Dijkstra's algorithm with negative weights

Question

Can we use Dijkstra\'s algorithm with negative weights?

STOP! Before you think \"lol nub you can just endlessly hop between two points and get an in

Answer 1

If you read the proof of optimality, one of the assumptions made is that all the weights are non-negative. So, no. As Bart recommends, use Bellman-Ford if there are no negative cycles in your graph.

You have to understand that a negative edge isn't just a negative number --- it implies a reduction in the cost of the path. If you add a negative edge to your path, you have reduced the cost of the path --- if you increment the weights so that this edge is now non-negative, it does not have that reducing property anymore and thus this is a different graph.

I encourage you to read the proof of optimality --- there you will see that the assumption that adding an edge to an existing path can only increase (or not affect) the cost of the path is critical.

Answer 2

There is actually an algorithm which uses Dijkstra's algorithm in a negative path environment; it does so by removing all the negative edges and rebalancing the graph first. This algorithm is called 'Johnson's Algorithm'.

The way it works is by adding a new node (lets say Q) which has 0 cost to traverse to every other node in the graph. It then runs Bellman-Ford on the graph from point Q, getting a cost for each node with respect to Q which we will call q[x], which will either be 0 or a negative number (as it used one of the negative paths).

E.g. a -> -3 -> b, therefore if we add a node Q which has 0 cost to all of these nodes, then q[a] = 0, q[b] = -3.

We then rebalance out the edges using the formula: weight + q[source] - q[destination], so the new weight of a->b is -3 + 0 - (-3) = 0. We do this for all other edges in the graph, then remove Q and its outgoing edges and voila! We now have a rebalanced graph with no negative edges to which we can run dijkstra's on!

The running time is O(nm) [bellman-ford] + n x O(m log n) [n Dijkstra's] + O(n^2) [weight computation] = O (nm log n) time

More info: http://joonki-jeong.blogspot.co.uk/2013/01/johnsons-algorithm.html

Answer 3

As long as the graph does not contain a negative cycle (a directed cycle whose edge weights have a negative sum), it will have a shortest path between any two points, but Dijkstra's algorithm is not designed to find them. The best-known algorithm for finding single-source shortest paths in a directed graph with negative edge weights is the Bellman-Ford algorithm. This comes at a cost, however: Bellman-Ford requires O(|V|·|E|) time, while Dijkstra's requires O(|E| + |V|log|V|) time, which is asymptotically faster for both sparse graphs (where E is O(|V|)) and dense graphs (where E is O(|V|^2)).

In your example of a mountainous terrain (necessarily a directed graph, since going up and down an incline have different weights) there is no possibility of a negative cycle, since this would imply leaving a point and then returning to it with a net energy gain - which could be used to create a perpetual motion machine.

Increasing all the weights by a constant value so that they are non-negative will not work. To see this, consider the graph where there are two paths from A to B, one traversing a single edge of length 2, and one traversing edges of length 1, 1, and -2. The second path is shorter, but if you increase all edge weights by 2, the first path now has length 4, and the second path has length 6, reversing the shortest paths. This tactic will only work if all possible paths between the two points use the same number of edges.

Answer 4

Yes, you could do that with adding one step at the end i.e.

            If v ∈ Q, Then Decrease-Key(Q, v, v.d)
            Else Insert(Q, v) and S = S \ {v}.

Answer 5

You can use Dijkstra's on a negative weighted graph but you first have to find the proper offset for each Vertex. That is essentially what Johnson's algorithm does. But that would be overkill since Johnson's uses Bellman-Ford to find the weight offset(s). Johnson's is designed to all shortest paths between pairs of Vertices.

http://en.wikipedia.org/wiki/Johnson%27s_algorithm

Answer 6

Actually I think it'll work to modify the edge weights. Not with an offset but with a factor. Assume instead of measuring the distance you are measuring the time required from point A to B.

weight = time = distance / velocity

You could even adapt velocity depending on the slope to use the physical one if your task is for real mountains and car/bike.