I'm curious about the following question:
Given a K3 surface, how does one proceed to compute its rank?
Of course the answer may depend on the form of the input, i.e. how the K3 is "given". So
For a given way of writing down a K3 surface, (e.g. quartics in $\mathbb{P}^3$)
How does one compute the Picard rank of the K3 surface?
(Aside: What I've seen people sometimes did is avoiding this question by nailing down a K3 surface $X$ with its $NS(X)$ together with the intersection form. Then find an embedding given by the ample class.)
Best Answer
There are some papers of van Luijk, where he computes the ranks of some K3s over number fields. The trick is to note that $NS(X) \hookrightarrow NS(X_p)$, where $X_p$ is the reduction of $X$ modulo a prime ideal $p$. One can determine the rank of $NS(X_p)$ by counting eigenvalues of Frobenius which differ from $q$ (the size of the residue field) by a root of unity. If you want to find rank 1 K3s, you can reduce modulo two different primes and hope to find rank 2 reductions which have lattices which are incompatible in some sense, forcing $NS(X)$ to be rank 1. (The issue here is that the rank of $NS(X_p)$ will always be even, so you can't win by using a single prime.)
I'm not sure how this works when you want to find K3s of larger rank though, unless you had a way of exhibiting linearly independent divisor classes. Anyhow, van Luijk uses this technique to find rank 1 quartics in $\mathbb{P}^3$ and I think others have done the same with genus 2 K3s defined over $\mathbb{Q}$.
I should add that the situation is much easier for Kummer surfaces. If I'm not mistaken, the rank of $X = K(A)$ ($A$ is an abelian surface) is 16 plus the Picard rank of $A$. The 16 comes from the 16 exceptional divisors you get when you blow up $A$ at its 2-torsion points. The rank of $A$ is usually not hard to figure out: a generic $A$ has rank 1, if $A$ is a product of elliptic curves then its rank is 2,3 or 4 depending on whether the curves are isogenous and whether they have CM or not, and there are a few other cases which one can probably figure out...