1. Add a constant $c\geq |w_{min}|$ to each edge of $G$, so that each edge now has non-negative weight.
  2. Run Dijkstra's algorithm

Can anyone tell me if this is viable or if it fails?


I'm afraid that this idea does not work (and I actually post this question as homework to my students, the reason being that at first glance it looks sound and complete).

Let $G(V, E)$ denote a graph with a cost function $c:e\in E\mapsto Z$, i.e., both positive and negative whole numbers, and no negative cycles. Let us assume there is an edge $e\langle v_i, v_j\rangle$ with a negative cost, i.e., $c(e_{ij}) <0$, so that $w_{min}=c(e_{ij})$.

Let $G'(V, E)$ denote a graph where the cost of each edge is now $c'(e_{ij})=c(e_{ij}) + w_{min}, \forall v_i, v_j\in V$. Certainly, there are no negative edge costs in $G'$.

Running Dijkstra on $G'$ to find the shortest path between two arbitrary vertices $s$ and $t$ would return a path $\pi:\langle s=v_0, v_1, v_2, ..., v_n=t\rangle$ with a cost equal to $\sum_{i=0}^{n-1}c'(e_{i, i+1})=\sum_{i=0}^{n-1}\left(c(e_{i,i+1})+w_{min}\right)=n\times w_{min}+\left(\sum_{i=0}^{n-1}c(e_{i, i+1})\right)$ ---please note the usage of $c'$ and $c$ in these expressions. In other words, the overall cost of paths in $G'$ penalizes larger paths by the constant $w_{min}$ and in the end, it might very easily return a different path.

It does not matter whether you correct now the cost of the path returned by running Dijkstra in $G'$, as it might be simply different.

Hope this helps,

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.