# Does this graph with one vertex have an Euler circuit?

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?

• The complete graph $K_1$ would just be the bare vertex, because $K_n$ is defined for simple graphs, so no loops allowed. – Joffan Mar 5 '17 at 22:30
• so there won't be any lines, it just be the vertex itself. So there won't be an Euler Circuit? – shawn edward Mar 5 '17 at 22:39
• The path of length zero still counts. – JMoravitz Mar 5 '17 at 22:40
• @JMoravitz since zero is a even number, therefore it is a Euler Circuit – shawn edward Mar 5 '17 at 22:42
• To trace the circuit, you put your pencil on the node and say "done" :-) – Joffan Mar 5 '17 at 22:46

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.
• Just to make things clear, a Hamiltonian circuit is when every vertex appears once, so like graph $K2$ would be Hamiltonian because every vertex appears once. – shawn edward Mar 5 '17 at 23:10
• No, $K_2$ is not Hamiltonian, because it has no Hamiltonian cycle, since cycle is a closed walk with no repeated edges. – Smylic Mar 5 '17 at 23:15