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You're exactly right - [argh] it's not at all the traveling salesman problem. Thanks for the correction. In any case, the thesis of my post was that it's a tot
by ratsbane 19y ago
You're exactly right - [argh] it's not at all the traveling salesman problem. Thanks for the correction.
In any case, the thesis of my post was that it's a totally solvable problem and that you could use other factors to reduce the number of nodes - like looking for shared group or network memberships.
I think this is the relavant bit? http://en.wikipedia.org/wiki/Dijkstra's_algorithm http://en.wikipedia.org/wiki/Dijkstra's_algorithm Except that the Facebook thing wouldn't be a weighted graph - unless you wanted to consider length of time people have been marked as friends, or count "worked together" more or less than some other form of association.
- eru 19y agoOr you just set the weights to one. You could even use an A*-Algorithm that uses geography as a hint.