Finding the shortest path in a graph without any negative prefixes

Question

Find the shortest path from source to destination in a directed graph with positive and negative edges, such that at no point in the path the sum of edges coming

Answer 1

Although people have shown that no fast solution exists (unless P=NP)..

I think for most graphs (95%+) you should be able to find a solution fairly quickly.

I take advantage of the fact that if there are cycles then there are usually many solutions and we only need to find one of them. There are probably some glaring holes in my ideas so please let me know.

Ideas:

1. find the negative cycle that is closest to the destination. denote the shortest distance between the cycle and destination as d(end,negC)

(I think this is possible, one way might be to use floyds to detect (i,j) with a negative cycle, and then breadth first search from the destination until you hit something that is connected to a negative cycle.)

2. find the closest positive cycle to the start node, denote the distance from the start as d(start,posC)

(I argue in 95% of graphs you can find these cycles easily)

    Now we have cases:
a) there is both the positive and negative cycles were found:
The answer is d(end,negC).

b) no cycles were found:
simply use shortest path algorithm?

c) Only one of the cycles was found. We note in both these cases the problem is the same due to symmetry (e.g. if we swap the weights and start/end we get the same problem). I'll just consider the case that there was a positive cycle found.

find the shortest path from start to end without going around the positive cycle. (perhaps using modified breadth first search or something). If no such path exists (without going positive).. then .. it gets a bit tricky.. we have to do laps of the positive cycle (and perhaps some percentage of a lap).
If you just want an approximate answer, work out shortest path from positive cycle to end node which should usually be some negative number. Calculate number of laps required to overcome this negative answer + the distance from the entry point to the cycle to the exit point of the cycle. Now to do better perhaps there was another node in the cycle you should have exited the cycle from... To do this you would need to calculate the smallest negative distance of every node in the cycle to the end node.. and then it sort of turns into a group theory/ random number generator type problem... do as many laps of the cycle as you want till you get just above one of these numbers.

good luck and hopefully my solutions would work for most cases.