Abstract
Finding a hamiltonian decomposition of G is one of the challenging problems in graph theory. We do not know for what classes of graphs G and H, their tensor product G x H is hamiltonian decomposable. In this paper, we have proved that, if G is a hamiltonian decomposable circulant graph with certain properties and H is a hamiltonian decomposable multigraph, then G x H is hamiltonian decomposable. In particular, tensor products of certain sparse hamiltonian decomposable circulant graphs are hamiltonian decomposable.
Highlights
A k-cycle is a subgraph of Kn with k distinct vertices x1, x2, x3, . . . , xk and k edges x1x2, x2x3, . . . , xk−1xk, xkx1
The wreath product of the graphs G and H, denoted by G ◦ H, has vertex set V (G) × V (H) in which (g1, h1)(g2, h2) is an edge whenever g1g2 is an edge in G, or g1 = g2 and h1h2 is an edge in H
The cartesian product of the graphs G and H, denoted by G H, has vertex set V (G) × V (H) in which (g1, h1)(g2, h2) is an edge whenever g1 = g2 and h1h2 is an edge in H, or h1 = h2 and g1g2 is an edge in G
Summary
A k-cycle is a subgraph of Kn with k distinct vertices x1, x2, x3, . . . , xk and k edges x1x2, x2x3, . . . , xk−1xk, xkx. Every circulant graph of order n is a Cayley graph with the underlying group being Zn. The problem of finding hamiltonian decompositions of product graphs is not new. Because of this, finding hamiltonian decompositions of the tensor products of hamiltonian decomposable graphs is considered to be difficult. Let G be a circulant graph with property Q and let H be any hamiltonian decomposable multigraph, G × H is hamiltonian decomposable After deleting suitable number of jumps of K4n+2 and K2m, the resulting graphs need not be dense but their tensor product is hamiltonian decomposable. This cannot be deduced from the existing results in this direction
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