Hamilton cycles in prisms

Abstract

Abstract:The prism over a graph G is the Cartesian product G K

Prism‐hamiltonicity of triangulations

“…Kaiser et al [9] attribute the following conjecture to Rosenfeld and Barnette [13]. It is implicit in [13] but does not seem to have been explicitly stated there.…”

Cycles in complementary prisms

“…Kaiser et al [10] trace this interest back to Barnette's famous open conjecture that the graphs of simple 4-polytopes are Hamiltonian. A related open conjecture posed by Rosenfeld and Barnette [15] claims that the prism of a 3-connected planar graph is Hamiltonian.…”

Closure for the property of having a hamiltonian prism

“…Recently, another property sandwiched between "having a 2-tree", i.e., a Hamilton path, and "having a 2-walk" has attracted attention of researchers [10]. This property is that the prism of a graph is hamiltonian.…”

“…We often identify one of the two copies of G in the prism with the graph G itself. It can be shown that if G has a Hamilton cycle, then its prism is hamiltonian and if its prism is hamiltonian, then G has a 2-walk [10]. Some old conjectures relaxed from "having a Hamilton cycle" to "having a hamiltonian prisms" become easy and some seem to remain still hard, e.g., it is not known whether there exists a constant k such that each k-tough graph has a hamiltonian prism (recall that a graph G is k-tough if, for every subset A of its vertices, G \ A is connected or has at most k|A| components).…”

“…We remark that a different kind of closures was developed by Ryjáček [11] for Hamilton cycles in the class of so-called clawfree graphs. All of these show a tight connection between hamiltonian problems and closures of graphs and thus the authors of [10] posed the following problem: Problem 1. Let G be a graph of order n and let x and y be two non-adjacent vertices such that the sum of their degrees is at least n. Is it true that G has a hamiltonian prism if and only if G + xy does?…”