I know that $P \subseteq NP$, but for a problem in $P$, e.g. MST in a graph, is it a correct statement to say that:

The MST problem belongs in NP-Class.

(I mean, i think it is correct, but could someone classify that as wrong because he would expect P instead of NP?)


1 Answer 1


The statement is correct for exactly the reason you started with: $\mathsf P$ is a subset of $\mathsf{NP}$ which means that every problem in $\mathsf P$ is also in $\mathsf{NP}$. You can also go through the definitions and find that a deterministic TM running in polynomial time is merely a special kind of nondeterministic TM running in polynomial time and take it from there.

However, while being true, the statement "MST is in $\mathsf{NP}$" holds less value than the statement "MST is in $\mathsf P$" as complexity theory is, at large, concerned with lower bounds. A decent analogy in my opinion would be that the statement $\pi \geq 3$ is stronger than the statement $\pi \geq 0$ eventhough both are true.


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