Why can't a Meredith Graph be Hamiltonian?

Question

I need to find out why Meredith Graph is not Hamiltonian

Meredith Graph is so complex to analyze in this regard. Can anyone help me wih their expertise?

Answer 1

The Meredith graph is roughly the Petersen graph where each vertex is replaced by a bipartite 'cluster' of 7 vertices: 4 'connecting points' which are connected to the rest of the graph and 3 'internal points' that have only edges inside the cluster.

Assume the Meredith graph has a Hamiltonian cycle. If this cycle visits each cluster only once, it would give rise to a Hamiltonian cycle for the Petersen graph, which is not possible.

So there must be at least one cluster that is visited twice. You enter a cluster at one of the four connecting points, then you can visit at most one internal point before you reach the next connecting point: then you already must leave the cluster. This leaves one internal point in this cluster that cannot be on the Hamiltonian cycle.