# Graph Theory: Directed Circuit Counterexample

I’m trying to answer the following:

Let P = v0,e1,v1,…,ek,vk be a path that is closed, v0,…,vkâˆ’1 are distinct, and k â‰¥ 3. Is P a circuit? Give either a proof, or a counterexample.

I don’t believe that this is a circuit as a circuit in my class is defined as being edge simple and closed… and this doesn’t make the claim that edge simplicity exists in this closed path; however, I’m not sure if my counterexample necessarily works the way I want it to. I think part of my problem is that I’m not sure if I completely understand what exactly constitutes edge simplicity. My book defines it as e1…ek are “distinct”, but what does distinct mean in this context? Is an edge not distinct if there are two edges connecting the same two nodes in an undirected graph? I’m pretty new to graph theory, so any intuition is super helpful. Thanks in advance!

=================

Different authors use “circuit” to mean either a cycle without repeated edged or a cycle without repeated vertices. So the right answer depends on which terminology the asker is expecting you to use.
– Henning Makholm
2 days ago

In our text, one of their criterion is that the vertices are distinct.
– too spicy
2 days ago

So, in your description of PP, the vertices are distinct, and that implies that no edge gets used more than once.
– Gerry Myerson
2 days ago

Is this because every edge is more or less just an ordered pair of vertices?
– too spicy
yesterday

Well, if that’s not what an edge is, then what is it?
– Gerry Myerson
yesterday

=================

=================