Finding paths in undirected graph with specific cost

Question

Suppose we have undirected, weighted graph. Our task is to find all paths beetween two vertices (source and destination) which total cost equal = N. I think it can be done with modified Dijkstra's algorithm combined with BFS or DFS, but I have no idea how implement such thing. Thanks for any help.

Equal, or at most equal? If it is the former, the problem is NP-hard via reduction to the hamiltonian path problem. — John Dvorak, Jan 24 '13 at 12:35
Then your problem is NP-hard (NP-complete or even worse), and I can't seem to find a NP solution to _count_ the paths. — John Dvorak, Jan 24 '13 at 12:38
Note there can be up to `(n-2)!` solutions (complete graph, all edges equal in length, a hamiltonian path is desired), so enumeration will be slow as hell. — John Dvorak, Jan 24 '13 at 12:41
my goal isnt count the path, is to count the total path's value that be equel to N — IlanMosheShimsoni, Jan 24 '13 at 12:43
You want to enumerate the paths whose length is `N`, right? There could be _many_ of them. — John Dvorak, Jan 24 '13 at 12:44
yes, this is my goal. (our graph has maximum 30 Vertex) thanks — IlanMosheShimsoni, Jan 24 '13 at 12:54

TemplateRex · Answer 1 · 2013-01-24T22:03:56.980

3

Assuming you have a framework / library to create a graph data structure and to traverse it, you could do a backtracking depth-first search with an early return if you cross your resource constraint. In pseudo-code:

void DFS(Vertex current, Vertex goal, List<Vertex> path, int money_left) {
  // oops
  if (money_left < 0) 
     return;

  // avoid cycles
  if (contains(path, current)
     return;

  // got it!
  if (current == goal)) {
     if (money_left == 0)
         print(path);
     return;
  }

  // keep looking
  children = successors(current); // optionally sorted from low to high cost
  for(child: children)          
      DFS(child, add_path(path, child), money_left - cost(child));      
}

and you can then call it as DFS(start, goal, List<Vertex>(empty), N)

edited Jan 24 '13 at 22:03

answered Jan 24 '13 at 21:43

TemplateRex

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Shouldn't you only print if money_left == 0? Looks good otherwise – Patashu Jan 24 '13 at 21:59
@Patashu tnx, updated. Without that clause, it would find all paths with `cost <= N` – TemplateRex Jan 24 '13 at 22:00
It should still return if current == goal but money_left != 0, so it doesn't go on a pointless search past the goal – Patashu Jan 24 '13 at 22:02
@Patashu good to see that in your time zone you are not almost asleep ;-) updated. – TemplateRex Jan 24 '13 at 22:06

Finding paths in undirected graph with specific cost

1 Answers1