Recognizing equistable graphs in FPT time^{†}^{†}thanks: This research is supported in part by “Agencija za raziskovalno dejavnost Republike Slovenije”, research program P– and research projects J, J, J, and BIFR/––Proteus–.
Abstract
A graph is called equistable if there exist a positive integer and a weight function such that is a maximal stable set of if and only if . Such a function is called an equistable function of . For a positive integer , a graph is said to be equistable if it admits an equistable function which is bounded by .
We prove that the problem of recognizing equistable graphs is fixed parameter tractable when parameterized by , affirmatively answering a question of Levit et al. In fact, the problem admits an vertex kernel that can be computed in linear time.
Keywords: equistable graphs, recognition algorithm, fixed parameter tractability.
1 Introduction
The main notion studied in this paper is the class of equistable graphs, introduced by Payan in 1980 [17] as a generalization of the well known and well studied class of threshold graphs [1, 9]. A stable (or independent) set in a (finite, simple, undirected) graph is a set of pairwise nonadjacent vertices. A maximal stable set is a stable set not contained in any other stable set. A graph is said to be equistable if there exists a function such that for every , set is a maximal stable set of if and only if . Equivalently, is equistable if and only if there exist a positive integer and a weight function such that is a maximal stable set of if and only if . Such a function is called an equistable function of , while the pair is called an equistable structure. Equistable graphs were studied in a series of papers [8, 13, 6, 7, 5, 18, 14, 15, 10, 17, 16]; besides threshold graphs and cographs, they also generalize the class of general partition graphs [11, 4, 13]. The complexity status of recognizing equistable graphs is open, and no combinatorial characterization of equistable graphs is known.
Levit et al. introduced in [8] the notion of equistable graphs. For a positive integer , a graph is said to be equistable if it admits an equistable function . Such a weight function is called a equistable function, and the corresponding structure is a equistable structure. We remark that there exist equistable graphs such that the smallest for which the graph is equistable is exponential in the number of vertices of [14].
For a positive integer , an equistable graph is said to be target equistable if it admits an equistable function with equistable structure . Clearly, every target equistable graph is also equistable (but not vice versa).
As mentioned above, the complexity of recognizing equistable graphs is open, but it seems plausible that the problem could be NPhard. It thus makes sense to search ways to simplify the recognition problem. To this end, we consider the following two parameterized problems related to equistability.
Equistability Input: A graph , a positive integer . Parameter: . Question: Is equistable?
Target Equistability Input: A graph , a positive integer . Parameter: . Question: Is target equistable?
Apart from being natural parameterizations of the equistability problem, the first problem has been tackled before (in a nonparameterized variant) in a paper by Levit et al. [8]. There they prove the following.
Theorem 1.1 (Levit et al. [8])
For every fixed , there is an algorithm to decide whether a given vertex graph is equistable. In case of a positive instance, the algorithm also produces a equistable structure of .
Also, the authors ask whether Theorem 1.1 can be strengthened in the sense that there is an FPTalgorithm for recognizing equistable graphs. We answer this question affirmatively.
More precisely, we prove the following results:

There is an vertex kernel for the Equistability problem that can be computed in linear time. This yields an FPT algorithm for the Equistability problem of running time , given a graph with vertices and edges. This affirmatively answers the question posed by Levit et al. [8].

The Target Equistability problem admits an vertex kernel, computable in linear time. Moreover, there is an time algorithm to solve the Target Equistability problem.
The first result we prove in Section 5, and the second in Section 4.
2 Preliminaries
2.1 Twin classes
Following [8], we say that vertices and of a graph are twins if they have exactly the same set of neighbors other than and . It is easy to verify that the twin relation is an equivalence relation. We recall some basic properties of the twin relation (see [8]):
Lemma 1
Let be a graph. The twin relation is an equivalence relation, and every equivalence class is either a clique or a stable set.
An equivalence class of the twin relation will be referred to as a twin class. Twin classes that are cliques will be referred to clique classes, and the remaining classes will be referred to as stable set classes. We say that two disjoint sets of vertices and in a graph see each other if every vertex of is adjacent to every vertex of , and they miss each other if every vertex of is nonadjacent to every vertex of . A vertex sees a set if the singleton sees , and similarly misses if misses . The set of all twin classes will be denoted by and referred to as the twin partition of . The number of twin classes of will be denoted by . The following observation is an immediate consequence of the fact that the twin classes are equivalence classes under the twin relation.
Observation 2.1
Every two distinct twin classes either see each other or miss each other.
By Observation 2.1, the quotient graph of , denoted , is thus well defined: Its vertex set is , and two twin classes are adjacent if and only if they see each other in . Given a graph , it is possible to find in linear time the twin partition , the quotient graph and , using any of the linear time algorithms for modular decomposition [19, 2, 12].
The following two lemmas due to Levit et al. [8] show why twin partitions are important in the study of equistable graphs.
Lemma 2
For every equistable function of and for every , every set of the form is a subset of a twin class of . In particular, if is a equistable graph, then .
Corollary 1
If is a target equistable graph, then .
Lemma 3
For every equistable function of an equistable graph and for every clique class there exists an such that .
2.2 Parameterized complexity
A decision problem parameterized by a problemspecific parameter is called fixedparameter tractable if there exists an algorithm that solves it in time , where is the instance size. The function is typically superpolynomial and depends only on . One of the main tools to design such algorithms is the kernelization technique. A kernelization is a polynomialtime algorithm which transforms an instance of a parameterized problem into an equivalent instance of the same problem such that the size of is bounded by for some computable function and is bounded by a function of . The instance is said to be a kernel of size . It is a folklore that a parameterized problem is fixedparameter tractable if and only if it admits a kernelization. In the remainder of this paper, the kernel size is expressed in terms of the number of vertices. For more background on parameterized complexity the reader is referred to Downey and Fellows [3].
3 A refined XPalgorithm for Equistability
In this section we propose a revised version of the algorithm of Levit et al. [8] for checking whether a given graph is equistable. We implement some speedups and give a more careful analysis of the running time. Let us remark that this improvement does not speed up the running time when is fixed, and it is thus not relevant for the main result of Levit et al. [8]. However, the improved running time is essential when the algorithm is applied to a kernelized instance, for the Equistability resp. Target Equistability problem. We refrain from formally restating the whole algorithm from [8] in order not to create redundancy.
Theorem 3.1
Let be a graph on vertices and edges, and let . Then there is an algorithm of running time to check whether is equistable. This algorithm computes a equistable structure, if one exists, and the same holds if a target is prescribed.
We emphasize that unlike in the statement of Theorem 1.1, the constant hidden in the notation in Theorem 3.1 does not depend on (in Theorem 3.1, is not restricted to be a constant).
Before we prove Theorem 3.1, we state the following observation.
Lemma 4
Let and let with . Then
Proof
If , the statement is immediate. So, let , and assume the statement is true for . Let with . We know that , and thus . A straightforward calculation shows that the right hand side is maximized (over ) for . Thus,
which completes the proof. ∎
We can now prove Theorem 3.1.
Proof (of Theorem 3.1)
Recall that by Lemma 2, any equistable weight function for assigns the same weight only to vertices of the same twin class. Following the algorithm of Levit et al. [8], we proceed as follows. First, we compute in time the twin partition of and the quotient graph (cf. Section 2.1). Fix any ordering such that vertices in each twin class appear consecutively in this ordering. Clearly, the permutation of the weights within a twin class produces an equivalent weight function, i.e., a weight function is an equistable function of if and only if after any permutation of the weights within a twin class we still have an equistable function. We aim to produce a family which contains all equistable functions up to permutations of the weights within a twin class. It suffices to produce all mappings such that the vertices in the set , , appear consecutively in the ordering of . Let be the number of partitions of into labeled intervals, where some of the intervals may be empty. It is straightforward to verify that is bounded by . A standard counting argument yields .
The set can be computed in time as follows. Generate all onetoone mappings from the set to an element set. Using the above ordering of , each such mapping determines a partition of . If the partition refines the twin partition of , then compute all the onetoone mappings from the resulting set of (at most ) nonempty intervals to the set . Each of these mappings specifies, in a natural way, a function in .
Let us now estimate more carefully the size of . Let . We have
implying . We thus obtain
Thus, we have to consider only many weight functions, which can be computed in time .
It remains to check if any of these weight functions in is an equistable function. For every weight function , the algorithm from [8] first computes the target value by evaluating the weight of an arbitrary (fixed) maximal stable set of (see [8] for details); in our setting, this computation can be implemented in time . The algorithm then computes the set of all dimensional vectors with integer coordinates such that for all . A vector represents the set of all subsets of such that the number of vertices of weight in the set equals .
Note that the number of vectors in is bounded by , which, by Lemma 4, is in turn bounded by , for each function . For each vector , the algorithm then checks whether the corresponding sets are of the right weight, that is, whether if and only if the vector encodes a set of maximal stable sets. This latter condition can be verified in time using the quotient graph (see [8] for details).
The running time of this algorithm is thus
This expression simplifies to , as desired.
We remark that, in case of a prescribed target value , the above algorithm can be modified in an obvious way to accept only those equistable functions under which all maximal stable sets have total weight . This completes the proof. ∎
4 An vertex kernel for the Target Equistability problem
Given a graph , the following reduction rule is specified by a positive integer as a parameter.

Clique Reduction. If a clique class contains more than vertices, delete from all but vertices.
The following lemma shows why Clique Reduction rule is safe for both problems, Target Equistability and Equistability.
Lemma 5
Let be a graph, a finite set, a clique class of with where , and a positive integer. Then, for every , graph is target equistable if and only if is target equistable, where is a graph obtained after the Clique Reduction rule has been applied to with respect to the clique class .
Proof
Let . First assume that is target equistable, say with a equistable structure . It is immediate that the restriction of to yields a equistable structure of . Therefore is target equistable.
Now assume that is target equistable, with a equistable structure . We define a function by extending to the set . Indeed, we simply put for all , and for all where . The choice of is arbitrary, since is constant on by Lemma 3.
We claim that is an equistable structure of . To show this, pick an arbitrary maximal stable set of . Then , and so we may assume that . Clearly is a maximal stable set of , and so . Therefore .
Conversely, let be a set with . Since , we have . As is constant on and , we may w.l.o.g. assume that . Hence, , and so is a maximal stable set of . Thus, is a maximal stable set of which completes the proof. ∎
In particular, Clique Reduction rule is safe for the Target Equistability problem. This is seen by putting and in the statement of Lemma 5.
Theorem 4.1
The Target Equistability problem admits a kernel of at most vertices, computable in linear time. Moreover, there is an time algorithm to solve the Target Equistability problem, given a graph with vertices and edges.
Proof
Let be a graph on vertices and edges. Using one of the linear time algorithms for modular decomposition [19, 2, 12], we can compute and in linear time. If , then we conclude that is not target equistable, by Corollary 1. Similarly, if there exists a stable set class with , then we conclude that is not target equistable. Also, we can apply Clique Reduction rule with parameter , to every clique class, in linear time. Afterward, the graph has at most vertices, which proves the first statement of the theorem.
Our FPT algorithm works as follows. First we compute in time a kernel with many vertices. Then we apply Theorem 3.1 to check whether is target equistable. For this, we can put and decide whether is equistable with target . We thus obtain a running time of . ∎
5 An vertex kernel for the Equistability problem
This section is devoted to the proof of the following result.
Theorem 5.1
The Equistability problem admits an vertex kernel, computable in linear time. Moreover, there is an time algorithm to solve the Equistability problem, given a graph with vertices and edges.
Proof
Let us first prove that the second statement follows from the first one. Assume that we can compute an vertex kernel for the Equistability problem in linear time. By Theorem 3.1, we can then decide whether this kernel is equistable in time .
We now turn to the construction of the vertex kernel. In case of a noinstance, our algorithm simply returns a nonequistable graph, say the vertex path . In what follows, we will assume that the input graph satisfies , since otherwise is not equistable, by Lemma 2. The following claim is the main step of our kernelization.
Claim 1
If there exist two distinct twin classes and such that one of them is a stable set and , then is not equistable.
Proof
Suppose for a contradiction that is equistable, with an equistable weight function , and that there exist two distinct twin classes and with such that is a stable set. If the set is contained in every maximal stable set of , then forms a twin class, a contradiction. Thus, we may assume without loss of generality that there exists a maximal stable set of such that and .
Recall that is either a clique class or a stable set class. Since every clique intersects every stable set in at most one vertex and every stable set class is either entirely contained in or disjoint from it, the fact that implies . Let be weights such that , and . Since , there exists a set of vertices in of weight . Since and , there exists a set of vertices in of weight such that . Then, the set is not a stable set, since otherwise by Observation 2.1 the set would be a stable set properly containing , contrary to the maximality of . Note that , contradicting the assumption that is an equistable weight function of . ∎
We consider the following two cases.
Case 1. Every twin class with is a clique class.
In this case, every stable set class has less than vertices, which implies that every maximal stable set of contains at most vertices from each twin class and is thus of total size at most . This implies that in every equistable structure of , we have .
We now perform Clique Reduction rule from Section 4 with .
By Lemma 5 applied with and , the application of Clique Reduction rule is safe.
When the rule can no more be applied, we have a graph with at most twin classes, each of size at most
.
We are done since .
Case 2. There exists a stable set twin class with .
By Claim 1, we may assume that is the unique twin class of size at least (since otherwise is not equistable).
Note that contains at most twin classes, each containing less than vertices, hence .
Suppose first that corresponds to an isolated vertex in the quotient graph . If , then and we are done.
So suppose that .
Claim 2
is equistable if and only if it admits a equistable function that is constant on .
Proof
The if part being trivial, assume that is equistable, and let be a equistable structure of . Let be such that , where . Now we define a weight function that equals outside , and is constantly on . We claim that is a equistable function of . Clearly, is bounded by . Under , all maximal stable sets of have weight .
The only possible problem is that for some vertex set that is not a maximal stable set of . In this case, we claim that . To see this, suppose . Since , we get . Therefore . But , since and . Thus , a contradiction.
So, , and since and is constant on we may assume that . But this yields
A contradiction. ∎
According to Claim 2, it suffices to test if is equistable by considering all possible functions that are constant on , and test for each of them whether it is a equistable function.
Before that, we reduce size of . For this, we compute a graph from by deleting all but many vertices from . Note that, since is a twin class, is unique up to isomorphism.
Claim 3
is equistable if and only if is equistable.
Proof
Let and .
First we assume that is equistable, say with an equistable structure . By Claim 2, we may assume that is constant on , say . We now consider the weight function with target value , and claim that is a equistable structure of . Since every maximal stable set of (resp., ) contains (resp., ) as a subset, it is straightforward that every maximal stable set of has weight . Suppose that there is a set with that is not a maximal stable set of . Then the set has total weight , but is not a maximal stable set of , a contradiction. This proves that is equistable.
Now we assume that is equistable, say with an equistable structure . By Claim 2 applied to , we may assume that is constant on , say . Consider the weight function defined as for all and for all with target value . We claim that is a equistable structure of . Again it is straightforward that any maximal stable set of has weight . Suppose that there is a set with that is not a maximal stable set of .
Recall that and consequently . If , we thus obtain
a contradiction. Thus, , and so we may assume that . Let . Then , but is not a maximal stable set of . This is contradictory, and so is equistable. ∎
By Claim 3, it suffices to check whether is equistable. Since , we are done.
Now, suppose that corresponds to a nonisolated vertex in the quotient graph . Then, there exists a twin class that sees . Let be a maximal stable set of containing a vertex of . Then, . Since , we have in particular that .
If , then for every equistable function of and every maximal stable set, say , such that , we have , hence is not equistable.
If , then .
Since it is clear that the above algorithm runs in time , the proof is complete. ∎
6 Future work
Several open problems surrounding our work remain, some of which we want to mention here in order to stimulate research on this topic.
Firstly, we believe it is NPhard to determine, given a graph and an integer , whether is equistable. It would be satisfying to see this proven, especially for the purpose of this paper. As mentioned in the introduction, the smallest such (if existing) might have to be exponential in the number of vertices of [14], which might serve as a hint for the hardness of this problem.
The analogous question is open also for the problem of Target Equistability: what is the computational complexity of determining, given a graph and an integer , whether is target equistable? Again, the smallest such (if existing) might have to be exponential in the number of vertices of the input graph [14].
A different computational problem in this context would be the following: given a graph and a number , does it admit an equistable weight function using at most different weights? Here, both the parameterized and classical complexity are unknown. Although we did not study this problem in depth, our impression is that it should be NPhard, but FPT when parameterized by . In view of the results of the present paper, there might very well be a polynomial kernel for this problem. Another problem that seems similar at first sight is whether equistability is FPT when parameterized by , the number of twinclasses of .
Apart from these recognition problems, it is apparently open whether the maximum stable set problem is FPT in the class of equistable graphs. Here we do at least know that this problem is APXhard in this class [14].
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