Algorithm to find diameter of a tree using BFS/DFS. Why does it work?

This link provides an algorithm for finding the diameter of an undirected tree using BFS/DFS. Summarizing:

Run BFS on any node s in the graph, remembering the node u discovered last. Run BFS from u remembering the node v discovered last. d(u,v) is the diameter of the tree.

Why does it work ?

Page 2 of this provides a reasoning, but it is confusing. I am quoting the initial portion of the proof:

Run BFS on any node s in the graph, remembering the node u discovered last. Run BFS from u remembering the node v discovered last. d(u,v) is the diameter of the tree.

Correctness: Let a and b be any two nodes such that d(a,b) is the diameter of the tree. There is a unique path from a to b. Let t be the first node on that path discovered by BFS. If the paths $p_1$ from s to u and $p_2$ from a to b do not share edges, then the path from t to u includes s. So

$d(t,u) \ge d(s,u)$

$d(t,u) \ge d(s,a)$

....(more inequalities follow ..)

The inequalities do not make sense to me.

All parts of proving the claim hinge on 2 crucial properties of trees with undirected edges:

• 1-connectedness (ie. between any 2 nodes in a tree there is exactly one path)

• any node can serve as the root of the tree.

Choose an arbitrary tree node $s$. Assume $u, v \in V(G)$ are nodes with $d(u,v) = diam(G)$. Assume further that the algorithm finds a node $x$ starting at $s$ first, some node $y$ starting at $x$ next. wlog $d(s,u) \geq d(s,v)$. note that $d(s,x) \geq d(s,y)$ must hold, unless the algorithm's first stage wouldn't end up at $x$. We will see that $d(x,y) = d(u,v)$.

The most general configuration of all nodes involved can be seen in the following pseudo-graphics ( possibly $s = z_{uv}$ or $s = z_{xy}$ or both ):

(u)                                            (x)
  \                                            /
   \                                          /
    \                                        /
     ( z_uv )---------( s )----------( z_xy )
    /                                        \
   /                                          \
  /                                            \
(v)                                            (y)

we know that:

• $d(z_{uv},y) \leq d(z_{uv},v)$. otherwise $d(u,v)

• $d(z_{uv},x) \leq d(z_{uv},u)$. otherwise $d(u,v)

• $d(s,z_{xy}) + d(z_{xy},x) \geq d(s,z_{uv}) + d(z_{uv},u)$, otherwise stage 1 of the algorithm wouldn't have stopped at $x$.

• $d(z_{xy},y) \geq d(v,z_{uv}) + d(z_{uv},z_{xy})$, otherwise stage 2 of the algorithm wouldn't have stopped at $y$.

1) and 2) imply $\, \\ d(u,v) = d(z_{uv},v) + d(z_{uv},u) \\ \qquad\geq d(z_{uv},x) + d(z_{uv},y) = d(x,y) + 2\, d(z_{uv}, z_{xy}) \\ \qquad\qquad\geq d(x,y)$.

3) and 4) imply $\, \\ d(z_{xy},y) + d(s,z_{xy}) + d(z_{xy},x) \\ \qquad\geq d(s,z_{uv}) + d(z_{uv},u) + d(v,z_{uv}) + d(z_{uv},z_{xy}) \qquad\qquad\qquad\qquad \\ \, $ equivalent to $\, \\ d(x,y) = d(z_{xy},y) + d(z_{xy},x) \\ \qquad\geq 2*\,d(s,z_{uv}) + d(v,z_{uv}) + d(u,z_{uv}) \\ \qquad\qquad\geq d(u,v)$.

therefore $d(u,v) = d(x,y)$.

analogue proofs hold for the alternative configurations

(u)                          (x)
                   \                          /
                    \                        /
                     \                      /
     ( s )---------( z_uv )----------( z_xy )
                     /                      \
                    /                        \
                   /                          \
                 (v)                          (y)

and

(x)        (u)  
                          /            \  
                         /              \ 
                        /                \
     ( s )---------( z_xy )----------( z_uv )
                        \                /          
                         \              /           
                          \            /            
                          (y)        (v)

these are all possible configurations. in particular, $x \not\in path(s,u), x \not\in path(s,v)$ due to the result of stage 1 of the algorithm and $y \not\in path(x,u), y \not\in path(x,v)$ due to stage 2.

StackExchange Computer Science Q#22855, answer score: 14