So suppose we have a connected graph G = (V, E), the following statements are equal:

- All vertices in G have even valence.
- G consist of the edges from a disjoint union of some cycles, and the vertices from these cycles.
- G has a Eulerian path.
- => 2. can be shown by induction on the number of cycles.
- => 3. can also be shown by induction on the number of cycles, and
- => 1. should be immediate.

There are no clear way known to see if a given graph has a Hamiltonian cycle, it mostly boils down to check if it has a lot of edges.

All graphs G=(V, E) with n = |V| >= 3 have a Hamiltonian cycle if and only if

v, u ∈ V : v != u => valence(v) + valence(u) >= n

(the valence of every two vertices which are not edge connected is more than or equal to the number of vertices in the graph).

It should, however, be noted, that this is only a condition that *ensure* that a Hamiltonian cycle exist, we cannot conclude the other way.
For an examply of a graph where this condition fails but which certainly have a Hamiltonian cycle, think of a cycle of more than 4 vertices (you can just travel around the edge, but the sum of every two non-edge-joined vertices is only 4).