So I'm kinda confused right now and I'm probably just confusing myself for no reason.
So let say we have a graph went just one vertex.
The complete graph of that one vertex will be a loop to itself.
It will have a Euler Circuit because it has a degree of two and starts and ends at the same point.
Am I right?
Also, I think it will have a Hamiltonian Circuit, right?
Best Answer
Complete graph $K_1$ has $\frac{0\cdot(-1)}{2} = 0$ edges and is Hamiltonian by convention. Also it is connected and all vertex degrees are even (I hope there is no surprise that 0 is even), therefore it is Eulerian.
If you want to consider pseudograph on 1 vertex then it is Hamiltonian and Eulerian, too, since addition/removal a loop doesn't change this properties.