I have doubt !

Wikipedia says :

An Eulerian graph is one in which all vertices have even degree; Eulerian graphs may be disconnected.

What I know :

Defitition of an euler graph

"An Euler circuit is a circuit that uses every edge of a graph exactly once. ▶ An Euler path starts and ends at different vertices. ▶ An Euler circuit starts and ends at the same vertex."

According to my little knowledge "An eluler graph should be degree of all vertices is even, and should be connected graph".

I am asking :Is it possible disconnected graph has euler circuit ? If it is possible show an example .

EDITED : Here is my suplimentary problem , I voted for the anwser .

Which of the following graphs has an Eulerian circuit?

  • $\begingroup$ Note that wikipedia says that Eulerian graphs may be disconnected not that every disconnected graph is Eulerian... $\endgroup$ – Michael Biro Aug 30 '15 at 13:13
  • $\begingroup$ bliendly ,I beilieved that eluer graph must be connected . That's wrong. $\endgroup$ – 1 0 Aug 30 '15 at 13:17

Here we go: $$\huge\cdot\qquad\cdot$$

remember that $0$ is even. The circuit is the "empty circuit"
Since the graph has no edges, we've already passed every edge if we don't even move :D

  • $\begingroup$ That means , Any k-regular graph where k is an even number has euler circuit ? $\endgroup$ – 1 0 Aug 30 '15 at 13:19
  • $\begingroup$ Suplimentay problem here $\endgroup$ – 1 0 Aug 30 '15 at 13:29
  • 1
    $\begingroup$ @user4791206 My bad, the wikipedia definition cited in your link was false. As such, not every $2n$-regular graph is eulerian and not every graph with even degrees is eulerian (see $\Delta\quad\Delta$). Instead, an eulerian graph can only be disconnected if there is exactly one connected component consisting of more than one vertex (i.e. all "disconnected vertices" must be singleton). $\endgroup$ – AlexR Aug 30 '15 at 13:38
  • $\begingroup$ yes , What I concluded now , need verification A graph contains eluer circuit if graph is : (1) null graph , (2) With k-component , any only one component contains eluer circuit and other (k-1) have no edges .i.e. other is only vertices , (3) otherwise graph must be connected with euler circuit . $\endgroup$ – 1 0 Aug 30 '15 at 13:43
  • $\begingroup$ @user4791206 I think you might think the correct thing, but I can't quite verify because your english doesn't quite get it across. $\endgroup$ – AlexR Aug 30 '15 at 13:48

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