Solving P vs. NP by writing a deductive solution-excluding algorithm

Question

I'm thinking about the P vs. NP problem. Take the Hamiltonian graph problem, for example. If I make a Hamiltonian graph-finding algorithm which goes through every edge of a given graph and deductively excludes as many possible Hamiltonian graphs as possible in each step, and then analyze its complexity with respect to the size of the input, could that be a solution to the P vs. NP problem? Or would a solution have to consider all possible elementary operations and show (logically/deductively) that no finite set of polynomial-time operations executed a polynomial number of times can solve the Hamiltonian graph decision problem?

Answer 1

If your algorithm can be proven to have polynomial-time complexity, then that would prove $P=NP$.

If your algorithm can be shown not to have polynomial-time complexity, that doesn't, in and of itself, prove $P \ne NP$, since there might be a better algorithm. However if you can also show your algorithm has complexity no worse than any other algorithm (good luck with that!), then that would prove $P \ne NP$.

Answer 2

If you want to use any problem $p$ to show that $\mathcal{P} \ne \mathcal{NP}$, you must show that

1. $p$ is $\mathcal{NP}$-complete.
2. There is no polynomial-time algorithm to solve $p$.