Let $p$ be prime and $\mathbb{F}_p$ the finite field with $p$ elements. Suppose we have a set $B\subseteq \mathbb{F}_p$ satisfying $|B|<p^{\alpha}$ for some $0<\alpha<1$ and there exists $A\subseteq \mathbb{F}_p$ such that $$A+A=B$$ where $A+A$ is the sumset $$A+A=\{ a_1+a_2 \ : \ a_1,a_2\in A\}.$$ Can we calculate $A$ in time subexponential in $|B|$? The $A$ may not be unique, so any such $A$ is sufficient.

Here's an easier version of the problem where this is possible: Let $0<\alpha<1/2$. Suppose $|B|<p^{\alpha}$ satisfies $$A+C=B$$ for some $|C|<p^{\alpha}$.

If we know $B,C$ then we can recover $A$ quickly with a high probability. This follows by grouping pairs into sets $D(\lambda)$ $$D(\lambda)=\{ (b,c)\in B\times C \ : b-c=\lambda\}.$$ We then try guesses for $A$ to be made up of values $\lambda$ where $D(\lambda)$ is approximately $|D(\lambda)|\sim |C|.$ This only works with a high probability (for example, no success if $|B+B|=O(|B|)$). So another question, instead of recovering $A$ with high probability, can we recover $A$ certainly?