[3] Other counterexamples were found later, in many cases based on Grinberg's theorem.

The "compulsory" edges of the fragments, that must be part of any Hamiltonian path through the fragment, are connected at the central vertex; because any cycle can use only two of these three edges, there can be no Hamiltonian cycle.

It can be realized geometrically from a tetrahedron (the faces of which correspond to the four large nine-sided faces in the drawing, three of which are between pairs of fragments and the fourth of which forms the exterior) by multiply truncating three of its vertices.

[7] Finally Holton and McKay showed there are exactly six 38-vertex non-Hamiltonian polyhedra that have nontrivial three-edge cuts.

They are formed by replacing two of the vertices of a pentagonal prism by the same fragment used in Tutte's example.