  resolves every pair in .

We are now ready to complete the proof of Theorem 1.

See 1

Proof

We may assume that . Let us first show the following claim: there exists a co-resolving set of of size at least if and only if there is a partition of such that contains at least equivalence classes of -equidistant vertices. Suppose that there exists a co-resolving set of of size at least . Then by Lemma 3, there is a set of size at most  that resolves . Let and for a partition . Then resolves and hence has at least equivalence classes of -equidistant vertices. Suppose now that there is a partition of such that has at least equivalence classes of -equidistant vertices. Choose a vertex from each equivalence class to form a set . Then is a co-resolving set of .

The above claim leads to the following randomized algorithm. Choose a natural number defined later on. Repeat times the following: uniformly at random partition the vertices of  into and , and derive equivalence classes of -equidistant vertices in . If the number of classes is at least , then conclude that  is a yes-instance and stop. If after all repetitions we do not conclude that is a yes-instance, then we conclude that is a no-instance.

Let us argue about the success probability of the randomized algorithm and how to choose

. The probability that for a random partition the vertices of  as , has at least equivalence classes of -equidistant vertices is at least the probability that and , where sets are as in Lemma 3, which is  Thus, is enough to achieve a constant success probability [5].

Observe that every loop in the randomized algorithm can be executed in polynomial time. Thus, the running time of the randomized algorithm is . The randomized algorithm can be derandomized using the standard -universal set technique [5], which brings an additional to the exponent of the running time.

4 Structural parameterizations for Metric Dimension

As Hartung and Nichterlein observed [17], Metric Dimension[VC] is trivially  by virtue of Observation 2: After reducing the size of each twin class to at most two, any instance with a vertex cover  of size  will have at most  vertices. In sparse graph classes, the twin reduction even results in a polynomial-size kernel: in classes of bounded expansion (planar graphs or graphs excluding a topological minor), the number of twin classes in  is bounded linearly in  and in nowhere dense classes by  (Lemma 4.3 and Corollary 4.4 in [13]). Furthermore, if the input graphs stem from a -degenerate class, the number of twin-classes and thus the number of vertices in the kernel is bounded by ; a fact that follows easily from the observation that in such a class at most  vertices in the independent set can have degree more than .

It is therefore natural to ask whether this structural parameterization allows a polynomial kernel in general graphs, a question we answer in the negative. We will use in the following that Hitting Set parameterized by the size of the universe plus the solution size does not admit a polynomial kernel unless the polynomial hierarchy collapses to the third level [7]

See 1

Proof

We provide a polynomial parameter transformation from Hitting Set[], i.e. parameterized by the size of the universe and the solution size, to Metric Dimension[]. Let  be a Hitting Set instance with  and . We construct a graph  as follows (Figure 1):

  1. Begin with the usual bipartite representation of , i.e., create a bipartite graph where for vertices and we have if and only if ;

  2. add vertices  to the graph and edges between  so that every vertex in  has a unique neighbourhood in  of size ;

  3. add  vertices  to the graph and edges between  and such that every vertex in  has a unique neighbourhood in  of size ;

  4. add three vertices  where , , and ;

  5. create true twin copies  of , and finally

  6. create false twin copies  of  but remove all edges from  to  afterwards. For simplicity, we will label the copy of any vertex  by .

Figure 1: A schematic of the reduction from a Hitting Set[] instance  to a Metric Dimension[] instance. The left drawing shows the basic construction, the right drawing the addition of false and true twins (an edge between a white set and its grey counterpart indicates that they are true twins, the absence of an edge that they are false twins). Note that the construction removes edges between the set and .

In summary, the sets  connect to  only, the sets  to  and , the edges between  encode the hitting set instance and the pairs , , and  are apices for the sets , and , respectively. Our construction concludes with  as the Metric Dimension[] instance with the vertex cover  and solution size .

Let us first show that if  is a yes-instance then so is . Suppose that  is a hitting set for  of size . We construct a landmark set  for  by setting ; let us now argue that is indeed a resolving set. First, note that the selected apices , , and  make sure that  is distinguished from  and  from . Since  and  are in , these sets are of course distinguished from their twin counterparts . By construction, every vertex in  has a unique neighbourhood in , hence all of  is resolved by . The same holds true for all pairs  as long as  and . The only pairs we have not yet shown to be resolved by  are of the form for  with its copy . Since  is a hitting set for , every set  is adjacent to at least one vertex in  while  has no neighbours at all in . Thus all such pairs are resolved by  and we conclude that  is a resolving set.

In the other direction, assume that  is a resolving set of size  for . Since for each pair of twins at least one vertex has to be in any resolving set, we may assume, without loss of generality, that . Let us call this collection of  vertices  and let us see what it resolves in . As argued above, every pair except those of the form , are certainly resolved. We first need to argue that  indeed does not resolve those pairs: this is immediately obvious for landmarks in  since  share the same neighbours inside this set. For landmarks in , note that all vertices in are at exactly distance two from every vertex in  via the apex vertex  (or ). Hence  cannot resolve any pair  and these pairs must then be resolved by the remaining  vertices in . All vertices outside of  are either selected or twins to selected vertices, hence we may assume that .

First, consider a potential landmark . Since  has distance exactly two to every vertex in  except itself, such a selection would only distinguish  from all other vertices and not resolve any other pair. Thus we can as well choose any vertex in  instead and potentially resolve more pairs, thus we may assume that .

Let us split  into  and . Again, only distinguishes  from the rest of . Thus  necessarily distinguishes all pairs  with  and therefore  hits all sets  for which . We finally construct a hitting set  of size  as follows: we take all vertices in  and for each pair  with  we select one (arbitrary) neighbour . By the previous observation, is a hitting set for  of size  and we conclude that  is a yes-instance.

This concludes the parameter preserving transformation. Let us conclude by checking that the parameter is polynomial in and :

where we used that .

We note that this reduction also gives evidence against a more general form of kernelization. Where a standard kernel can be understood as a many-one reduction from a problem to itself, with output size bounded by a function of the parameter, a Turing kernel is the corresponding Turing reduction notion. In other words, informally, a Turing kernel is a polynomial-time procedure that solves a parameterized problem, with access to an oracle for the problem but with a bound on the maximum length of the questions it may ask of the oracle. A polynomial Turing kernel is a Turing kernel with a bound on the question size. For a more formal definition, see [18, 5]. It is known that there are parameterized problems that do not allow a polynomial kernel unless the polynomial hierarchy collapses, but which do allow polynomial Turing kernels; cf. [18, 26, 20].

Although we do not have a framework for excluding polynomial Turing kernels that is as powerful as that for excluding standard polynomial kernels, Hermelin  [18] defined a hierarchy of complexity classes, conjectured to represent problems that do not allow polynomial Turing kernels. The most basic and most common of these hardness classes is WK[1], which is in turn contained in a larger class MK[2]. It is conjectured in [18] that no WK[1]-hard problem has a polynomial Turing kernel. Since Hitting Set[] is known to be MK[2]-hard [18], the above reduction gives the following.

Corollary.

Metric Dimension[] is MK[2]-hard (hence also WK[1]-hard) under polynomial parameter transformations, and does not allow a polynomial Turing kernel unless CNF-SAT and every other problem in MK[2] does.

5 Conclusion

We initiated the study of the parameterized complexity of the dual of the classic Metric Dimension problem and obtained a polynomial kernel as well as a single-exponential algorithm. To the best of our knowledge, this is the first non-trivial parameterization for Metric Dimension which leads to a polynomial kernel. Since our focus in this article was on obtaining new classification results, we leave the improvement of the kernel size or a potential proof of a lower bound on the bitsize of our kernel, to future work.

In addition, we note that it remains open whether Metric Dimension is polynomial time solvable even on series-parallel graphs. Since series- parallel graphs are precisely the graphs of treewidth 2, a negative answer would also imply that there is no XP algorithm for Metric Dimension parameterized by the treewidth. Consequently, a natural starting point of enquiry towards addressing this question could be the study of the parameterized complexity of Metric Dimension parameterized by treewidth.

Acknowledgement

Gutin was partially supported by Royal Society Wolfson Research Merit Award.

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