2

I was trying to come up with a solution for finding the single-source shortest path algorithm for an undirected weighted graph using BFS.

I came up with a solution to convert every edge weight say x into x edges between the vertices each new edge with weight 1 and then run the BFS. I would get a new BFS tree and since it is a tree there exists only 1 path from the root node to every other vertex.

The problem I am having with is trying to come up with the analysis of the following algorithm. Every edge needs to be visited once and then be split into the corresponding number of edges according to its weight. Then we need to find the BFS of the new graph.

The cost for visiting every edge is O(m) where m is the number of edges as every edge is visited once to split it. Suppose the new graph has km edges (say m'). The time complexity of BFS is O (n + m') = O(n + km) = O(n + m) i.e the Time complexity remains unchanged. Is the given proof correct?

I'm aware that I could use Dijkstra's algorithm here, but I'm specifically interested in analyzing this BFS-based algorithm.

templatetypedef
  • 328,018
  • 92
  • 813
  • 992
Prateek Agrawal
  • 33
  • 5

1 Answers1

1

The analysis you have included here is close but not correct. If you assume that every edge's cost is at most k, then your new graph will have O(kn) nodes (there are extra nodes added per edge) and O(km) edges, so the runtime would be O(kn + km). However, you can't assume that k is a constant here. After all, if I increase the weight on the edges, I will indeed increase the amount of time that your code takes to run. So overall, you could give a runtime of O(kn + km).

Note that k here is a separate parameter to the runtime, the same way that m and n are. And that makes sense - larger weights give you larger runtimes.

(As a note, this is not considered a polynomial-time algorithm. Rather, it's a pseudopolynomial-time algorithm because the number of bits required to write out the weight k is O(log k).)

templatetypedef
  • 328,018
  • 92
  • 813
  • 992