The problem is this:
Can every closed triangle mesh (an approximation of a 3d object using triangles, eg. a tetrahedron) be 'peeled like an orange', that is, can we find a sequence of triangles such that consecutive triangles share edges, and all triangles in the mesh are included once (and only once) in our sequence.
This is my logic so far:
Model the mesh as its dual graph, where vertices are triangles, and edges join triangles that share edges, then the problem boils down to finding a Hamiltonian Path in the graph.
Each vertex will have degree three; this makes the graph by definition cubic.
I did some googling; I did find that Tait's conjecture, that every cubic polyhedral graph has a Hamiltonian cycle, is false. I guess then that every cubic polyhedral graph has a Hamiltonian path (otherwise a counterexample would disprove Tait's conjecture). Cubic polyhedral graphs are planar and 3-vertex-connected so if my guess is correct, we can 'peel' all convex polyhedra, but can we always 'peel' a torus?
I believe that if our 3d mesh is an object without holes, then the graph is always planar.
So my question is:
When are Hamiltonian paths guaranteed to exist in cubic graphs? Particularly those cubic graphs that correspond to physical triangle meshes.
In my quest for a counterexample I constructed a genus-2 surface out of triangles, but it was fairly easy for my computer to find a 'peeling', so I am now a little more convinced that everything can be peeled.