Hamiltonian Cycles in T-Graphs

There is only one finite, 2-connected, linearly convex graph in the Archimedean triangular tiling that does not have a Hamiltonian cycle. The vertices and polygonal edges of the planar Archimedean tilings 44 and 36 of the plane, partially shown in Figs. 1 and 2, respectively, are called the square tiling graph (STG) and the triangular tiling graph (TTG). (See [1].) A subgraph G of TTG is linearly convex if, for every line L which contains an edge of TTG, the set L ∩ G is a (possibly degenerate or empty) line segment. Such a line L is called a grid line. Linearly convex subgraphs of STG are defined similarly. A T-graph (respectively, S-graph) is any nontrivial, finite, linearly convex, 2-connected subgraph of TTG (respectively, STG). For example, the graph G shown in Fig. 3 is linearly convex even though it has three components including an isolated vertex v, and G has vertices x and y whose midpoint z is a vertex of TTG but not of G. (Each component of G is 2-connected, and the two nontrivial components are each T-graphs.) If a nontrivial graph G is Hamiltonian (i.e., has a Hamiltonian cycle), then it is clearly 2-connected. Zamfirescu and Zamfirescu [5] investigated which S-graphs have a Hamiltonian cycle. The situation is much easier for T-graphs, as we will show. With only one exception, any T-graph is Hamiltonian. Let D denote the T-graph shown in Fig. 4—the linearly-convex hull of the Star of David. Even though D is 2-connected and linearly convex, it is clearly not Hamiltonian. Theorem. Every T-graph, other than D, is Hamiltonian. Proof. Let G be any T-graph. The boundary of G, denoted ∂G, is the boundary of the unbounded component of the complement of G in the plane. It is clear that ∂G is a cycle 498 J. R. Reay and T. Zamfirescu