Token Sliding on Split Graphs

07/14/2018
by   Rémy Belmonte, et al.
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We show that the independent set reconfiguration problem on split graphs under the token sliding rule is PSPACE-complete.

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1 Introduction

A reconfiguration problem is a problem of the following type: we are given an instance of a decision problem, two feasible solutions , and a local modification rule. The question is whether can be transformed to by repeated applications of the modification rule in a way that maintains the solution feasible at all times. Due to their numerous applications, reconfiguration problems have attracted much interest in the literature, and reconfiguration versions of standard problems (such as Satisfiability, Dominating Set, and Independent Set) have been widely studied (see the surveys [10, 19] and the references therein).

Among reconfiguration problems on graphs, Independent Set Reconfiguration is certainly the most well-studied. The complexity of this problem depends heavily on the rule specifying the allowed reconfiguration moves. The main reconfiguration rules that have been studied for Independent Set Reconfiguration are Token Addition & Removal (TAR) [16, 18], Token Jumping (TJ) [2, 3, 12, 13, 14], and Token Sliding (TS) [1, 5, 6, 8, 11, 17]. In all rules, we are required to keep the current set independent at all times. TAR allows us to add or remove any vertex in the current set, as long as the set’s size is always higher than a predetermined threshold. TJ allows to exchange any vertex in the set with any vertex outside it (thus keeping the size of the set constant at all times). Finally, under TS, we are allowed to exchange a vertex in the current independent set with one of its neighbors, that is, we are allowed to perform a TJ move only if the two involved vertices are adjacent.

The Independent Set Reconfiguration problem has been intensively studied under all three rules. Because the problem is PSPACE-complete in general for all three rules [16], this has motivated the study of its complexity in restricted classes of graphs, with an emphasis on graphs where Independent Set is polynomial-time solvable, such as chordal graphs and bipartite graphs. By now, many results of this type have been discovered (see Table 1 for a summary).

Our first, and main, focus of this paper is to concentrate on a case of this problem which has so far remained elusive, namely, the complexity of Independent Set Reconfiguration on chordal graphs under the TS rule. This case is of particular interest because it is one of the few cases where the problem is known to be tractable under both TAR and TJ. Indeed, Kamiński, Medvedev, and Milanič [16] showed that under these two rules Independent Set Reconfiguration is polynomial-time solvable on even-hole-free graphs, a class that contains chordal graphs. In the same paper they explicitly asked as an open question if the same problem is tractable on even-hole-free graphs under TS ([16, Question 2]).

This question was then taken up by Bonamy and Bousquet [1] who made some progress by showing that Independent Set Reconfiguration under TS is polynomial-time solvable on interval graphs, an important subclass of chordal graphs. They also gave some first evidence that it may be hard to obtain a similarly positive result for chordal graphs by showing that a related problem, the problem of determining if all independent sets of the same size can be transformed to each other under TS, is coNP-hard on split graphs, another subclass of chordal graphs. Note, however, that this is a problem that is clearly distinct from the more common reconfiguration problem (which asks if two specific sets are reachable from each other), and that the coNP-hardness is not tight, since the best known upper bound for this problem is also PSPACE.

The complexity of Independent Set Reconfiguration under TS on split and chordal graphs has thus remained as an open problem. Our first, and main, contribution in this paper is to settle this problem by showing that the problem is PSPACE-complete already on split graphs (Theorem 3), and therefore also on chordal and even-hole-free graphs.

Independent Set Reconfiguration
TS TJ/TAR
perfect PSPACE-complete [16]
even-hole-free PSPACE-complete (Theorem 3) P [16]
chordal PSPACE-complete (Theorem 3) P (even-hole-free)
split PSPACE-complete (Theorem 3) P (even-hole-free)
interval P [1] P (even-hole-free)
bipartite PSPACE-complete [17] NP-complete [17]
Table 1: Complexity of Independent Set Reconfiguration on some graph classes.

-Colorable Reconfiguration

A natural generalization of Independent Set Reconfiguration was recently introduced in [15]: in -Colorable Reconfiguration we are given a graph and two sets , both of which induce a -colorable graph. The question is whether can be transformed to (under any of the previously mentioned rules) in a way that maintains a -colorable graph at all times. Clearly, is the case of Independent Set Reconfiguration. It was shown in [15] that this problem is already PSPACE-complete on split graphs under all three rules, when is part of the input. It was thus posed as an open question what is the complexity of the same problem when is fixed. Some first results in this direction were given in the form of an (XP) algorithm that works for split graphs under the TAR and TJ rules (but not TS). Motivated by this work, the second area of focus of this paper is to investigate how the hardness of -Colorable Reconfiguration for split graphs established in Theorem 3 extends to larger, but fixed .

Our first contribution in this direction is to show that, for chordal graphs, -Colorable Reconfiguration under TS is PSPACE-complete for any fixed . This is, of course, not surprising, as the problem is PSPACE-complete for ; indeed, the reduction we present in Theorem 4 is a tweak of the construction of Theorem 3 that increases .

What is perhaps more surprising is that we show (under standard assumptions) that, even though Theorem 3 establishes hardness for on split graphs, a similar tweak cannot establish hardness for higher on the same class for TS. Indeed, we provide an algorithm which solves TS -Colorable Reconfiguration in split graphs in time for any except . Thus, Independent Set Reconfiguration turns out to be the only hard case of -Colorable Reconfiguration for split graphs under TS. Since the algorithm of [15] for TAR/TJ reconfiguration of split graphs works for all fixed , it thus seems that this anomalous behavior is peculiar to the Token Sliding rule.

Finally, we address the natural question of whether one can improve this algorithm, by showing that the problem is W[2]-hard parameterized by and the length of the solution for all three rules. This is in a sense doubly tight, since in addition to our algorithm and the algorithm of [15] which run in , it also matches the trivial algorithm which tries out all solutions of length . More strongly, under the ETH our reduction implies that the problem cannot be solved in meaning that these algorithms are in a sense “optimal”.

2 Definitions

We use standard graph-theoretic terminology. For a graph and a set we use to denote the graph induced by . A graph is chordal if it does not contain a -vertex cycle as an induced subgraph for any . A graph is split if its vertex set can be partitioned into two sets such that induces a clique and induces an independent set. It is a well-known fact that split graphs are chordal, and it is easy to see that both classes are closed under induced subgraphs. We use to denote the chromatic number and maximum clique size of a graph respectively. It is known that, because chordal graphs are perfect, if is chordal then [21]. We also recall that a graph is chordal if and only if every induced subgraph of contains a simplicial vertex, where a vertex is simplicial if its neighborhood is a clique.

Let be a graph and an integer. Given two sets such that , we say that can be -transformed into by one token sliding (TS) move if and there exist with such that , . One easy way to think of TS moves is by picturing the elements of the current set as tokens placed on the vertices of the graph, and a single move as “sliding” a token along an edge (hence the name Token Sliding).

We say that is -reachable from , or that can be -transformed into , by a sequence of TS moves if there exists a sequence of sets , with and for each , and can be -transformed into by one TS move. We will simply say that can be transformed into or that is reachable from , if are independent sets and can be -transformed into . We focus on the following problems.

In -Colorable Reconfiguration we are given a graph and two sets with and . We are asked if can be -transformed into . Independent Set Reconfiguration is the special case of -Colorable Reconfiguration where .

In addition to TS moves we will consider Token Jumping (TJ) and Token Addition & Removal (TAR) moves. A TJ move is the same as a TS move except that the two vertices are not required to be adjacent. Two -colorable sets are reachable with one TAR move with threshold if and . We note here that, because our main focus in this paper is the TS rule, whenever we refer to a transformation without explicitly specifying under which rule this transformation is performed the reader may assume that we are referring to the TS rule.

We assume that the reader is familiar with basic complexity notions such as the class PSPACE [20], as well as basic notions in parameterized complexity, such as the class W[2] (see e.g. [4]). In Theorem 3 we will perform a reduction from the PSPACE-complete NCL (non-deterministic constraint logic) reconfiguration problem introduced by Demaine and Hearn in [8] (see also [7, 9]). Let us recall this problem. In the NCL reconfiguration problem we are given as input a graph , whose edge set is partitioned into two sets, (red) and (blue). We consider blue edges as edges of weight and red edges as edges of weight . A valid configuration of is an orientation of all the edges with the property that all vertices have weighted in-degree at least . In the NCL configuration-to-configuration problem we are given two valid orientations of , and , and are asked if there is a sequence of valid orientations such that and for all we have that agree on all edges except one. We recall the following theorem:

[Corollary 6 of [8]] The NCL configuration-to-configuration problem is PSPACE-complete even if all vertices of have degree exactly three and, moreover, even if all vertices belong in one of the following two types: OR vertices, which are vertices incident on exactly three blue edges and no red edges; and AND vertices which are vertices incident on two red edges and one blue edge.

3 Token Sliding on Split Graphs is PSPACE-complete

The main result of this section is that Independent Set Reconfiguration is PSPACE-complete under the TS rule when restricted to split graphs.

Overview of the proof

Our proof is a reduction from the NCL (non-deterministic constraint logic) reconfiguration problem of Theorem 2. The first step of our proof is a relatively straightforward reduction from the NCL reconfiguration problem to token sliding on split graphs. Its main idea is roughly as follows: for each edge of the original graph we construct two selection vertices in the independent set of our split graph. The idea is that at each point exactly one of the two will contain a token (i.e. will belong in the current independent set), hence our independent set will in a natural way represent an orientation of the original graph. In order to allow a single reconfiguration step to take place we add for each pair of selection vertices one or two “gate” vertices (depending on the color of ), which are common neighbors of and belong in the clique. The idea is that a single re-orientation step would, for example, take a token from , slide it to a gate vertex connected to the pair , and then slide it to : this sequence would represent re-orienting from to . In order to simulate the in-degree constraint we add edges between each selection vertex and gate vertices corresponding to edges incident on the other endpoint of , since keeping a token on represents an orientation of towards , which makes it harder to re-orient the edges incident on the other endpoint of .

The above sketch captures the basic idea of our reduction, except for one significant obstacle. The correspondence between orientations and independent sets is only valid if we can guarantee that no intermediate independent set will “cheat” by, for example, placing tokens on both and . Since we have added edges from to gate vertices that correspond to other edges (in order to simulate the interaction between edges in the NCL instance), nothing prevents a reconfiguration solution from using these edges to slide a token from one selection pair to another. The main problem thus becomes enforcing consistency, or in other words forcing the solution sequence to only use the appropriate gate vertices to slide tokens as intended. This is handled in the second step of our reduction which, given the split graph construction sketched above, makes a large number of copies and connects them appropriately in a way that the only feasible token sliding solutions are indeed those that correspond to valid orientations of the original graph.

In the remainder of this section we use the following notation: , where , is the graph supplied with the initial NCL reconfiguration instance and are the initial and target orientations; is the “basic” split graph of our construction in the first step and the independent sets of for which we need to decide reachability; and is the split graph of our final token sliding instance with being its corresponding independent sets.

Before we proceed, let us first slightly edit our given NCL reconfiguration instance. We will now allow some vertices to have degree two and call these vertices COPY vertices. Using these we can force the OR vertices to become an independent set.

NCL reconfiguration remains PSPACE-complete on graphs where (i) all vertices are either AND vertices (two incident red edges, one incident blue edge), OR vertices (three incident blue edges), or COPY vertices (two incident blue edges) (ii) every blue edge is incident on exactly one COPY vertex.

Proof.

For every blue edge in the original graph we delete this edge from the graph, introduce a new COPY vertex , and connect to with blue edges. It is not hard to see that this transformation does not change the type of any original vertex or the answer to the reconfiguration problem. ∎

First Step of the Construction

We assume (Lemma 3) that in the given graph we have three types of vertices (AND, OR, COPY) and that each blue edge is incident on one COPY vertex. Let us now describe the construction of .

  1. For each we construct two selector vertices and one gate vertex .

  2. For each we construct two selector vertices and two gate vertices .

  3. For each edge we connect to both . For each edge we connect both to both . We call the edges added in this step gate edges.

  4. For each AND vertex , such that and we add the following edges: (see Figure 1). In other words, for each edge involved in this part we connect the selector which represents its other endpoint (not ) to the gate vertices of edges that should be unmovable if this edge is not oriented towards .

  5. For each OR vertex such that we add the following edges: . In other words, we connect the selector vertex for each to a distinct gate of the edges , for distinct. Informally, this makes sure that if two of the edges are oriented away from the third edge is stuck, but if at most one is oriented away from the other edges have a free gate.

  6. For each COPY vertex such that we add the following edges: . In other words, we connect the selector vertex for in a way that blocks the movement of the token from , and similarly for .

  7. We connect all gate vertices into a clique to obtain a split graph. Note that the remaining vertices (that is, the selector vertices ) form an independent set.

We now construct two independent sets of in the natural way: given an orientation , for each we place in if and only if orients towards ; we construct from in the same way. This completes the basic construction.

Before proceeding, let us make some basic observations regarding the neighborhoods of gate vertices of the graph . We have the following:

  • If , let be vertices of such that , (that is, are the second endpoints of the blue edges incident on ). We have that .

  • If , is a COPY vertex and is an AND vertex, let be the other edge incident on , and be the other two edges incident on . Then .

  • If , is a COPY vertex and is an OR vertex, let be the other edge incident on , and be the other two edges incident on . Then one of the vertices has neighbors and the other has neighbors .





Figure 1: Construction when is an AND vertex (top) or an OR vertex (bottom). In both cases is a COPY vertex. The part of the construction corresponding to is not drawn: would be a common neighbor of and would be a common neighbor of . Edges connecting selector vertices to their corresponding gates are drawn thinner for readability. On the right, black (gate) vertices are connected in a clique.

We are now ready to show that if we only consider “consistent” configurations in , then the new instance simulates the original NCL reconfiguration problem.

There is a valid reconfiguration of the NCL instance given by if and only if there exists a valid reconfiguration under the TS rule from to in such that no independent set of the reconfiguration sequence contains both for any .

Proof.

Since is a split graph, any independent set contains at most one vertex from the clique made up of the gate vertices. We will call an independent set that contains no gate vertices a “main” configuration. Furthermore, for main configurations that also obey the restrictions of the lemma (i.e. do not contain both for any ), we observe that there is a natural one-to-one correspondence with the set of orientations of : an edge is oriented towards if and only if is in the independent set. (We implicitly use the fact that the number of tokens is , therefore for each pair exactly one vertex has a token in such a main configuration).

Suppose now that we have two consecutive valid orientations in the reconfiguration sequence of such that differ only on the edge , which orients towards . We want to show that the sets obtained using the correspondence above from can be obtained from each other with a pair of sliding token moves. Indeed, the sets are identical except that and . We would like to slide the token from to using a gate vertex adjacent to both vertices.

First, assume that , so there exists a single gate vertex . Furthermore, are both AND vertices. Since both are valid configurations, in both configurations the blue edges incident on are oriented towards these two vertices. As a result has no neighbor in .

Second, suppose and one of is a COPY vertex. If is incident on an AND vertex, because both are valid and agree on all edges except we have that both red edges incident on the AND vertex are oriented towards it in both configurations. Similarly, the second blue edge incident on the COPY endpoint of is oriented towards it in both configurations. We therefore observe that neither , nor has a neighbor in except , so we can safely slide .

Similarly, for the last case, suppose that and one of the endpoints of is an OR vertex, while the other is a COPY vertex. Again, because are both valid and only disagree on , at least one of the blue edges incident on the OR vertex (other than ) is oriented towards it in both configurations. As before, the second blue edge incident on the COPY vertex is oriented towards it in both configurations. Therefore, one of has no neighbor in except , so we can safely slide the token from to with two moves.

To complete the proof, we need to show that if we have a valid token sliding reconfiguration sequence, this gives a valid reorientation sequence for . The main observation now is that in a shortest token sliding solution that obeys the properties of the lemma, a token that slides out of must necessarily in the next move slide into , where . To see this, observe that because of the requirement that the set does not contain both selector vertices of any edge, the tokens found on other selector vertices dominate all gate vertices except those corresponding to . Since we can neither repeat configurations, nor add a second token to the clique made up of gate vertices, the next move must slide the token to the other selector vertex.

To see that the orientation sequence obtained through the natural translation of main configurations is valid, consider two consecutive main configurations in the token sliding solution, such that the corresponding orientations are , and is valid. We will show that is also valid. Suppose that differs from in the edge which is oriented towards in (it is not hard to see that cannot differ in more than one edge). Thus, is transformable in two moves to by sliding to a gate corresponding to and then to . If is a red edge, this means that in both blue edges incident on are directed towards , so the reorientation is valid. If is blue, we first assume that is a COPY vertex. Since a gate corresponding to is free, the other blue edge incident on is oriented towards in and we have a valid move. Finally, if is blue and is an OR vertex, we conclude that, since at least one gate from is available in , at least one of the two other blue edges incident on is directed towards in and we have a valid move. ∎

Second Step: Enforcing Consistency

We will now construct a graph that will function in a way similar to the graph we have already constructed but in a way that enforces consistency. Let be the graph constructed in the first step of our reduction, and let be the set of gate edges, that is, the set of edges that connect the selector vertices for an edge to the corresponding gate(s).

Let and . We first take disjoint copies of and for a vertex we will use the notation , where to denote the vertex corresponding to in the -th copy. Then, for every edge (every non-gate edge) and for all with , we add the edge . This completes the construction of and it is not hard to see that the graph is split, as the copies of the clique of form a larger clique. To complete our instance let us explain how to translate an independent set of that contains no vertices of the clique to an independent set of : we do this in the natural way by including in the new independent set all copies of vertices of the original independent set. Since both the initial and final independent sets in our first construction use no vertices in the clique, we have in this way two independent sets of size in the new graph, and thus a valid Token Sliding instance. Let be the two independent sets of we are asked to transform and the corresponding independent sets of .

We first show that if we have a solution for reconfiguration in then we have a solution for reconfiguring the sets in the new graph.

Let be two independent sets of of size that use no vertices of the clique, respect the conditions of Lemma 3, and can be transformed to one another by two sliding moves. Then the independent sets which are obtained in by including all copies of vertices of respectively can be transformed into one another by a sequence of TS moves.

Proof.

Each of uses exactly one of the vertices , for each edge , because of their size, the fact that they contain no vertex of the clique, and the fact that neither contains both for any edge (this is the condition of Lemma 3). If can be transformed into with two sliding moves, the first move takes a token from an independent set vertex, say and moves it to the clique and the second moves the same token to . Since contains a token on each pair of selector vertices, the only vertex of the clique on which the token can be moved is a gate vertex corresponding to , say (if is red) or (if is blue). We now observe that if (or similarly ) is available in (that is, it has no neighbors in besides ), then the same is true for for all in . To see this, note that the neighbors of are, , and, for each all the vertices for . Since none of the neighbors of is in , is available. We therefore slide, one by one, a token from to and then to , for all . ∎

Now, for the more involved direction of the reduction we first observe that it is impossible for a reconfiguration to arrive at a situation where the solution is highly irregular, in the sense that, for an edge we have multiple tokens on copies of both and .

Let be the initial independent set constructed in our instance and be an independent set which for some and for some with has . Then is not reachable with TS moves from .

Proof.

Let be an independent set that satisfies the conditions of the lemma but is reachable from with the minimum number of token sliding moves. Consider a sequence that transforms to , and let be the independent set immediately before in this sequence. contains exactly three of the vertices . Without loss of generality say . Therefore, the move that transforms to slides a token into from one of the neighbors of this vertex. We now observe that contains copies of each neighbor of in , plus the gate vertices corresponding to in the -th copy of . However, the copies of the neighbors of are also neighbors of , hence a token cannot slide through these vertices. Furthermore, the gate vertices of are also neighbors of . We therefore have a contradiction. ∎

We now use Lemma 3 to show that for each original edge, the graph contains some non-trivial number of tokens on the selector vertices of that edge.

Let be the initial independent set constructed in our instance and be an independent set which for some has . Then is unreachable from .

Proof.

Suppose is reachable. Then by Lemma 3, for each edge we have , because otherwise there would exist (by pigeonhole principle) . We now use a simple counting argument. The total number of tokens is , while for any edge we have . However, , where we use the fact that . As a result for any edge , as the independent set uses at most one vertex from the clique. ∎

We are now ready to establish the final lemma that gives a mapping from a sliding token reconfiguration in to one in .

If there exists a reconfiguration from to in under the TS rule then there exists a reconfiguration from to in under the TS rule which for each edge contains at most one of the vertices in every independent set in the sequence.

Proof.

Take a configuration of , that is an independent set in the supposed sequence from to . We map this independent set to an independent set of as follows: for each edge , we set if and only if . Informally, this means that we take the majority setting from . We note that this always gives an independent set that contains exactly one vertex from for each .

Our main argument now is to show that if are two consecutive independent sets of the solution for , then the sets which are obtained in the way described above in are either identical or can be obtained from one another with two sliding moves. If are not identical, they may differ in at most two vertices corresponding to an edge , say and . This is not hard to see, since is obtained from with one sliding move, and this move can only affect the majority opinion for at most one edge.

Now we would like to argue that it is possible to slide to a gate vertex associated to and then to in . Consider the transition from to . This move either slides a token from some to the clique, or slides a token from the clique to some (because the majority opinion changed from to ). Because of Lemma 3, both and contain at least four vertices in some copies of . Hence, since at least half of these vertices are in copies of in , there exists some . Similarly, there exists some . Consider now a gate vertex in the clique of such that is not associated with . If has an edge to in , then all copies of in have an edge to , therefore cannot belong in either set. As a result, the clique vertex that is used in the transition from to is a copy of a gate vertex associated with (either , or one of , depending on the color of ). This gate vertex copy therefore has no neighbor in . From this we conclude that the same gate vertex in also has no neighbor in , as the majority opinion only changed for . It is therefore legal to slide from to this gate vertex and then to . ∎

Sliding Token Reconfiguration is PSPACE-complete for split graphs.

Proof.

We begin with an instance of the PSPACE-complete NCL reconfiguration problem, as given in Lemma 3. We construct the instance of Sliding Token Reconfiguration on split graphs as described (it’s clear that this can be done in polynomial time). If the NCL reconfiguration instance is a YES instance, then by Lemma 3 there exists a sliding token reconfiguration of , and by repeated applications of Lemma 3 to independent sets that do not contain clique vertices in the reconfiguration of there exists a sliding token reconfiguration of . If on the other hand there exists a sliding token reconfiguration on , then by Lemma 3 there exists a reconfiguration that satisfies the condition of Lemma 3 on , hence the original NCL instance is a YES instance. ∎

4 PSPACE-completeness for Chordal Graphs for

In this section, we build upon the PSPACE-completeness result from Section 3 to show that -Colorable Set Reconfiguration is PSPACE-complete, for every , when the input graph is restricted to be chordal.

For every , the -Colorable Set Reconfiguration problem under the TS rule is PSPACE-complete, even when the input graph is restricted to be chordal.

Proof.

We provide a reduction from Independent Set Reconfiguration where the input graph is restricted to be a split graph, which we proved to be PSPACE-complete in Theorem 3. Let be an input split graph for Independent Set Reconfiguration. We construct a chordal graph as follows, starting from a graph isomorphic to and two non-empty independents set of the same size. For every edge , we add sets of new vertices , such that induces a clique for every , and every vertex of is made adjacent to both and , for every . In addition, we create a new set and a set . In other words, we append disjoint cliques of size to every edge of , and add all those newly created vertices to and to . The chordality of follows from the fact that the new vertices of the sets are all simplicial in , hence is chordal if and only if is chordal as well (and is split).

We now claim the following: given in independent set of , the instance of Independent Set Reconfiguration is a YES-instance if and only if the instance of -Colorable Set Reconfiguration is a YES-instance as well. Observe that, by the construction, and are -colorable because the maximum clique in contains at most one vertex of and at most the vertices of a clique .

The forward direction of the previous claim follows easily: performing the same moves as those of a reconfiguration sequence from to in , starting from , yields a reconfiguration sequence where every step preserves -colorability, and produces the desired set .

For the backwards direction, we claim that, for any -colorable set reachable from , it holds that the vertices of are pairwise non-adjacent. In other words, the tokens placed on original vertices of form an independent set. Indeed, observe that the number of vertices of that do not belong to satisfies . This immediately implies that for any set and edge , we have , and therefore contains a clique of size as an induced subgraph, i.e., one of the sets is completely contained in . This implies that, for every edge of , we have , i.e., the vertices of are pairwise non-adjacent, as desired. ∎

5 XP-time Algorithm on Split Graphs for fixed

In this section we present an algorithm for -Colorable Reconfiguration under the TS rule, on split graphs, for . Recall that a split graph is a graph whose vertex set is partitioned into a clique and an independent set . An input instance consists of a split graph , and two -colorable sets .

Before proceeding, let us give some high-level ideas as well as some intuition why this problem, which is PSPACE-complete for (Theorem 3), admits such an algorithm for larger . Our algorithm consists of two parts: a rigid and a non-rigid reconfiguration part. In the rigid reconfiguration part the algorithm decides if two sets are reachable by using moves that never slide tokens into or out of . Because of this restriction and the fact that the sets are -colorable, the total number of possible configurations is , so this part can be solved with exhaustive search (this is similar to the algorithm of [15] for TJ/TAR). In the non-rigid part we assume we are given two sets which, in addition to being -colorable, have . The main insight is now that any two such sets are reachable via TS moves (Lemma 5 below). Informally, the algorithm guesses a partition of the optimal reconfiguration into a rigid prefix, a rigid suffix, and a non-rigid middle, and uses the two parts to calculate each independently.

The intuitive reason that our algorithm cannot work for is the non-rigid part. The crucial Lemma 5 on which this part is based fails for : for instance, if is a star with three leaves and are two distinct sets each containing two leaves, then satisfy all the conditions for , but are not reachable from each other with TS moves. Such counterexamples do not, however, exist for higher , because for sets that satisfy the conditions of Lemma 5 we know we can always freely move tokens around inside the clique (and without loss of generality, such tokens exist). Note also, that this difficulty is specific to the TS rule: the algorithm of [15] implicitly uses the fact that any two sets with tokens in the clique are always reachable, as this is an almost trivial fact if one is allowed to use TJ moves. Thus, Lemma 5 is the main new ingredient that makes our algorithm work.

Let us now proceed with a detailed description of the algorithm. First, let us fix some notation. For a vertex set , we write the subsets and as and respectively. Throughout this section, we assume that input graph is connected (and thus each vertex in has a neighbor in ); otherwise we can consider instances induced by each component separately.

Let be a split graph, , and be two -colorable sets such that . Then is -reachable from . Furthermore, a reconfiguration sequence from to can be produced in polynomial time.

Proof.

We first observe that if , then there is an easy optimal -transformation. By making one TS move from to , one can -transform to with sliding moves (thus yielding an optimal reconfiguration sequence). It is clear that all the sets resulting from these TS moves are -colorable because each of them has at most vertices in .

Therefore, it suffices to show that there is always a -transformation of which decrease as long as . Note that we can assume that there exists (otherwise we exchange the roles of and ). In the case when , one can transform to with TS moves from a vertex of to . Trivially this is a -transformation, and it holds that . (Note that this argument would not be valid if ). If , then one can make at most two TS moves from a vertex of to . Because has at most vertices and these TS moves maintain at most vertices in , -colorability of is preserved. Moreover, the new set has at most vertices in while its intersection with in is strictly larger. This completes the proof of the first statement. The proof is constructive and easily translates to a polynomial-time algorithm. ∎

Let us now introduce a notion that will be useful in our algorithm. For two -colorable sets with we say that has a rigid -transformation to if there exists a valid -transformation from to with TS moves which also has the property that every -colorable set of the transformation has .

Given a split graph , with , and two -colorable sets with , there is an algorithm that decides if there exists a rigid -transformation of to in time .

Proof.

The main observation is that since all intermediate sets must have , we are only allowed to slide tokens inside . However, contains at most vertices (as it is -colorable), therefore, there are at most potentially reachable sets: one for each collection of vertices of the clique.

We now construct a secondary graph with a node for each subset of that contains vertices of and the vertices of , and connect two such nodes if their corresponding sets are reachable with a single TS move in . In this graph we check if there is a path from the node that represents to the one that represents and if yes output the sets corresponding to the nodes of the path as our rigid reconfiguration sequence. ∎

There is an algorithm that decides -Colorable Reconfiguration on split graphs under the TS rule in time , for .

Proof.

We distinguish the following cases: (i) , (ii) and , (iii) . This covers all cases since are -colorable and we can assume without loss of generality that .

For case (i) we invoke Lemma 5. The answer is always Yes, and the algorithm of the lemma produces a feasible reconfiguration sequence.

For case (ii), suppose there exists a reconfiguration sequence from to , call it . Let be the smallest index such that . Clearly such an index exists, since . We now guess the configuration and the configuration (that is, we branch into all possibilities). Observe that there are at most choices for as we have and . Furthermore, once we have selected a , there are possibilities for , as is reachable from with one TS move.

We observe that if we guessed correctly, then there exists a rigid -transformation from to (by the minimality of and the fact that ); we use the algorithm of Lemma 5 to check this. Furthermore, the configuration is always transformable to by Lemma 5. Therefore, if the algorithm of Lemma 5 returns a solution, then we have a -transformation from to . Conversely, if a -transformation from to exists, since we tried all possibilities for , one of the branches will find it.

Finally, for case (iii), if we first use Lemma 5 to check if there is a rigid -transformation from to . If one is found, we are done. If not, or if we observe that, similarly to case (ii), in any feasible transformation , there exists an such that (otherwise the transformation would be rigid). Pick the minimum such . We now guess the configurations (as before, there are possibilities) and use Lemma 5 to verify that is reachable from . If is reachable from , we need to verify that is reachable from . However, we observe that this reduces to case (ii), because , so we proceed as above. If the algorithm returns a valid sequence we accept, while we know that if a valid sequence exists, then there exists a correct guess for that we consider. ∎

6 W-hardness for Split Graphs

In this section we show that -Colorable Reconfiguration on split graphs is W[2]-hard parameterized by and the length of the reconfiguration sequence under all three reconfiguration rules (TAR, TJ, and TS). In this sense, this section complements Section 5 by showing that the algorithm that we presented for -Colorable Reconfiguration on split graphs cannot be significantly improved under standard assumptions.

We will rely on known results on the hardness of Dominating Set Reconfiguration. We recall that in this problem we are given a graph , two dominating sets of size at most and are asked if we can transform into by a series of TAR operations while keeping the size of the current set at most at all times. More formally, we are asked if there exists a sequence such that for each , , is a dominating set of , and .

[[18]] Dominating Set Reconfiguration is W[2]-hard parameterized by the maximum size of the allowed dominating sets and the length of the reconfiguration sequence under the TAR rule.

Before proceeding, let us make two remarks on Theorem 6: first, because the reduction of [18] is linear in the parameters, it is not hard to see that it also implies a tight ETH-based lower bound based on known results for Dominating Set; second, using an argument similar to that of Theorem 1 of [16], the same hardness can be obtained for the TJ rule.

Dominating Set Reconfiguration is W[2]-hard parameterized by the maximum size of the allowed dominating sets and the length of the reconfiguration sequence under the TAR, or TJ rule. Furthermore, the pro