1 Introduction
Throughout the paper, we consider only simple and finite graphs. For a positive integer , let and respectively denote the chordless path and the complete graph on vertices. For a positive integer , is the chordless cycle on vertices. A diamond is the fourvertex complete graph minus an edge, and a gem is the graph consisting of a (say ) plus a vertex which is adjacent to all the vertices of . Given a graph , we say that a graph is free if it does not contain an induced subgraph isomorphic to . Given a set of graphs, we say that is free if is free, for each .
A clique (stable set) in a graph is a set of mutually adjacent (nonadjacent) vertices in . A split graph is a graph whose vertex set can be partitioned into a stable set and a clique. A complete split graph is a split graph where every vertex of the stable set is adjacent to every vertex of the clique.
Domination and vertex colorings are two of the major topics extensively studied in graph theory. It is reflected in large number of books, monographs, and periodic surveys. In a graph , a subset of is a dominating set if every vertex in is adjacent to some vertex in . The dominating induced subgraph of a graph is the subgraph induced by a dominating set in . For any integer , a coloring of a graph is a mapping such that any two adjacent vertices in satisfy . The chromatic number of a graph is the smallest integer such that admits a coloring. It is interesting to note that in many proof techniques employed to obtain optimal coloring of graphs, the structure of dominating sets is exploited; see [9] for examples. We refer to a classical work of Bacsó [2] for a theoretical foundation of structural domination.
Wolk [21] showed that every connected induced subgraph of a graph is dominated by the graph if and only if is (, )free. Bacsó and Tuza [4], and Cozzens and Kellehar [11] independently showed that every connected induced subgraph of a graph is dominated by a complete graph if and only if is (, )free. Pim van’t Hof and Paulusma [18] showed that every connected induced subgraph of a graph is dominated by a complete bipartite graph (not necessarily induced) if and only if is (, )free. Bacsó, Michalak, and Tuza [3] characterized the classes of graphs which are dominated by bipartite graphs, cycles, stars, and complete partite graphs. For other related results, we refer to [3, 6, 18] and the references therein. In this paper, by employing similar techniques as in [3], we show that every connected induced subgraph of a graph is dominated by an induced connected split graph if and only if is free, where is a set of six graphs which includes and , and each containing an induced . A similar characterisation is shown for the class of graphs which are dominated by induced complete split graphs. Motivated by these results, we study the structural descriptions of some classes of free graphs. In particular, we give structural descriptions for the class of (,,, gem)free graphs and for the class of (,,, diamond)free graphs. Using these results, we show that every (,,, gem)free graph satisfies , and that every (,,, diamond)free graph satisfies . These two upper bounds are tight for any subgraph of the Petersen graph containing a .
A family of graphs is bounded [15] with binding function if holds whenever and is an induced subgraph of . Thus the class of (,,, gem)free graphs, and the class of (,,, diamond)free graphs are bounded. In this respect, we note the following existing results which are relevant to this paper, and we refer to a recent extensive survey [19] for several families of graphs which admit a binding function.

Every (, diamond)free graph satisfies [7].

Every (, gem)free graph satisfies [9].

Every (, )free graph satisfies [16].

Every ()free graph satisfies [8].

Since the class of free graphs admits a binding function () [15], the class of free graphs is bounded. Also we note that the problem of obtaining a polynomial binding function for the class of free graphs is open, and is open even for the class of ()free graphs; the best known binding function for such class of graphs is [10].

Since the class of free graphs does not admit a linear binding function [14], it follows that the class of free graphs too does not admit a linear binding function.
Furthermore, the class of free graphs are of particular interest in algorithmic graph theory as well since the computational complexity of Minimum Dominating Set Problem, Maximum Independent Set Problem, and Colorability Problem (for ) are unknown for the class of free graphs.
2 Notation and terminology
We follow West [20] for standard terminology and notation.
A hole in a graph is an induced subgraph which is a cycle of length at least four. A hole is called even if it has an even number of vertices. An evenholefree graph is a graph with no even holes.
If and are two vertex disjoint graphs, then is the graph with vertex set and the edge set . For any two disjoint subsets and of , we denote by , the set of edges with one end in and other end in . We say that is complete to or is complete if every vertex in is adjacent to every vertex in ; and is anticomplete to if . If is singleton, say , we simply write is complete (anticomplete) to . A cliquecutset of a graph is a clique in such that has more connected components than .
In a graph , the neighborhood of a vertex is the set ; we drop the subscript when there is no ambiguity. The neighborhood of a subset is the set . A vertex of a graph is bisimplicial if its neighborhood is the union of two cliques (not necessarily disjoint). Given a set , denote the subgraph of induced by in , and for any , we denote the set by . If , then a vertex is a private neighbor of some if is adjacent to and is nonadjacent to every vertex in . Given a connected graph , a vertex of is a cutvertex of if is disconnected. Given a graph , attaching a leaf to a vertex of means adding a vertex and an edge to .
A blowup of a graph with vertices is any graph such that can be partitioned into (not necessarily nonempty) cliques , , such that is complete to if , and is anticomplete to if . A blowup of a graph is a maxblowup if is maximal for each . In a blowup of a graph , if is notempty for some , then for convenience, we call one vertex of as .
3 Graphs dominated by split graphs
A is a graph on four vertices and , and four edges and . Let be the graph obtained from the paw by adding the vertex and an edge , and be the graph obtained from the by adding an edge .
Let and let .
For the proof of our main result of this section, we rewrite a characterisation of split graphs where every forbidden graph is connected. A well known result of Földes and Hammer [13] states that a graph is a split graph if and only if it is ()free. A recent result proved in [12] states that a connected graph is free if and only if it is free. Thus we have the following lemma.
Lemma 3.1
A connected graph is a split graph if and only if it is free.
Lemma 3.2
A connected graph is a complete split graph if and only if it is free.
Proof. Using a result in [17], it is easy to deduce that a graph is a complete split graph if and only if it is free. So it is sufficient to show that a connected graph is free if and only if is (, paw)free. If is free, then obviously is (, paw)free as and paw both contain a . Conversely, let be a (, paw)free graph. Suppose to the contrary that there exists a in with vertices and such that . Then since is connected there exists a shortest path of length at least 2 between and . Now since is free, is of length , and thus there exists a vertex such that is adjacent to both and . But then induces either a or a paw, a contradiction. So is free and the lemma follows.
Let and be five graphs as shown in Figure 1.
Let and .
Theorem 3.1
Every connected induced subgraph of a graph is dominated by an induced connected split graph if and only if is free.
Proof. If every connected induced subgraph of a graph is dominated by an induced connected split graph, then is free as none of the graphs in is dominated by a connected split graph. To prove the converse, we may assume that is connected and it is sufficient to prove that is dominated by an induced connected split graph. In view of Lemma 3.1, we prove that there exists a connected dominating induced subgraph of , where is free.
Let be the set of all subgraphs of such that is a connected dominating induced subgraph of . Then we claim the following.
(1) 
Proof of : Suppose to the contrary that every graph in contains a . Choose such that contains minimum number of 4cycles with as many leaves as possible attached to a 4cycle, say . We prove that each vertex has a leaf attached to it. If is a cutvertex of , then there exists a leaf attached to it. If is a non cutvertex of , then there exists a private neighbor of in (otherwise, is a connected dominating subgraph with at least one less than which is a contradiction). But then is a connected dominating subgraph of with the same number of 4cycles as , but with more leaves attached to which is a contradiction. Hence every vertex in has a leaf attached to it. This implies that contains , a contradiction to the fact that is free. So holds.
Let be the set is free. Then by , is nonempty. Then a similar proof as in shows that the following holds.
(2) 
Let be the set is free. Then by , is nonempty. Then we claim the following.
(3) 
Proof of : Suppose to the contrary that every graph in contains a . Let us choose containing minimum number of copies of with as many leaves as possible attached to non cut vertices of a ; and we call this copy of as . We prove that each non cutvertex has a leaf attached to it. If is a cutvertex of , then there exists a leaf attached to it. If is a non cutvertex of , then there exists a private neighbour of in (otherwise, with less number of copies of than which is a contradiction). Moreover, as adding to does not induce a or a . But then contains the same number of copies of as with more leaves attached to which is a contradiction. Hence every non cutvertex in has a leaf attached to it. This implies that contains , a contradiction to the fact that is free. So holds.
Let be the set is free. Then by , is nonempty. Then we claim the following.
(4) 
Proof of : Suppose to the contrary that every graph in contains a . Let us choose containing minimum number of copies of with as many leaves as possible attached to non cut vertices of a ; and we call this copy of as . Then we shall prove that each non cutvertex has a leaf attached to it. If is a cutvertex of , then there exists a leaf attached to it. If is a non cutvertex of , then there exists a private neighbour of in ; for otherwise, contains less number of copies of than , a contradiction. Also we see that as adding to does not induce a or a or a . But then will have same number of copies of as , but with more leaves attached to , a contradiction. Hence every non cutvertex in has a leaf attached to it. This implies that contains , a contradiction to the fact that is free. So holds.
Since is ()free, we see that every minimal dominating subgraph of is free (see also [5]). So there exists a minimal dominating subgraph which is free. This completes the proof.
Theorem 3.2
Every connected induced subgraph of a graph is dominated by an induced complete split graph if and only if is free.
Proof. If every connected induced subgraph of is dominated by an induced complete split graph, then is free as none of the graphs in is dominated by a complete split graph. To prove the converse, we may assume that is connected and it is sufficient to prove that is dominated by an induced complete split graph. In view of Lemma 3.2, we prove that there exists a connected dominating induced subgraph of , where is free.
Let be the set of all subgraphs of such that is a connected dominating induced subgraph of . Let be the set is free. A proof similar to that of Theorem 3.1 proves that is not empty. Then we claim there exists a graph in which is pawfree. Suppose not. Choose containing a minimum number of copies of paw with as many leaves as possible attached to non cut vertices of a paw; and we call this copy of paw as . We prove that each non cutvertex of has a leaf attached to it. If is a cutvertex of , then there exists a leaf attached to it. If is a non cutvertex of , then there exists a private neighbour of in ; for otherwise, belongs to , containing less number of copies of paw than , a contradiction. Moreover, as adding to does not induce a . But then contains same number of copies of paw as with more leaves attached to , a contradiction. Hence every non cutvertex in has a leaf attached to it. This implies that contains , a contradiction to the fact that is free.
Since is ()free, we see that every minimal dominating subgraph of is free (see also [5]). So there exists a minimal dominating subgraph which is free. This completes the proof.
4 The class of (, , , gem)free graphs
Let . In this section, we give a structural description of free graphs, and show that the class of free graphs is bounded.
Let be a .
Let be the graph obtained from by adding vertices , and edges and .
Let be the graph obtained from by adding vertices , and edges and .
Throughout this section, we follow the convention that if the set of vertices induces a gem, then  is an induced path in the neighborhood of .
Theorem 4.1
Let be a connected free graph that contains an induced , say . Let be the vertex set of a maxblowup of contained in such that . Let has no neighbor in . For each , mod , let:
Let , , , and . Then the following properties hold for all :

and hence .

is complete to .

, , are empty.

One of and is empty, one of and is empty, and one of and is empty.

If is not empty, then (i) is empty, and (ii) or .

If is free, then is empty.

If is free, then is empty.
Proof.
(a) Consider any . For each , let
be a neighbor of in (if any such vertex exists) and
be a nonneighbor of in (if any exists). Let
exists. Then since . If for some , then since
 is a , does not exist and hence is complete to , and so .
Now up to symmetry is one of the following sets (1)–(4). In each case we make an observation.
(1) or for some . Then .
(2) for some . Then .
(3) for some . Then is complete to as proved above. Moreover, is complete to , for otherwise or induces a gem.
So can be added to , contradicting the
maximality of .
(4) for some or . Then induces
a gem for some .
So we conclude that , and hence (a) holds.
Note that for each , by the symmetry of and is mod 6. Also by (b), any is complete to . From now on, we use the following notation for each . For a vertex , we let be a neighbor of in . Also, for a vertex , we let and be neighbors of , respectively, in and . We prove the statements (c)–(g) for ; for other cases of the proof is similar.
(c) If and , then induces a gem. If and , then  or  or  is a . If and , then  is a . If and , then  or  is a . If and , then  is a . If and , then  or  is a . If and , then  is a . If belongs to one of the sets listed in (c), then it is symmetric to one of the cases showed above. This proves (c).
(d) It is sufficient to prove or is empty. If there are vertices and , then has a neighbor , has a neighbor . By (c), is not adjacent to . But then  is a . So (d) holds.
(e) Let and be adjacent. Suppose to the contrary that , and let . Then by (c), is anticomplete to . But now  is a or  is a , a contradiction.
If , then by (e:(i)), , and by (d), . So . If , then by (e:(i)), , and by (d), . So . Thus (e) holds.
(f) Suppose that there is an edge with and . Let be a neighbor of in , and be a neighbor of in . Now induces . So (f) holds.
(g) Suppose, up to symmetry, that there are adjacent vertices and . Since is complete to (by (b)), we have is complete to . But then induces . So (g) holds.
This completes the proof of the theorem.
Theorem 4.2
Let be a connected free graph. Then the following hold:

If contains an , then has a bisimplicial vertex.

If contains an , then either has a clique cutset or has a bisimplicial vertex.
Proof. We may assume that contains an or an with the same vertexset and edgeset as defined earlier. Since both and contain an induced , let be the vertex set of a maxblowup of contained in such that . Now we partition the vertexset of as in Theorem 4.1, and we use the properties in Theorem 4.1 with the same notation.
Proof of : Suppose that contains . Note that and . Then by Theorem 4.1(d), is empty, and one of and is empty. We may assume, up to symmetry, that . Now we claim that . Suppose there is a vertex , then has a neighbor and , and since is anticomplete to (by Theorem 4.1(c)), we have  a , which is a contradiction. So . Now we see that is a union of two cliques and , and hence is a bisimplicial vertex. This proves (i).
Proof of : Suppose that contains . Assuming that has no clique cutset, we show that has a bisimplicial vertex. Note that and . Then by Theorem 4.1(b), is complete to , and by Theorem 4.1(d), is empty. Moreover, by the definition of , has a neighbor in , say . Now here too we claim that . Suppose there is a vertex , then has a neighbor and , and since is anticomplete to (by Theorem 4.1(c)), we have  a , which is a contradiction. So is empty. Next we claim that:
(1) 
Proof of : Suppose that there is an edge with and . Then by Theorem 4.1 (c), we see that , and hence has a neighbor . Since is complete to (by Theorem 4.1 (b)), . From Theorem 4.1 (c) by choosing the appropriate sets, we see that is anticomplete to , and . Hence  is a . Also is complete to (by Theorem 4.1 (b)), and since can not induce a gem, we have . But now we see that  is a , which is a contradiction. This proves (1).
Next we claim that:
(2) 
Proof of : Suppose . Let be a connected component in containing . Then by (1), we see that which is a clique. This implies that is a clique cutset separating from , which is a contradiction. This proves (2).
Now by the above observations we note that is a union of two cliques and , and hence is a bisimplicial vertex. This proves (ii).
Thus the proof of Theorem 4.2 is complete.
Let be the graph obtained from by adding vertices and , and edges and .
Theorem 4.3
Let be a connected free graph that contains an induced . Then has a cliquecutset or has a bisimplicial vertex or is a blowup of the Petersen graph.
Proof. First note that if contains either or , then by Theorem 4.2, has a clique cutset or has a bisimplicial vertex, and the theorem holds. So we may assume that is (, )free. Moreover, we may assume that has no clique cutset. We may assume that contains an with the same vertexset and edgeset as defined earlier. Since contains an induced , let be the vertex set of a maxblowup of contained in such that . Now we partition the vertexset of as in Theorem 4.1, and we use the properties in Theorem 4.1 with the same notation. Note that and .
Now if is not empty, for some , then, by (e) (of Theorem 4.1), either or . Hence either or is a bisimplicial vertex, and we conclude the theorem. So we may assume that: , for all . Then by (g) and (c), we see that is empty, for all . Hence = is a clique, for all . Since has no clique cutset, it follows that:
is empty, for every .  (1) 
Moreover, we claim that:
(2) 
Proof of (2): Clearly, . If , then is a bisimplicial vertex, and we conclude the theorem. If and , then by (1) and (c) we see that is empty. Since is not a clique cutset, it follows that . So there is an edge, say with and . Then by the definition of ’s, has a neighbor in , say , and has a neighbor in , say . But now induces an . So we conclude that is not empty. Likewise, is not empty. This proves (2), by the symmetry of .
We say a vertex is pure if is complete to , and is good if has a neighbor in .
Now we claim that, for each :
(3) 
Proof of : We prove for . Let be a good vertex in . So there exists a neighbor of in , say . If has a nonneighbor in , say , then since has a neighbor in , say , we see that  is a . So is complete to . Likewise, is complete to . This proves (3).
Next we claim that, for each
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