2

Another heuristic idea: Find a long shortest path, and pick the vertex halfway along it. Pick a vertex and run BFS from it. For some small $k$, take the $k$ furthest vertices from the original vertex that the BFS determines, and repeat the process on each of them, keeping the $k$ overall furthest vertices each time. Repeat a few times. If the graph is a ...


1

If you consider the format of the TCP-IP datagram. Source Address: The 32-bit IP address of the originator of the datagram. Note that even though intermediate devices such as routers may handle the datagram, they do not normally put their address into this field—it is always the device that originally sent the datagram. Destination Address: The 32-bit IP ...


1

After a bit of reading through literature I've come upon "closeness centrality" which is the reciprocal of what I'm calculating (mean distance, which they call "farness" in the article). But I still haven't found any algorithms for finding the "closeness center" (node with maximum closeness centrality) that is faster than $O(N^2)$. As a heuristic, I have ...


Only top voted, non community-wiki answers of a minimum length are eligible