Recall that if all edge weights are non-negative, Dijkstra's algorithm computes the shortest path distances from a given source vertex S to all other vertices.
All of the algorithms considered here will work even if there are negative-weight edges in the graph (as long as there are no negative-cost cycles).
This algorithm computes shortest path distances from a given source vertex S to all other vertices in the digraph, assuming the digraph is acyclic.
Order the vertices v1,v2,...,vn in topological order.
Define
Since all paths in the graph go forward in the topological ordering, the following recurrence holds:
Base case:
Algorithm:
1. For i = 1,2,...,i0-1 2. D[i] = ∞ 3. D[i0] = 0 4. For i = i0+1, i0+2,...,n 5. D[i] = min { D[j] + WT(j,i) : (j,i) ∈ E, j < i } 6. Return D
This algorithm runs in O(n+m) time, since each edge is considered once in line 5.
Here is another algorithm that computes shortest path distances from a given source vertex S to all other vertices. This one is based on dynamic programming and works even if the graph has negative-weight edges, as long as the graph has no negative-weight cycles.
Order the vertices v1,v2,...,vn in some arbitrary way. Define
Here are the base cases for the recurrence relation:
Here is the recurrence relation. For k > 0,
This gives the following algorithm:
1. for each vertex i, set D[i] = ∞ 2. D[S] = 0 3. for k = 1,2,...,n-1 4. D[i] = min { D[i], min { D[j] + WT(j,i) : (j,i) ∈ E } }. 5. return D.
This algorithm runs in O(nm) time, since line 4 can be implemented in O(m) time.
How can we modify this algorithm to detect if the graph has a negative-weight cycle?
The next algorithm computes shortest path distances between all pairs of vertices. It also works as long as there are no negative-weight cycles in the graph.
The algorithm is based on TransitiveClosureByDP algorithm.
Order the vertices v1,v2,...,vn in some arbitrary way.
Define D[i,j,k] to be the length of the shortest path from i to j that goes through only the first k vertices v1,...,vk. (That is, i and j need not be in the set of k vertices, but all the other vertices on the path must be.)
Claim: Assume the graph has no negative-weight cycles. Then, for k > 0, D[i,j,k] = min { D[i,j,k-1], D[i,k,k-1] + D[k,j,k-1] }
To see this, consider the diagram:
A path is simple if it contains no cycles. If the graph has no negative-weight cycles, then between any two vertices there is always a shortest path that is simple (as long as there is some path).
Any simple path from i to j going only through vertices in the set {1,2,...,k} is of one of the following two forms:
The base case for our recurrence relation is
1. Initialize D[i,j] = ∞ for all i,j. 2. Set D[i,j] = WT(i,j) for each edge (i,j) in E. 3. For k = 1,2,...,n 4. For i = 1,2,...,n 5. For j = 1,2,...,n 6. D[i,j] = min { D[i,j], D[i,k] + D[k,j] } 7. Return D.
The algorithm runs in O(n3) time.
Exercise: how do we detect it if the graph does have a negative cycle? (In this case, shortest paths are not well defined, because by going around the cycle repeatedly, we get paths with arbitrarily negative weight.)