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On an undirected and unweighted graph, how is it possible to enumerate all the groups of connected nodes having a length of 1,2,..,n (n is a user defined value)?

This question is similar to this one; with this difference: for n=3; I also require to find the path: A-B-C and C-E-F.

If n is 4, then the paths should also include:

A-B-C-D

A-B-C-E

A-B-C-F

A-C-E-F

I guess this is a problem of something like; "all pairs - all paths", where each path can contain at most n nodes. Would you also please tell the methods computational complexity?

My thinking is that I need to use both DFS and BFS simultaneously, but I am not sure whether this is efficient?

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banbar
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1 Answers1

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You can basically use DFS with an extra variable that is passed down the recursion of length, which will be reduced at every iteration. The stop condition will be when this extra variable reached 0.

Something along the lines of:

DFS(source,length,path):
   print path //this is always done, because we want all paths up to n
   if (length == 0): //stop clause
      return
   for each (source,u) is an edge:
       path.append(u)
       DFS(u,length-1,path)
       path.removeLast() //clean up environment

Another (less efficient, but might be more elegant) is doing an Iterative Deepening DFS, with length=1,2,...,n (and put the print in the stop clause only)

amit
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  • Thanks for the reply. Could you please tell what do you mean by path.removeLast()? – banbar Jan 09 '13 at 16:49
  • @user1959766: In here `path` is a list, and `path.removeLast()` removes the last element from it. `path.append(u)` adds `u` to the end of the list – amit Jan 09 '13 at 17:34
  • Thanks. However, I think the problem persists: as the last element is removed, we also lose some of the sets of nodes. In the example given, how is it possible for the group of nodes A-B-C to be detected, if we remove B? – banbar Jan 09 '13 at 20:56
  • @user1959766: You don't remove anything, you remove it from the *current* path only, after you have finished exploring it as much as you can. r am I misunderstanding you? – amit Jan 09 '13 at 21:35
  • Firstly, I think I wrongly posed the question initially (sorry the inconvenience it caused), better way is: finding the set of connected groups of nodes (rather than paths), where each groups size is <= n. When I trace your pseudo-code for length = 2 (i.e. n=3) in the example I gave, I think the resulting groups of nodes look like this: - A - AB - ABD - AC - ACE - ACF. Please correct me if I am wrong. – banbar Jan 10 '13 at 00:20