A cycle plus triangles graph is a 4-regular graph $G$ with a Hamiltonian circuit $C$ and such that the chords of $C$ induce a set of disjoint triangles (3-circuits). A 4-regular graph $G$ has a Hamilton decomposition if its set of edges can be partitioned into two sets that induce Hamiltonian circuits in $G$. A cycle plus triangles graph is edge 3-connected if it doesn’t have any edge 2-cuts (sets of two edges which when removed disconnect the graph).
Our question is the following
Question Let $G$ be a cycle plus triangles graph which is edge 3-connected. Is it true then that $G$ has a Hamiltonian decomposition, or is there a counterexample that might be known?
All edge 3-connected cycle plus triangles graph with 18 or fewer vertices have Hamiltonian decompositions. However, the graph in the picture below with 21 vertices is edge 3-connected and does not have a hamiltonian decomposition. It got derived from a counterexample to another hamiltonicity conjecture made some years ago about a class of cubic graphs (MO-graphs, cubic graphs with a chordless dominating circuit); the example was found by Brendan McKay - the derivation entails replacing dominated vertices by triangles. From 21 vertices onwards there seem to be counterexamples of every possible order. There would seem to be a relation between hamiltonian MO-graphs and cycle plus triangles which have a hamiltonian decomposition, but we are unable to establish it at present or be more specific about it.