# 3-SAT complexity

I'm trying to understand how $$3$$-SAT problems are assigned complexity to try and get a better understanding of the P vs NP problem.

Would a polynomial-time solution to an increasing number of clauses but constant number of distinct literals resolve P vs NP?

Or, would an algorithm have to work in polynomial time in both the number of distinct literals and the number of clauses?

It seems like a 2-dimensional problem and I'm not sure in which 'direction' the P vs NP problem lies.

There is a way to reduce SAT to $$3$$-SAT. This means that any polynomial algorithm for $$3$$-SAT can be used to construct a polynomial algorithm for SAT. This means any polynomial algorithm for $$3$$-SAT would prove P=NP.
The input for $$3$$-SAT consists $$n$$ variables and $$k$$ clauses consisting of $$3$$ literals each. A literal is of the form $$v$$ or $$\neg v$$ for some variable $$v$$. A clause is of the form $$a\lor b\lor c$$ for some literals $$a,b,c$$. So the size of the input of a $$3$$-SAT instance is $$n+3k$$.
What we mean when we talk about a polynomial algorithm for $$3$$-SAT is an algorithm with running time $$\mathcal{O}(p(n+3k))$$ for some polynomial $$p$$. This means that the algorithm has to run in polynomial time in the number of variables, but also in the number of clauses and literals.