1 Introduction
Recently, Combinatorial reconfiguration [12] has been extensively studied in the field of theoretical computer science. (See, e.g., surveys [10, 19].) A reconfiguration problem is generally defined as follows: we are given two feasible solutions of a combinatorial search problem, and asked to determine whether we can transform one into the other via feasible solutions so that all intermediate solutions are obtained from the previous one by applying the specified reconfiguration rule. This framework is applied to several wellstudied combinatorial search problems; for example, Independent Set [3, 9, 11, 14, 15], Vertex Cover [17, 18], Dominating Set [8, 16, 18, 20], and so on.
The Dominating Set Reconfiguration problem is one of the wellstudied reconfiguration problems. For a graph , a vertex subset is called a dominating set of if contains at least one vertex in the closed neighborhood of each vertex in . Figure 1 illustrates four dominating sets of the same graph. Suppose that we are given two dominating sets and of a graph whose cardinalities are at most a given upper bound . Then the Dominating Set Reconfiguration problem asks to determine whether we can transform into via dominating sets of cardinalities at most such that all intermediate ones are obtained from the previous one by adding or removing exactly one vertex. Note that this reconfiguration rule, i.e. adding or removing exactly one vertex while keeping the cardinality constraint, is called the token addition and removal (TAR) rule. Figure 1 illustrates an example of transformation between two dominating sets and for an upper bound .
Combinatorial reconfiguration models “dynamic” transformations of systems, where we wish to transform the current configuration of a system into a more desirable one by a stepbystep transformation. In the current framework of combinatorial reconfiguration, we need to have in advance a target (a more desirable) configuration. However, it is sometimes hard to decide a target configuration, because there may exist exponentially many desirable configurations. Based on this situation, Ito et al. introduced the new framework of reconfiguration problems, called optimization variant [13]. In this variant, we are given a single solution as a current configuration, and asked for a more “desirable” solution reachable from the given one. This variant was introduced very recently, hence it has only been applied to Independent Set Reconfiguration to the best of our knowledge. Therefore and since Dominating Set Reconfiguration is one of the wellstudied reconfiguration problems as we already said, we focus on this problem and study it under this framework.
1.1 Our problem
In this paper, we study the optimization variant of Dominating Set Reconfiguration (denoted by OPTDSR); to avoid the confusion, we call the original Dominating Set Reconfiguration the reachability variant (denoted by REACHDSR). Suppose that we are given a graph , two integers , and a dominating set of whose cardinality is at most ; we call an upper bound and a solution size. Then OPTDSR asks for a dominating set satisfying the following two conditions: (a) the cardinality of is at most , and (b) can be transformed from under the TAR rule with upper bound . For example, if we are given a dominating set in Figure 1 and two integers and , then one of the solutions is , because can be transformed from and holds.
1.2 Related results
Although OPTDSR is being introduced in this paper, some results for REACHDSR relate to OPTDSR in the sense that the techniques to show the computational hardness or construct an algorithm will be used in our proof for OPTDSR. We thus list such results for REACHDSR in the following.
There are several results for the polynomialtime solvability of REACHDSR. Haddadan et al. [8] showed that REACHDSR under TAR rule is PSPACEcomplete for split graphs, for bipartite graphs, and for planar graphs, while lineartime solvable for interval graphs, for cographs, and for forests. REACHDSR is also studied well from the viewpoint of fixedparameter (in)tractability. Mouawad et al. [18] showed that REACHDSR under TAR is W[2]hard when parameterized by an upper bound . As a positive result, Lokshtanov et al. [16] gave a fixedparameter algorithm with respect to for graphs that exclude as a subgraph.
1.3 Our results
In this paper, we study OPTDSR from the viewpoint of the polynomialtime (in)tractability and fixedparameter (in)tractability.
We first study the polynomialtime solvability of OPTDSR with respect to graph classes (See Figure 2). Specifically, we show that the problem is PSPACEcomplete even for split graphs, for bipartite graphs, and for bounded pathwidth graphs, and NPhard for planar graphs with bounded maximum degree. On the other hand, the problem is lineartime solvable for cographs, trees and interval graphs. The inclusions of these graph classes are represented in Figure 2.
We next study the fixedparameter (in)tractability of OPTDSR. We first focus on the following four graph parameters: the degeneracy , the maximum degree , the pathwidth , and the vertex cover number (that is the size of minimum vertex cover). Figure 3(a) illustrates the relationship between these parameters, where A B means that the parameter A is bounded by some function of B. This relation implies that if we have a result stating that OPTDSR is fixedparameter tractable for A then the tractability for B follows, while if we have a negative (i.e. intractability) result for B then it extends to A. From results for polynomialtime solvability, we show the PSPACEcompleteness for fixed and NPhardness for fixed , and hence the problem is fixedparameter intractable for each parameter , and under P PSPACE or P NP. As a positive result, we give an FPT algorithm for , hence the problem is fixedparameter tractable for . We then consider two input parameters: the solution size and the upper bound . (See Figure 3(c).) We show that OPTDSR is W[2]hard when parameterized by . We note that we can assume without loss of generality that holds, as explained in Section 2. Therefore, it immediately implies W[2]hardness for . Most single parameters (except for ) cause a negative (intractability) result. We thus finally consider combinations of one graph parameter and one input parameter. We give an FPT algorithm with respect to . (See Figure 3(b).) In the end, we can conclude from the discussion above that for any combination of a graph parameter and an input parameter , OPTDSR is fixedparameter tractable when parameterized by . Due to space limitations, proofs of statements marked with (*) have been moved to Appendix.
2 Preliminaries
For a graph , we denote by and the vertex set of and edge set of , respectively. For a vertex , we let and ; we call a vertex in a neighbor of in . For a vertex subset , we let . If there is no confusion, we sometimes omit from the notation.
2.1 Optimization variant of Dominating Set Reconfiguration
For a graph , a vertex subset is a dominating set of if . For a dominating set , we say that dominates if holds. We say that a vertex has a private neighbor in if there exists a vertex such that . In other words, the vertex is dominated only by in . Note that the private neighbor of a vertex can be itself. A dominating set is (inclusionwise) minimal if and only if each of its vertices has a private neighbor, and minimum if and only if the cardinality is minimum among all dominating sets. Notice that any minimum dominating set is minimal. For a graph , we denote by the dominating number of defined as the cardinality of a minimum dominating set of ; if it is clear from the context that is an input graph, then we just write instead of .
Let and be two dominating sets of . We say that and are adjacent if , where and we denote this by . Let us now assume that both and are both of size at most , for some given . Then, a reconfiguration sequence between and under the TAR rule (or sometimes called a TARsequence) is a sequence of dominating sets of such that:

for each , is a dominating set of such that ; and

for each , holds.
Considering a reconfiguration sequence under the TAR rule, we sometimes write TAR() instead of TAR to emphasize the upper bound on the size of a solution. We say that is reachable from if there exists a reconfiguration sequence between and ; since a reconfiguration sequence is reversible, if is reachable from , then is also reachable from . We write if (resp. ) is reachable from (resp. ). Then, the optimization variant of the Dominating Set Reconfiguration problem (OPTDSR) is defined as follows:
We denote by a 4tuple an instance of OPTDSR.
2.2 Observations
From the definition of OPTDSR, we have the following observation.
Observation .
Let be an instance of OPTDSR. If and violate the inequality , then is a solution of the instance.
Proof.
By the definition of , we know . Therefore if the inequality is violated, we have or . In both cases, holds, and hence is a solution. ∎
It is observed that the condition in Observation 2.2 can be checked in linear time. Therefore, we sometimes assume without loss of generality that holds. Then, another observation follows.
Observation .
Let be an instance of OPTDSR such that holds. If is minimal and holds, then the instance has no solution.
Proof.
Since , we cannot add any vertex to without exceeding the threshold . Besides, since is minimal, we cannot remove any vertex while maintaining the domination property. As a result, there is no dominating set of size at most reachable from i.e. does not hold for any dominating set such that . ∎
Again, the conditions in Observation 2.2 can be checked in linear time, and hence we can assume without loss of generality that is not minimal or holds. Suppose that is not minimal. Then we can always obtain a dominating set of size less than by removing some vertex without private neighbor from , that is, we have a dominating set with and . Note that has a solution if and only if does. Therefore, it suffices to consider the case where holds. Combining it with Observation 2.2, we sometimes assume without loss of generality that holds.
3 Polynomialtime (in)tractability
In this section, we give some results for the polynomialtime solvability of OPTDSR with respect to graph classes. In Subsection 3.1, we show the NPhardness for the case where the input graph has maximum degree 3, or is planar with maximum degree 4. In Subsection 3.2, we show the PSPACEcompleteness for bounded pathwidth graphs, for split graphs, and for bipartite graphs. In Subsection 3.3, we give polynomialtime algorithms for cographs, trees and interval graphs.
3.1 NPhardness for planar graphs with bounded maximum degree
To show the NPhardness, we will use the following observation:
Observation .
Let be a graph and be an integer. Then an instance of OPTDSR is equivalent to finding a dominating set of with size at most .
Proof.
We claim that any dominating set of cardinality at most is a solution to the instance . Suppose that has a dominating set of cardinality at most and let be one of such dominating sets. Then, we can transform the input dominating set into by removing vertices in one by one; the observation follows. ∎
Observation 3.1 implies that results for the classical Dominating Set problem can be applied to OPTDSR. Recall that given a graph and an integer , the Dominating Set problem consists in deciding whether admits a dominating set of size at most . This problem is known to be NPhard even for the case where the input graph has maximum degree 3, or is planar with maximum degree 4 [7]. We thus obtain the following theorem. OPTDSR is NPhard even for the case where the input graph has maximum degree 3, or is planar with maximum degree 4.
3.2 PSPACEcompleteness for several graph classes
The following is the main theorem in this subsection. OPTDSR is PSPACEcomplete even for bounded pathwidth graphs, for split graphs, and for bipartite graphs.
First, observe that OPTDSR is in PSPACE. Indeed, when we are given a dominating set as a solution for some instance of OPTDSR, we can check in polynomial time whether it has size at most or not. Furthermore, since REACHDSR is in PSPACE, we can check in polynomial space whether it is reachable from the original dominating set . Therefore, we can conclude that OPTDSR is in PSPACE.
In the rest of this subsection, we thus show the PSPACEhardness for bounded pathwidth, split and bipartite graphs, respectively. To this end, we give polynomialtime reductions from the optimization variant of Vertex Cover Reconfiguration (denoted by OPTVCR). We note that all reductions are almost identical to the ones of PSPACEhardness for REACHDSR [8].
We now give the definition of OPTVCR. For a graph , a vertex subset is called a vertex cover if contains at least one endpoint of each edge in . Suppose that we are given a graph , two integers , and a vertex cover of whose cardinality is at most . Then OPTVCR asks for a vertex cover satisfying the following two conditions: (a) the cardinality of is at most , and (b) can be transformed from via vertex covers of size at most such that each intermediate one can be obtained from the previous one by adding or removing exactly one vertex. The problem is known to be PSPACEcomplete even for bounded pathwidth graphs [13] ^{1}^{1}1In [13], Ito et al. actually showed the PSPACEcompleteness for the optimization variant of Independent Set Reconfiguration. However, the result can easily be converted to OPTVCR from the observation that any vertex cover of a graph is the complement of an independent set..
We first consider bounded pathwidth graphs. The pathwidth of a graph is defined as follows. A path decomposition of is a sequence , where each , for each , satisfies the following properties:

each vertex is contained in (at least) one bag ;

each edge is contained in (at least) one bag i.e. there exists such that ;

for every three indices , .
The width of a given path decomposition is one less than the size of its largest bag, that is . Finally, the pathwidth of , denoted by , is the minimum width of any path decomposition of . Then the following lemma completes the proof of PSPACEcompleteness for bounded pathwidth graphs. OPTDSR is PSPACEhard even for bounded pathwidth graphs.
Proof.
Our reduction follows from the classical reduction from Vertex Cover to Dominating Set [7]. Let be an instance of OPTVCR. Let be the graph constructed from as follows: for each edge , we add a new vertex and join it with both of and by edges (see Fig. 4). Then let be the corresponding instance of OPTDSR. This construction can clearly be done in polynomial time.
It remains to prove that is a yesinstance for OPTVCR if and only if is a yesinstance for OPTDSR.
Suppose that is a yesinstance and let be a vertex cover of size at most reachable from under the TAR() rule, by a sequence . Since any vertex cover of is a dominating set of and , then the sequence yields a reconfiguration sequence from to . Thus, is a yesinstance.
We now prove the other direction. Suppose that is a yesinstance and let be a TAR() sequence of dominating sets of starting at and reaching a dominating set that satisfies . Recall that does not contain any newly added vertex in . We want a sequence that does not touch any newly added vertex . To this end, we proceed by eliminating them one by one from the sequence. Let be a vertex, and be the associated newly added vertex. Whenever a dominating set contains , we instead consider the set , obtainable from in one step under TAR() rule by simply ignoring the addition of and maybe adding . It is still a dominating set since . The resulting sequence does not touch , hence by repeating the operation on all vertices of we obtain a sequence that does not touch any of them. In this way, we can obtain a reconfiguration sequence of vertex covers in between and as needed.
Since OPTVCR is PSPACEcomplete for bounded pathwidth graphs, the reduction above implies PSPACEhardness on bounded pathwidth graphs. ∎
We next consider the class of split graphs. A graph is a split graph if its vertex set can be partitioned into a clique and an independent set. Then the following lemma completes the proof of PSPACEcompleteness for split graphs. [*] OPTDSR is PSPACEhard even for split graphs.
We finally consider the class of bipartite graphs. [*] OPTDSR is PSPACEhard even for bipartite graphs.
3.3 Lineartime algorithms
In this subsection, we show that OPTDSR can be solved in linear time for several graph classes. To this end, we deal with the concept of a canonical dominating set. We say that a dominating set is canonical if is a minimum dominating set which is reachable from any dominating set under the TAR() rule. Then we have the following theorem. Let be a class of graphs such that any graph has a canonical dominating set and we can compute it in linear time. Then OPTDSR can be solvable in linear time on .
Proof.
Let be an instance of OPTDSR, where . Recall that we can assume without loss of generality that ; we can check in linear time whether the inequality is satisfied or not, and if it is violated, then we know from Observation 2.2 and 2.2 that it is a trivial instance. Since , admits a canonical dominating set and we can compute in linear time an actual one. Let be such a canonical dominating set. Then it follows from the definition that is reachable from under the TAR() rule since . Since is a minimum dominating set, we can output it if holds, and noinstance otherwise. All processes can be done in linear time, and hence the theorem follows. ∎
Haddadan et al. showed in [8] that every cographs, trees (actually, forests), and interval graphs admit a canonical dominating set. Their proofs are constructive, and hence we can find an actual canonical dominating set. It is observed that the constructions on cographs and trees can be done in linear time. The construction on interval graphs can also be done in linear time with a nontrivial adaptation by using an appropriate data structure. Therefore, we have the following lineartime solvability of OPTDSR. OPTDSR can be solved in linear time on cographs, trees, and interval graphs.
4 Fixedparameter (in)tractability
In this section, we study the fixedparameter complexity of OPTDSR with respect to several graph parameters: the upper bound , solution size , minimum size of a vertex cover and degeneracy .
More precisely, we first show that OPTDSR is W[2]hard when parameterized by the upper bound . To prove it, we use the idea of the reduction constructed by Mouawad et al. in [18] to show the W[2]hardness of REACHDSR.
[*] OPTDSR is W[2]hard when parameterized by the upper bound .
On the other hand, we give FPT algorithms with respect to the combination of the solution size and the degeneracy in Subsection 4.1 and the vertex cover number in Subsection 4.2.
4.1 FPT algorithm for degeneracy and solution size
The following is the main theorem in this subsection. OPTDSR is fixedparameter tractable when parameterized by . To prove the theorem, we give an FPT algorithm with respect to . Note that our algorithm uses the idea of an FPT algorithm solving the reachability variant of Dominating Set Reconfiguration, developed by Lokshtanov et al. [16]. Their algorithm uses the concept of domination core; for a graph , a domination core of is a vertex subset such that any vertex subset is a dominating set of if and only if [6].
Suppose that we are given an instance of OPTDSR where is a degenerate graph. By Observation 2.2, we can assume without loss of generality that . We first check whether has a dominating set of size at most : this can be done in FPT() time for degenerate graphs [1]. If does not have it, then we can instantly conclude that this is a noinstance.
In the remainder of this subsection, we assume that has a dominating set of size at most . In this case, we kernelize the instance: we shrink by removing some vertices while keeping the existence of a solution until the size of the graph only depends on and . To this end, we use the concept of domination core. [Lokshtanov et al. [16]] If is a degenerate graph and has a dominating set of size at most , then has a domination core of size at most and we can find it in FPT time.
Therefore, one can compute a domination core of of size at most in FPT time by Lemma 4.1. In order to shrink , we use the reduction rule R1: if there exists two vertices such that , we remove . We need to prove that R1 is “safe”, that is, we can remove from without changing the existence of a solution. However, if the input dominating set contains , we cannot do it immediately. Therefore, we first remove from .
[*] Let be a dominating set such that both and hold. Then there exists such that and , and can be computed in linear time.
We can now redefine as a dominating set which does not contain . We then consider removing from . Let . The following lemma ensures that removing keeps the existence of a solution.
Let be an instance where . Then, has a solution if and only if has a solution.
Proof.
We first prove the if direction. Suppose that has a solution . Then there exists a reconfiguration sequence of dominating sets of . It suffices to show that any dominating set of in is also a dominating set of . Since is a dominating set of and , we have . By the definition of domination core, we know that is also a dominating set of .
We then prove onlyif direction. Suppose that has a solution . Then there exists a reconfiguration sequence of dominating sets of . Based on , we construct another sequence of vertex sets of , where
for each . Notice that any vertex subset in does not contain . Our claim is that is a solution of . To prove it, we show the following two statements:

for each , is a dominating set of (and hence of ); and

for each , holds i.e. we have .
Then the sequence obtained by removing redundant ones from is a reconfiguration sequence from to .
We first show the statement i. Let be any dominating set in . If , then the statement clearly holds. Thus we consider the other case where . Since is a dominating set of , we know . Furthermore, since , we have . By the definition of domination core, is a dominating set of , and hence the statement i follows.
We then show the statement ii. Let and be any two consecutive dominating sets in . Then, we know . We assume without loss of generality that ; otherwise the proof is symmetric. We prove the statement in the following three cases:

Case 1: both and hold;

Case 2: either or holds (but not both); and

Case 3: both and hold.
In Case 1, we know that , and hence the statement clearly holds. We then consider Case 2. In this case, since , we observe that and , and hence Therefore, . Thus we can conclude that , and hence the statement follows. We finally deal with Case 3. In this case, we have . Therefore, holds, and hence the statement follows. In this way, we can conclude that is a solution of . This concludes the proof. ∎
We exhaustively apply the reduction rule R1 to shrink . Let and be the resulting graph and dominating set, respectively. Then, any two vertices satisfy (more precisely, ). Then the following lemma completes the proof of Theorem 4.1. can be solved in FPT time.
Proof.
We first show that the size of the vertex set of is at most . Since , it suffices to show that holds. Recall that any two vertices satisfy . Then since the number of combination of vertices in is at most , we have the desired upper bound .
We now prove that can be solved in FPT time. To this end, we construct an auxiliary graph , where the vertex set of is the set of all dominating sets of , and any two nodes (that correspond to dominating sets of ) and in are adjacent if and only if holds. Let and . Then the number of candidate nodes in (vertex subsets of ) is bounded by . For each candidate, we can check in time if it forms a dominating set. Thus we can construct the vertex set of in time. We then construct the edge set of . There are at most pairs of nodes in . For each pair of nodes, we can check in time if their corresponding dominating sets differ in exactly one vertex. Therefore we can construct the edge set of in time, and hence the total time to construct is time. We finally search a solution by running a breadthfirst search algorithm from on in time.
We can conclude that our algorithm runs in time in total. Since and , this is an FPT time algorithm. ∎
4.2 FPT algorithm for vertex cover number
Let be an instance of OPTDSR. As in the previous section, we may first assume by Observation 2.2 that . We recall that is the size of a minimum vertex cover of . In order to lighten notations, we simply denote by the vertex cover number of the input graph. Then, we have the following:
OPTDSR is fixedparameter tractable when parameterized by .
We first establish the following fact that is going to be useful later.
Observation .
If is degenerate, then .
Proof.
Let be a graph, a minimum vertex oover of and be any subgraph of . If contains a vertex outside , then has a degree at most in and therefore in . Otherwise, is a subraph of and thus has at most vertices. Hence all vertices of have degree at most in . Therefore, since any subgraph of contains a vertex of degree at most , is degenerate. ∎
We are now able to get down to the proof of Theorem 4.2, by providing an algorithm that solves OPTDSR and runs in time FPT. We first compute a minimum vertex cover of in time FPT() [4]. We partition the vertices of into two components, the vertex cover and the remaining vertices . By definition of vertex cover, no edge can have both ends outside , therefore is an independent set. Note that if , then by Observation 4.2 we have , where is the degeneracy of . In this case we are able to use the algorithm of the last section, that runs in time FPT.
We may therefore assume . In the remainder of the proof, we assume that the graph has no isolated vertex since an isolated vertex must belong to any dominating set of . We now prove that is always a yesinstance i.e. there exists a dominating set of size at most that is reachable from under the TAR() rule.
We associate to every vertex a special neighbor among its neighbors that dominate it (which can be either in or ) i.e. we pick arbitrarily a vertex in . We denote this special neighbor . Let be the set of special neighbors i.e. . This corresponds to the set of vertices that are used to dominate the vertices in that do not belong to . Note that .
We are now able to describe the algorithm we use to output , the target dominating set. It consists in exhaustively applying the two following rules on the vertices of that belong to the current dominating set:

if there is a vertex in but not in that is already dominated by another vertex, then we remove from the dominating set; and

if there is a vertex in but not in that is dominated only by itself, then we add any one of its neighbors to the dominating set, and then remove . The vertex does not need a special neighbor anymore, since it now belongs to the dominating set. We thus update the set by only keeping the special neighbors of vertices that are still in .
We first prove that these two rules are safe i.e. we do not break the domination property at any step. Since Rule i removes a vertex that is not required to dominate itself or another vertex (because it has not been chosen in ), we can safely remove it. In Rule ii, after adding a neighbor of to the dominating set, is not required to dominate itself anymore. Since is not in , we can now apply Rule i which is safe.
Recall that . Then, each dominating set obtained after applying one of these rules is of size at most since Rule i only removes vertices and Rule ii consists in an addition immediately followed by a removal.
Now, let be the dominating set obtained once we cannot apply Rule i and Rule ii anymore (see Figure 5 for an example). All remaining vertices in now belong to . By definition of , each vertex in has (exactly) one neighbor in (but they are not necessarily distinct). Therefore, . As a result, . Since , the size of is at most , as desired.
Finally, we focus on the complexity of this algorithm. As we already said, we first compute a minimum vertex cover of in time FPT(). If , we run the FPT algorithm of Section 4.1. Otherwise, we compute the set and we run the subroutine that exhaustively applies the two aforementioned rules. Since these rules only apply to vertices in and whenever one is applied, exactly one vertex in is removed (and none is added), these rules are applied at most times. Therefore, the subroutine runs in polynomial time and produces the desired dominating set . As a result, this algorithm is FPT with respect to . This concludes the proof.
5 Conclusion
In this paper, we have studied a new variant of combinatorial reconfiguration recently introduced by Ito et al. in [13] and we have applied it to the wellstudied Dominating Set Reconfiguration problem. We have tackled this problem from a complexity perspective with respect to some graph parameters or graph classes. More precisely, we have shown that OPTDSR is PSPACEcomplete, even when restricted to bounded pathwidth graphs, split graphs or bipartite graphs. On the other hand, we have shown that the problem is lineartime tractable on cographs, trees, and interval graphs. These results highlight the frontier between hardness and tractability since the problem is PSPACEhard for bipartite graphs but linear for trees. We have also studied the problem from a parameterized complexity viewpoint and we have showed that OPTDSR is fixedparameter tractable when parameterized by the minimum size of a vertex cover or by the degeneracy and the size of the desired dominating set.
References
 [1] N. Alon and S. Gutner. Linear time algorithms for finding a dominating set of fixed size in degenerated graphs. Algorithmica, 54(4):544, 2008.
 [2] A.A. Bertossi. Dominating sets for split and bipartite graphs. Information Processing Letters, 19(1):37–40, 1984.
 [3] M. Bonamy and N. Bousquet. Token sliding on chordal graphs. In Proceedings of the 43rd International Workshop on GraphTheoretic Concepts in Computer Science (WG 2017), pages 127–139, 2017.
 [4] J. Chen, I.A. Kanj, and G. Xia. Improved upper bounds for vertex cover. Theoretical Computer Science, 411(40):3736 – 3756, 2010.
 [5] R.G. Downey and M.R. Fellows. Parameterized Complexity. Springer, 1999.
 [6] P. Drange, M.S. Dregi, F.V. Fomin, S. Kreutzer, D. Lokshtanov, M. Pilipczuk, M. Pilipczuk, F., F. Villaamil, S. Saurabh, S. Siebertz, and S.Sikdar. Kernelization and sparseness: the case of dominating set. In 33rd Symposium on Theoretical Aspects of Computer Science (STACS 2016), pages 31:1–31:14, 2016.
 [7] M.R. Garey and D.S. Johnson. Computers and Intractability: A Guide to the Theory of NPCompleteness. Freeman, San Francisco, CA, 1979.
 [8] A. Haddadan, T. Ito, A.E. Mouawad, N. Nishimura, H. Ono, A. Suzuki, and Y. Tebbal. The complexity of dominating set reconfiguration. Theoretical Computer Science, 651:37–49, 2016.
 [9] R.A. Hearn and E.D. Demaine. PSPACEcompleteness of slidingblock puzzles and other problems through the nondeterministic constraint logic model of computation. Theoretical Computer Science, 343(1–2):72–96, 2005.
 [10] J. van den Heuvel. The complexity of change. In Surveys in Combinatorics 2013, volume 409 of London Mathematical Society Lecture Note Series, pages 127–160. Cambridge University Press, 2013.
 [11] D.A. Hoang and R. Uehara. Sliding tokens on a cactus. In Proceedings of the 27th International Symposium on Algorithms and Computation (ISAAC 2016), pages 37:1–37:26, 2016.
 [12] T. Ito, E.D. Demaine, N.J.A. Harvey, C.H. Papadimitriou, M. Sideri, R. Uehara, and Y. Uno. On the complexity of reconfiguration problems. Theoretical Computer Science, 412(12–14):1054–1065, 2011.
 [13] T. Ito, H. Mizuta, N. Nishimura, and A. Suzuki. Incremental optimization of independent sets under reachability constraints. In Proceedings of the 25th International Computing and Combinatorics Conference (COCOON 2019), 2019. to appear.
 [14] M. Kamiński, P. Medvedev, and M. Milani. Complexity of independent set reconfigurability problems. Theoretical Computer Science, 439:9–15, 2012.
 [15] D. Lokshtanov and A.E. Mouawad. The complexity of independent set reconfiguration on bipartite graphs. In Proceedings of the 29th Annual ACMSIAM Symposium on Discrete Algorithms (SODA 2018), pages 7:1–7:19, 2019.
 [16] D. Lokshtanov, A.E. Mouawad, F. Panolan, M.S. Ramanujan, and S. Saurabh. Reconfiguration on sparse graphs. J. Comput. Syst. Sci., 95:122–131, 2018.
 [17] A.E. Mouawad, N. Nishimura, and V. Raman. Vertex cover reconfiguration and beyond. In Proceedings of the 25th International Symposium on Algorithms and Computation (ISAAC 2014), pages 452–463, 2014.
 [18] A.E. Mouawad, N. Nishimura, V. Raman, N. Simjour, and A. Suzuki. On the parameterized complexity of reconfiguration problems. Algorithmica, 78(1):274–297, 2017. doi:10.1007/s0045301601592.
 [19] N. Nishimura. Introduction to reconfiguration. Algorithms, 11(4):52, 2018.

[20]
A. Suzuki, A.E. Mouawad, and N. Nishimura.
Reconfiguration of dominating sets.
Journal of Combinatorial Optimization
, 32(4):1182–1195, 2016.
Appendix A Omitted proof for Lemma 3.2
Proof.
We again give a polynomialtime reduction from OPTVCR. We extend the idea developed for the NPhardness proof of Dominating Set problem on split graphs [2].
Let be an instance of OPTVCR, where and . We construct the corresponding split graph as follows (see also Figure 6.) Let , where and ; the vertex corresponds to the edge . We join all pairs of vertices in so that forms a clique in . In addition, for each edge in , we join with each of and . Let be the resulting graph, and let be the corresponding instance of OPTDSR (we will prove later that is a dominating set of ). Clearly, this instance can be constructed in polynomial time. It remains to prove that is a yesinstance if and only if is a yesinstance.
We first prove the onlyif direction. Suppose that is a yesinstance. Then, there exists a vertex cover of size at most reachable from under the TAR(’) rule. Since , and both problems employ the same reconfiguration rule, it suffices to prove that any vertex cover of is a dominating set of . Since and is a clique, all vertices in are dominated by the vertices in . Thus, consider a vertex , which corresponds to the edge in . Then, since is a vertex cover of , at least one of and must be contained in . This means that is dominated by the endpoint or in . Therefore, each vertex cover in the reconfiguration sequence between and is a dominating set of (including and ) and thus, is a yesinstance.
We now focus on the if direction. Suppose that is a yesinstance. Then, there exists a dominating set of of size at most reachable under the TAR() rule by a sequence , with . Recall that and thus is a vertex cover of . We want to produce a sequence of dominating sets that are subsets of . To this end, in the same spirit as in the previous proof, we eliminate the vertices of one by one. If a contains a vertex associated to the edge , then we replace by , which is also a dominating set and is reachable in one step from . Thus, the resulting sequence does not touch , and by repeating the operation to all vertices of , we obtain the wanted TAR() sequence of subets of . Observe that any dominating set of such that forms a vertex cover of , because each vertex is dominated by at least one vertex in . Therefore, is a yesinstance. The conclusion follows. ∎
Appendix B Omitted proof of Lemma 3.2
Proof.
We give a polynomialtime reduction from OPTDSR on split graphs to the same problem restricted to bipartite graphs. The same idea is used in the NPhardness proof of Dominating Set problem on bipartite graphs [2].
Let be an instance of OPTDSR, where is a split graph. Then can be partitioned into two subsets and which form a clique and an independent set in , respectively. Furthermore, by the reduction given in the proof of Lemma 3.2, the problem on split graph remains PSPACEcomplete even if the given dominating set consists of vertices only in . We thus assume that holds.
We now construct the corresponding bipartite graph , as follows. First, we delete any edge joining two vertices in so that forms an independent set. Then, we add a new edge consisting of two new vertices and , and join with each vertex in . The resulting graph is bipartite (see Fig. 6 for an example). Let , and . Then we obtain the corresponding OPTDSR instance where is bipartite (here again, we will prove later that is dominating set of ). Clearly, this instance can be constructed in polynomial time. We then prove that is a yesinstance if and only if is a yesinstance.
We first prove the onlyif direction. Suppose that there exists a dominating set of such that and . Consider any dominating set of . Then, holds because and we have deleted only the edges which have both endpoints in . Since , we can conclude that is a dominating set of . Furthermore, . Thus there exists a dominating set of such that and , as desired.
We then prove the if direction. Suppose that there exists a dominating set of , of size at most and reachable from by a TAR() sequence , with . Recall that , and notice that any dominating set of contains at least one of and . Since , we can assume that contains . Therefore, we can also assume that is contained in every dominating set of the reconfiguration sequence. Recall that the assumption holds. As in the previous proof, we can produce an equivalent sequence that does not touch any vertex of . Again, if a dominating set touches a vertex associated to the edge , we replace by . We repeat the operation for all and obtain the wanted sequence.
Consider any dominating set of in such a reconfiguration sequence. Since , we have . Furthermore, since and forms a clique in , we have . Since there is no edge joining and a vertex in , each vertex in is dominated by some vertex in . Therefore, is a dominating set of with cardinality at most , and hence there exists a dominating set of such that and . ∎
Appendix C Omitted proof of Theorem 4
Proof.
We give an FPTreduction from the (original) Dominating Set problem that is W[2]hard when parameterized by its natural parameter [5].
Let be an instance of Dominating Set, where and . Then we construct the corresponding instance of OPTDSR, as follows. We first describe the construction of . Let be the graph obtained by adding a universal vertex to , and be copies of . The vertex set of consists of . For any and , we use to denote the vertex in corresponding to in . Then, for each vertex in except for , we connect by new edges to all vertices in in each ; formally, the edge set of consists of . This completes the construction of ; see Figure 7 for an example of this reduction. However, for readability purposes, we do not draw all the edges between the vertices in and those of , for . The only such drawn edges are the dotted ones (in gray) that are incident to the vertices and . We set , , and ; notice that has vertices. In this way, we constructed the corresponding instance . Then our claim is that is a yesinstance if and only if has a solution.
We first prove the onlyif direction. Suppose that is a yesinstance, hence there exists a dominating set of of size at most . Then by the construction of , we know that is also a dominating set of (if we identify the vertices of with those of ).
Thus it suffices to show that , since has at most vertices. We first add vertices in
Comments
There are no comments yet.