Calculating “Kevin Bacon” Numbers

Question

I\'ve been playing around with some things and thought up the idea of trying to figure out Kevin Bacon numbers. I have data for a site that for this purpose we can consider

Answer 1

Sounds like a job for Dijkstra's algorithm.

ED: Eh, I shouldn't have pulled the trigger so fast. Dijkstra's (and Bellman-Ford) reduces to a breadth-first search when the weights are 1, so this isn't too useful. Oh well.

The A* algorithm, mentioned by tvanfosson, may be ideal for this. The idea is that instead of searching and recursing in whatever order the elements are in each level of the tree (rooted on your start- or end-point), you use some heuristic to determine which element you are going to try first. In your case a good bet would probably be the degree of a node (number of "friends"), but you could possibly want to use the number of people within some arbitrary number of degrees of a given person (i.e., the guy who has has three friends who each have 100 friends is likely to be a better node than the guy who has 20 friends in a clique that shuns outsiders). There's all sorts of other things you could use as a heuristic (friends get 2 points, friends-of-friends get 1 point; whatever, experiment).

Combine this with a depth limit (cut off after 6 degrees of separation, or whatever), and you can vastly improve your average case (worst case is still the same as basic BFS).

Answer 2

This is a standard shortest path problem. There are lots of solutions, including Dijkstra's algorithm and Bellman-Ford. You may be particularly interested in looking at the A* algorithm and seeing how it would perform with the cost function relative to the inverse of any particular node's degree. The idea would be to visit more popular nodes (those with higher degree) first.

Answer 3

run a breadth-first search in both directions (from each endpoint) and stop when you have a connection or reach your depth limit

Answer 4

This one might be better overall Floyd-Warshall the all pairs shortest distance.