Bellman Ford and One Olympiad Questions?

Question

I took an Olympiad Exam three days ago. I ran into a nice question as follows.

We know the bellman-ford algorithm checks all edges in each step, and for each edge if,

Answer 1

1 is incorrect. First of all, we always find shortest paths with k edges, if any. But also, if we happen to relax the arcs in the topological order of a shortest path tree, then we converge in one iteration, despite the fact that the shortest path tree may be arbitrarily deep.

s --> t --> u --> v

Relax s->t, t->u, u->v; shortest path from s->v is three hops,
but B--F has made two iterations.

2 is incorrect, because who knows what the weights are?

  100
s --> t

Relax s->t; weight is 100, but B--F has made two iterations.

3 is correct, because by an averaging argument at least one arc of a negative cycle would be unrelaxed. Let v1, ..., vn be a cycle. Since the arcs are relaxed, we have d(vi) + len(vi->vi+1) - d(vi+1) >= 0. Sum the inequalities for all i and telescope the d terms to simplify to sum_i len(vi->vi+1) >= 0, which says that the cycle is nonnegative.