Lets say we are trying to solve some algorithmic problem $A$ that is dependent on input of size $n$. We say algorithm $B$ that runs in time $T(n)$, is asymptotically better than algorithm $C$ which runs in time $G(n)$ if we have: $G(n) = O(T(n))$, but $T(n)$ is not $O(G(n))$.

My question is related to the asymptotic running time of graph algorithms, which is usually dependent on $|V|$ and $|E|$. Specifically I want to focus on Prim's algorithm. If we implement the priority queue with a binary heap the run-time would be $O(E\log V)$. With Fibonacci heap we could get a run-time of $O(V\log V + E)$.

My question is do we say that $O(V\log V + E)$ is asymptotically better than $O(E\log V)$?

Let me clarify: I know that if the graph is dense the answer is yes. But if $E=O(V)$ both of the solutions are the same. I am more interested in what is usually defined as an asymptotic improvement in the case we have more than one variable, and even worse - the variables are not independent ($V-1\le E<V^2$, since we assume the graph is connected for Prim's algorithm).



1 Answer 1


The most permissive definition is as follows.

Suppose that $f(V,E),g(V,E)$ are two running times of graph algorithms solving the same problem.

We say that $f$ is an asymptotic improvement in some regime on $g$ if there exists a sequence $V_n,E_n$ with $V_n \to \infty$ such that $$ \lim_{n\to\infty} \frac{f(V_n,E_n)}{g(V_n,E_n)} = 0. $$

Sometimes the regime is not deemed interesting, but that's a more subjective matter.

Note also that the problem already manifests itself for one-variable functions. Consider $$ f(n) = n^2, \qquad g(n) = \begin{cases} 2^n &\text{if } n = 2^m, \\ n & \text{otherwise}. \end{cases} $$ Is $f$ an asymptotic improvement over $g$? It is for inputs of a certain length, and is indeed so under our permissive definition above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.