 # On the Maximum Weight Independent Set Problem in graphs without induced cycles of length at least five

A hole in a graph is an induced cycle of length at least 4, and an antihole is the complement of an induced cycle of length at least 4. A hole or antihole is long if its length is at least 5. For an integer k, the k-prism is the graph consisting of two cliques of size k joined by a matching. The complexity of Maximum (Weight) Independent Set (MWIS) in long-hole-free graphs remains an important open problem. In this paper we give a polynomial time algorithm to solve MWIS in long-hole-free graphs with no k-prism (for any fixed integer k), and a subexponential algorithm for MWIS in long-hole-free graphs in general. As a special case this gives a polynomial time algorithm to find a maximum weight clique in perfect graphs with no long antihole, and no hole of length 6. The algorithms use the framework of minimal chordal completions and potential maximal cliques.

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## 1 Introduction

All graphs in this paper are finite and simple. A clique in a graph is a set of pairwise adjacent vertices, and an independent set (or a stable set) is a set of pairwise non-adjacent vertices. A graph is perfect if every induced subgraph of satisfies , where is the chromatic number of and is the maximum clique size in . In a graph , a hole is an induced cycle with at least vertices and an antihole is the complement of a hole. The length of a hole or an antihole is the number of vertices in it. A hole or antihole is long if it has length at least .

For two graphs and we say that contains if is isomorphic to an induced subgraph of . A graph is -free if it does not contain , and for a family of graphs , is -free if is -free for every . The class of perfect graphs was introduced by Claude Berge 

, and became a class of central importance in graph theory. Berge conjectured that a graph is perfect if and only if it does not contain an odd hole or an odd antihole. This question (the Strong Perfect Graph Conjecture) was solved by Chudnovsky, Robertson, Seymour and Thomas

. Moreover, Chudnovsky, Cornuéjols, Liu, Seymour and Vušković  devised a polynomial-time algorithm that determines if a graph is perfect.

The Maximum Independent Set (MIS) is the problem of finidng an independent set of maximum cardinality in a graph, and the Maximum Clique (MC) is the problem of finding a clique of maximum cardinality. Similarly, given a graph with non-negative weights on its vertices, Maximum Weight Independent Set (MWIS) is the problem of finding an independet set of maximum total weight, and Maximum Weight Clique (MWC) is the problem if finding a clique on maximum total weight.

It is known that the Maximum Independent Set (MIS), Maximum Weight Independent Set (MWIS), Maximum Clique (MC), and Maximum Weight Clique (MWC) problems can be solved in polynomial time on perfect graphs using the algorithm of Grötschel, Lovász and Schrijver 

. This algorithm however is not combinatorial and uses the ellipsoid method. Here by a “combinatorial algorithm” we mean an algorithm that can be described entirely in terms of the graph in question. No combinatorial polynomial-time algorithm is known for any of the above problems in perfect graphs; finding one is a major open problem in the field. At the moment we do not even have a polynomial-time combinatorial algorithm to solve

MIS in perfect graphs with no hole of length four. Another important special case is the MC problem for perfect graphs with no long antiholes; again no polynomial-time combinatorial algorithm is known. By taking complements, the latter question is a special case of solving MIS in the class of long-hole-free graphs.

We denote by is the path on vertices. Recently significant progress on the question of the complexity of MWIS was made using the approach of “potential maximal cliques “ (PMCs) that was originally developed by Bouchitté and Todinca [3, 4]. A milestone result was obtained in 2014 by Lokshtanov, Vatshelle, and Villanger  who designed a polynomial-time algorithm for MWIS in -free graphs. Within the same framework, recently Grzesik et al.  showed polynomial-time algorithm for MWIS in -free graphs.

The starting point of this paper was to try to apply this powerful technique to various subclasses of perfect graphs. However, our main results are about a class of graphs that includes both perfect and imperfect graphs, and contains an interesting subclass of perfect graphs, as follows. For an integer the -prism is the graph consisting of two cliques of size , and a -edge matching between them. More precisely, the -prism has vertex set ; each of the sets and is a clique, for every , and there are no other edges in . Our first result is the following:

###### Theorem 1.1.

For every integer the Maximum Weighted Independent Set problem in a (long-hole, -prism)-free -vertex graph can be solved in time .

Since -prisms come up naturally in the context of perfect graphs, the following corollary, obtained by taking complements, is of interest:

###### Theorem 1.2.

Let be an integer. Let be an -vertex perfect graph with no long antihole, and such that the complement of does not contain the -prism. Then the Maximum Weighted Clique problem in can be solved in time . In particular, the Maximum Weighted Clique problem in a perfect -vertex graph with no long antihole and no hole of length can be solved in time .

The last statement of Theorem 1.2 follows from the fact that the complement of a the -prism is the cycle of length .

The algorithm of Theorem 1.1 easily implies a subexponential algorithm for MWIS in long-hole-free graphs, as we now explain.

###### Theorem 1.3.

The Maximum Weighted Independent Set problem in a long-hole -vertex graph can be solved in time .

###### Proof.

Set and check (by exhaustive enumeration) if contains the -prism as an induced subgraph. If such a prism has been found, then branch into subcases guessing for the sought optimum independent set (since consists of two cliques, it intersects with any independent set in at most two vertices). In every branch, delete from the graph for the guessed value of and recurse; since , the number of vertices in the graph drops by at least . Otherwise, if no such is found, apply the algorithm of Theorem 1.1, which now runs in time . Standard analysis shows that this algorithm has running time bound . ∎

Recently, two groups of authors [1, 7] reported a subexponential-time algorithm for MWIS in a related class of -free graphs for every fixed . Their result depends heavily on the notion of “bounded balanced separators”, which we explain next. A balanced separator for a graph and a weight function is a set of vertices such that every connected component of has total weight (w.r.t. ) at most half of the total weight of . We say that a graph class has balanced separators bounded by if for every and every weight function there exists a balanced separator for and of size at most . The main technical statement of  is that a -free graph admits balanced separator of size bouned by where is the maximum degree in . Our second result is a similar statement for long-hole-free graphs.

###### Theorem 1.4.

For every long-hole-free graph and every weight function there exists a balanced separator of and of size at most .

Standard arguments (see e.g. ) show that if a graph class has balanced separators bounded by then the treewidth of a graph is bounded by and, if a balanced separator of size at most for given and can be found in polynomial time, so can a tree decomposition of width . In  a subexponential algorithm for MWIS in a -free -vertex graph with running time bound is obtained by first setting a threshold , branching exhaustively on vertices of degree at least and, once the maximum degree drops below this threshold, by computing a tree decomposition of width and solving MWIS by a dynamic programming algorithm on this tree decomposition. Following exactly the same strategy with threshold we obtain the following.

###### Theorem 1.5.

The Maximum Weighted Independent Set problem on a long-hole -vertex graph can be solved in time .

### Organization

In Section 2 we explain the general framework of potential maximal cliques. In Section 3 we prove Theorem 1.1, and finally in Section 4 we prove Theorem 1.4.

## 2 Separators and potential maximal cliques

Let be a graph and let . We denote by the sugbraph of induced by and by the graph . A component of (or of ) is the vertex set of a maximal connected subgraph of . We write to mean the set of connected components of . We denote by the set of vertices of with a neighbor in , and write . When we use the notation (or ) instead of (or ). For the graph has vertex set and edge set .

A graph is chordal if it has no holes. A set is a fill-in or a chordal completion (of ) if is a chordal graph. A fill-in is minimal if it is inclusion-wise minimal.

Let . For , we say that is an -separator if and lie in different connected components of . An -separator is a minimal -separator if it is an inclusion-wise minimal -separator. is said to be a minimal separator if there exist such that is a minimal -separator in . We say that is a full component for if . It is easy to see that is a minimal separator if and only at least two members of are full components.

An important property of minimal separators is that no new minimal separator appears when a minimal fill-in is added to a graph. More precisely:

###### Proposition 2.1 ().

Let be a graph and let be a minimal fill-in for . If is a minimal separator of , then is a minimal separator of . Furthermore, .

A set is a potential maximal clique (PMC) if there exists a minimal fill-in of such that is a maximal (inclusion-wise) clique of . A PMC is surrounded by minimal separators in the following sense:

###### Proposition 2.2 ().

Let be a graph, let be a PMC of , and let . Then is a minimal separator of and is a full component for .

Next we state an important characterization of PMCs in graphs.

###### Theorem 2.3 ().

A set is a PMC in if and only if the following two conditions hold:

1. for every we have ;

2. for every either , , or there exist with .

In the second condition of Theorem 2.3, we say that a component covers the nonedge .

Our main algorithmic engine is the following.

###### Theorem 2.4 ().

Given a graph with vertex weights and a family that contains all PMCs of , one can solve MWIS in in time polynomial in the size of and .

Thus it is enough to construct a family as in Theorem 2.4. However, it turns out that instead of constructing a family of PMCs, it is easier to construct a family of components that result from deleting PMCs. This approach was taken in [3, 10, 9], and is justified by the following result:

###### Theorem 2.5 ().

Given a graph and a family of vertex sets of connected induced subgraphs of such that for every potential maximal clique of we have , one can compute the family of all potential maximal cliques of . The running time of the algorithm and the size of the family is bounded polynomially in the size of and .

Our final observation is the following

###### Theorem 2.6.

Given a graph and a family of all minimal separators of , one can compute the family of subsets of such that for every potential maximal clique of we have . The running time of the algorithm and the size of the family is bounded polynomially in the size of and .

###### Proof.

For every we can compute in polynomial time the set . Let ; we claim that is the desired family. To see this, let be a potential maximal clique of and let . By Proposition 2.2, is a minimal separator of , and therefore . ∎

We remark that all minimal separators in a graph can be enumerated in time polynomial in the graph size and the number of output minimal separators . In view of Theorem 2.6 from now on we focus on studying minimal separators.

## 3 k-prism and minimal separators

The goal of this section is to prove Theorem 1.1. We say that a graph class has the polynomial separator property if there exists such that every has at most minimal separators. In view of the results of Section 2, MWIS can be solved in polynomial time in any graph class with polynomial separator property. It is easy to see that the -prism has minimal separators while being long-hole-free, and therefore the class of long-hole-free graphs does not have the polynomial separator property. In this section we prove that in long-hole-free graphs -prisms are the only reason the property is violated.

We show:

###### Theorem 3.1.

Let be an integer and let be a long-hole-free graph that does not contain a -prism. Then has at most minimal separators.

###### Lemma 3.2.

Let be an integer and let be a long-hole-free graph that does not contain a -prism. Let be a minimal separator in and let with . Assume that there exists such that for every we have . Then there exists a set such that , and .

###### Proof.

Let . Let and be two distinct connected components of .

Assume that for there exist , that is, the neighborhoods of the components and are incomparable inside . Let be a shortest path from to via , and let be a shortest path from to via (possibly consists of one edge if and are adjacent). Note that and are of length at least two as , while is of length at least one. Hence, the concatenation of , , and is a hole of length at least five, a contradiction.

We deduce that we can enumerate the components of as such that

 N(A1)∩S′⊇N(A2)∩S′⊇…⊇N(Am)∩S′.

In particular, since for no component we have , there exist with . Since , we have and .

Assume there exists such that a shortest path from to via is of length at least . Let be a shortest path from to via . Then, the concatenation of , , and the edge is a hole of length at least five, a contradiction. We deduce that for every , there exists a vertex with . In particular, .

Consider now an inclusion-wise minimal set with . By minimality, for every pick ; clearly, . Pick two distinct . Let be a shortest path from to via . Let equal the edge if it is present, or the concatenation of edges and if . Then, the concatenation of , , and edges for is a hole of length at least five unless both and .

Write . We conclude that both and are cliques of . Since does not contain a -prism, it follows that , and therefore the has the desired properties. This completes the proof. ∎

Now we can prove Theorem 3.1.

###### Proof.

For a set , we define . Note that if and only if is a minimal separator. By induction on the number of vertices of , we show that

 ∑S⊆V(G)ζG(S)≤|V(G)|k+2. (1)

The statement is straightforward for .

Pick arbitrary and let . To show (1), it suffices show that

 ∑S⊆V(G)ζG(S)−∑S⊆V(G′)ζG′(S)≤|V(G)|k+1. (2)

Let be such that , that is, is a minimal separator in . We consider two cases.

We say that is special if

• ,

• if is the connected component of that contains , then , and

• for every we have .

If is special, then by Lemma 3.2 there exists of size at most such that . Thus, every connected component distinct from with is a connected component of . Since there are at most choices for , we infer that the contribution to the sum from the sets that are special is at most .

Define if is special and otherwise.

If is a minimal separator that is not special, then either or and, if is the connected component of that contains , then either or still one connected component of satisfies . In both options is a minimal separator in . Hence, to show (2) it suffices to show that for every minimal separator in it holds that:

 ζG′(S′)≥ζ′G(S′)+ζG(S′∪{v}). (3)

Let and let . Clearly,

 ζG′(S′)=|A|−1.

If , then there exists a single connected component of that contains and all connected components of . Hence,

 ζG(S′∪{v}) =|B|−1, ζ′G(S′)≤ζG(S′) =|A|−|B|.

This proves (3) in the case . Otherwise, if , then is not a minimal separator in and . If the connected component of that contains satisfies , then is special in , , and (3) is proven. Otherwise, we observe that and we are done. This completes the proof. ∎

Finally, we prove Theorem 1.1.

###### Proof of Theorem 1.1..

Since is -prism-free, Theorem 3.1 implies that the number of minimal separators in is at most . By a result of , all minimal separators of can be enumerated in time . Now Theorem 1.1 follows from Theorems 2.62.5, and 2.4. ∎

## 4 Dominating a PMC with three vertices

In this section we prove Theorem 1.4. To this end, we show that in a long-hole-free graph every PMC is contained in a neighborhood of at most three vertices of . This is done by a sequence of structural lemmas that follows next.

###### Lemma 4.1.

Let be a long-hole-free graph, let be a minimal separator in , and let satisfy . Then for every independent set there exist a vertex with .

###### Proof.

For every let . Suppose that for every . Then there are two vertices in with and such that both and are maximal (inclusion-wise). Choose with the following properties:

• is maximal;

• ; and

• subject to the first two conditions, the length of a shortest path from to in is the smallest possible.

Let . By the maximality of , there exists . Let be a shortest path from to in . Then and for every we have that . In particular, no vertex of is adjacent to . Let be the neighbor of closest to along . Then . Now is an induced path with at least vertices. Let be a full component for , and let be a path with endpoints and and all internal vertices in . By concatenating and we get a hole of length at least five, a contradiction. ∎

###### Lemma 4.2.

Let be long-hole-free graph and let be a minimal separator in . Let with . Then for every there exist and with .

###### Proof.

For each , let . Let be the set of all for which is inclusion-wise minimal among , and let be an inclusion-wise maximal independent subset of . Since , the set is independent as well. By Lemma 4.1, there exists and with , , , and .

We claim that

 Z0⊆N(a)∪N(b). (4)

Assume the contrary: there exists . By the maximality of , there exists . By the choice of and , we have . By the definition of , there exists a vertex . By symmetry, assume . Then is a long hole in , a contradiction. This finishes the proof of (4).

By the definition of , (4) implies . This finishes the proof of the lemma. ∎

###### Lemma 4.3.

Let be a long-hole-free graph, let be a PMC in , and let be an independent set. Then either and , or there exists with .

###### Proof.

Assume that the first alternative does not hold. That is, if then . Then, by Theorem 2.3, for every there exists a component covering a nonedge from to some other vertex of . Let maximize . Suppose, contrary to the second alternative, that and let . Denote ; note that . For every let . Since is an independent set, for every there exists satisfying . In particular, . On the other hand, the maximality of implies that for every . Consequently, there exist two components with inclusion-wise incomparable and . Let for and let be a shortest path from to via . Let be a shortest path from to via . Then is a long hole in , a contradiction. This finishes the proof of the lemma. ∎

###### Lemma 4.4.

Let be a long-hole-free graph, let be a PMC in , and let . Then either or there exists with (in particular, ).

###### Proof.

Suppose that but there is no component as in the statement of the lemma. Let be inclusion-wise minimal such that there is no component with . Observe that is non-empty, because there is at least one non-edge within with one endpoint , so in particular there is at least one component satisfying . If is an independent set, then so is , and then Lemma 4.3 contradicts the choice of . Hence, there exists an edge . By the minimality of , there exist a component with and a component with . By the choice of , we have and . Let be a shortest path from to via and similarly define in . But then is a long hole in , a contradiction. ∎

###### Lemma 4.5.

Let be a long-hole-free graph, let be a PMC in , and let be arbitrary. Then either or there exist two vertices and such that .

###### Proof.

Assume that . Let be a component with (it exists by Lemma 4.4). Recall that is a minimal separator and is a full component for . Let be another full component for . By Lemma 4.2, there exist and with . Since , we have as desired. ∎

We can now deduce the following:

###### Theorem 4.6.

For every long-hole-free graph and for every potential maximal clique in there exists a set of size at most such that .

Theorem 4.6 immediately implies Theorem 1.4 using standard techniques.

###### Proof of Theorem 1.4..

Let and be as in the statement of the theorem. By Theorem 4.6, it suffices to show that there is a potential maximal clique in that is a balanced separator with respect to . To this end, let be a minimal chordal completion of . A folklore result (see e.g. ) is that admits a tree decomposition where the bags are exactly the maximal cliques of . Let be the tree of the decomposition. For every edge , let and be the two components of and for let be the union of all the bags of . Orient the edge from the endpoint in with smaller weight of to the one with the larger weight, breaking ties arbitrarily. Let be a node of zero outdegree. Then it can be easily checked that the bag at is a balanced separator and we are done. ∎

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