#### Abstract

In 1974, Goodman and Hedetniemi proved that every 2-connected -free graph is hamiltonian. This result gave rise many other conditions for Hamilton cycles concerning various pairs and triples of forbidden connected subgraphs under additional connectivity conditions. In this paper we investigate analogous problems when forbidden subgraphs are disconnected which affects more global structures in graphs such as tough structures instead of traditional connectivity structures. In 1997, it was proved that a single forbidden connected subgraph in 2-connected graphs can create only a trivial class of hamiltonian graphs (complete graphs) with . In this paper we prove that a single forbidden subgraph can create a non trivial class of hamiltonian graphs if is disconnected: every -free graph either is hamiltonian or belongs to a well defined class of non hamiltonian graphs; every 1-tough -free graph is hamiltonian. We conjecture that every 1-tough -free graph is hamiltonian and every 1-tough -free graph is hamiltonian.

#### 1. Introduction

Only finpite undiprected graphs without loops or multiple edges are considered. Let be a graph with vertex set and edge set . We write for the subgraph of induced by . The neighborhood of a vertex will be denoted by . We use to denote the number of vertices (order) of . The independence number of , denoted by , is the maximum size of an mutually nonadjacent vertices in . If are graphs then a graph is said to be —free if contains no copy of any of the graphs as an induced subgraph; the graphs will be also referred to in this context as forbidden subgraphs. We denote by and the path and the cycle on vertices. Further, denotes the complete graph of order , and denotes the complete bipartite graph with partite sets of cardinalities and . Specifically, the graph will be called a claw. The graph is obtained from by adding an edge. Let be the graph which is obtained by identifying each vertex of a triangle with an end vertex of one of three vertex-disjoint paths of lengths . Let denote the number of components of a graph . A graph is -tough if for every subset of the vertex set with . The toughness of , denoted , is the maximum value of for which is -tough (taking for all ). A good reference for any undefined terms is [1].

The first sufficient condition for hamiltonicity of a graph in terms of forbidden subgraphs is by Goodman and Hedetniemi [2].

Theorem A (see [2]). *Every 2-connected -free graph is hamiltonian.*

This result gave rise to many other hamiltonicity conditions for various pairs and triples of forbidden connected subgraphs under additional connectivity conditions.

In 1997, Faudree and Gould [3] proved that a single forbidden connected subgraph in 2-connected graphs can create only a trivial class (complete graphs) of hamiltonian graphs with .

Theorem B (see [3]). *Let be a connected graph and let be a 2-connected graph. Then is -free implies is hamiltonian if and only if . **The following theorem presents the forbidden pairs version of Theorem B.*

Theorem C (see [3]). *Let and be connected graphs and be a 2-connected graph of order . Then is -free implies is Hamiltonian if and only if and is one of the graphs:
*

The forbidden triple version of Theorems B and C still is not known.

In this paper we investigate analogous problems when forbidden subgraphs are disconnected which affects more global structures in graphs such as tough structures instead of traditional connectivity structures.

First of all, we prove that a single forbidden subgraph can create a non trivial class of hamiltonian graphs if is disconnected. As starting point, we consider and as disconnected sugraphs consisting of two and three isolated vertices, respectively. In other words, is a compliment of and is a complement of (a triangle). The first proposition follows immediately.

Proposition 1. *Every -free graph is complete and therefore is hamiltonian. *

The next proposition shows that -free graphs are hamiltonian if and only if they are 2-connected.

Proposition 2. *Every 2-connected -free graph is hamiltonian.*

Observe that is the minimal forbidden disconnected subgraph containing at least one edge. To describe the hamiltonian graphs with forbidden subgraph , we need the following recursive definition.

*Definition 3. *We say that if and only if either is independent set of vertices or is complete graph or there is a bipartition such that(1) for each ,(2). By the Definition, if is a complete bipartite graph then .

Theorem 4. *If is a –free graph then either is Hamiltonian or with .*

If is a 1-tough graph then by the definition, . Hence, the following holds from Theorem 4 immediately.

Corollary 5. *Every 1-tough -free graph is hamiltonian.*

This result can be slightly improved.

Theorem 6. *Every 1-tough -free graph is hamiltonian.*

Theorem 6 is sharp in some respects. Indeed, is a nonhamiltonian -free (even -free) graph with and , implying that the condition “ is 1-tough” in Theorem 6 cannot be removed or replaced by “ is 2-connected.” Now form a graph, denoted by , by adding a new vertex to and new edges . Since is a nonhamiltonian -free graph with , we can claim that the condition “ is -free” in Theorem 6 cannot be relaxed to “ is -free.” Finally, is a -free graph and hence the condition “ is -free” in Theorem 6 cannot be relaxed to “ is -free.” So, Theorem 6 is best possible in many respects. The condition “ is -free” in Theorem 6 perhaps can be relaxed to “ is -free.”

Conjecture 7. *Every 1-tough -free graph is hamiltonian.*

Conjecture 7 (if true) is best possible in all respects. Indeed, the above constructed graph shows that the condition “ is -free” in Conjecture 7 cannot be replaced by “ is -free.”

Moreover, the following seems reasonable as well.

Conjecture 8. *Every 1-tough -free graph is hamiltonian.*

For more than one tough graphs, we conjecture the following.

Conjecture 9. *Every -free graph with is hamiltonian.*

The Petersen graph shows that the condition “ is -free” in Conjecture 9 cannot be replaced by “ is -free”.

The next conjecture concerns -free graphs.

Conjecture 10. *Every -free graph with is hamiltonian.**The graph (see the graph examples concerning the sharpness of Theorem 6) shows that the condition in Conjecture 10 can not be replaced by . Further, the Petersen graph shows that the condition “ is -free” in Conjecture 10 can not be replaced by “ is -free”.*

The next conjecture is based on a forbidden subgraph containing the claw as a substructure.

Conjecture 11. *Every -free graph with is hamiltonian.*

The Petersen graph shows that the condition in Conjecture 11 can not be replaced by .

#### 2. Additional Notation and Preliminaries

For a subset of , we denote by the subgraph of induced by . For a subgraph of we use short for . Furthermore, for a subgraph of and , we define .

A simple cycle (or just a cycle) of length is a sequence of distinct vertices with for each , where . When , the cycle on two vertices , coincides with the edge . A graph is hamiltonian if contains a Hamilton cycle, that is, a cycle of length . A cycle in is dominating if is edgeless.

Paths and cycles in a graph are considered as subgraphs of . If is a path or a cycle, then the length of , denoted by , is . We write with a given orientation by . For , we denote by the subpath of in the chosen direction from to . The expression means that are the endvertices of . For , we denote the th successor and the th predecessor of on by and , respectively. We abbreviate by . For each , we define .

To prove Proposition 2 and Theorems 4, 6, we need the following useful lemma.

Lemma 12. *Let be a 2-connected graph, a longest cycle in with a given orientation and . Let be two internally disjoint paths with given orientations starting at and terminating on at different vertices . Then .*

#### 3. Proofs

*Proof of Lemma 12. *Assume the contrary, that is . Then
is longer than , contradicting the fact that is a longest cycle in . Hence, . Lemma 12 is proved.

*Proof of Proposition 2. *Since is 2-connected graph, it contains a cycle. Let be a longest cycle in with a given orientation. If then is a Hamilton cycle and we are done. Let and let . Since is 2-connected, there are two internally disjoint paths with given orientations starting at and terminating on at different vertices . If then is longer than , contradicting the fact that is a longest cycle. Hence, . Analogously, . Furthermore, by Lemma 12, , implying that is an independent set of vertices which contradicts the fact that is -free graph. This final contradiction proves that is hamiltonian. Proposition 2 is proved.

*Proof of Theorem 4. *Let be a -free graph. If is independent set of vertices, then by the definition, with and we are done. Let contain at least one edge. Next, if is disconnected, then clearly contains as an induced subgraph, contradicting the hypothesis. Let be connected. Further, if is a tree, then clearly either is a star (i.e., is a complete bipartite graph and hence with ) or contains as an induced subgraph, again contradicting the hypothesis. Now assume that is not a tree and let be a longest cycle in with a given orientation. If , then is a Hamilton cycle and we are done. Let . Since is connected, we have for some and . Assume without loss of generality that . Since is a longest cycle, we have . If , then and form an induced subgraph , contradicting the hypothesis. Hence . By a similar argument, for each and for each where is even. Further, since is a longest cycle, by Lemma 12, is an independent set of vertices. Moreover, for each , there is no path connecting and and passing through .

To prove that , assume the contrary, that is for some . Recalling that is a longest cycle, we conclude that and . Therefore, and form a disconnected subgraph which contradicts the fact that is -free. Hence . But then it is not hard to see that is an independent set of vertices.

Now let us prove that for each . Assume that . Since is an independent set of vertices, we have . if , then as above , implying that
is longer than , a contradiction. Let . But then (as above) , implying that . Put
Clearly Since is an independent set of vertices and , we can state that and thus is not hamiltonian. It remains to prove that . If is independent or is complete, then and we are done. Otherwise denote by a largest independent subset in and put . Clearly, and . Let be any vertex in . Since is a largest independent set, there exists a vertex such that . Let . If , then and form an induced subgraph , contradicting the hypothesis. Hence implying that for each . This means also that each vertex in is adjacent to all vertices in . Applying the same arguments to instead of , we conclude that .

* Proof of Theorem 6. *Let be a 1-tough -free graph. In particular, is 2-connected graph and therefore contains a cycle. Let be a longest cycle in with a given orientation. If then is a Hamilton cycle and we are done. Let and let be a connected component of of maximum order. Denote by the elements of occurring on in a consecutive order. Since is 1-tough, we have . Set
where . Since is a longest cycle in , we have and for each . The segments are called elementary segments on induced by . We call a path an intermediate path between two distinct elementary segments and if

Define to be the set of all intermediate paths between elementary segments . If , then has at least connected components, contradicting the fact that is 1-tough. Otherwise for some distinct . Choose a path in such that and . If , then each vertex with the subpath of of length 2 forms an induced subgraph , contradicting the hypothesis. Let , that is, and hence . Put
Assume without loss of generality that is chosen from such that is minimum. Since is a longest cycle in , by Lemma 12, is an independent set of vertices, implying that either or , say . Since is minimum, we have , that is, forms an induced subgraph . But then each vertex with forms an induced , again contradicting the hypothesis. Theorem 6 is proved.

#### Acknowledgments

The author would like to thank the anonymous referees for careful reading of the manuscript and for their valuable suggestions and corrections.