How to set exactly one edge to zero in directed weighted graph in order to find shortest path?

Question

The following is the question I am working on:

Consider a directed, weighted graph G where all edge weights are positive. The goal of this problem is to find the shortest path in G between two pre-specified vertices s and t , but with an added twist: you are allowed to change the weight of exactly one edge (of your choosing) to zero.

In other words, you must pick an edge in G to set to zero that minimizes the shortest path between s and t . Give an efficient algorithm to achieve this goal in O ( E lg V ) time and analyze your algorithm’s running time. Sub-optimal solutions will receive less credit.

Hint: You may have to reverse the edges, run a familiar algorithm a number of times, plus do some extra work

So I have tried running Dijkstra's from s to all other nodes and then I have tried reversing the edges and running it again from s to all other nodes. However, I found out that we have to run Dijskstra's from s to all other nodes and then reverse the edges and then run Dijkstra's from all other nodes to t. I am not exactly sure how this helps us to find the edge to set to zero. By my intuition I thought that we would simply set the maximum weight edge to zero. What is the point of reversing the edges?

Answer 1

We need to run Dijkstra's algorithm twice - once for the original graph with s as the source vertex, and once with the reversed graph and t as the source vertex. We'll denote the distance we get between vertex s and i from the first run as D(i) and the distance we get between vertex t and i second run D_rev(i).

Note that we can go follow the reversed edges backwards (i.e., follow them in the original direction), thus D_rev(i) is actually the shortest distance from vertex i to t. Similarly, D(i) is the shortest distance from vertex s to i following Dijkstra's algorithm.

We can now loop through all the edges, and for each edge e which connects v1 and v2, add up D(v1) and D_rev(v2), which corresponds to the weight of the path s -> v1 -> v2 -> t with e being the zero edge, since we can go from s to v1 with a distance of D(v1), set e to 0, go from v1 to v2, and then go from v2 to t with a distance of D_rev(v2). The minimum over these is the answer.

A rough proof sketch (and also a restatement) : if we set an edge e to 0, but don't use it in the path, we can be better off setting an edge that's in the path to 0. Thus, we need only consider paths that includes the zeroed edge. The shortest path through a zeroed edge e is to first take the shortest path from s to v1, and then take the shortest path from v2 to t, which are exactly what were computed using the Dijkstra algorithm, i.e., D and D_rev.

Hope this answer helps!