New Sufficient Conditions for Hamiltonian and Traceable Graphs
Citation: “New Sufficient Conditions for Hamiltonian and Traceable Graphs”. American Research Journal of Mathematics; V3, I1; pp:12.
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Abstract:
In this note, we present new sufficient conditions for Hamiltonian and traceable graphs.
Keywords: Hamiltonian graph, traceable graph
Description:
1 INTRODUCTION
We consider only finite undirected graphs without loops or multiple edges. Notation and terminology not defined here follow those in [1]. For a graph G = (V, E), we use n and e to denote its order V and size E, respectively. The connectivity of the graph G is denoted by k(G). For disjoint subsets S, T of the vertex set V(G) of a graph G, let E(S, T) be the set of the edges in G that join a vertex in S and a vertex in T. We use G˅H to denote the join of two disjoint graphs G and H. The graph consists of p isolated vertices is denoted by Ep. A cycle C in a graph G is called a Hamiltonian cycle of G if C contains all the vertices of G. A graph G is called Hamiltonian if G has a Hamiltonian cycle. A path P in a graph G is called a Hamiltonian path of G if P contains all the vertices of G. A graph G is called traceable if G has a Hamiltonian path. In this note, we present new sufficient conditions for Hamiltonian and traceable graphs. The main results are as follows.
Theorem 1. Let G be a graph of order n ≥ 3, e edges, and connectivity k ≥ 2. If
then G is Hamiltonian or Kk ˅Ek+ 1, where n = 2k + 1.
Theorem 2. Let G be a graph of order n ≥ 2, e edges, and connectivity k ≥ 1. If
then G is traceable or Kk ˅Ek+ 2, where n = 2k + 2.
Next, we will prove Theorem 1 and Theorem 2.
2 PROOFS
Proof of Theorem 1.
Let G be a graph satisfying the conditions in Theorem 1. If G has a Hamiltonian cycle, then the proof is finished. Now we assume that G is not Hamiltonian. Choose a longest cycle C in G and give an orientation on C. Since G is not Hamiltonian, there exists a vertex x0 in V(G)  V(C). By Menger’s theorem, we can find s (s ≥ k) pairwise disjoint (except for x0 ) paths P1 , P2 , ...,Ps between x0 and V(C). Let ui be the end vertex of Pi on C, where 1 ≤ i ≤ s. We, without loss of generality, assume that the appearance of u1 , u2 , ...,us on C agrees with the given orientation of C. We use ui + to denote the successor of ui along the given orientation of C, where 1 ≤ i ≤ s. Then a standard proof in Hamiltonian graph theory yields that S :={x0, u1 + , u2 + , ..., uk + } is independent (otherwise G would have cycles which are longer than C). Thus
Therefore S = k + 1, xy is in E for any vertex x in S and for any vertex y in V  S, and G[V  S)] is complete.
Let H be the component of G[V  V(C)] containing x0 . Since u1 + is adjacent to any vertex in V  S, H must consist of a singleton x0 (otherwise G would have a cycle which is longer than C). Since x0 is adjacent to any vertex in V  S, H must be the only component of G[V  V(C)] (otherwise x0 would be adjacent to a vertex in another component of G[V  V(C)], which is a contradiction). Again, since x0 is adjacent to any vertex in V  S, the segment from ui + to ui + 1 along the given orientation of C, for each i with 1 ≤ i ≤ s and s + 1 is regarded as 1, must consist of only ui + and ui + 1 (otherwise G would have a cycle which is longer than C). Hence G is
This completes the proof of Theorem 1. QED
Proof of Theorem 2. Let G be a graph satisfying the conditions in Theorem 2. If G has a Hamiltonian path, then the proof is finished. Now we assume that G is not traceable. Choose a longest path P in G and give an orientation on P. Let y and z be the two end vertices of P. We assume that the appearance of y and z on P agrees with the given orientation of P. Since G is not traceable, there exists a vertex x0 in V(G)  V(P). By Menger’s theorem, we can find s (s ≥ k) pairwise disjoint (except for x0 ) paths P1 , P2 , ...,Ps between x0 and V(P). Let ui be the end vertex of Pi on P, where 1 ≤ i ≤ s. We, without loss of generality, assume that the appearance of u1 , u2 , ...,us on P agrees with the given orientation of P. Since P is a longest path in G, y ≠ui and z ≠ ui , for each i with 1 ≤ i ≤ s, otherwise G would have paths which are longer than P. We use ui + to denote the successor of ui along the given orientation of P, where 1 ≤ i ≤ s. Then a standard proof in Hamiltonian graph theory yields that S :={x0 , y, u1 + , u2 + , ..., uk + } is independent (otherwise G would have paths which are longer than P). Thus
Therefore S = k + 2, xy in E for any vertex x in S and for any vertex y in V  S, and G[V  S)] is complete.
Let H be the component of G[V  V(P)] containing x0 . Since u1 + is adjacent to any vertex in V  S, H must consist of a singleton x0 (otherwise G would have a path which is longer than P). Since x0 is adjacent to any vertex in V  S, H must be the only component of G[V  V(P)](otherwise x0 would be adjacent to a vertex in another component of G[V  V(P)], which is a contradiction). Again, since x0 is adjacent to any vertex in V  S, the segment from ui + to ui + 1 along the given orientation of P, for each i with 1 ≤ i ≤ s–1, must consist of only ui + and ui + 1 (otherwise G would have a path which is longer than P). Moreover, the segment from y to u1 along the given orientation of P must consist of only y and u1 and the segment from us + to z along the given orientation of P must consist of only us + . Hence G is
This completes the proof of Theorem 2. QED
References
1. J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, Macmillan, London and Elsevier, New York (1976).
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