MDST

There is a spoj problem where you need to find the diameter of the MDST for a given graph. The graph is undirected, connected and unweighted. There are up to 1000 nodes, but no bound is given on the number of edges. (the solution I found online looped through all the edges and than all the nodes, and got AC. I think its thus safe to assume the number of edges is O(n))

From what I've read online here's the logic of the solution:

1. find the shortest distance dist[u][v] from each node to each other node (done in O(n^2), with bfs from each node)

2. brute force the center of the MDST

   1. If the center of the MDST is a node v, I find the node u such that dist[v][u] is maximal. -> The diameter if v is center will be 2*dist[v][u] I keep a variable mdst as the minimum diameter I have found so far.

   mdst = INF;
   for(v = 0 to N){
      mx = 0
      for(u = 0 to N) 
          mx = max(mx, dist[v][u])
      mdst = min(mdst, mx)
   }
   

   2. If the center of the MDST is an edge e = (u, v) I find the node x such that min(dist[u][x], dist[v][x]) is maximal. -> the diameter if e is the center will be 2*min(dist[u][x], dist[v][x]) + 1

   for((u, v) : edges){
      mx = 0
      for(x = 0 to N) 
          mx = max(mx, min(dist[u][x], dist[v][x]))
      mdst = min(mdst, mx)
   }
   

In either of the above cases, if the chosen node/edge is not the center of the MDST, the resulting diameter I find will be strictly greater than that of the actual MDST. On the other hand if the chosen node/edge is the center of the MDST, I should find the correct diameter.

Above I give an explanation to why this solution works, but it relies on the fact : if the chosen node/edge is not the center of the MDST, the resulting diameter I find will be strictly greater than that of the actual MDST which I didn't prove.

I have worked out a proof on paper for why this is true, but I'm not sure how to write it down formally here so I will not try to do so. I recommend you try to prove it yourself, if you are struggling write a message to me and I will try to help with what I can.