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Hi I'm seeking for a formal proof of the question listed in the title.

I'm a new learner that do not have extensive knowledge for that, So if any of you could be so kind to tell me and explain to me the steps of proof.

Thanks a lot!!

Shore
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  • Consider any partition of the nodes of the graph into two pieces, say A and B. Because of weak connectedness, without loss of generality there is an edge from A to B . Next, use the in degree = out degree to count the edges and find that there must be an edge from B to A. You can first prove a lemma saying for each subset the total in degree is the total out degree. – Elle Najt Nov 15 '19 at 15:28
  • It's not really a duplicate question, but it's proved in my answer here. – Misha Lavrov Nov 15 '19 at 17:15
  • Hi thanks, for both of you to be so kind. But may I further ask that in what way I could separate the graph in 2 partition?? – Shore Nov 15 '19 at 23:11

1 Answers1

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If each vertex has in degree equal to our degree, then there is an Eulerian cycle in which each edge is visited exactly once. Since this cycle exists, any vertex can use the cycle to get to any other vertex. As there is a directed path from each starting vertex to each destination vertex, the digraph is strongly connected.

Heather Guarnera
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  • Hmmm, I actually asked this question due to the reason of proving eulerian cycle. Wiki said it has to be strongly connected while my text book said it only need weakly. However, I actually find a prove here that proved weakly connect. And as you said I can actually using this to prove the title..... – Shore Nov 18 '19 at 03:24