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Given n by n points and the distance d between these points, I need to find the corresponding undirected weighted graph that would result in these distances. I tried using Prim's algorithm to find the MST, however this set is of size n-1 and does not include the n needed edges. E.g. given n by n distances

0 3 5
3 0 4
5 4 0

I need to find the corresponding edges:

1 - 2 = 3
1 - 3 = 5
2 - 3 = 4

Which results in the graph:

      3
1 --------- 2
 \         /
  \5      /4
   \     /
    \   /
      3

However Prim's would return only the first 2 edges since a MST doesn't contain any cycles.

MGG
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1 Answers1

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One graph that would result in these distances is the graph that has an edge from every node to every other node and the length of that edge is the distance according to the matrix. (I'm not sure what you mean by unweighted directed because the example you give appears to be undirected and I'm not sure what the difference is between weights and lengths here).

Another option would be to consider the distances in increasing order, as you have done with Prim's algorithm, and, as well as checking to see if the edge is required to connect its two ends, check to see if the minimum distance between those ends in the graph reconstructed so far is the same as the distance in the matrix. If it is not, add the edge even if the ends are connected in the graph so far.

mcdowella
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  • As asked in the question I need to find the undirected weighted graph with n edges that results in these shortest distances. In the example of n = 3 constructing every edge according to the matrix works, but if n = 8 e.g. this would not satisfy the question. In this case Prim's would return 7 of these edges, how would I go about finding the last one as mentioned in your second paragraph? I implemented Prim's in Java: https://pastebin.com/Zmfpm3D1 – MGG Nov 13 '17 at 15:47
  • If you know there are n edges I would use Prim to get the first n-1 edges and then go through the unused edges in increasing order of length, keeping the first one that is shorter than the path through the tree formed by the existing n-1 edges. You may be able to save some time computing distances by taking advantage of the fact that you are computing distances on a tree. – mcdowella Nov 13 '17 at 18:19