0
$\begingroup$

Let $G$ be a directed Graph with $c: E(G) \to \mathbb{R}$.

$G$ has no negative circle $\Leftrightarrow$ there is a function $\pi:V(G) \to \mathbb{R}$ so that for all $e=(v, w) \in E(G)$ apply $\pi(v) + c(e) - \pi(w) \ge 0$

Proof:
"$\Rightarrow$": We define new vertics $s$ and new edges $(s, t)$ with $c(s, t) = 0$ for all $t \in V(G)$. We also define $\pi:V(G) \to \mathbb{R}$ with $\pi(t) \mapsto dist(s, t)$
For all $e=(v, w) \in E(G): \pi(w) - \pi(v) = dist(s, w) - dist(s, v) \le c(e)$
$\Leftrightarrow \pi(v) + c(e) - \pi(w) \ge 0$
$dist(s, w) \ge c(e) + dist(s, v)$ $(2)$
Proof: $dist(s, v)$ is a $s$-$v$-path with the weight of $dist(s, v) + c(e)$. But since in G are no negative circles there is also a shortest path so $(2)$ is true.

"$\Leftarrow$": Let $C = (v_0, e_1, v_1, ..., e_m, v_m)$ be a circle with $v_0 = v_m$. Then is true that:
$$c(C) = \sum_{i=1}^m c(e_i) \ge \sum_{i=1}^m(\pi(v_i)-\pi(v_{i-1}))=0$$

So the statement is true.
Can please anyone look over this and tell me if I did any mistakes?

Regards,
Chiray

  • 0
    What is $dust(s,v)$? How is it different from $dist(s,v)$? – 2017-01-19
  • 0
    sorry, my mistake, should have been $dist(s, v)$, I will change it – 2017-01-19
  • 0
    Thus $\pi$ is a kind of potential... or "infra-potential"... – 2017-01-19
  • 0
    @JeanMarie what is that? And does it make the proof incorrect? – 2017-01-19
  • 1
    My remark does no invalidate the proof at all! I thought you had realized that, in case of equality, your characterization become: $c(e)=\pi(w)-\pi(v)$ but this is a difference of potential (think to electrical networks). I think that this analogy can be helpful; ask your lecturer: he/she will probably say: "oh yes of course...". – 2017-01-19
  • 0
    @JeanMarie ok tank you, I will read into it! – 2017-01-19

0 Answers 0