HiveBrain v1.2.0
Get Started
← Back to all entries
patternMinor

Find the lightest weight of a path between $u$ to $v$ that contains no more then $k$ edges (If there's one)

Submitted by: @import:stackexchange-cs··
0
Viewed 0 times
paththeedgesmoreweightonebetweenthatcontainsthen

Problem

Given a directed and strongly connected graph $G=(V,E)$, weight function $w: E \to \mathbb{R}$ and two distinct vertices $u,v \in V$. We know that there aren't negative cycles.
I need to find algorithm, efficient as possible,such that for every value of $k$, $2 \leq k \leq |V|-1$, it will find the lightest weight of a path between $u$ to $v$ that contains no more then $k$ edges (If there's one).

I don't know what to do. I want to use Bellman Ford somehow, I just don't sure how.

Thanks a lot.

Solution

Consider the dynamic programming formulation of Bellman-Ford:

$\qquad \displaystyle \mathrm{bf}(i,j) = \begin{cases}
0 &, i = 0 \land j = s \\
\infty &, i = 0 \land j \neq s \\
\min\limits_{k \in V}\, \big(\, \mathrm{bf}(i-1,k) + c(k,j)\,\big) &, \text{else}
\end{cases}$

What meaning does $\mathrm{bf}(i,j)$ have, that is what does DP-matrix entry $\mathrm{bf}[i,j]$ contain?


If you unfold the definition (i.e. proof by induction) you see that $\mathrm{bf}(i,j)$ is the weight of the shortest path from starting node $s$ to $j$ with at most $i$ edges -- exactly what you are looking for.

Context

StackExchange Computer Science Q#3352, answer score: 4

Revisions (0)

No revisions yet.