7
$\begingroup$

I read that a directed graph has an Eulerian cycle if and only if every vertex has equal in degree and out degree, and all of its vertices with nonzero degree belong to a single strongly connected component.

But I think that if every vertex has equal in degree and out degree, then we just need weak connected component, which mean that all of its vertices with nonzero degree belong to a single connected component of the underlying undirected graph.

Can you show me that if a digraph that has: - Every vertex has equal in degree and out degree. - All of its vertices with nonzero degree belong to a single connected component of the underlying undirected graph. - This digraph is not strongly connected component. Then this can not has Eulerian cycle?

$\endgroup$

2 Answers 2

8
$\begingroup$

I think this question is best clarified with the following observation:

Once we assume that in-degrees and out-degrees of every vertex are equal, weakly connected and strongly connected components become one and the same.

So it is correct to say that an Eulerian cycle in a digraph requires the graph to be strongly connected, but it is also correct to say that being weakly connected is enough.


Here is a proof of the observation. Suppose we have a digraph $D$ which is weakly connected, but not strongly connected. Then $D$ can be decomposed into two parts, $D_1$ and $D_2$, which are connected by some $k>0$ edges oriented from $D_1$ and $D_2$, but with no edges oriented from $D_2$ and $D_1$.

If there are $m$ edges internal to $D_1$, then the sum of the out-degrees of all the vertices of $D_1$ is $m+k$, but the sum of the in-degrees of all the vertices of $D_1$ is only $m$. These are not equal, so there must be some vertex of $D_1$ where the out-degree is not equal to the in-degree. (In fact, there must be some vertex of $D_1$ where the out-degree exceeds the in-degree.)

Conversely, if out-degrees and in-degrees are equal at each vertex, then such a decomposition cannot exist: if $D$ is weakly connected, it must be strongly connected.

(More generally, if $D$ is not necessarily connected, but satisfies the degree condition at each vertex, then this argument applies to make each of its weak components strong.)

$\endgroup$
0
$\begingroup$

I saw that statement on the Wikipedia page and was curious too.

The vertices with nonzero degree of a directed graph containing an Eulerian cycle are strongly connected.

Proof

Suppose $G$ is not strongly connected. Let $S$ and $T$ be two strongly connected components of $G$ (comprised of vertices with degree $> 0$). Since $G$ has an Eulerian cycle, there must be an edge connecting $S$ and $T$. Label this edge $uv$, where to be clear $u \in S$ and $v \in T$. Let $u_1$ be another vertex in $S$. Since $S$ is strongly connected there exists a path from $u$ to $u_1$ and from $u_1$ back to $u$. However, then $u$ has an unequal number of incoming and outgoing edges. Since $u_1$ was arbitrary we have a contradiction. $\Box$

Let me know if I'm wrong in my reasoning somewhere.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .