2

I'm trying to implement graph power calculation using BFS. According to Wikipedia, BFS with depth limit k will suffice (I'm using adjacency list representation, my graphs are sparse, so adjacency matrix power is not optimal).

Complexity of BFS is O(V+E), running it for every vertex is trivially O(V*(V+E)). I also have depth limit k here and Wikipedia claims (in the page linked above) that running BFS with this constraint for every vertex will be O(V*E). Why is that? In the worst case when every vertex is in the "range" of k from every other vertex I'll have to search the entire graph V times, after all.

qalis
  • 169
  • 5

1 Answers1

2

There are two cases.

First. $V-1 \leq E$. Then $O(V \cdot (V + E))$ and $O(V \cdot E)$ are equivalent.

Second. $E < V-1$. Then graph is disconnected and the biggest component has size at most $E + 1$. The whole algorithm will consist of $V$ BFS runs each on the component which has at most $E + 1$ vertices and at most $E$ edges. So each BFS run will take $O(E)$ time and the whole algorithm - $O(VE)$.

Note: this reasoning does not depend on $k$.

Vladislav Bezhentsev
  • 489
  • 2
  • 8