Coloring graph classes with no induced fork via perfect divisibility

04/06/2021 ∙ by T. Karthick, et al. ∙ 0

For a graph G, χ(G) will denote its chromatic number, and ω(G) its clique number. A graph G is said to be perfectly divisible if for all induced subgraphs H of G, V(H) can be partitioned into two sets A, B such that H[A] is perfect and ω(H[B]) < ω(H). An integer-valued function f is called a χ-binding function for a hereditary class of graphs C if χ(G) ≤ f(ω(G)) for every graph G∈ C. The fork is the graph obtained from the complete bipartite graph K_1,3 by subdividing an edge once. The problem of finding a polynomial χ-binding function for the class of fork-free graphs is open. In this paper, we study the structure of some classes of fork-free graphs; in particular, we study the class of (fork,F)-free graphs G in the context of perfect divisibility, where F is a graph on five vertices with a stable set of size three, and show that every G∈ G satisfies χ(G)≤ω(G)^2. We also note that the class G does not admit a linear χ-binding function.

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1 Introduction

For a positive integer , will denote the complete graph on vertices, and will denote the path on vertices. For an integer , will denote the cycle on vertices. A hole in a graph is an induced cycle with ; an antihole is the complement of a hole. A hole or antihole is odd (even) if

is odd (even). The

union of two vertex-disjoint graphs and , denoted by , is the graph with vertex set and edge set . The union of copies of the same graph will be denoted by . A stable set (or an independent set) is a set of vertices that are pairwise nonadjacent.

A class of graphs is hereditary if every induced subgraph of every graph in is also in . An important and well studied type of hereditary class of graphs is the class of graphs which are defined by forbidden induced subgraphs. Given a graph , we say that a graph is -free if has no induced subgraph that is isomorphic to . Given a class of graphs , we say that a graph is -free if is -free for every .

For a graph , will denote its chromatic number, and its clique number. For every graph , . A graph is called perfect if for every induced subgraph of , .

A graph is said to be perfectly divisible if for all induced subgraphs of , can be partitioned into two sets , such that is perfect and . Perfectly divisible graphs were introduced by Hoáng [12], and can be thought of as a generalization of perfect graphs in the sense that perfect graphs are perfectly divisible. However not all perfectly divisible graphs are perfect. For example, the vertex set of an odd hole can be partitioned into two sets such that the first set induces a perfect graph and the other is a stable set. So an odd hole is perfectly divisible but not perfect. Hoáng [13] observed that the class of -free graphs is perfectly divisible, and in [12] he showed that the class of (banner, odd hole)-free graphs is perfectly divisible. Chudnovsky and the third author showed that the class of (, bull)-free graphs is perfectly divisible [6]. See Sections 2 and 4 for more on perfect divisibility.

A hereditary class of graphs is -bounded if there is a function (called a -binding function) such that for every graph . In addition, if is a polynomial function then the class is polynomially -bounded. It has long been known that there are hereditary graph classes that are not -bounded (see [24] for examples) but it is not known whether there is a hereditary graph class that is -bounded but not polynomially -bounded. A recent survey of Scott and Seymour [24] gives a detailed overview of this area of research.

The class of claw-free graphs (or -free graphs) is widely studied in a variety of contexts and has a vast literature; see [10] for a survey. A detailed and complete structural classification of claw-free graphs has been given by Chudnovsky and Seymour; see [9]. A result of Gyárfás [11] together with a result of Kim [16] show that the class of claw-free graphs is -bounded, and that every such graph satisfies . It is also known that there is no linear -binding function even for a very special class of claw-free graphs; see [3]. Chudnovsky and Seymour [8] showed that every connected claw-free graph with a stable set of size at least satisfies .

The fork is the graph obtained from the complete bipartite graph by subdividing an edge once. The class of claw-free graphs is a subclass of the class of fork-free graphs. It is a natural line of research to see what properties of claw-free graphs are also enjoyed by fork-free graphs. A classic example is the polynomial-time solvability of the (weighted) stable set problem in the class of fork-free graphs [1, 18], generalizing the result for claw-free graphs [19, 21]. It has long been known that the class of fork-free graphs is -bounded [15], and it not known whether the class of fork-free graphs is polynomially -bounded or not. Indeed, Randerath and Schiermeyer [23] asked the following interesting question.

Problem 1

Does there exist a polynomial -binding function for the class of fork-free graphs?

The third author (unpublished) has conjectured that the class of fork-free graphs is perfectly divisible which in turn will yield a quadratic -binding function. In this paper, we are interested in polynomial -binding functions for some classes of fork-free graphs, namely (fork, )-free graphs , where is any nontrivial graph on at most five vertices, and we give below some known results in this direction.

  • If , then it is observed in [23] that every satisfies . Moreover if is connected, then equality holds if and only if is an odd hole.

  • If , then clearly every is perfect.

  • If , then is linearly -bounded; see [5] and the reference therein.

  • It follows from a result of Wagon [25] that, if , then every satisfies . Further, it is known that does not admit a linear -binding function; see [3].

  • If , then every satisfies , and does not admit a linear -binding function; see [11, 16].

  • If , then it follows from Theorem 18 of [22] that is quadratically -bounded.

  • Randerath [20] showed that, if , then every satisfies .

  • Recently, Chudnovsky et al [4] proved a structure theorem for the class of (fork, antifork)-free graphs, and used it to prove that every (fork, antifork)-free graph satisfies . (Here, an antifork is the complement graph of a fork.)

Thus if , then the class of (fork, )-free graphs is known to be quadratically -bounded except when , and not much is known when . Here, we study the class of (fork, )-free graphs, where is a graph on five vertices with a stable set of size three. More precisely, we consider the class of (fork, )-free graphs , where is one of the following graphs: , dart, co-dart, co-cricket, banner, and bull, and show that the following hold:

  • Every is perfectly divisible, when .

  • Every is either claw-free or perfectly divisible, when {dart, banner, co-cricket}.

  • Every satisfies .

  • Since the class of -free graphs does not admit a linear -binding function [3], and since each graph and the fork has a stable set of size , it follows that the class does not admit a linear -binding function.

2 Preliminaries

We follow West [26] for standard notation and terminology used here, and we refer to the website ‘https://www.graphclasses.org/smallgraphs.html’ for some special graphs used in this paper. For a vertex in a graph , is the set of vertices adjacent to , is the set , and is the set . Given a subset , is the set , and is the set . We drop the subscript in the above notations if there is no ambiguity. For a vertex set , denotes the subgraph of induced by . We say that a graph contains a graph if is an induced subgraph of . The complement of a graph will be denoted . Given disjoint vertex sets , we say that is complete to if every vertex in is adjacent to every vertex in ; we say is anticomplete to if every vertex in is nonadjacent to every vertex in ; and we say is mixed on if is not complete or anticomplete to . When has a single vertex, say , we can instead say that is complete to, anticomplete to, or mixed on . A vertex in is universal if it is complete to . A set is a homogeneous set if and for every , is either complete or anticomplete to . We say that a graph admits a homogeneous set decomposition if has a homogeneous set. The independence number of a graph is the size of a largest stable set in . A triad in a graph is a stable set of size .

For a vertex subset of , we write ---- to denote an induced cycle in with vertex set and edge set , and we write --- to denote an induced path in with vertex set and edge set .

We say that a vertex is a center of a claw in a graph , if has neighbors such that is a triad; and we call the vertices , , the leaves of the claw.

The paw is the graph that consists of a with a pendant vertex attached to it. The diamond is the graph .

We say that a vertex set induces:

  • a fork if induces a claw with center , and is a leaf adjacent to .

  • a dart if induces a claw with center , and is adjacent to and but not to .

  • a banner if induces a claw with center , and is adjacent to but not to .

  • a co-dart if induces a paw, and is anticomplete to .

  • a co-cricket if induces a diamond, and is anticomplete to .

  • a bull if --- is a path, and is adjacent to but not to .

Note that a co-cricket , and a co-dart and is the complement graph of a dart.

We use the following known results. The class of perfect graphs admits a forbidden induced subgraph characterization, namely, the strong perfect graph theorem, given below.

Theorem 1 ([7])

A graph is perfect if and only if it has no odd hole or odd antihole.

Chudnovsky and Seymour [8] give a simple proof of the following -bound for claw-free graphs in general.

Theorem 2 ([8])

Every claw-free graph satisfies . Moreover, the bound is asymptotically tight.

Although perfect divisibility is a structural property, it immediately implies a quadratic -binding function. Indeed, we have the following (see also [6]).

Lemma 1 ([12])

Every perfectly divisible graph satisfies .

A graph is said to be perfectly weight divisible if for every nonnegative integer weight function on , there is a partition of into two sets and such that is perfect and the maximum weight of a clique in is smaller than the maximum weight of a clique in . We will also use the following results.

Theorem 3 ([6])

A minimal non-perfectly weight divisible graph does not admit a homogeneous set decomposition.

The proof of the following theorem is similar to the proof of Theorem 3.7 in [6], and we give it here for completeness.

Theorem 4

Let be a hereditary class of graphs. Suppose that every graph has a vertex such that is perfect. Then every is perfectly weight divisible, and hence perfectly divisible.

Proof. Let be a minimal counterexample to the theorem. Then there is a nonnegative integer weight function on for which there is no partition of as in the definition of perfectly weight divisibility. Let be the set , and let . Since is hereditary, , and so by the hypothesis of the theorem, has a vertex such that is perfect. But now, since , if we let and , then we get a partition of as in the definition of perfectly weight divisibility, a contradiction. This proves the theorem.

3 Classes of fork-free graphs

3.1 The class of (fork, )-free graphs

In this section we prove that (fork, )-free graphs are perfectly divisible, and hence the class of (fork, )-free graphs is quadratically -bounded. A vertex set in a graph is said to be anticonnected if the subgraph induced by in is connected. Also, a vertex is an anticenter for a vertex set if . An antipath in is the complement of the path --- in , for some .

Theorem 5

Let be a (fork, )-free graph which is not perfectly divisible. Then has a homogeneous set.

Proof. Since is not perfectly divisible, given , is not perfect, so it contains an odd hole or an odd antihole , by Theorem 1. Note that , and since is -free, does not contain for . So contains an odd antihole induced by with anticenter . We construct a sequence of vertex sets such that for each , is obtained from by adding one vertex that has a neighbor and a nonneighbor in . Let be the maximal vertex set obtained this way. By maximality, is a homogeneous set. We show that ; in particular, we show that does not intersect the set of anticenters for .

5.1

is complete to .

Let be the vertices of , with edges whenever (indices are modulo ). Suppose to the contrary that there exists a vertex that has a nonneighbor in . By assumption, has a neighbor in , and a neighbor . Suppose that has two consecutive neighbors in . Then there is some such that and . But then induces a fork. So we may assume that does not have two consecutive neighbors in . Then since is odd, there must exist some such that and . But now ----- is a which is a contradiction. This proves 5.1.

5.2

For each , is complete to , and does not intersect .

We prove the assertion by induction on . By 5.1, we may assume that . Suppose to the contrary that has a nonneighbor ; then is a center for . Since has a neighbor in and , we have . Since has a nonneighbor in and is complete to , we have . Since , has a neighbor in , say . Note that is anticonnected for each . Then if is a shortest antipath from to , contains at least three vertices; label its vertices --- in order. Since , induce a fork which is a contradiction. So has no nonneighbors in . This proves 5.2.

Now by 5.2, it follows that the vertex set does not intersect . Since , we have , as desired.

Corollary 5.1

Every (fork, )-free graph is perfectly divisible.

Proof. This follows from Theorems 3 and 5.

We immediately have the following corollary which generalizes the result that the class of -free graphs is perfectly divisible [13].

Corollary 5.2

Every -free graph is perfectly divisible.

Corollary 5.3

Every (fork, )-free graph satisfies .

Proof. This follows from Lemma 1.

3.2 The class of (fork, dart)-free graphs

In this section, we prove that the class of (fork, dart)-free graphs is quadratically -bounded.

Theorem 6

Let be a connected (fork, dart)-free graph. If contains a claw with center , then is perfect.

Proof. Let be a (fork, dart)-free graph containing a claw with center . Let be the set of leaves of claws in with center . Then has a triad, and so . Let denote the set , and denote the set . Then is the set .

6.1

If has a neighbor in a triad in , then it has at least two neighbors in . Moreover, every has a neighbor in a triad , so has two nonadjacent neighbors in .

Let be a triad in , and suppose that is adjacent to . Then since does not induce a fork, we see that is adjacent to or . This prove the first assertion of 6.1. Note that by definition, any in has a neighbor ; then there exist such that is a triad, and hence is also adjacent to or . This proves 6.1.

6.2

If , then is -free.

If there is a , say --, in , then induces a dart, a contradiction. This proves 6.2.

6.3

is complete to .

Suppose to the contrary that has a nonneighbor . Then since , there are vertices and in such that is a claw. Now, if is not adjacent to , then is a leaf of the claw induced by , a contradiction to our assumption that . So is adjacent to . Likewise, is adjacent to . But then induces a dart which is a contradiction. This proves 6.3.

6.4

is anticomplete to .

Suppose to the contrary that there exists a vertex such that has a neighbor, say . Since is anticomplete to (by the definition), we may assume that . By 6.3, is complete to . Then since , there exist such that induces a dart which is a contradiction. This proves 6.4.

6.5

Let and let be a component of . Then is not mixed on .

Suppose not. We may assume that is mixed on an edge in . Then since , by 6.1, there exist nonadjacent vertices such that . But then induces a fork, a contradiction. This proves 6.5.

6.6

is complete to .

By 6.4, . Since is connected, it follows from 6.5 that every vertex in has a neighbor in . Suppose to the contrary that is mixed on . Let and let . By 6.1, let and be the neighbors of in . Recall that, by 6.4, is anticomplete to . Suppose that is adjacent to . Then since does not induce a fork, is adjacent to . Likewise, is adjacent to . But then induces a dart, a contradiction. So we may assume that is not adjacent to . Then since does not induce a fork, is not adjacent to . Likewise, is not adjacent to . Then by 6.1, has nonadjacent neighbors . Then since does not induce a fork or a dart, we may assume that is adjacent to , but not to . Then since does not induce a dart, is not adjacent to . Now, induces a fork which is a contradiction. This proves 6.6.

6.7

is a clique.

Suppose not. Let and be two nonadjacent vertices in . By 6.4, . Now choose any and any adjacent to . Then, by 6.4 and 6.6, induces a fork which is a contradiction. This proves 6.7.

6.8

If is an odd hole or an odd antihole in , then .

By 6.6 and 6.7, every vertex in is universal in . Since odd holes and odd antiholes have no universal vertices, we see that . So . This proves 6.8.

6.9

Let ---- be an odd hole in . Then every vertex in which has a neighbor in is adjacent to exactly two consecutive vertices of .

Let . We may assume that is a neighbor of in . Then since does not induce a fork or a dart, we may assume that is adjacent to , and is nonadjacent to . Then since does not induce a dart, is not adjacent to . If has a neighbor in , say with the largest index , then . By the choice of , we have , and then induces a fork. Thus is anticomplete to . Hence every vertex in which has a neighbor in is adjacent to exactly two consecutive vertices of . This proves 6.9.

6.10

is -free, where .

Suppose to the contrary that contains an odd hole, say ----. By 6.8, . By 6.1, let be a triad in , and let and be the neighbors of in . Then by 6.9, is anticomplete to , and we may assume that . If is adjacent to , then induces a dart. So we may assume that is not adjacent to . Then by 6.1, is a neighbor of . As earlier, we see that is not adjacent to , and hence by 6.9, is adjacent to . But now induces a fork which is a contradiction. This proves 6.10.

6.11

is -free, where .

Suppose to the contrary that contains a , say with vertices and edges whenever (indices are modulo ). By 6.8, . Let be a neighbor of . Consider any consecutive pair of vertices . Then since does not induce a fork or a dart, is adjacent to exactly one of , . Therefore, if is adjacent to , then is adjacent to precisely the vertices with even index in , and if is not adjacent to , then is adjacent to precisely the vertices with odd index in . We may assume that is adjacent to . Then is not adjacent to . Then since -- is a , by 6.2, is not adjacent to . But then induces a fork which is a contradiction. This proves 6.11.

Now by 6.10 and 6.11, and by Theorem 1, we conclude that is perfect. This completes the proof.

Theorem 7

Let be a connected (fork, dart)-free graph. Then is either claw-free or for any claw in with center, say , is perfect.

Proof. This follows from Theorem 6.

The following corollary generalizes the result known for the class of claw-free graphs (Theorem 2).

Corollary 7.1

Every (fork, dart)-free graph satisfies .

Proof. Let be a (fork, dart)-free graph. We may assume that is connected. If is claw-free, then the desired result follows from Theorem 2. So let us assume that contains a claw with center, say . Then by Theorem 7, is perfect. Now since and since is perfect, we see that is perfectly divisible, and hence the result follows from Lemma 1.

3.3 The class of (fork, co-dart)-free graphs

In this section, we prove that (fork, co-dart)-free graphs are perfectly divisible, and hence the class of (fork, co-dart)-free graphs is quadratically -bounded.

Theorem 8

Let be a connected (fork, co-dart)-free graph. Then either admits a homogeneous set decomposition or for each vertex in , is perfect.

Proof. Suppose to the contrary that does not admit a homogeneous set decomposition and that there is a vertex in such that is not perfect. So by Theorem 1, contains an odd hole or an odd antihole. Since has no co-dart, is paw-free, and so has no odd antiholes except . So suppose that contains an odd hole. Let ---- be a shortest odd hole in for some with vertex set .

8.1

If , then .

Suppose to the contrary that is nonadjacent to . First suppose that has two adjacent neighbours in . We may assume that . Then since does not induce a co-dart, we see that is adjacent to . Then by similar arguments, we conclude that . But now induces a co-dart, a contradiction. So suppose that is nonadjacent to any two consecutive vertices in . Since has a neighbor in , we may assume that is adjacent to . Then is nonadjacent to both and . Then since and do not induce forks, is adjacent to both and . Since is nonadjacent to two consecutive vertices in , this implies that , and is nonadjacent to . But now induces a fork, a contradiction. This proves 8.1.

8.2

Any vertex in is complete to .

Let be a vertex in . Then by 8.1, is adjacent to . We may assume that is adjacent to . Then since does not induce a fork, is adjacent to either or . We may assume that is adjacent to . Then for , since does not induce a co-dart, is complete to . Now suppose to the contrary that is nonadjacent to one of or , say . Then is adjacent to . For, otherwise if , then induces a fork, and if , then induces a co-dart which are contradictions. But then induces a co-dart, a contradiction. So is adjacent to both and , and hence is complete to . This proves 8.2.

By 8.2, we see that is a homogenous set, a contradiction. This proves Theorem 8.

Corollary 8.1

Every (fork, co-dart)-free graph is perfectly weight divisible, and hence perfectly divisible.

Proof. Let be a minimal counterexample to the theorem. Then, by Theorem 3, does not admit a homogeneous set decomposition. So, by Theorem 8, there is a vertex in such that is perfect. Then, by Theorem 4, it follows that is perfectly weight divisible, and hence perfectly divisible, a contradiction. This proves Corollary 8.1.

Corollary 8.2

Every (fork, co-dart)-free graph satisfies .

Proof. This follows from Corollary 8.1, and from Lemma 1.

3.4 The class of (fork, banner)-free graphs

In this section, we prove that (fork, banner)-free graphs are either claw-free or perfectly divisible, and hence the class of (fork, banner)-free graphs is quadratically -bounded. We use the following lemma.

Lemma 2 ([2])

If is a banner-free graph that does not admit a homogeneous set decomposition, then is -free.

Theorem 9

Let be a (fork, banner)-free graph that contains a claw. Then either admits a homogeneous set decomposition or there is a vertex in such that is perfect.

Proof. Let be a (fork, banner)-free graph that contains a claw. Suppose that does not admit a homogeneous set decomposition. Then is connected, and, by Lemma 2, we may assume that is -free. Let be a vertex in such that is maximized. Let be a maximum stable set in , and let denote the set . Since contains a claw, we see that and so has a triad.

9.1

is anticomplete to .

Suppose has a neighbor in a triad . Then since does not induce an or a banner, is not adjacent to and . But then induces a fork, a contradiction. This proves 9.1.

9.2

is a stable set.

Suppose to the contrary that has a component, say with more than one vertex. Then, by 9.1, is anticomplete to . Let be neighbors, and suppose is adjacent to . Then is adjacent to at least two vertices in any given triad (otherwise, contains a fork, a contradiction). We may assume . Then since does not induce a fork, is adjacent to . Thus we conclude that every vertex in is either complete or anticomplete to , and so is a homogeneous set, a contradiction to our assumption. This proves 9.2.

Now it follows from 9.2 that is perfect. This completes the proof.

Corollary 9.1

Let be a (fork, banner)-free graph. Then either is claw-free or admits a homogeneous set decomposition or there is a vertex in such that is perfect.

Proof. This follows from Theorem 9.

Corollary 9.2

Let be a (fork, banner)-free graph. Then either is claw-free or is perfectly weight divisible, and hence perfectly divisible.

Proof. This follows from Theorems 3 and 4, and from Corollary 9.1.

Corollary 9.3

Every (fork, banner)-free graph satisfies .

Proof. This follows from Corollary 9.2, Theorem 2, and from Lemma 1.

3.5 The class of (fork, co-cricket)-free graphs

In this section, we prove that (fork, co-cricket)-free graphs are either claw-free or perfectly divisible, and hence the class of (fork, co-cricket)-free graphs is quadratically -bounded.

Theorem 10

Let be a (fork, co-cricket)-free graph. Then either is claw-free or admits a homogeneous set decomposition or for each vertex in , is perfect.

Proof. Let be a (fork, co-cricket)-free graph. Suppose to the contrary that none of the assertions hold. We may assume that is connected. Let be a vertex in such that is not perfect. Since is co-cricket-free, we see that does not contain a diamond, and hence does not contain an odd antihole except . So by Theorem 1, contains an odd hole. Let ---- be a shortest odd hole in for some , and let denote the vertex set of .

10.1

If is a vertex in which has three consecutive neighbors in , then is complete to .

We may assume that is adjacent to the vertices , and . Now if there is a vertex in that is nonadjacent to , say , then induces a co-cricket, a contradiction. So is complete to . In particular, is adjacent to . Next suppose to the contrary that is not complete to . Let be the least positive integer such that is adjacent to , and is nonadjacent to . Now if , then induces a fork, and if , then induces a fork, a contradiction. So is complete to . This proves 10.1.

10.2

Let be a vertex in which has a neighbor in . Then the following hold:

  1. If , then is either or or , for some mod .

  2. If , then is either or , for some , mod .

If has three consecutive vertices of as neighbors, then by 10.1, , and we conclude the proof. So we may assume that no three consecutive vertices of are neighbors of .

Now suppose that 10.2: does not hold. So by our assumption, there is an index , mod such that is adjacent to , and is anticomplete to . Then induces a fork, a contradiction. So 10.2: holds.

Next suppose that 10.2: does not hold. First let us assume that no two consecutive vertices of are neighbors of . Since has a neighbor in , we may assume that is adjacent to . By assumption, is not adjacent to both and . Then since does not induce a fork, is adjacent to . Likewise, is adjacent to . Also by our assumption, is not adjacent to . Now induces a fork, a contradiction. So we may assume that there is an index , mod such that is adjacent to both and , say . Moreover, by our contrary assumption, has a neighbor in . Also by our earlier arguments, is nonadjacent to both and . Suppose that has a neighbor in ; let be the least possible integer in such that is adjacent to . Now since does not induce a fork, . Then since ----- is not an odd hole in which is shorter than , we see that is odd. Also since does not induce a fork, is adjacent to . So by our assumption, is nonadjacent to . Moreover, since ------ is not an odd hole in which is shorter than , has a neighbor in , say . But now induces a fork, a contradiction. Thus, by using symmetry, we may assume that has no neighbor in . So by our contrary assumption, is adjacent to . But then induces a fork, a contradiction. So 10.2:(b) holds. This proves 10.2.

Let be the set , and let be the set . Moreover, if , then let be the set