Proof of correctness: Algorithm for diameter of a tree in graph theory

Let's call the endpoint found by the first BFS x. The crucial step is proving that the x found in this first step always "works" -- that is, that it is always at one end of some longest path. (Note that in general there can be more than one equally-longest path.) If we can establish this, it's straightforward to see that a BFS rooted at x will find some node as far as possible from x, which must therefore be an overall longest path.

Hint: Suppose (to the contrary) that there is a longer path between two vertices u and v, neither of which is x.

Observe that, on the unique path between u and v, there must be some highest (closest to the root) vertex h. There are two possibilities: either h is on the path from the root of the BFS to x, or it is not. Show a contradiction by showing that in both cases, the u-v path can be made at least as long by replacing some path segment in it with a path to x.

[EDIT] Actually, it may not be necessary to treat the 2 cases separately after all. But I often find it easier to break a configuration into several (or even many) cases, and treat each one separately. Here, the case where h is on the path from the BFS root to x is easier to handle, and gives a clue for the other case.

[EDIT 2] Coming back to this later, it now seems to me that the two cases that need to be considered are (i) the u-v path intersects the path from the root to x (at some vertex y, not necessarily at the u-v path's highest point h); and (ii) it doesn't. We still need h to prove each case.

I'm going to work out j_random_hacker's hint. Let s, t be a maximally distant pair. Let u be the arbitrary vertex. We have a schematic like

    u
    |
    |
    |
    x
   / \
  /   \
 /     \
s       t ,

where x is the junction of s, t, u (i.e. the unique vertex that lies on each of the three paths between these vertices).

Suppose that v is a vertex maximally distant from u. If the schematic now looks like

    u
    |
    |
    |
    x   v
   / \ /
  /   *
 /     \
s       t ,

then

d(s, t) = d(s, x) + d(x, t) <= d(s, x) + d(x, v) = d(s, v),

where the inequality holds because d(u, t) = d(u, x) + d(x, t) and d(u, v) = d(u, x) + d(x, v). There is a symmetric case where v attaches between s and x instead of between x and t.

The other case looks like

    u
    |
    *---v
    |
    x
   / \
  /   \
 /     \
s       t .

Now,

d(u, s) <= d(u, v) <= d(u, x) + d(x, v)
d(u, t) <= d(u, v) <= d(u, x) + d(x, v)

d(s, t)  = d(s, x) + d(x, t)
         = d(u, s) + d(u, t) - 2 d(u, x)
        <= 2 d(x, v)

2 d(s, t) <= d(s, t) + 2 d(x, v)
           = d(s, x) + d(x, v) + d(v, x) + d(x, t)
           = d(v, s) + d(v, t),

so max(d(v, s), d(v, t)) >= d(s, t) by an averaging argument, and v belongs to a maximally distant pair.