1 Introduction
Recently, the reconfiguration framework has been applied to several search problems. In a reconfiguration problem, we are given two feasible solutions of a search problem and are asked to determine whether we can modify one to the other by repeatedly applying prescribed reconfiguration rules while keeping the feasibility (see [17, 27, 24]). Studying such a problem is important for understanding the structure of the solution space of the underlying problem. Computational complexity of reconfiguration problems has been studied intensively. For example, the Independent Set Reconfiguration problem under the reconfiguration rules [16], [17], and [18] is studied for several graph classes such as planar graphs [16], perfect graphs [18], clawfree graphs [5], trees [8], interval graphs [4], and bipartite graphs [20].
In this paper, we initiate the study on problems of reconfiguring colorable sets, which generalizes Independent Set Reconfiguration. For a graph and an integer , a vertex set is colorable if the subgraph induced by admits a proper coloring. For example, the colorable sets in a graph are exactly the independent sets of the graph. Recently, colorable sets have been studied from the viewpoint of wireless network optimization (see [2, 3] and the references therein). The Colorable Set Reconfiguration problem asks given two colorable sets and in a graph , whether we can reach from to by repeatedly applying local changes allowed. We consider the following three local change operations (see Section 2 for formal definitions):

: either adding or removing one vertex while keeping the size of the set at least a given threshold .

: swap one member for one nonmember.

: swap one member for one nonmember along an edge.
In perfect graphs, being colorable is equivalent to having no clique of size more than . This property often makes problems related to coloring tractable. Thus, to understand this very general problem, we start the study on Colorable Set Reconfiguration with classes of perfect graphs. Figure 1 shows the graph classes studied in this paper and the inclusion relationships (see Section 2.2 for definitions).
Our contribution
Before we start our investigation on the reconfiguration problem, we first fill a gap in the complexity landscape of the search problem Colorable Set that asks for finding a large colorable set. When , Colorable Set is equivalent to the classical problem of finding a large independent set that can be solved in polynomial time for perfect graphs. For larger , it was only known that the case is NPcomplete for perfect graphs [1]. To make the complexity status of Colorable Set for perfect graphs complete, we show that it is NPcomplete for any fixed (Theorem 3.1).
We then show complexity divergences among the classes of perfect graphs in Figure 1, in particular under and . See Table 1 for a summary of our results. Our results basically say that the problem under and is tractable on interval graphs but further generalization is not quite possible.
Colorable Set Reconfiguration under  

fixed  arbitrary  
perfect  PSPACEc  
cocomparability  PSPACEc [18]^{2}^{2}2The reduction in [18] outputs cocomparability graphs. See also Theorem 6.1 in this paper.  PSPACEc (Thm 6.1)  
chordal  P [18]  ?  PSPACEc 
split  P  P (Thm 5.4)  PSPACEc (Thm 5.5) 
interval  P  P (Thm 4.11)  
bipartite  NPc [20]  Trivial if 
More specifically, we first study the problem on interval graphs and show that a shortest reconfiguration sequence under can be found in linear time (Theorem 4.11). This implies the same result under . Next we study the problem on split graphs. We show that the complexity depends on . When is a fixed constant, the problem is polynomialtime solvable under and (Theorem 5.4). If is a part of input, then we can show that the problem is PSPACEcomplete under all rules, including (Theorem 5.5). While the hardness result applies also to chordal graphs, it is unclear whether a similar positive result for chordal graphs can be obtained when is a fixed constant. We only know that the case of under and is polynomialtime solvable as chordal graphs are evenholefree [18]. We finally show that for every fixed the problem is PSPACEcomplete for cocomparability graphs under all rules (Theorem 6.1). Thus, our results are in some sense tight since the interval graphs are exactly the chordal cocomparability graphs and split graphs are chordal graphs (see Figure 1).
As a byproduct of Theorems 4.11 and 5.4, the Feedback Vertex Set Reconfiguration problem [23] turns out to be polynomialtime solvable for split graphs and interval graphs under and . These are the first polynomialtime solvable cases for Feedback Vertex Set Reconfiguration. To see the polynomialtime solvability, observe that the complements of colorable sets in a chordal graph are exactly the feedback vertex sets in the graph^{3}^{3}3
Each induced cycle in a chordal graph is a triangle, and thus 2colorable (or equivalently, odd cycle free) chordal graphs are forests.
and reconfigurations of the complements are equivalent to reconfigurations of the original vertex sets under and .2 Preliminaries
We say, as usual, that an algorithm for a graph runs in linear time if the running time of the algorithm is .
A proper coloring of a graph assigns a color from to each vertex in such a way that adjacent vertices have different colors. Given a graph and an integer , Graph Coloring asks whether admits a proper coloring. This problem is NPcomplete even if is fixed to 3 [12]. The minimum such that a graph admits a proper coloring is its chromatic number.
The Colorable Set problem is a generalization of Graph Coloring where we find a large induced subgraph of the input graph that admits a proper coloring. Let be a graph. For a set of vertices , we denote by the subgraph induced by . A vertex set is colorable in if has a proper coloring. Now the problem is defined as follows:

Problem: Colorable Set

Input: A graph and integers and .

Question: Does have a colorable set of size at least ?
The problem of finding a large colorable set is studied for a few important classes of perfect graphs (see Figure 1 and Table 1). For the class of perfect graphs, it is known that a maximum colorable set (that is, a maximum independent set) can be found in polynomial time [15]. Parameterized complexity [19] and approximation [9] of Colorable Set on perfect graphs are also studied.
2.1 Reconfiguration of colorable sets
Let and be colorable sets in a graph . Then, under for a nonnegative integer if and , where denotes the symmetric difference . Here means that and can be reconfigured to each other in one step and stands for “token addition & removal.” A sequence of colorable sets in is a reconfiguration sequence of length between and under if holds under for all . A reconfiguration sequence under is simply called a sequence. We write under if there exists a sequence between and . Note that every reconfiguration sequence is reversible, that is, if and only if . Now the problem we are going to consider is formalized as follows:

Problem: Colorable Set Reconfiguration under ( for short)

Input: A graph , integers and , and colorable sets and of .

Question: Does under hold?
We denote by an instance of . We assume that both and hold; otherwise it is trivially a noinstance. Note that the lower bound guarantees that none of the sets in the reconfiguration sequence is too small. Without the lower bound, the reachability problem becomes trivial as can always reach via .
For a instance , we denote by the length of a shortest sequence in between and ; if there is no such a sequence, then we set .
We note that is a decision problem and hence does not require the specification of an actual sequence. Similarly, the shortest variant of simply requires to output the value of .
Other reconfiguration rules
Although the rule is our main target, we also study two other wellknown rules (token jumping) and (token sliding). Let and be colorable sets in a graph . For and , we additionally assume that because these rules do not change the size of a set. Now the rules are defined as follows:

under if ;

under if and the two vertices in are adjacent in .
Reconfiguration sequences under and as well as the reconfiguration problems and are defined analogously. An instance of or is represented as , and and are defined in the same way.
The following relation can be shown in almost the same way as Theorem 1 in [18] and means that is not harder than in the sense of Karp reductions.
Lemma 2.1.
Let and be colorable sets of size in a graph . Then, under if and only if under . Furthermore, holds.
To make the presentation easier, we often use the shorthands for and for . For a vertex of a graph , we denote the neighborhood of in by .
2.2 Graph classes
A clique in a graph is a set of pairwise adjacent vertices. A graph is perfect if the chromatic number equals the maximum clique size for every induced subgraph [14]. The following fact follows directly from the definition of perfect graphs and will be used throughout this paper.
Proposition 2.2.
A vertex set of a perfect graph is colorable if and only if has no clique of size more than .
There are many graph classes of perfect graphs. Chordal graphs form one of the most wellknown subclasses of perfect graphs, where a graph is chordal if it contains no induced cycle of length greater than 3.
Cocomparability graphs form another large class of perfect graphs. A graph is a cocomparability graph if there is a linear ordering on such that and imply or . Although they are less known than chordal graphs, cocomparability graphs generalize several important graph classes such as interval graphs, permutation graphs, trapezoid graphs, and cobipartite graphs (see [14, 25]).
The classes of chordal graphs and cocomparability graphs are incomparable.^{4}^{4}4A cycle of four vertices is a cocomparability graph but not chordal. The net graph obtained by attaching a pendant vertex to each vertex of a triangle is chordal but not a cocomparability graph. It is known that the class of interval graphs characterizes their intersection; namely, a graph is an interval graph if and only if it is a cocomparability graph and chordal [13]. Recall that a graph is an interval graph if it is the intersection graph of closed intervals on the real line.
Another wellstudied subclass of chordal graphs (and hence of perfect graphs) is the class of split graphs. A graph is a split graph if can be partitioned into a clique and an independent set . To emphasize that is a split graph, we write . The classes of interval graphs and split graphs are incomparable.^{5}^{5}5A path with five or more vertices is an interval graph but not a split graph. The net graph is a split graph but not an interval graph.
3 NPhardness of Colorable Set on perfect graphs for fixed
It is known that if is unbounded, Colorable Set is polynomialtime solvable for interval graphs [28, 22] and more generally for cocomparability graphs [10], while it is NPcomplete for split graphs (and thus for chordal graphs) [28, 7]. On the other hand, if is a fixed constant, Colorable Set is polynomialtime solvable even for chordal graphs [28, 7].
For perfect graphs, the case of is solvable in polynomial time [15], while the case of is NPcomplete [1]. Here we demonstrate that the problem is hard for any fixed .
Theorem 3.1.
Colorable Set is NPcomplete on perfect graphs for every fixed .
In [1], the problem actually studied was the dual of our problem. An odd cycle is a cycle of odd length. An odd cycle transversal of a graph is a set of vertices that intersects every cycle of odd length in . In other words, is an odd cycle transversal if and only if is bipartite. They study the following problem of finding a small odd cycle transversal:

Problem: Odd Cycle Transversal (OCT)

Input: A graph and an integer .

Question: Does have an odd cycle transversal of size at most ?
Proposition 3.2 ([1]).
OCT is NPcomplete for perfect graphs.
The join of two disjoint graphs and is the graph . That is, is obtained from the disjoint union of and by adding all possible edges between and .
Lemma 3.3.
The class of perfect graphs is closed under join. That is, if two disjoint graphs are perfect, then so is their join.
Proof.
An odd hole is an induced odd cycle of length at least 5. Let and be disjoint perfect graphs. By the strong perfect graph theorem [6], it suffices to show that none of and its complement contains an odd hole.
Suppose to the contrary that contains an odd hole . Since and are perfect, intersects both and . Moreover, as , one of and has at least three vertices in . Therefore, contains a vertex of degree at least 3. This contradicts that is an induced cycle and .
Observe that is the disjoint union of the complements of and of . By the (weak) perfect graph theorem [21], and are perfect and thus have no odd hole. Hence, has no odd hole. ∎
Now we are ready for proving the main claim of this section.
Proof of Theorem 3.1.
Let be an instance of OCT for perfect graphs. Let be the disjoint union of cliques of size . Let . By Lemma 3.3, is perfect (as is clearly perfect). Now it suffices to show that for any , has an odd cycle transversal of size if and only if has a colorable set of size .
To show the onlyif part, let be an odd cycle transversal of size in . Let . Clearly, . Since contains no clique of size 3, the maximum clique size of is at most . Since is perfect, is colorable.
To show the if part, let be a colorable set of size in . Observe that contains at least one entire clique of size in since otherwise . Let . Suppose to the contrary that is not bipartite and thus contains a clique of size 3. By the definition, . Thus, contains the clique of size . This contradicts the colorability of . ∎
4 Shortest reconfiguration in interval graphs
In this section, we show that for interval graphs can be solved in linear time. Our result is actually stronger and says that an actual shortest sequence can be found in linear time, if any. By Lemma 2.1, the same result is obtained for . We first give a characterization of the distance between two colorable sets in an interval graph (Section 4.1). This characterization says that a shortest sequence has length linear in the number of vertices of the graph. We then show that the distance can be computed in linear time (Section 4.2). We finally present a lineartime algorithm for finding a shortest sequence (Section 4.3).
It is known that a graph is an interval graph if and only if its maximal cliques can be ordered so that each vertex appears consecutively in that ordering [13, 11]. We call a list of the maximal cliques ordered in such a way a clique path. Let be an interval graph and be a clique path of ; that is, for each vertex , there are indices and such that if and only if . Given an interval graph, a clique path and the indices and for all vertices can be computed in linear time [26]. Hence we can assume that we are additionally given such information. Note that is an interval representation of . Namely, if and only .
Let be a clique in an interval graph . By the Helly property of intervals, the intersection of all intervals in is nonempty; that is, (see [25]). A point in the intersection is a clique point of .
4.1 The distance between colorable sets
Let be an instance of . The set is locked in if is a maximal colorable set in and . The following lemma follows immediately from the definition.
Lemma 4.1.
Let be a graph, and let and be distinct colorable sets of size at least in . If or is locked in , then .
Proof.
Assume without loss of generality that is locked in . If there is a colorable set in such that , then as . This contradicts the maximality of . Since , we can conclude that . ∎
The rest of this subsection is dedicated to a proof of the following theorem, which implies that the converse of the lemma above also holds for interval graphs.
Theorem 4.2.
Let be an interval graph, and let and be distinct colorable sets of size at least in . If and are not locked in , then the distance is determined as follows.

If and are not locked in , then .

If exactly one of and is locked in , then .

If and are locked in , then we have the following two cases.

If there is such that both and are colorable in , then .

Otherwise, .

Corollary 4.3.
For , if and only if none of and is locked in .
Observe that for any pair of colorable sets and in . We use this fact implicitly in the following arguments.
Lemma 4.4 (Theorem 4.2 (1)).
Let be an interval graph, and let and be colorable sets of size at least in . If and are not locked in , then .
Proof.
We proceed by induction on . The base case of is trivial. Assume that and that the statement is true if the symmetric difference is smaller.
We first consider the case where . Since is not locked in , is not maximal in . Thus there is a vertex such that is colorable. The set is not locked in , , and . By the induction hypothesis, . Hence, we have . If , we can apply the same argument.
In the following, we assume that and . If , then we can add the elements of onebyone in an arbitrary order to get a shortest reconfiguration sequence of length . The case where is the same.
We now consider the case where and . Let and be vertices with the smallest rightend in each set. That is, and . By symmetry, assume that . Let be a vertex that minimizes . (Note that and may be the same.) Now we have . We set and . Clearly, . To apply induction hypothesis, it suffices to show that is not locked in . To this end, we prove that . Suppose to the contrary that is not colorable; that is, contains a clique of size . Since does not contain such a large clique, must include . Let be a clique point of . If , then includes no vertex in as has the minimum in . This contradicts the colorability of and thus . This implies that is a clique of size , a contradiction. Therefore, we can conclude that is colorable. Now, by the induction hypothesis, , and thus . ∎
Lemma 4.5 (Theorem 4.2 (2)).
Let be an interval graph, and let and be distinct colorable sets of size at least in . If and are not locked in , and exactly one of and is locked in , then .
Proof.
Without loss of generality, assume that is locked in . This implies that . Since is not locked in , is not maximal in . Hence, there is a vertex such that is a colorable set of . Observe that and are not locked in . Thus, by Theorem 4.2 (1), it holds that .
On the other hand, since is locked in , every colorable set of with contains a vertex in . Thus holds. This implies that . ∎
Lemma 4.6 (Theorem 4.2 (3a)).
Let be an interval graph, and let and be distinct colorable sets of size at least in . Assume and are locked in but not in . If there is a vertex such that both and are colorable in , then .
Proof.
Let be a vertex such that both and are colorable in . We have and . Since and are not locked in , Theorem 4.2 (1) implies that .
The lower bound can be shown in exactly the same way as in the proof of Lemma 4.5. ∎
4.2 Computing the distance in linear time
We here explain how to check which case of Theorem 4.2 applies to a given instance in linear time.
Lemma 4.8.
Given an interval graph and colorable sets and in , one can either find a vertex such that and are colorable or decide that no such vertex exists in linear time.
Proof.
Let be a clique path of . Recall that are the maximal cliques of . Thus, for every , the maximum clique size of is equal to .
We compute for as follows. Initialize all to ; for each , add to all with . In the same way, we compute for . From the observation above, we can conclude that for each vertex , and are colorable if and only if for .
The initialization and the test for all nonmembers of can be done in time . It suffices to show that . Since , there is a vertex with . Thus . By induction on the number of vertices, our claim holds. ∎
By setting in the lemma above, we have the following lemma.
Lemma 4.9.
Given an interval graph and a colorable set in , one can either find a vertex such that is colorable or decide that is maximal in linear time.
Corollary 4.10.
Given an interval graph and colorable sets and in , the distance can be computed in linear time.
Proof.
We first check whether or is locked in . If so, the distance is . Otherwise, we check whether and are locked in . If not both of them are locked in , then we can apply Theorem 4.2 (1) or (2) and determine the distance. If both and are locked in , we find a vertex such that both and are colorable in . Everything can be done in linear time by Lemmas 4.8 and 4.9. ∎
4.3 Finding a shortest reconfiguration sequence in linear time
Here we describe how we find an actual shortest reconfiguration sequence in linear time. To this end, we need to be careful about the representation of a reconfiguration sequence. If we always output the whole set, the total running time cannot be smaller than . However, this product can be quadratic. To avoid this blow up, we output only the difference from the previous set. That is, if the current set is and the next set is (), we output (, resp.). We also fully use the reversible property of reconfiguration sequences and output them sometimes from left to right and sometimes from right to left. For example, we may output a reconfiguration sequence as first , next , then . It is straightforward to output the sequence from left to right by using a linearsize buffer.
Theorem 4.11.
Given an interval graph and colorable sets and in , a sequence of length can be computed in linear time.
Proof.
We first test which case of Theorem 4.2 applies to the given instance. This can be done in linear time as shown in the proof of Corollary 4.10. We reduce Cases (2) and (3) to Case (1). The reductions below can be done in linear time by using Lemmas 4.8 and 4.9.
Assume first that Case (2) applies; that is, is locked but is not in . We find a vertex such that is a colorable set of . We then add to . As we saw in the proof of Lemma 4.5, this is a valid step in a shortest reconfiguration sequence. Furthermore, after this step, and are not locked in .
Next assume that Case (3) applies; that is, both and are locked in . We find vertices such that and are colorable in . In Case (3a), we further ask that . We then add to and to . The proofs of Lemmas 4.6 and 4.7 imply that these are valid steps in a shortest reconfiguration sequence, and that and are no longer locked in after these steps.
We now handle Case (1), where and are not locked in . Assume that and since otherwise finding a shortest sequence is trivial. We first compute two orderings of the vertices in : nondecreasing orderings of leftends and of rightends . Such orderings can be constructed in linear time from a clique path. We maintain information for each vertex whether , , or . Using this information, we can also maintain vertices of the smallest leftend and of the smallest rightend in each of and .
Let and be vertices with the smallest rightend in each set. By symmetry, assume that . Let be a vertex that minimizes . As shown in the proof of Lemma 4.4, under and . We output the two steps and .
We then set and update the information as anymore. We also have to maintain the vertices of the smallest left and rightends in each and . Let be a vertex with the smallest rightend. The vertex can be found by sweeping the nondecreasing ordering of the rightends from the position of to the right. The vertex with the smallest leftend can be found in an analogous way. Although a single update can take super constant steps, it sums up to a linear number of steps in total since it can be seen as a single lefttoright scan of each nondecreasing ordering. Therefore, the total running time is linear. ∎
5 Split graphs
For split graphs, we consider two cases. In the first case, we assume that is a fixed constant, and show that the problem under (and ) can be solved in time. The second case is the general problem having as a part of input. We show that in this case the problem is PSPACEcomplete under all reconfiguration rules.
5.1 Polynomialtime algorithm for fixed
Let be a split graph, where is a clique and is an independent set. For with , we define as follows:
We can see that is colorable for every with as follows. Every clique includes at most vertices in and at most one vertex in . Since a vertex in has fewer than neighbors in , the maximum clique size of is at most .
Lemma 5.1.
If is a colorable set of with , then under .
Proof.
Note that is colorable since is colorable and thus . We now show that , which implies .
If , then , and thus . If , then since is colorable. Thus it holds that . ∎
By the reversibility of reconfiguration sequences, we can reduce the problem as follows.
Corollary 5.2.
If and are colorable sets of with and , then under if and only if under .
Now we state the crucial lemma for solving the reduced problem.
Lemma 5.3.
Let and . If and are colorable sets of size at least , then under if and only if .
Proof.
To prove the if part, assume that . Then, . Since , it holds that by Lemma 5.1. Thus we have .
To prove the onlyif part, assume that . If , then is a maximal colorable set and no other colorable set of size at least can be reached from . Assume that , and hence . Then, and . Therefore, we have , a contradiction. ∎
Combining the arguments in this subsection, we are now ready to present a polynomialtime algorithm.
Theorem 5.4.
Given an vertex split graph and colorable sets and of size at least in , it can be decided whether under in time .
Proof.
We construct a graph from as follows:
For each with , the size can be computed in constant time (assuming that we know the size in advance). If