什么是O（n * log m）+ O（m）？

Question

I am confused about addition with the big O notation.

I'm to create an algorithm to find a MST for a graph with some other requirements for a school problem. It's time complexity is to be in O(E * log V) , where E is the number of edges and V the number of vertices in the graph. I have arrived at a solution that is in O(E * log V) + O(V) .

Does it hold that O(E * log V) + O(V) = O(E * log V) ?

Thank you for all the answers! I am assuming this complexity on connected graphs, on graphs that are not connected, my algorithm works in O(E * log V) .

Answer 1

For any x, you can make a graph with x edges and 2ˣ (mostly disconnected) vertices.

For such a graph, E log V = x², so (V + E log V)/(E log V) = (2ˣ+x²)/x².

This grows without bound as x increases, so O(E log V) + O(V) is NOT the same as O(E log V), even for graphs.

HOWEVER, if you specify connected graphs, then you have V < E. In that case, as long as V>=2, you have V + E log V < E + E log V <= 2(E log V)

So O(E log V) = O(E log V) + O(V) for connected graphs.

Answer 2

O(ElogV+V) is not the same as O(ElogV). In general V can be arbitrarily larger than ElogV, which makes the two complexity classes different.

But, assuming you have an O(ElogV + V) time algorithm for finding an MST if one exists, you can turn it into a guaranteed O(ElogV) time algorithm, assuming the graph is represented in adjacency list form.

We can determine, in O(E) time, if E>=V/2. Go through the vertices of the graph, and see if there's any edges adjacent to that vertex. If you find a vertex with no adjacent edges, the graph clearly has no MST since that vertex is not connected to the rest of the graph. If you have gone through all vertices, you know that E>=V/2. If you find a vertex with no adjacent edges after n steps, you know you have at least (n-1)/2 edges in the graph, so this procedure takes O(E) time (even though naively it looks like it's O(V) time).

If E is less than V/2, the graph is disconnected (since in a connected graph, E>=V-1), and there's no MST.

So: check if E>=V/2 and only if so, run your MST algorithm.

This takes O(E + ElogV + V) = O(E + ElogV + 2E) = O(ElogV) time.