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?

  • $\begingroup$ The complete graph $K_1$ would just be the bare vertex, because $K_n$ is defined for simple graphs, so no loops allowed. $\endgroup$ – Joffan Mar 5 '17 at 22:30
  • $\begingroup$ so there won't be any lines, it just be the vertex itself. So there won't be an Euler Circuit? $\endgroup$ – shawn edward Mar 5 '17 at 22:39
  • $\begingroup$ The path of length zero still counts. $\endgroup$ – JMoravitz Mar 5 '17 at 22:40
  • $\begingroup$ @JMoravitz since zero is a even number, therefore it is a Euler Circuit $\endgroup$ – shawn edward Mar 5 '17 at 22:42
  • 1
    $\begingroup$ To trace the circuit, you put your pencil on the node and say "done" :-) $\endgroup$ – 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.

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.

  • $\begingroup$ 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. $\endgroup$ – shawn edward Mar 5 '17 at 23:10
  • $\begingroup$ No, $K_2$ is not Hamiltonian, because it has no Hamiltonian cycle, since cycle is a closed walk with no repeated edges. $\endgroup$ – Smylic Mar 5 '17 at 23:15
  • $\begingroup$ So it has to have repeated edges to be Hamiltonian? $\endgroup$ – shawn edward Mar 5 '17 at 23:19

You are right. As there is only one vertex in this graph, and depending on what the graph looks like (a single vertex without an edge or a single vertex with a loop), we find that every top has even degree. It is also trivial to notice that this is a connected graph, so we deduce, by a theorem proven by Euler, that this graph contains an eulerian cyclus. Also, draw both cases and apply your definition of Eulerian cyclus to it! Convince yourself the definition applies here.


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