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BFSexample

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The other answers here do a great job showing how BFS runs and how to analyze it. I wanted to revisit your original mathematical analysis to show where, specifically, your reasoning gives you a lower estimate than the true value.

The rest of the vertex will not be evaluated for the segment 1 and would just add V-1 times of processing (since they are already visited in segment 2 which is O(V).

You are very close to having the right estimate here. The question is where the missing V term comes from. The issue here is that, weirdly enough, you can't say that O(V) · O(E / V) = O(E).

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BFSalgorithm

I hope this is helpful to anybody having trouble understanding computational time complexity for Breadth First Search a.k.a BFS.

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What's actually happening here is that no matter how many edges there are in the graph, there's some baseline amount of work you have to do for each node independently of those edges. That's the setup to do things like run the core if statements, set up local variables, etc.

You are saying that total complexity should be O(V*N)=O(E). Suppose there is no edge between any pair of vertices i.e. Adj[v] is empty for all vertex v. Will BFS take a constant time in this case? Answer is no. It will take O(V) time(more accurately θ(V)). Even if Adj[v] is empty, running the line where you check Adj[v] will itself take some constant time for each vertex. So running time of BFS is O(V+E) which means O(max(V,E)).

The answers I've seen here are correct, but there's a catch. Saying that BFS (or DFS) complexity is O(|V|+|E|) is only true if you need to traverse the whole graph, visiting each of its connectivity components. In this case, you can't take |V| out of the equation since you may need to start your BFS algorithms more than once if the graph is disconnected. Keeping in mind that when we compute complexity, we take into account all the atomic operations and treat them as O(1), your algorithm will have to start O(|V|) times.

Case 1: Consider a graph with only vertices and a few edges, sparsely connected graph (100 vertices and 2 edges). In that case, the segment 1 would dominate the course of traversal. Hence making, O(V) as the time complexity as segment 1 checks all vertices in graph space once.

BFSgraph

You are totally correct that the average work per node is O(E / V). That means that the total work done asympotically is bounded from above by some multiple of E / V. If we think about what BFS is actually doing, the work done per node probably looks more like c1 + c2E / V, since there's some baseline amount of work done per node (setting up loops, checking basic conditions, etc.), which is what's accounted for by the c1 term, plus some amount of work proportional to the number of edges visited (E / V, times the work done per edge). If we multiply this by V, we get that

However, if you only need to traverse one connectivity component, then the number of nodes in it is limited by |E| + 1, which gives us |V|=O(|E|), and the eventual complexity for this case is O(|E|).

What's happening here is that those lovely lower-order terms that big-O so conveniently lets us ignore are actually important here, so we can't easily discard them. So that's mathematically at least what's going on.

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I would just like to add to above answers that if we are using an adjacency matrix instead of a adjacency list, the time complexity will be O(V^2), as we will have to go through a complete row for each vertex to check which nodes are adjacent.

The time complexity to go over each adjacent edge of a vertex is, say, O(N), where N is number of adjacent edges. So, for V numbers of vertices the time complexity becomes O(V*N) = O(E), where E is the total number of edges in the graph. Since removing and adding a vertex from/to a queue is O(1), why is it added to the overall time complexity of BFS as O(V+E)?

Performing an O(1) operation L times, results to O(L) complexity. Thus, removing and adding a vertex from/to the Queue is O(1), but when you do that for V vertices, you get O(V) complexity. Therefore, O(V) + O(E) = O(V+E)

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It would be more proper to look at all the steps of the algorithm and count how many times it performs the atomic operations to get the O(|E|) result for the single connectivity component case, but the explanation above gives the general idea of why it's true.

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Here the segment 2 will dominate as the number of edges are more and the segment 2 gets evaluated 2|E| times for an undirected graph.

One of the ways that I grasped the intuition of the time complexity O ( V + E) is that when we traverse the graph (let's take BFS pseudocode in Java):

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