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[Solved]: Dijkstra's algorithm to compute shortest paths using k edges?

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Problem Detail: 

I am aware of using Bellman-Ford on a graph $G=(V,E)$ with no negative cycles to find the single-source single-destination shortest paths from source $s$ to target $t$ (both in $V$) using at most $k$ edges. Assuming we have no negative edge weights at all, can we use Dijkstra's algorithm for the same?

My thoughts/algorithm: I was wondering if instead of having a $dist[u$] array (storing the best known distance from s to u), we could use a $dist[u][k]$ table to store the best known distance from $s$ to $u$ using at most $k$ edges (dynamic programming maybe?), and similarly have the priority queue with $(u,n)$ tuples as keys. We can then terminate the algorithm when the tuple popped off the priority queue is $(t,n)$ where t is the target destination and $n <= k$?

Asked By : Mathguy

Answered By : D.W.

If the graph has no negative edges, the problem can be solved in $O(k \cdot (|V|+|E|) \lg |E|)$ time using Dijkstra's algorithm combined with a product construction. We construct a new graph $G'=(V',E')$ with vertex set $V' = V \times \{0,1,2,\dots,k\}$ and edge set

$$E' = \{((v,i), (w,i+1)) : (v, w) \in E\}.$$

In other words, for each edge $v \to w$ in $G$, we have edge $(v,i) \to (w,i+1)$ for all $i$ in $G'$.

Now use Dijkstra's algorithm to find the shortest path in $G'$ from $(s,0)$ to a vertex of the form $(t,i)$ where $i \le k$. This will be the shortest path in $G$ that uses at most $k$ edges.

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