Forbidden Induced Subgraphs for Star-Free Graphs

Forbidden induced subgraphs for star-free graphs

Jun Fujisawa1∗ Katsuhiro Ota2 Kenta Ozeki3† Gabriel Sueiro2

1Department of Applied Science, Kochi University

2Department of Mathematics, Keio University

3National Institute of Informatics

Abstract Let H be a family of connected graphs. A graph G is said to be H-free if G is H-free for every graph H in H. In [1] it was pointed that there is a family of connected graphs H not containing any induced subgraph of the claw having the property that the set of H-free connected graphs containing a claw is ﬁnite, provided also that those graphs have minimum degree at least two and maximum degree at least three. In the same work, it was also asked whether there are other families with the same property. In this paper we answer this question by solving a wider problem. We consider not only the claw-free graphs but the more general class of star-free graphs. Concretely, given t ≥ 3, we characterize all the graph families H such that every large enough H-free connected graph is K1,t-free. Additionally, for the case t = 3 we show the families that one gets when adding the condition |H| ≤ k for each positive integer k.

Keywords: star-free, claw-free, forbidden subgraph

1 Introduction

In this paper we only consider simple ﬁnite graphs. Let G be a connected graph. Given a connected graph H, G is said to be H-free if G does not contain H as an induced subgraph. Given a family of connected graphs H, G is said to be H-free if G is H-free for all H ∈ H. K1,3-free graphs are also called claw-free graphs. Claw-free graphs have been widely studied in the literature (see for example [6] for a survey on claw-free graphs). The search for forbidden induced subgraphs implying some property on a graph has received a lot of attention in recent years. See for example [2, 3, 8, 9, 10, 11, 12, 13]. The following result was proved in [1] (the graphs Y4,Z2,3, W3,2 and K2,3 will be deﬁned in Section 2 and Section 4).

Theorem 1 ([1]). Let G be a connected graph with minimum degree at least two and maximum degree at least three. If G is {Y4,Z2,3,W3,2,K2,3}-free and |V (G)| ≥ 9, then G is K1,3-free.

∗This work was supported by JSPS Grant-in-Aid for Young Scientists (Start-up). †Research Fellow of the Japan Society for the Promotion of Science.

1 The interesting point about this result is that even though no graph of the family H = 3 {Y4,Z2,3,W2 ,K2,3} is an induced subgraph of K1,3, when considering the H-free graphs under certain conditions, the graph K1,3 is also forbidden. In the same work [1], the following theorem was also proved. Theorem 2 ([1]). Let G be a connected graph with minimum degree at least two and maximum degree at least three. Let t ≥ 3 and let H0 be the set of connected graphs of order 0 t + 2 containing an induced K1,t. If G is (H − {K1,t+1})-free and |V (G)| ≥ t + 2, then G is K1,t-free. Under the view of these results, it is natural to ask what other families of graphs satisfy the same property. We remove the conditions on the minimum and maximum degree since those are necessary conditions related to the problem studied in [1]. We can state our problem as follows. Given t ≥ 3, characterize all the families of connected graphs H such that every large enough H-free connected graph is K1,t-free. In this paper, we solve this problem for all t ≥ 3. The rest of the paper is organized as follows. In Section 2, we make all needed deﬁnitions and present our main results. In Section 3, we do the corresponding proofs. In Section 4, we consider restricting the size of the family of forbidden subgraphs for the case t = 3. Concretely, we characterize the families of forbidden subgraphs H with |H| ≤ k for each k ≤ 1. See Section 4 for a formal statement of the problem. In Section 5, we give the corresponding proofs. Finally, in Section 6 we make some discussion, propose some open questions and comment on the cases t = 1 and t = 2.

2 Deﬁnitions and main results

If H1 and H2 are two connected graphs, we write H1 H2 to indicate that H1 is an induced subgraph of H2. We say that a family of connected graphs H is redundant if there are two different graphs H1,H2 ∈ H such that H1 H2. It is easy to see that we can restrict our problem to considering only non-redundant families. Define G as the set of all non-redundant families of connected graphs. Let t ≥ 3 and define H(t) as the set of families H ∈ G such that there is a constant n0 = n0(t, H) with the property that all H-free connected graphs G with |V (G)| ≥ n0 are K1,t-free. Then, our problem is reduced to finding all the elements in the set H(t). We define a binary relation “≤” in G as follows. For H1, H2 ∈ G, we say that H1 ≤ H2 if for each H2 ∈ H2, there is an H1 ∈ H1 such that H1 H2. It is easy to see that the relation “≤” defines a partial order in G. Furthermore, if H1 ≤ H2 then any H1-free graph is also an H2-free graph (see for example Lemma 3 of [7]). Kn is the complete graph on n vertices. Pn is the path on n vertices. Kn,m is the complete bipartite graph with partite sets on n and m vertices.

Let t ≥ 2. To state our results we deﬁne the following graphs (see Figure 1).

t • Ym is a path on m vertices with t − 1 extra vertices attached to the ﬁrst vertex of the t path. The last vertex of the path is called the tail of Ym.(m ≥ 1) t • Ys,m is the graph obtained by joining s degree one vertices of a K1,t with the ﬁrst t vertex of the path on m vertices. The last vertex of the path is called the tail of Ys,m. (m ≥ 1, 1 ≤ s ≤ t)

t • Ybs,m is the graph obtained by joining s degree one vertices of a K1,t with the ﬁrst vertex of the path on m vertices and adding the edge between the center of the K1,t t and the ﬁrst vertex of the path. The last vertex of the path is called the tail of Ybs,m. (m ≥ 1, 1 ≤ s ≤ t)

2 t • Wq is the graph obtained by joining a Kq with t independent vertices. (q ≥ 1) t • Ts,q is the graph obtained by joining s degree one vertices of a K1,t with all the vertices of a Kq.(q ≥ 1, 1 ≤ s ≤ t)

t • Ds,q is the graph obtained by joining s degree one vertices and the center of a K1,t with all the vertices of a Kq.(q ≥ 1, 0 ≤ s ≤ t)

t t t • Zm,r,Zs,m,r and Zbs,m,r are the graphs obtained by identifying a vertex of a Kr with t t t the tail of a Ym,Ys,m and Ybs,m, respectively. (m ≥ 1, r ≥ 1, 1 ≤ s ≤ t)

t l

t K1,l Wq

Kq s t − s s Kq+1 t − s

t t Ts,q Ds,q

t − 1 Kr t − 1

m m t t Ym Zm,r

s t − s Kr s t − s

m m t t Ys,m Zs,m,r

s t − s Kr s t − s

m m t t Ybs,m Zbs,m,r Figure 1: Some forbidden subgraphs

3 For t ≥ 3, deﬁne the following families of graphs. t t •T (q) = { Ts,q: 2 ≤ s ≤ t − 1 }. t t •D (q) = { Ds,q: 2 ≤ s ≤ t − 1 }. t t •Y (m) = { Ys,m: 2 ≤ s ≤ t − 2 }. t t •Z (m, r) = { Zs,m,r: 2 ≤ s ≤ t − 2 }.

t t • Yb (m) = { Ybs,m: 2 ≤ s ≤ t − 2 }.

t t • Zb (m, r) = { Zbs,m,r: 2 ≤ s ≤ t − 2 }.

•YZ t(m, r) = Yt(m+2)∪Zt(1, r)∪...∪Zt(m, r)∪Ybt(m+2)∪Zbt(1, r)∪...∪Zbt(m, r). t t t t t t t t •H (m, l, q, r) = {K1,l,Wq } ∪ {Ym+2,Z1,r,...,Zm,r} ∪ T (q) ∪ D (q) ∪ YZ (m, r). Notice that for the case t = 3, Yt(m), Zt(m, r), Ybt(m) and Zbt(m, r) are empty and both T t(q) and Dt(q) have only one element.

For t ≥ 3, define the following subset of G. • F(t) = { H ∈ G: H ≤ Ht(m, l, q, r) for some m ≥ 1, l ≥ t + 1, q ≥ 2, r ≥ 3}. Our main result in this paper is the following theorem. It gives the characterization of families of forbidden subgraphs for star-free graphs we described in Section 1. Theorem 3. Let t ≥ 3, then H(t) = F(t). For our proofs we need the following definitions. For terminology and notation not defined in this paper, we refer the reader to [4]. i Let G be a connected graph. For v ∈ V (G), define NG(v) = {w ∈ V (G): the distance 0 1 from v to w is exactly i}. Notice that NG(v) = {v} and NG(v) = NG(v). If the graph G is i i obvious from the context, we sometimes write N (v) for NG(v). A clique of a graph is a set of pairwise adjacent vertices, and an independent set is a set of pairwise nonadjacent vertices. For two positive integers l and r, the Ramsey number R(l, r) is the minimum positive integer R such that any graph of order at least R contains either an independent set of cardinality l or a clique of cardinality r. The Ramsey number R(l, r) exists for every positive integers l and r (see for example [4]). 0 0 0 If G is a graph and S ⊆ V (G), for S ⊆ S, define BS(S ) = {v ∈ V (G): N(v) ∩ S = S }. Observation: if for some N ⊆ V (G), there is a constant k such that for every S0 ⊆ S, 0 |S| |N ∩ BS(S )| ≤ k, then |N| ≤ 2 · k (remember that the number of subsets of a set S is 2|S|). We will implicitly use this fact in the proofs of several lemmas in section 3.

3 Proof of Theorem 3

First, we will prove the following theorem that shows that forbidding some family of F(t) is enough to imply that the graph is star-free provided it is large enough. We prove this fact in the following theorem. Theorem 4. Let t ≥ 3. Then F(t) ⊆ H(t). Before giving the proof, we would like to comment on non-redundancy of the family Ht(m, l, q, r). It is not diﬃcult to check that the family Ht(m, l, q, r) is non-redundant for the parameters used in the deﬁnition of F(t)(m ≥ 1, l ≥ t + 1, q ≥ 2, r ≥ 3). These conditions were chosen so that Ht(m, l, q, r) is not redundant nor it contains any induced subgraph of K1,t. Moreover, reducing by 1 any of the constants in the condition would make t H (m, l, q, r) either redundant or contain an induced subgraph of K1,t.

4 We divide the proof of Theorem 4 in several lemmas that we state and prove bellow.

Lemma 5. Let t ≥ 3 and let G be a graph with an induced K1,t of center x0. If G is t t t m+1 ({Ym} ∪ Y (m) ∪ Yb (m))-free for some m ≥ 3, then N (x0) = ∅.

Proof. Let Y ⊆ V (G) with |Y | = t, such that {x0} ∪ Y is an induced K1,t in G. Suppose m+1 t t that N (x0) 6= ∅. We will show that G contains a Ym, some graph of Y (m) or some graph of Ybt(m), which is a contradiction. i Let k = m + 1 and let P = x0x1 ··· xk be an induced path of G with xi ∈ N (x0) for j all 0 ≤ i ≤ k. Notice that N (x0) ∩ N(Y ) = ∅ for all 3 ≤ j ≤ k. Otherwise, an element j v ∈ N (x0) ∩ N(Y ) would have a path of length 2 to x0, passing through some element of j Y , contradicting that v ∈ N (x0). Then N(Y ) ∩ P ⊆ {x0, x1, x2}. Let Y1 = N(x1) ∩ Y and Y2 = N(x2) ∩ Y . If |Y2| ≥ t − 1, then Y2 ∪ {x2, . . . , xm+1} t t contains a Ym. If 2 ≤ |Y2| ≤ t − 2, then (Y − Y2) ∪ {x0} ∪ Y2 ∪ {x2, . . . , xm+1} is a Ys,m, t where s = |Y2|. If |Y2| = 1, then (Y − Y2) ∪ {x0} ∪ Y2 ∪ {x2, . . . , xm−1} is a Ym. Suppose now that |Y2| = 0, that is N(x2) ∩ Y = ∅. If |Y1| ≥ t − 1, then Y1 ∪ {x1, . . . , xm} t t contains a Ym. If 2 ≤ |Y1| ≤ t − 2, then (Y − Y1) ∪ {x0} ∪ Y1 ∪ {x1, . . . , xm} is a Ybs,m, where t s = |Y1|. If |Y1| ≤ 1, then (Y − Y1) ∪ {x0, . . . , xm−1} contains a Ym.

Lemma 6. Let t ≥ 3 and let G be a graph with an induced K1,t of center x0. Suppose t t t that G is ({K1,l,Z1,r,Wq } ∪ D (q))-free for some l ≥ t + 1, r ≥ 3, q ≥ 2. Then |N(x0)| < 2t · R(l, max(r, q)).

Proof. Let Y ⊆ V (G) with |Y | = t, such that {x0} ∪ Y is an induced K1,t in G. Let 0 0 Y ⊆ Y . We will show that |N(x0) ∩ BY (Y )| < R(l, max(r, q)), and since |Y | = t we get t that |N(x0)| < 2 · R(l, max(r, q)). 0 0 0 If |Y | ≤ 1, then |Y − Y | ≥ t − 1 and so |N(x0) ∩ BY (Y )| < R(l, r), since otherwise 0 0 t (Y − Y ) ∪ {x0} ∪ (N(x0) ∩ BY (Y )) contains a Z1,r or a K1,l. 0 0 0 0 If 2 ≤ |Y | ≤ t − 1, then |N(x0) ∩ BY (Y )| < R(l, q), since otherwise Y ∪ (Y − Y ) ∪ 0 t 0 {x0} ∪ (N(x0) ∩ BY (Y )) contains a Ds,q or a K1,l, where s = |Y |. 0 0 0 0 If |Y | = t, then |N(x0) ∩ BY (Y )| < R(l, q), since otherwise Y ∪ (N(x0) ∩ BY (Y )) t contains a Wq or a K1,l.

Lemma 7. Let t ≥ 3 and let G be a graph with an induced K1,t of center x0. Suppose that t t t t t G is ({K1,l,Z1,r,Z2,r,Wq } ∪ Zb (1, r) ∪ T (q))-free for some l ≥ t + 1, r ≥ 3, q ≥ 2. Then 2 t |N (x0)| < 2 · R(l, max(r, q)) · |N(x0)|.

Proof. Let Y ⊆ V (G) with |Y | = t, such that {x0} ∪ Y is an induced K1,t in G. Let 0 2 0 x1 ∈ N(x0). Let Y ⊆ Y . Let N = N (x0) ∩ N(x1). It suﬃces to show that |N ∩ BY (Y )| < R(l, max(r, q)). 0 0 0 If |Y | = 1, then |Y − Y | = t − 1 and so |N ∩ BY (Y )| < R(l, r), since otherwise 0 0 0 t (Y − Y ) ∪ {x0} ∪ Y ∪ (N ∩ BY (Y )) contains a Z2,r or a K1,l. 0 0 0 0 If 2 ≤ |Y | ≤ t − 1, then |N ∩ BY (Y )| < R(l, q), since otherwise (Y − Y ) ∪ {x0} ∪ Y ∪ 0 t 0 (N ∩ BY (Y )) contains a Ts,q or a K1,l, where s = |Y |. 0 0 0 0 If |Y | = t, then |N ∩ BY (Y )| < R(l, q), since otherwise Y ∪ (N ∩ BY (Y )) contains a t Wq or a K1,l. 0 0 Suppose now that |Y | = 0, that is N ∩ BY (Y ) ∩ N(Y ) = ∅. Notice that if x1 ∈ Y , then 0 N ∩ BY (Y ) = ∅. Then we may suppose that x1 ∈/ Y . Let Y1 = Y ∩ N(x1). 0 0 If |Y1| ≥ t − 1, then |N ∩ BY (Y )| < R(l, r), since otherwise Y1 ∪ {x1} ∪ (N ∩ BY (Y )) t contains a Z1,r or a K1,l. 0 If 2 ≤ |Y1| ≤ t − 2, then |N ∩ BY (Y )| < R(l, r), since otherwise (Y − Y1) ∪ {x0} ∪ Y1 ∪ 0 t {x1} ∪ (N ∩ BY (Y )) contains a Zbs,1,r or a K1,l, where s = |Y1|. 0 If |Y1| ≤ 1, then |Y − Y1| ≥ t − 1 and so |N ∩ BY (Y )| < R(l, r), since otherwise 0 t (Y − Y1) ∪ {x0, x1} ∪ (N ∩ BY (Y )) contains a Z2,r or a K1,l.

5 Lemma 8. Let t ≥ 3 and let G be a graph with an induced K1,t of center x0. Let i ≥ 2 and t t t t t suppose that G is ({K1,l,Zi−1,r,Zi,r,Zi+1,r} ∪ Z (i − 1, r) ∪ Zb (i, r))-free for some l ≥ t + 1 i+1 i and r ≥ 3. Then |N (x0)| < R(l, r) · |N (x0)|.

Proof. Let Y ⊆ V (G) with |Y | = t, such that {x0} ∪ Y is an induced K1,t in G. Let i j xi ∈ N (x) and let x0x1 ··· xi be an induced path with xj ∈ N (x) for all 0 ≤ j ≤ i. Let i+1 N = N (x0) ∩ N(xi). It suﬃces to show that |N| < R(l, r). Let Y1 = Y ∩N(x1) and Y2 = Y ∩N(x2). As in the proof of Lemma 5, for all 3 ≤ j ≤ i+1, N j(x) ∩ N(Y ) = ∅. t If |Y2| ≥ t − 1, then |N| < R(l, r), since otherwise Y2 ∪ {x2, . . . , xi} ∪ N contains a Zi−1,r or a K1,l. If 2 ≤ |Y2| ≤ t−2, then |N| < R(l, r), since otherwise (Y −Y2)∪{x0}∪Y2∪{x2, . . . , xi}∪N t contains a Zs,i−1,r or a K1,l, where s = |Y2|. If |Y2| = 1, then |N| < R(l, r), since otherwise (Y − Y2) ∪ {x0} ∪ Y2 ∪ {x2, . . . , xi} ∪ N t contains a Zi+1,r or a K1,l. Suppose now that |Y2| = 0, that is N(x2) ∩ Y = ∅. t If |Y1| ≥ t − 1, then |N| < R(l, r), since otherwise Y1 ∪ {x1, . . . , xi} ∪ N contains a Zi,r or a K1,l. If 2 ≤ |Y1| ≤ t−2, then |N| < R(l, r), since otherwise (Y −Y1)∪{x0}∪Y1∪{x1, . . . , xi}∪N t contains a Zbs,i,r or a K1,l, where s = |Y1|. If |Y1| ≤ 1, then |Y − Y1| ≥ t − 1 and so |N| < R(l, r), since otherwise (Y − Y1) ∪ t {x0, . . . , xi} ∪ N contains a Zi+1,r or a K1,l. We use the above lemmas to prove Theorem 4.

Proof of Theorem 4. Let H ∈ F(t). Let m ≥ 1, l ≥ t + 1, q ≥ 2 and r ≥ 3 such that H ≤ Ht(m, l, q, r). Let G be an H-free connected graph. Suppose that there is an induced K1,t of center x0. We will show that |V (G)| is bounded by a function depending only on t, l, m, q and r. t t Notice that since G is Ym+2-free, then G is Zi,r-free for all i ≥ m+1. Since we also know t t that G is Zi,r-free for all 1 ≤ i ≤ m, we conclude that G is Zi,r-free for all i ≥ 1. Using a similar argument, we have that G is Zt(i, r)-free and Zbt(i, r)-free for all i ≥ 1. Thus, G satisﬁes all the conditions of Lemmas 5, 6, 7 and 8. By Lemma 5, N m+1(x) = ∅. Then we only need to show that N i(x) is bounded for all 1 ≤ i ≤ m. By Lemmas 6 and 7, N(x) and N 2(x) are bounded. By Lemma 8, |N i+1(x)| < R(l, r) · |N i(x)| for all 2 ≤ i ≤ m − 1. Using an inductive argument we get that |N i(x)| < R(l, r)i−2 · |N 2(x)| for all 3 ≤ i ≤ m. We conclude that |N i(x)| < R(l, r)m−2 · |N 2(x)| for all 3 ≤ i ≤ m.

Finally, we prove our main theorem.

Proof of Theorem 3. By Theorem 4, we already know that every family of graphs in F(t) is also in H(t). It remains to show that every family of graphs in H(t) is also in F(t). Let H ∈ H(t). Then there is a positive integer n0 such that every H-free connected graph of order at least n0 is K1,t-free. Let n be an integer such that n ≥ max(n0, t + 1). Consider the family H0 = Ht(n, n, n, n). All the graphs in H0 are connected graphs of 0 order at least n0 containing an induced K1,t. Then it must be that no graph of H is H-free. In other words, for each H0 ∈ H0, there is an H ∈ H such that H H0. But this is exactly the deﬁnition of H ≤ H0. Then since H0 is in F(t), we conclude that H is also in F(t).

6 4 Restricting the size of the family of forbidden subgraphs

In this section, we consider restricting the size of the family of forbidden subgraphs. Con- cretely, we add the condition |H| ≤ k to some family H ∈ H(t) for some positive integer k. We restrict ourselves to the case t = 3 (claw-free graphs) which was our original motivation. We were able to characterize such families for each k ≥ 1. In other words, for each k ≥ 1, we characterized the families H ∈ H(3) such that |H| ≤ k. The result is expressed in Theorem 9.

To state and prove the result, we deﬁne some additional graphs.

3 • Ym is Ym. 3 • Zm,r is Zm,r. − • Zm,r is the graph obtained by identifying a vertex of a Kr with the end vertex of a path on m + 1 vertices.

3 • Dq is D2,q. 3 • Tq is T2,q. − • Tq is Tq with the only vertex of degree one of Tq removed. Some of these deﬁnitions may be not necessary as they just rename some graphs, but they help making the statement of the result easier to read and the proof easier to understand, as we reduce the number of subindices and superindices.

Deﬁne the following families of graphs.

A 2 •H i (l, q, r) = {K1,l,Yi+2,Wq ,Z1,r,...,Zi,r} (for i ≥ 1).

B 2 − •H i (l, m, q, r) = {K1,l,Ym,Wq ,Z1,r,...,Zi−1,r,Zi,r} (for i ≥ 2).

C 3 •H i (l, q, r) = {K1,l,Yi+2,Wq ,Dq,Tq,Z1,r,...,Zi,r} (for i ≥ 1).

D 3 − •H i (l, m, q, r) = {K1,l,Ym,Wq ,Dq,Tq,Z1,r,...,Zi−1,r,Zi,r} (for i ≥ 3). Deﬁne the following subsets of G.

• F1 = { H ∈ G : H ≤ {K1,3}}.

• F3 = { H ∈ G : H ≤ {K1,l,Ym,Kr} for some l ≥ 4, m ≥ 3 and r ≥ 3}.

3 − • F4 = { H ∈ G : H ≤ {K1,l,Ym,Wq ,Z1,r} for some l ≥ 4, m ≥ 3, q ≥ 2 and r ≥ 3}.

3 • F5 = { H ∈ G : H ≤ {K1,l,P4,Wq ,Dq,Z1,r} for some l ≥ 4, q ≥ 2 and r ≥ 3}.

3 − • F6 = { H ∈ G : H ≤ {K1,l,Ym,Wq ,Dq,Z1,r,Z2,r} for some l ≥ 4, m ≥ 4, q ≥ 2 and r ≥ 3}.

A A • Fi = { H ∈ G: H ≤ Hi (l, q, r) for some l ≥ 4, q ≥ 2, r ≥ 3} (i ≥ 1). B B • Fi = { H ∈ G: H ≤ Hi (l, m, q, r) for some l ≥ 4, m ≥ i + 3, q ≥ 2, r ≥ 3} (i ≥ 2). C C • Fi = { H ∈ G: H ≤ Hi (l, q, r) for some l ≥ 4, q ≥ 2, r ≥ 3} (i ≥ 1). D D • Fi = { H ∈ G : H ≤ Hi (l, m, q, r) for some l ≥ 4, m ≥ i + 3, q ≥ 2, r ≥ 3} (i ≥ 3).

7 The following is the main theorem of this section.

Theorem 9. Let k ≥ 1 be an integer and let H ∈ H(3) with |H| ≤ k. Then

•H∈ Fi for some i ∈ {1, 3, 4, 5, 6} with i ≤ k or

A •H∈ Fi for some 1 ≤ i ≤ k − 3 or B •H∈ Fi for some 2 ≤ i ≤ k − 3 or C •H∈ Fi for some 1 ≤ i ≤ k − 5 or D •H∈ Fi for some 3 ≤ i ≤ k − 5. S C Notice that F(3) = i≥1 Fi . Moreover, all the families in the other sets mentioned in Theorem 9 are also in F(3), and so they are all in H(3). This fact can by derived from the following lemma. Lemma 10. The following statements hold:

C C D (1) F1 ⊆ F1 , F3 ⊆ F4, F4 ⊆ F6, F5 ⊆ F1 and F6 ⊆ F3 . A C (2) Let i ≥ 1, then Fi ⊆ Fi . B A (3) Let i ≥ 2, then Fi ⊆ Fj for some j ≥ 1.

D C (4) Let i ≥ 3, then Fi ⊆ Fj for some j ≥ 1. Proof. Statements (1) and (2) are easy to verify. B B Proof of (3): Let i ≥ 2 and H ∈ Fi . Since H ≤ Hi (l, m, q, r) for some l ≥ 4, m ≥ i + 3, − q ≥ 2 and r ≥ 3, we have that H ≤ {Ym} for some m ≥ i + 3. Since Zi,r Zh,r for all h ≥ i A and all r ≥ 3, then H ∈ Fm−2. D D Proof of (4): Let i ≥ 3 and H ∈ Fi . Since H ≤ Hi (l, m, q, r) for some l ≥ 4, m ≥ i + 3, − q ≥ 2 and r ≥ 3, we have that H ≤ {Ym} for some m ≥ i + 3. Since Zi,r Zh,r for all h ≥ i C and all r ≥ 3, then H ∈ Fm−2.

5 Proof of Theorem 9

First, we prove two lemmas that deal with the inductive part of the proof of Theorem 9. Lemma 11. Let k ≥ 4 be an integer and let H ∈ H(3) with |H| ≤ k. Suppose that A B H {K1,3}, H ∈/ Fj for all 1 ≤ j ≤ k − 3 and H ∈/ Fj for all 2 ≤ j ≤ k − 3. Suppose also that there are graphs B1,B2,B3,H1 ∈ H such that

• B1 = K1,l for some l ≥ 4.

• B2 = Pm+1 or B2 = Ym for some m ≥ 3.

2 • B3 = Wq for some q ≥ 2.

• H1 = Z1,r1 for some r1 ≥ 3.

Then there are graphs H2,...,Hk−3 in H and integers r2, . . . , rk−3 such that for all 2 ≤ i ≤ k − 3, Hi = Zi,ri and ri ≥ 3. Additionally, m ≥ k. Proof. The proof is by induction on i. Let 2 ≤ i ≤ k − 3 and suppose that there are graphs H1,...,Hi−1 in H such that

Hj = Zj,rj for some rj ≥ 3 and all 1 ≤ j ≤ i−1. We will prove that there is a graph Hi ∈ H such that Hi = Zi,ri for some ri ≥ 3.

8 0 2 A Let r = max(r1, . . . , ri−1). Since H ≤ {K1,l,Wq ,Z1,r0 ,...,Zi−1,r0 } and H ∈/ Fi−1, then H {Yi+1}. In particular, B2 = Pm+1 or B2 = Ym for some m ≥ i + 2. 2 B − Since H ≤ {K1,l,Ym,Wq ,Z1,r0 ,...,Zi−1,r0 } and H ∈/ Fi , then H {Zi,r} for all r ≥ 3. Since H ∈ H(3), there is a positive integer n0 = n0(H) such that every H-free connected graph of order at least n0 is claw-free. Let n = max(n0, 3). Consider G = Zi,n. Since G contains an induced claw, G must contain some graph in 2 H as an induced subgraph. Since G contains neither K1,4, Pi+3 nor W2 then Bj G for all j ∈ {1, 2, 3}. Furthermore, since Zj,3 G for all 1 ≤ j ≤ i − 1, then Hj G for all 1 ≤ j ≤ i − 1. Then there must be some other graph Hi ∈ H such that Hi G. − Since Hi K1,3, Hi Yi+1 and that Hi Zi,r for all r ≥ 3, then Hi = Zi,ri for some ri ≥ 3. Notice that if ri = 2, then it would contradict that Hi Yi+1. This concludes the inductive proof. We now prove that m ≥ k. Let i = k − 3. Let 2 r = max(r1, . . . , ri). Suppose that H ≤ {Yi+2}. Then H ≤ {K1,l,Yi+2,Wq ,Z1,r,...,Zi,r}, A and hence H ≤ Hi (l, q, r) (with i = k−3), a contradiction. We conclude that H {Yi+2} = {Yk−1} and so B2 = Pm+1 or B2 = Ym for some m ≥ k. Lemma 12. Let k ≥ 7 be an integer and let H ∈ H(3) with |H| ≤ k. Suppose that C D H {K1,3}, H ∈/ Fj for all 1 ≤ j ≤ k − 5 and H ∈/ Fj for all 3 ≤ j ≤ k − 5. Suppose also that there are graphs B1,...,B5,H1,H2 ∈ H such that

• B1 = K1,l for some l ≥ 4.

• B2 = Pm+1 or B2 = Ym for some m ≥ 3. • B = W 3 for some q ≥ 2. 3 q1 1

• B4 = Dq2 for some q2 ≥ 2. • B = T − or H = T for some q ≥ 1 and 5 q3 5 q3 3

• H1 = Z1,r1 for some r1 ≥ 3.

• H2 = Z2,r1 for some r2 ≥ 3.

Then there are graphs H3,...,Hk−5 in H and integers r3, . . . , rk−5 such that for all 3 ≤ i ≤ k − 5, Hi = Zi,ri and ri ≥ 3. Additionally, m ≥ k − 2. Proof. The proof of this lemma is essentially the same as Lemma 11. The proof is by induction on i. Let 3 ≤ i ≤ k − 5 and suppose that there are graphs H1,...,Hi−1 in H such that

Hj = Zj,rj for some rj ≥ 3 and all 1 ≤ j ≤ i−1. We will prove that there is a graph Hi ∈ H such that Hi = Zi,ri for some ri ≥ 3. 0 3 Let r = max(r , . . . , r ). Since H ≤ {K ,W ,D ,T ,Z 0 ,...,Z 0 } and H ∈/ 1 i−1 1,l q1 q2 q3 1,r i−1,r C Fi−1, then H {Yi+1}. In particular, B2 = Pm+1 or B2 = Ym for some m ≥ i + 2. 3 D − Since H ≤ {K ,Y ,W ,D ,T ,Z 0 ,...,Z 0 } and H ∈/ F , then H {Z } for 1,l m q1 q2 q3 1,r i−1,r i i,r all r ≥ 3. Let n0 be as in Lemma 11. Let n = max(n0, 3). Consider G = Zi,n. Since G contains 3 − neither K1,4, Pi+3, W2 , D2, T1 then Bj G for all j ∈ {1, 2, 3, 4, 5}. Furthermore, since Zj,3 G for all 1 ≤ j ≤ i − 1, then Hj G for all 1 ≤ j ≤ i − 1. Then there must be some other graph Hi ∈ H such that Hi G. − Since Hi K1,3, Hi Yi+1 and that Hi Zi,r for all r ≥ 3, then Hi = Zi,ri for some ri ≥ 3. Notice that if ri = 2, then it would contradict that Hi Yi+1. This concludes the inductive proof. We now prove that m ≥ k − 2. Let i = k − 5. Let r = max(r1, . . . , ri). Suppose that H ≤ {Y }. Then H ≤ {K ,Y ,W 3 ,D ,T ,Z ,...,Z }, and hence H ≤ HC (l, q, r) i+2 1,l i+2 q1 q2 q3 1,r i,r i (with i = k−5), a contradiction. We conclude that H {Yi+2} = {Yk−3} and so B2 = Pm+1 or B2 = Ym for some m ≥ k − 2.

9 Proof of Theorem 9. Suppose that H ∈ H(3) and |H| ≤ k. Suppose contrary to the theorem that

•H ∈/ Fi for all i ∈ {1, 3, 4, 5, 6} with i ≤ k,

A •H ∈/ Fi for all 1 ≤ i ≤ k − 3, B •H ∈/ Fi for all 2 ≤ i ≤ k − 3, C •H ∈/ Fi for all 1 ≤ i ≤ k − 5 and D •H ∈/ Fi for all 3 ≤ i ≤ k − 5.

Since H ∈ H(3), there is a positive integer n0 = n0(H) such that every H-free connected graph of order at least n0 is claw-free. Let n = max(n0, 3). We will consider several connected graphs G of order at least n containing an induced claw. By the deﬁnition of H(3), there will be some H ∈ H such that H G. Consider G = K1,n. Then there is a graph B1 ∈ H such that B1 G. Since H ∈/ F1, then H {K1,3}, and so B1 K1,3. We conclude that

• B1 = K1,l for some l ≥ 4.

Consider G = Yn. Since G contains no K1,4, then B1 G. Then k ≥ 2 and there is a graph B2 ∈ H such that B2 G. Since B2 K1,3 then

• B2 = Pm+1 or B2 = Ym for some m ≥ 3.

3 Consider G = Wn . Since G contains neither K1,4 nor P4, then B1 G and B2 G. Then k ≥ 3 and there is a graph B3 ∈ H such that B3 G. Since H ∈/ F3, then H {Kr} for all r ≥ 3. Since B3 K1,3 and B3 Kr for all r ≥ 3, then • B = W 2 or B = W 3 for some q ≥ 2. 3 q1 3 q1 1 2 Consider G = Z1,n. Since G contains neither K1,4, P4 nor W2 , then Bi G for all i ∈ {1, 2, 3}. Then k ≥ 4 and there is a graph H1 ∈ H such that H1 G (the name H1 − will be better understand later in the proof). Since H ∈/ F4, then H {Z1,r} for all r ≥ 3. − Since H1 K1,3 and H1 Z1,r for all r ≥ 3, then

• H1 = Z1,r1 for some r1 ≥ 3.

2 Case 1 : H ≤ {Wq } for some q ≥ 2.

2 0 0 2 Since H ≤ {Wq } for some q ≥ 2 then there is a graph B in H such that B Wq for 0 0 0 some q ≥ 2. Notice it may be that B = B3 or not. Since B K1,3 and B Kr for all 0 2 r ≥ 3, then B = Wq for some q ≥ 2.

By Lemma 11, there are graphs H2,...,Hk−3 in H such that Hi = Zi,ri for some ri ≥ 3 and all 2 ≤ i ≤ k − 3. From the same lemma, we have that m ≥ k and so B2 = Pm+1 or 0 B2 = Ym for some m ≥ k. Notice that {B1,B2,B ,H1,...,Hk−3} ⊆ H. Since |H| ≤ k, then 0 B = B3 and H has no other graphs, namely, H = {B1,B2,B3,H1,...,Hk−3}. 2 Consider G = Zk−2,n. Since G contains neither K1,4, Pk+1 nor W2 then Bi G for all i ∈ {1, 2, 3}. Furthermore, since Zi,3 G for all 1 ≤ i ≤ k − 3, then Hi G for all 1 ≤ i ≤ k − 3. Then G contains no graph of H, which is a contradiction.

2 Case 2 : H {Wq } for all q ≥ 2.

2 Since H {Wq } for all q ≥ 2, then • B = W 3 for some q ≥ 2. 3 q1 1

10 3 Consider G = Dn. Since G contains neither K1,4, P4, W2 nor Z1,3, then Bi G for all i ∈ {1, 2, 3} and H1 G. Then k ≥ 5 and there is a graph B4 ∈ H such that B4 G. Since 2 − B4 K1,3, B4 Kr for all r ≥ 3, B4 Wq for all q ≥ 2 and B4 Z1,r for all r ≥ 3, then

• B4 = Dq2 for some q2 ≥ 2.

Since H ∈/ F5, then H {P4}. Then B2 = Pm for some m ≥ 5, or B2 = Ym for some m ≥ 3. 3 Consider G = Tn. Since G contains neither K1,4, P5, Y3, W2 , D2 nor Z1,3, then Bi G for all 1 ≤ i ≤ 4 and H1 G. Then k ≥ 6 and there is a graph B5 ∈ H such that B5 G. − 2 Since H ∈/ F6, then H {Z2,r} for all r ≥ 3. Since B5 K1,3, B5 Wq for all q ≥ 2, and − that B5 Zj,r for j ∈ {1, 2} and all r ≥ 3, then • B = T − or B = T for some q ≥ 1. 5 q3 5 q3 3

Suppose that H ≤ {Y3}. Since H ≤ {K ,Y ,W 3 ,D ,T ,Z }, then H ≤ HC (l, max(q , q , q ), r ), a contradic- 1,l 3 q1 q2 q3 1,r1 1 1 2 3 1 tion (since 1 ≤ k−5). Then we may suppose that H {Y3} and so B2 = Pm+1 , or B2 = Ym for some m ≥ 4. 3 − Consider G = Z2,n. Since G contains neither K1,4, P5, W2 , D2, T1 nor Z1,3, then Bi G for all i ∈ {1, 2, 3, 4, 5} and H1 G. Then k ≥ 7 and there is a graph H2 ∈ H such − that H2 G. Since H2 K1,3, H2 Y3 and H2 Zj,r for j ∈ {1, 2} and all r ≥ 3, then

• H2 = Z2,r2 for some r2 ≥ 3.

By Lemma 12, there are graphs H1 ...Hk−5 in H such that Hi = Zi,ri for some ri ≥ 3 and all 1 ≤ i ≤ k − 5. From the same lemma, we have that m ≥ k − 2 and so B2 = Pm+1 or B2 = Ym for some m ≥ k − 2. Notice that {B1,...,B5,H1,...,Hk−5} ⊆ H. Since |H| ≤ k, then H has no other graphs, namely, H = {B1,...,B5,H1,...,Hk−5}. 3 − Consider G = Zk−4,n.OPSmatching Since G contains neither K1,4, Pk−1, W2 , D2 nor T1 then Bi G for all i ∈ {1, 2, 3, 4, 5}. Furthermore, since Zi,3 G for all 1 ≤ i ≤ k − 5, then Hi G for all 1 ≤ i ≤ k − 5. Then G contains no graph of H, which is a contradiction.

6 Conclusions

The characterization we were looking for is given by Theorem 3. We have solved the problem of characterizing H(t) for any t ≥ 3, but it is also possible to consider t = 1 and t = 2. It is not diﬃcult to see that F(t) is also the solution for t = 1 and t = 2. In these cases, the corresponding families H1(m, l, q, r) and H2(m, l, q, r) after removing redundant graphs are as follows.

1 •H (m, l, q, r) = {K1,l,Kq,Pm}

2 2 2 2 2 •H (m, l, q, r) = {K1,l,Wq ,Ym,Z1,r,...,Zm−2,r} The case t = 1 is an easy proposition that can also be found in [5, Proposition 9.4.1]. In this paper we considered connected graphs with no degree conditions. But it is also possible to consider graphs with higher connectivity or with some minimum degree condition. Problem 1. Let t ≥ 3 and k ≥ 1. Characterize all the families of connected graphs H satisfying the following property. Every large enough H-free k-connected graph is K1,t-free. In this paper we were able to resolve Problem 1 for the case k = 1. Problem 2. Let t ≥ 3 and d ≥ 2. Characterize all the families of connected graphs H satisfying the following property. Every H-free large enough connected graph with minimum degree at least d is K1,t-free. Even a combination of Problems 1 and 2 is possible.

11 References

[1] R. Aldred, J. Fujisawa, and A. Saito. Forbidden subgraphs and the existence of a 2-factor. J. Graph Theory, 64(3):250 – 266, 2010.

[2] H. Broersma, R. J. Faudree, A. Huck, H. Trommel, and H. J. Veldman. Forbidden subgraphs that imply hamiltonian-connectedness. J. Graph Theory, 40(2):104 – 119, 2002. [3] J. Brousek. Forbidden triples for hamiltonicity. Discrete Math., 251(1):71–76, 2002.

[4] G. Chartrand and L. Lesniak. Graphs & Digraphs. Wadsworth & Brooks/Coles, Mon- terey, CA, 3rd edition, 1996. [5] R. Diestel. Graph Theory (Graduate Texts in Mathematics). Springer, 2nd edition, 2000.

[6] R. Faudree, E. Flandrin, and Z. Ryj´aˇcek.Claw-free graphs – a survey. Discrete Math., 164(1-3):87 – 147, 1997. [7] S. Fujita, K. Kawarabayashi, C. L. Lucchesi, K. Ota, M. D. Plummer, and A. Saito. A pair of forbidden subgraphs and perfect matchings. J. Combin. Theory, Ser. B, 96(3):315 – 324, 2006.

[8] R. J. Gould and M. S. Jacobson. Forbidden subgraphs and hamiltonian properties of graphs. Discrete Math., 42(2-3):189 – 196, 1982. [9] R. J. Gould, T.Luczak,and F. Pfender. Pancyclicity of 3-connected graphs: Pairs of forbidden subgraphs. J. Graph Theory, 47(3):183 – 202, 2004.

[10] T.Luczakand F. Pfender. Claw-free 3-connected P11-free graphs are hamiltonian. J. Graph Theory, 47(2):111 – 121, 2004. [11] K. Ota and G. Sueiro. Forbidden induced subgraphs for perfect matchings. submitted. [12] M. D. Plummer and A. Saito. Forbidden subgraphs and bounds on the size of a maximum matching. J. Graph Theory, 50(1):1–12, 2005.

[13] Z. Ryjcek. Closure and forbidden pairs for hamiltonicity. J. Combin. Theory, Ser. B, 86(2):331 – 346, 2002.