Source is in Mathworld, but it has no proof. Also, I am not completely sure what "unbalanced" means here. Non-equal number of even and uneven degrees?

What I am interested in, are there some universal properties of (connected) graphs that prevent hamiltonicity? Something to do with number of vertices versus edges, or average degrees or such? Something that is easy to demonstrate?

I will (hopefully) be using them as counterexamples of hamiltonian graphs in my thesis.


1 Answer 1


A bipartite graph has unbalanced parity if the two vertex sets are not the same size. Such a graph cannot be Hamiltonian, because a Hamilton circuit must alternate between the two vertex sets.

Added: A bit of searching turned up Grinberg’s theorem, which gives a necessary condition for a graph to be Hamiltonian and hence a sufficient condition for a graph to be non-Hamiltonian. This paper might be of use, but I know nothing about it beyond the title. This answer on CSTheory.SE gives another necessary condition for Hamiltonicity and hence sufficient condition for non-Hamiltonicity.

  • $\begingroup$ Grinberg's theorem is about planar graphs, so it is somewhat less than what I was hoping for. The C. Hoede's paper could be interesting, but not enough for the price, unfortunately. The last example is almost good enough, it's only lacking the property of "easy to demonstrate" :) $\endgroup$ Jan 16, 2013 at 11:52

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