What's the simplest algorithm/solution for a single pair shortest path through a real-weighted undirected graph?

Question

I need to find a shortest path through an undirected graph whose nodes are real (positive and negative) weighted. These weights are like resources which you can gain or loose by entering the node.

The total cost (resource sum) of the path isn't very important, but it must be more than 0, and length has to be the shortest possible.

For example consider a graph like so:

A-start node; D-end node

A(+10)--B( 0 )--C(-5 )
     \     |    /
       \   |  /
D(-5 )--E(-5 )--F(+10)

The shortest path would be A-E-F-E-D

Dijkstra's algorithm alone doesn't do the trick, because it can't handle negative values. So, I thought about a few solutions:

First one uses Dijkstra's algorithm to calculate the length of a shortest path from each node to the exit node, not considering the weights. This can be used like some sort of heuristics value like in A*. I'm not sure if this solution could work, and also it's very costly. I also thought about implement Floyd–Warshall's algorithm, but I'm not sure how.

Another solution was to calculate the shortest path with Dijkstra's algorithm not considering the weights, and if after calculating the path's resource sum it's less than zero, go through each node to find a neighbouring node which could quickly increase the resource sum, and add it to the path(several times if needed). This solution won't work if there is a node that could be enough to increase the resource sum, but farther away from the calculated shortest path.

For example:

A- start node; E- end node
A(+10)--B(-5 )--C(+40)
      \
        D(-5 )--E(-5 )

Could You help me solve this problem?

EDIT: If when calculating the shortest path, you reach a point where the sum of the resources is equal to zero, that path is not valid, since you can't go on if there's no more petrol.

Answer 1

Edit: I didn't read the question well enough; the problem is more advanced than a regular single-source shortest path problem. I'm leaving this post up for now just to give you another algorithm that you might find useful.

The Bellman-Ford algorithm solves the single-source shortest-path problem, even in the presence of edges with negative weight. However, it does not handle negative cycles (a circular path in the graph whose weight sum is negative). If your graph contains negative cycles, you are probably in trouble, because I believe that that makes the problem NP-complete (because it corresponds to the longest simple path problem).

Answer 2

This doesn't seem like an elegant solution, but given the ability to create cyclic paths I don't see a way around it. But I would just solve it iteratively. Using the second example - Start with a point at A, give it A's value. Move one 'turn' - now I have two points, one at B with a value of 5, and one at D also with a value of 5. Move again - now I have 4 points to track. C: 45, A: 15, A: 15, and E: 0. It might be that the one at E can oscillate and become valid so we can't toss it out yet. Move and accumulate, etc. The first time you reach the end node with a positive value you are done (though there may be additional equivalent paths that come in on the same turn)

Obviously problematic in that the number of points to track will rise pretty quickly, and I assume your actual graph is much more complex than the example.

Answer 3

I would do it similarly to what Mikeb suggested: do a breadth-first search over the graph of possible states, i.e. (position, fuel-left)-pairs.

Using your example graph:

• Octagons: Ran out of fuel
• Boxes: Child nodes omitted for space reasons

Searching this graph breadth-first is guaranteed to give you the shortest route that actually reaches the goal if such a route exists. If it does not, you will have to give up after a while (after x nodes searched, or maybe when you reach a node with a score greater than the absolute value of all negative scores combined), as the graph can contain infinite loops.

You have to make sure not to abort immediately on finding the goal if you want to find the cheapest path (fuel wise) too, because you might find more than one path of the same length, but with different costs.