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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
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1 Answers1

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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
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  • Simple and elegant, thank you. – qalis Apr 19 '20 at 20:59
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    I like this but don't understand how you got to the intermediate claim of $O(E^2)$ -- after bounding the size of each component, doesn't $O(VE)$ follow directly from the fact that we perform $V$ BFS operations, each on a (sub)graph of size at most $E+1$ vertices and $E-2$ edges? The $O(E^2)$ claim seems to imply that arbitrarily many isolated vertices can be processed/ignored in constant time. – j_random_hacker Apr 20 '20 at 17:35
  • @j_random_hacker Thank you, fixed it. Obviously, there can't be $O(E^2)$ because for graph consisting of only isolated vertices it will imply $O(1)$ algorithm. – Vladislav Bezhentsev Apr 20 '20 at 18:30
  • Thanks, +1. Must try to remember this component size upper bound for the sparse case, feels like it could be useful. – j_random_hacker Apr 20 '20 at 19:34