In Goldberg-Tarjan's algorithm for finding the max flow, why is $d(v)$ a lower bound for the shortest path?

Question

I hope my question is precise enough. In Goldberg-Tarjan's algorithm (also push-relabel algorithm, and other names) a distance (or label- as it's called in my lecture) function $d(v)$ is needed. This function is defined as

$d: V \to \mathbb{N}_0:$
$\forall (v,w) \in E: d(v) \le d(w) + 1$
For the start of the flow $s: d(s) = n = |V|$
For the end of the flow $t: d(t) = 0$

Now for the proof of the correctness of this algorithm the next step (in our lecture) is to show that

$d_x(v,t) \ge d(v)$

with $d_x(v,t)$ being the length of a shortest (number of edges) v-w-path in $G$. This means, that $d(v)$ is a lower bound for the amount of edges in a shortes path from $v$ to $t$.

Now comes the question: Why is it a lower bound? If $n$ is the amount of edges in the graph, and $d(s)=n$, isn't the shortest path from $s$ to $t$ bound to have less than or equal to $n$ edges?

I'm sorry for not providing any sources to this, but none of my sources are in English, and you can easily look up ones of your own on wikipedia for example.

Answer 1

The confusion comes from your condition $\forall (v,w)\in E$. The labels must be respected in the residual graph $G_f$, hence the correct condition is $\forall (v,w)\in E_f: d(v)\leq d(w)+1$.

In particular, the algorithm starts with a flow that removes all outgoing edges from $s$, so actually $t$ is not reachable from $t$ (again, in $G_f$).

For a clear exposition, I would recommend the book Algorithm Design (Kleinberg-Tardos), they go in detail in chapter 7.4.

Cheers