Can top SAT-solvers factor easy numbers?

Question

Modern SAT-solvers are very good at solving many real-world examples of SAT instances. However, we know how to generate hard ones: for instance use a reduction from factoring to SAT and give the RSA numbers as input.

This raises the question: what if I take an easy example of factoring. Instead of taking two large primes on $n/2$ bits, what if I take a prime $p$ on $\log n$ bits and a prime q on $n/\log n$ bits, let $N = pq$ and the encode $\mathrm{FACTOR}(N)$ as a SAT instance. $N$ would be an easy number to factor by brute-force search or sieve methods since one of the factors in so small; does a modern SAT-solver with some standard reduction from factoring to SAT also pick up on this structure?

Can top SAT-solvers factor $N = pq$ where $|p| = \log n$ quickly?

Answer 1

There are other much simpler instances which we provably know that the current algorithms cannot solve in sub-exponential time. These algorithms are incapable of counting (almost all of them are improvements of DPLL which correspond to Resolution propositional proof system).

Unfortunately such examples are unsatisfiable instances. The question about finding natural satisfiable hard instances for these algorithms is an interesting research problem (Russell Impagliazzo mentioned this during the proof complexity workshop last year at Banff). There are satisfiable instances that we provably know the algorithms fail badly if there is any such instance, but they are not very natural (they are based on formulas expressing the soundness of algorithms).

Regarding factoring, if the size of numbers are small (e.g. logarithmic as in your case, i.e. the numbers are given in unary), then theoretically there is no result that says there cannot be solved by current algorithms, and in fact we can write simple polynomial time algorithms that factor these numbers. So whether a particular SAT solver program can solve them might depend on the particular algorithm.