Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
432 views
in Technique[技术] by (71.8m points)

python - All minimum spanning trees implementation

I've been looking for an implementation (I'm using networkx library.) that will find all the minimum spanning trees (MST) of an undirected weighted graph.

I can only find implementations for Kruskal's Algorithm and Prim's Algorithm both of which will only return a single MST.

I've seen papers that address this problem (such as Representing all minimum spanning trees with applications to counting and generation) but my head tends to explode someway through trying to think how to translate it to code.

In fact i've not been able to find an implementation in any language!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I don't know if this is the solution, but it's a solution (it's the graph version of a brute force, I would say):

  1. Find the MST of the graph using kruskal's or prim's algorithm. This should be O(E log V).
  2. Generate all spanning trees. This can be done in O(Elog(V) + V + n) for n = number of spanning trees, as I understand from 2 minutes's worth of google, can possibly be improved.
  3. Filter the list generated in step #2 by the tree's weight being equal to the MST's weight. This should be O(n) for n as the number of trees generated in step #2.

Note: Do this lazily! Generating all possible trees and then filtering the results will take O(V^2) memory, and polynomial space requirements are evil - Generate a tree, examine it's weight, if it's an MST add it to a result list, if not - discard it.
Overall time complexity: O(Elog(V) + V + n) for G(V,E) with n spanning trees


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...