Disconnected Forbidden Subgraphs, Toughness and Hamilton Cycles

In 1974, Goodman and Hedetniemi proved that every 2-connected $(K_{1,3},K_{1,3}+e)$-free graph is hamiltonian. This result gave rise many other hamiltonicity conditions for various pairs and triples of forbidden connected subgraphs under additional connectivity conditions. In 1997, it was proved that a single forbidden connected subgraph $R$ in 2-connected graphs can create only a trivial class of hamiltonian graphs (complete graphs) with $R=P_3$. In this paper we prove that a single forbidden subgraph $R$ can create a non trivial class of hamiltonian graphs if $R$ is disconnected: $(\ast1)$ every $(K_1\cup P_2)$-free graph either is hamiltonian or belongs to a well defined class of non hamiltonian graphs; $(\ast2)$ every 1-tough $(K_1\cup P_3)$-free graph is hamiltonian. We conjecure that every 1-tough $(K_1\cup P_4)$-free graph is hamiltonian and every 1-tough $P_4$-free graph is hamiltonian


Introduction
Only finite undirected graphs without loops or multiple edges are considered. We denote by n and α the order and the independent number of a graph, respectively. If H 1 , ..., H t (t ≥ 1) are graphs then a graph G is said to be (H 1 , ..., H t )-free if G contains no copy of any of the graphs H 1 , ..., H t . Further, we denote by P i and C i the path and the cycle on i vertices. A good reference for any undefined terms is [1].
The first sufficient condition for hamiltonicity of a graph in terms of forbidden subgraphs is due to Goodman and Hedetniemi [3].
This result gave rise many other hamiltonicity conditions for various pairs and triples of forbidden connected subgraphs under additional connectivity conditions.
In 1997, Faudree and Gould [2] proved that a single forbidden connected subgraph R in 2-connected graphs can create only a trivial class (complete graphs) of hamiltonian graphs with R = P 3 .
Theorem B. Let R be a connected graph and G be a 2-connected graph. Then G is R-free implies G is hamiltonian if and only if R = P 3 .
In this paper we prove that a single forbidden subgraph R can create a non trivial class of hamiltonian graphs if R is disconnected. First of all, observe the following. Proposition 1. Every (K 1 ∪ K 1 )-free graph is complete and therefore is hamiltonian.
It is not hard to see that every (K 1 ∪ K 1 ∪ K 1 )-free graph either is hamiltonian or consists of two complete graphs having at most one vertex in common. In other words, we have the following.
Observe that K 1 ∪ P 2 is the minimum forbidden disconnected subgraph containing at least one edge. To describe the hamiltonian graphs in (K 1 ∪ P 2 )-free graphs, we need the following recursive definition.
Definition. We say that G ∈ ℵ if and only if either V (G) is independent set of vertices or G is complete graph or there is a bipartition V = V 1 ∪ V 2 such that 1) V 1 is an independent set of vertices, Theorem 1. Every (K 1 ∪ P 2 )-free graph G either is hamiltonian or G ∈ ℵ with α(G) > n/2.
The following corollary follows immediately.
Further relaxing of the condition "G is (K 1 ∪ P 2 )-free" implies the following.
Examples for sharpness. Clearly, K 2,3 is a non hamiltonian (K 1 ∪ P 3 )-free (even (K 1 ∪ P 2 )-free) graph with τ (K 2,3 ) = 2/3 and κ(K 2,3 ) = 2, implying that the condition "G is 1-tough" in Theorem 2 can not be removed or replaced by "G is 2-connected". Now form a graph, denoted by H 1 , by adding a new vertex x 7 to C 6 = x 1 x 2 x 3 x 4 x 5 x 6 x 1 and new edges x 7 x 1 , x 7 x 4 , x 2 x 6 . Since H 2 is a non hamiltonian (K 2 ∪ P 3 )-free graph with τ (H 2 ) = 1, we can claim that the condition "G is (K 1 ∪ P 3 )-free" in Theorem 2 cannot be relaxed to "G is (K 2 ∪ P 3 )-free". Finally, H 2 is a (K 1 ∪ K 1,3 )-free graph and hence the condition "G is (K 1 ∪ P 3 )-free" in Theorem 2 cannot be relaxed to "G is (K 1 ∪ K 1,3 )free". So, Theorem 2 is best possible in many respects. The condition "G is (K 1 ∪ P 3 )-free" in Theorem 2 perhaps can be relaxed to "G is (K 1 ∪ P 4 )-free".
Example for sharpness. The graph H 1 (see the sharpness examples for Theorem 2) shows that the condition "G is (K 1 ∪ P 4 )-free" in Conjecture 1 (if true) can not be replaced by "G is (K 1 ∪ P 5 )-free".
For more than one tough graphs, the following is reasonable.
Example for sharpness. The Petersen graph shows that the condition "G is (K 1 ∪ P 5 )-free" in Conjecture 2 can not be replaced by "G is (K 1 ∪ P 6 )-free".
Examples for sharpness. The graph H 1 (see the sharpness examples for Theorem 2) shows that the condition τ > 1 in Conjecture 3 can not be replaced by τ = 1. Further, the Petersen graph shows that the condition "G is (K 2 ∪K 2 )free" in Conjecture 3 can not be replaced by "G is ( The next conjecture is based on K 1,3 (Claw).
Example for sharpness. The Petersen graph shows that the condition τ > 4/3 in Conjecture 4 can not be replaced by τ = 4/3.
Finally, we hope that the following is true.

Notations and preliminaries
The set of vertices of a graph G is denoted by V (G) and the set of edges by E(G). For S a subset of V (G), we denote by G\S the maximum subgraph of G with vertex set V (G)\S. We write G[S] for the subgraph of G induced by S. For a subgraph H of G we use G\H short for G\V (H). The neighborhood of a vertex x ∈ V (G) will be denoted by N (x). Furthermore, for a subgraph H of G and x ∈ V (G), we define N H (x) = N (x) ∩ V (H). Let s(G) denote the number of components of a graph G. A graph G is t-tough if |S| ≥ ts(G\S) for every subset S of the vertex set V (G) with s(G\S) > 1. The toughness of G, denoted τ (G), is the maximum value of t for which G is t-tough (taking τ (K n ) = ∞ for all n ≥ 1).
A simple cycle (or just a cycle) C of length t is a sequence When t = 2, the cycle C = v 1 v 2 v 1 on two vertices v 1 , v 2 coincides with the edge v 1 v 2 , and when t = 1, the cycle C = v 1 coincides with the vertex v 1 . So, all vertices and edges in a graph can be considered as cycles of lengths 1 and 2, respectively. A graph G is hamiltonian if G contains a Hamilton cycle, i.e. a cycle of length n. A cycle C in G is dominating if G\C is edgeless.
Paths and cycles in a graph G are considered as subgraphs of G. If Q is a path or a cycle, then the length of Q, denoted by |Q|, is |E(Q)|. We write Q with a given orientation by − → Q . For x, y ∈ V (Q), we denote by x − → Q y the subpath of Q in the chosen direction from x to y. For x ∈ V (C), we denote the h-th successor and the h-th predecessor of x on − → C by x +h and x −h , respectively. We abbreviate x +1 by x + . For each X ⊂ V (C), we define X +h = {x +h |x ∈ X}.

Proofs
Proof of Theorem 1. Let G be a (K 1 ∪P 2 )-free graph. If V (G) is independent then G ∈ ℵ and we are done. Let G contains at least one edge. Next, if G is disconnected then clearly G contains K 1 ∪ P 2 as in induced subgraph, contradicting the hypothesis. Let G is connected. Further, if G is a tree then clearly G either is a star (that is G is a complete bipartite graph and hence G ∈ ℵ) or contains K 1 ∪ P 2 as in induced subgraph, again contradicting the hypothesis. Now let G is not a tree, that is contains a cycle, and let C = v 1 v 2 ...v t v 1 be a longest cycle in G. If V (G\C) = ∅ then C is a Hamilton cycle and we are done. Let V (G\C) = ∅. It follows that xy ∈ E(G) for some x ∈ V (G\C) and y ∈ V (C). Assume w.l.o.g. that y = v 1 . Since C is extreme, we have xv 2 ∈ E(G). If xv 3 ∈ E(G) then x and v 2 v 3 form an induced subgraph K 1 ∪P 2 , contradicting the hypothesis. Hence xv 3 ∈ E(G). By a similar argument, xv i ∈ E(G) for each i = 1, 3, 5, ...t − 1 and t is even. Further, since C is extreme, {x, v 2 , v 4 , ..., v t } is an independent set of vertices. Moreover, for each u, v ∈ {x, v 2 , v 4 , ..., v t }, there is no a path connecting u and v and passing then recalling that C is extreme, we conclude that zv 2 ∈ E(G) and zv 3 ∈ E(G), contradicting the fact that G is K 1 ∪ P 2 -free. Hence V (G\C) is an independent set of vertices. If zv 2 ∈ E(G) then as above, zv 4 ∈ E(G) and hence is longer than C, a contradiction. This means that for each v ∈ V (G\C), we have Since V 1 is independent and |V 1 | > |V 2 |, G is not hamiltonian and α(G) > n/2. If V 2 is independent or G[V 2 ] is complete then G ∈ ℵ and we are done. Otherwise denote by V 3 a largest independent subset in V 2 and put V 4 = V 2 \V 3 . Let w 1 ∈ V 3 and w 2 ∈ V 4 . Clearly w 2 w 3 ∈ E(G) for some w 3 ∈ V 3 \{w 1 }, since otherwise V 3 ∪ {w 2 } is an independent set of vertices, contradicting the maximality of V 3 . If w 1 w 2 ∈ E(G) then w 1 and w 2 w 3 form an induced subgraph K 1 ∪ P 2 , contradicting the hypothesis. Hence w 1 w 2 ∈ E(G) implying that N (v) = V 4 for each v ∈ V 3 . Applying the same arguments to V 4 instead of V 2 , we conclude that G ∈ ℵ.
Proof of Theorem 2. Let G be a 1-tough (K 1 ∪ P 3 )-free graph. Since G is 1-tough, it contains a cycle. Let C be a longest cycle in G and H a connected component of G\C of maximum order. If V (H) = ∅ then C is a Hamilton cycle and we are done. Let V (H) = ∅ and let ξ 1 , ..., ξ s be the elements of N C (H) occuring on C in a consecutive order. Since G is 1-tough, we have s ≥ 2. Set w}. Define Υ(I i1 , I i2 , ..., I it ) to be the set of all intermediate paths between elementary segments I i1 , I i2 , ..., I it . If Υ(I 1 , I 2 , ..., I s ) = ∅ then G\{ξ 1 , ..., ξ s } has at least s+1 connected components, contradicting the fact that G is 1-tough. Otherwise Υ(I a , I b ) = ∅ for some distinct a, b ∈ {1, ..., s}. Choose a path L = x − → L y in Υ(I a , I b ) such that x ∈ V (I * a ) and y ∈ V (I * b ). If |V (L)| ≥ 3 then each vertex v ∈ V (H) with x − → L x ++ forms an induced subgraph K 1 ∪ P 3 , contradicting the hypothesis. Let |V (L)| = 2, i.e. L = xy. Put Assume without loss of generality that L is chosen from Υ(I a , I b ) such that |V (Q)| is minimum. Since C is extreme, by standard arguments, {ξ 1 , ..., ξ s } + is an independent set of vertices, implying that either x = ξ + a or y = ξ + b , say x = ξ + a . Since |V (Q)| is minimum, we have x − y ∈ E(G), that is x − xy forms an induced subgraph P 3 . But then each vertex v ∈ V (H) with x − xy forms an induced K 1 ∪ P 3 , again contradicting the hypothesis. Theorem 2 is proved.