The Bellman-Ford algorithm checks all edges in each step, and if for each edge the following: $d(v)>d(u)+w(u,v)$ holds, then $d(v)$ will be updated. $w(u,v)$ is the weight of edge $(u, v)$ and $d(u)$ is the length of best path was found for vertex $u$.
If in any step there is no update for all vertexes, the algorithms terminate.
If Bellman-Ford will be used for finding all shortest paths from vertex $s$ in graph $G$ with $n$ vertex, terminate after $k < n$ iteration then following is True:
number of edges in all shortest paths from $s$ is at most $k-1$
I think some times this is true and sometimes is false (when we check all edges simultaneously... maybe this is wrong), but I need a clear definition for the above sentence or a small example about logic. I need verification of the above fact. is it always true? why?
The above statement is dependent on the implementation of your algorithm (at least I think).
Which implementation makes it true and under which condition it becomes false...
I see 4 version of Belman Ford on This Jeff Erricson Algorithm Book on Page 295.
Can you please add…?is to be interpreted literally as a question (instead of the prompt I seem to have failed to communicate politely), it remains unanswered: I don't know whether you can name source or originator (judging from the material linked, the latter may be difficult) - you did not update your question (as of now). $\endgroup$