Graph which is Bipartite, has an Euler circuit, but not a Hamiltonian circuit

Is there a graph which is bipartite, has an Euler circuit, but not a Hamiltonian circuit?

I know the answer is yes, but if you consider something like this:

I don't think this would be bipartite, considering that $$1$$ and $$5$$ are both connected to themselves? It should still have an Euler circuit and no Hamiltonian circuit.

If we seperated the vertices into sets by color:

$$A = \{1,3,5\}$$

$$B = \{2,4\}$$

$$1$$ and $$5$$ would not have to be in both sets, but they are still connected to themselves. So if you have a loop at any vertex in a graph, it is automatically not considered bipartite?

• This graph also doesn't have an Euler circuit, since $1$ and $5$ have odd degree. – Misha Lavrov Apr 8 at 1:42

You are correct. One of the equivalences of $$G$$ being bipartite is that $$G$$ has no odd cycles, and a loop is a cycle of size 1.