The Structure of Graphs with Forbidden Induced $C_4$, $\Overline{C}_4$, $C_5$, $S_3$, Chair and Co-chair

We find the structure of graphs that have no C4, $\overline{C}_4$, C5, S3, chair and co-chair as induced subgraphs. Then we deduce the structure of the graphs having no induced C4, $\overline{C_4}$, S3, chair and co-chair and the structure of the graphs G having no induced C4, $\overline{C_4}$ and such that every induced P4 of G is contained in an induced C5 of G.


Introduction
In this paper, graphs are finite and simple. The vertex set and edge set of a graph G are denoted by V (G) and E(G) respectively. Two edges of a graph G are said to be adjacent if they have a common endpoint and two vertices x and y are said to be adjacent if xy is an edge of G. The neighborhood of a vertex v in a graph G, denoted by N G (v), is the set of all vertices adjacent to v and its degree is d G (v) = |N G (v)|. We omit the subscript if the graph is clear from the context. For two set of vertices U and W of a graph G, let E[U, W ] denote the set of all edges in the graph G that joins a vertex in U to a vertex in W . A graph is empty if it has no edges. For A ⊆ V (G), G[A] denotes the sub-graph of G induced by A. If G[A] is an empty graph, then A is called a stable set. While, if G[A] is a complete graph, then A is called a clique set, that is any two distinct vertices in A are adjacent. The complement graph of G is denoted by G and defined as follows: V (G) = V (G) and xy ∈ E(G) if and only if xy / ∈ E(G). A graph H is called a forbidden subgraph of G if H is not (isomorphic to) an induced subgraph of G.
A cycle on n vertices is denoted by C n = v 1 v 2 ...v n v 1 while a path on n vertices is denoted by P n = v 1 v 2 ...v n . A chair is any graph on 5 distinct vertices x, y, z, t, v with exactly 5 edges xy, yz, zt and zv. The co-chair or chair is the complement of a chair. S 3 is the graph on 6 vertices as indicated in Figure 1. Many graphs encountered in the study of graph theory are characterized by configurations or subgraphs they contain. However, there are occasions where it is easier to characterize graphs by sub-graphs or induced sub-graphs they do not contain. For example, trees are the connected graph without (induced) cycles. Bipartite graphs are those without (induced) odd cycles ( [1]). Split graphs are those without induced C 4 , C 4 and C 5 . Line graphs are characterized by the absence of only nine particular graphs as induced sub-graph (see [2]). Perfect graphs are characterized by C 2n+1 and C 2n+1 being forbidden, for all n ≥ 2 (see [3]). The purpose of this paper is to find the structure of graphs such that C 4 , C 4 , C 5 , S 3 chair and co-chair are forbidden subgraphs. These graphs will be called generalized combs and they are generalization of generalized stars and generalization of combs (See [6,8]). Seymour's Second Neighborhood Conjecture (see [9]) is proved for orientation of graphs obtained from the complete graph by deleting the edges of a generalized star and for those obtained by deleting the edges of a comb [6,8]. Generalized stars (also called threshold graphs) are the graphs with C 4 , C 4 and P 4 forbidden. Finding the structure of the generalized comb, might give a clearer vision for an attempt to prove Seymour's conjecture for oriented graphs obtained from the complete graph by deleting the edges of a generalized comb.

Preliminary Definitions and Theorems
Definition 1. A graph G is a called a split graph if its vertex set is the disjoint union of a stable set S and a clique set K. In this case, G is called an {S, K}-split graph.
If G is an {S, K}-split graph and ∀s ∈ S, ∀x ∈ K we have sx ∈ E(G), then G is called a complete split graph.
If G is an {S, K}-split graph and E[S, K] forms a perfect matching of G, then G is called a perfect split graph.
where the A i 's and X i 's are pair-wisely disjoint sets.
X i is a clique and the X i 's are nonempty, except possibly X n+1 .

3) S := n i=0
A i is a stable set and the A i 's are nonempty, except possibly A 0 .
is a complete split graph.

5)
The only edges of G are the edges of the subgraphs mentioned above.
In this case, G is called an {S, K}-threshold graph.
In fact, threshold graphs are exactly the generalized stars defined in [6].
is empty or isomorphic to the cycle C 5 ; 3) every vertex in C is adjacent to every vertex in K but to no vertex in S.

Main Results
Lemma 3.1. Suppose that C 4 , C 4 , C 5 , chair and co-chair are forbidden subgraphs of G. If the path mbb ′ m ′ is an induced subgraph of G, then: Proof. Since C 4 , C 4 and C 5 are forbidden, then G is an {S, K}-split graph for some stable set S and a clique set Since xm is an edge of G and S is stable, then we must have x ∈ K. But K is a clique, then x is adjacent to b and b ′ . Thus Proof. We prove this by induction on the number of vertices of G. This is clearly true for small graphs. Suppose that P 4 is a forbidden subgraph of an {S, K}-split graph G. It is clear that G is a threshold graph. We have to prove that G is {S, K}-threshold graph. Let x ∈ K be a vertex with minimum degree in G, that is d G (x) = min{d G (y); y ∈ K} and G ′ := G − x be the graph induced by the vertices of G except x (If K = φ, then the statement is true). Then P 4 is a forbidden subgraph of the {S, K − {x}}-split graph G ′ . By the induction hypothesis, G ′ is an {S, K − {x}}threshold graph. We follow the notations in Definition 2. Assume that ∃a ∈ S − A n such that ax ∈ E(G). Let x n ∈ X n . Since d(x n ) ≥ d(x), then there is a n ∈ A n such that a n x n ∈ E(G) but a n x / ∈ E(G). Then axx n a n is an induced P 4 in G. Contradiction. Thus we may suppose that In this case we do the following: (These sets are called the sets of the generalized comb G). www.ejgta.org The structure of graphs with forbidden induced C 4 , C 4 , C 5 , S 3 , chair and co-chair | S. Ghazal and A 0 are the only possibly empty sets among the X ′ i s, Y ′ i s, A ′ i s.

9)
The only edges of G are the edges of the subgraphs mentioned above.
In this case, we say that G is an {S, K}-generalized comb. Note, that we may assume that no two consecutive sets M i and M i+1 are both empty. We use this assumption in the rest. It is clear that the comb defined in [8] is a particular case of the generalized comb (see Figure  3). Moreover, we have the following:   Assume that ∃i ≥ 2 such that b ∈ X i . Let x ∈ X 1 and a ∈ A 1 . Then mbxa is induced P 4 in G, a contradiction. So b ∈ X 1 . Then also b ′ ∈ X 1 , because b and b ′ have the same neighborhood in H.
Otherwise, G has at least two induced P 4 . Let m be a vertex of G such that d(m) = min{d(z); z is a leaf of an induced P 4 in G} and let P = mbb ′ m ′ be an induced P 4 . Note that d(m) = d(m ′ ).
. We may assume without loss of generality that m / ∈ {u, u ′ } and let Suppose first that m ′ ∈ {u, u ′ } and assume without loss of generality that m ′ = u ′ . Assume that b ′ = d ′ . If b = d, then by using Lemma 3.1 repeatedly, we can prove easily that ∈ E(G). By applying Lemma 3.1 repeatedly, we have the following: , a contradiction. Since udd ′ u ′ = udb ′ m ′ is an induced path of length four of G, then by Lemma 3.1 also udbm is an induced path of G and thus of H. Then, by the definition of the generalized comb H, ∃i; u, m ∈ M i (We follow the notations of definition 3.). In this case we add m ′ to M i and b ′ to Y i . This shows that G is an {S, K}−generalized comb. Now, suppose that m ′ / ∈ {u, u ′ }. Assume that m ∈ A. By definition of the generalized comb H and since udd ′ u ′ is an induced P 4 of H, we get that . Then b / ∈ N (u) and thus by Lemma 3.1, we , which is a contradiction to the choice of m. Hence, b ∈ N H (u) and so, by Lemma 3.1, So m ∈ M . Let l be the greatest such that M l = φ. Suppose that m / ∈ M l . Let m ′′ ∈ M l and b ′′ ∈ Y l be its neighbor. ∃i < l such that m ∈ M i . Then b ′′ m ∈ E(G) and N H (m ′′ ) ⊆ N H (m). Let c ∈ Y i be the neighbor of m. Let k be the smallest such that k > i and M k = φ (Note that k exists and i < k ≤ l, moreover we may assume k = i + 1 or k = i + 2).
Suppose b ∈ N (m ′′ ). Then also b ′ ∈ N (m ′′ ). If b = b ′′ , then ∃j > k such that b ∈ Y j . Then by using Lemma 3.1, we can prove easily that G[{m, m ′ , m ′′ , b, b ′ , c}] is an induced S 3 of G, a contradiction. However, if b = b ′ , then also by using Lemma 3.1, we can observe that Therefore m ∈ M l . Let Y l ∩ N (m) = {c}. If b = c, then we add b ′ to Y l and m ′ to M l and thus G is {S, K}-generalized comb. Now suppose that b = c. Suppose that c is not the only vertex in Y l and thus there is an induced path mcc ′′ m ′′ with c, c ′′ ∈ Y l and m ′′ ∈ Y l . By using Lemma 3.1, we can prove easily that G[{b, b ′ , c, m, m ′ , m ′′ }] is an induced S 3 of G a contradiction. Therefoe c is the only vertex in Y l . Since bm ∈ E(H), then b ∈ Y l+1 . We do the following: To Y l+1 add c and remove b and to Y l add b and remove c. Then we add b ′ to Y l and m ′ to M l (as in the case b = c) and this shows that G is an {S, K}-generalized comb. Proof. The necessary condition is obvious by the definition of the generalized comb. For the sufficient condition it is enough to note that the statement C 4 , C 4 , C 5 , S 3 , chair and co-chair are forbidden subgraphs of G is equivalent to the statement that G is a split graph and S 3 , chair and co-chair are forbidden subgraphs of G.

Corollary 3.2. G is a generalized comb if and only if every induced subgraph of G is a generalized comb.
Proof. Let G ′ be an induced subgraph of a generalized comb G. It is clear that G ′ contains no induced C 4 , C 4 , C 5 , chair and co-chair. Thus G ′ is a generalized comb. The sufficient condition is clear. 3) every vertex in C is adjacent to every vertex in K but to no vertex in S.
Proof. The sufficient condition is clear by construction of G. We prove the necessary condition. Suppose that C 4 , C 4 , S 3 , chair and co-chair are forbidden subgraphs of a graph G. Then by Theorem 2.3, V (G) is disjoint union of three sets S, K and C such that:

1) G[S ∪ K] is an {S, K}-split graph;
2) G[C] is empty or isomorphic to the cycle C 5 ; 3) every vertex in C is adjacent to every vertex in K but to no vertex in S.  3) every vertex in C is adjacent to every vertex in K but to no vertex in S.
Proof. The sufficient condition is clear by construction of G. We prove the necessary condition. Suppose that C 4 , C 4 are forbidden subgraphs of a graph G and every induced P 4 of G is contained in an induced C 5 of G. Then by Theorem 2.3, V (G) is disjoint union of three sets S, K and C such that: 2) G[C] is empty or isomorphic to the cycle C 5 ; 3) every vertex in C is adjacent to every vertex in K but to no vertex in S.
Then G[C] is the unique induced C 5 of G or G has no induced C 5 . Then C 4 , C 4 , P 4 are forbidden subgraphs of G[S ∪ K]. Thus G[S ∪ K] is an {S, K}-threshold graph.