Can there be any relations regarding the number of nodes available in a digraph so that to qualify it as NP-Complete problem. If we consider this problem for instance:Input: A digraph $G=(V,E)$ and two nodes $u,v \in V$ Question: Is there a path in $G$ from $u$ to $v$? Can we say this problem is NP-Complete problem since the digraph have only two nodes that they have a path from one another and this makes it be a Hamiltonian Path. Hints are appreciated! 
Relation between digraph and NP-Complete problem
complexity theory;graph theory;np complete;hamiltonian path
When a problem is NP-complete, it doesn't necessarily mean that all instances are hard to solve. It just means that enough instances are hard to solve that, if you could solve this problem efficiently, you could do the same for all problems in NP. Enough, in this case, does mean infinitely many1 and that should lead you towards looking at large inputs if you want to find hard cases.2For example, the fact that 3-SAT is NP-complete doesn't mean that you would have any difficulty telling me whether the formula $(X\vee Y\vee Z)$ is satisfiable. In your example, you've observed that, for a graph with a single edge, that edge is a Hamiltonian path. That doesn't mean that it's hard to tell whether a graph has a single edge, or that it's easy to compute Hamiltonian paths on all graphs.1 Probably infinitely many, anyway. If P = NP, there wouldn't be any hard instances of NP-complete problems.2 Note that, strictly speaking, there's no such thing as a single hard case since, if we restrict to that single input, the problem can be solved in constant time. Properly, I mean a whole infinite class of instances such that it's hard to write an algorithm that answers the problem on that whole class of inputs.