Q:

Bellman-Ford - is number of interations greater than diameter?

Diameter of a connected, undirected graph is the smallest natural
  number d, so that between any two vertices of the graph exist path
  of length at most d.
Prove or disprove: in Bellman-Ford is the number of iterations always
  equal or lower than d.

I'm trying to solve this issue. What I tried was sketching a lot of graphs, however I have failed to find a single graph where the number of iterations would be higher than the diameter. 
The only graph where the number of iterations wouldn't be <= than diameter would be a graph with negative edges, however I found out that in undirected graph there can't be any negative edges, otherwhere there would be a negative cycle.
So, AFAIK the statement is correct. However, how would I prove such a statement? I don't even know how to start. Thanks for any help

A:

In the Bellman–Ford algorithm, after $t$ iterations the array contains, for any two nodes, the minimal walk of length at most $t$ connecting them. Assuming your graph doesn't have negative weights, the shortest walk between any two vertices will be a path, and so its length would be at most the diameter. Therefore there is no need to run the algorithm beyond $d$ iterations.

