Average distance between two nodes in a weighted undirected tree

Question

Hello I'm trying to figure out how to calculate the average distance between two nodes in a weighted undirected graph. Furthermore, this graph is a tree so it has V - 1 edges.

I thought about using Floyd Warshall to compute all-shortest paths and then calculate the average. But that would turn out to be in O(E^3) time complexity. And it really isn't enough.

I've also been thinking of using Dynamic Programming to solve it but I can't really see how... Can anyone give me a few pointers please? I don't want a direct answer, just a few tips so that I can keep thinking about it :)

if you found it in a programming site, can you please give it's link? We some friends found this to be interesting too. — Shihab Shahriar Khan, Nov 02 '17 at 05:52
Hello, I found it on a french website so it won't be in English. Apparently, there is a solution in O(V) with a single DFS. — Noé Weeks, Nov 02 '17 at 12:16
Me and a friend of mine today accidentally found a O(V) solution. So we were looking forward to submitting that — Shihab Shahriar Khan, Nov 02 '17 at 12:18
Well done, what is your strategy? I though about calculating for each vertice , in how many paths it is used. But I have trouble implementing it — Noé Weeks, Nov 02 '17 at 12:59
Think about every edge, and yes, how many times it's used.If a tree is of size N, and for an edge connecting a parent and child,the size of child subtree is P, then this edge is used P*(N-P) times — Shihab Shahriar Khan, Nov 02 '17 at 13:02
Yup, this sounds right, I'll try to implement it and I'll tell you if it works ! — Noé Weeks, Nov 02 '17 at 13:04

score 0 · Answer 1 · answered Nov 01 '17 at 21:22

Build and maintain a table of all shortest paths (no surprise). Each entry includes the lowest-known cost and the path.

Start the table by filling in all the edges (only paths of length 1); the other entries are inf.

Iterate through the table, considering two cases: add a node on the left, and add a node on the right. At each end, consider only the nodes reachable by a single edge from the end node. Is that path shorter than the existing entry?

For instance, assume we have an edge between nodes I and J; call it IJ, we also have edge JK. Is the cost of IJ + JK less than the current entry for edge IK? If so, replace it and record the path IJK; if not, go to the next edge/node/path.

You can iterate through all existing paths until there are no updates. Perhaps better, index the paths in which each shortest path is used; that allows you to update all paths whenever a particular sub-path is improved.

Continue the process until you make a full pass over the graph with no updates. If you maintain that sub-path index, I think you can do this in a only one or two passes over the graph. I believe that this would make the process O(E*N^2) (on edges and nodes).

Hello, this should work but doesn't have a good enough complexity :/ — Noé Weeks, Nov 02 '17 at 12:17

Average distance between two nodes in a weighted undirected tree

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