Fix a field $K$ and consider the boolean function $f : \{0, 1\}^n \rightarrow \{0, 1\}$, $f : (x_0, \ldots , x_{n - 1}) \mapsto (\overline{x_0} \vee \overline{x_1}) \wedge \cdots \wedge (\overline{x_{n - 2}} \vee \overline{x_{n - 1}})$, where $n$ is even. Fix an affine hyperplane $A \subset K^n$.

Conjecture. If $A \cap \{0, 1\}^n \subset f^{-1}(1)$, then $\left| A \cap \{0, 1\}^n \right| \leq 2^{\frac{n}{2}}$.

Notes.

• Because $\left| f^{-1}(1) \right| = 3^{\frac{n}{2}}$, the claimed bound on $\left| A \cap \{0, 1\}^n \right|$ is much smaller than the a priori maximum.
• The claimed upper bound can easily be attained: indeed, set $K := \mathbb{Q}$ (e.g.) and set $A$ as the kernel of the functional $A : (x_0, \ldots , x_{n - 1}) \mapsto x_1 + x_3 + \cdots + x_{n - 1}$. It's easy to check that $A \cap \{0, 1\}^n \subset f^{-1}(1)$ and $\left| A \cap \{0, 1\}^n \right| = 2^{\frac{n}{2}}$. Thus the claim is that this is the best you can do.
• The problem can also be phrased "dually" in terms of subset sums. It says that if an array of field elements $a_0, \ldots , a_{n - 1}, a$'s subset sums "lacks adjacent elements", in the sense that every $\{i_0, \ldots , i_{k - 1}\} \subset \{0, \ldots , n -1 \}$ such that $\sum_{j = 0}^{k - 1} a_{i_j} = a$ also satisfies $\{2 j, 2 j + 1 \} \not \subset \{i_0, \ldots , i_{k - 1}\}$ for each $j \in \{0, \ldots , \frac{n}{2} - 1\}$, then there can be at most $2^{\frac{n}{2}}$ subsets $\{i_0, \ldots , i_{k - 1}\} \subset \{0, \ldots , n - 1\}$ such that $\sum_{j = 0}^{k - 1} a_{i_j} = a$.