Question 99
Let $S$ be the set $\{-1,1\}^n$, that is, $n$-tuples such that each coordinate is either $-1$ or $1$. For
$$
s = (s_1, s_2, \ldots, s_n),\ t = (t_1, t_2, \ldots, t_n) \in \{-1,1\}^n
$$
define $s \odot t = (s_1 t_1, s_2 t_2, \ldots, s_n t_n)$.
Let $c$ be a positive constant, and let $f: S \rightarrow \{-1,1\}$ be a function such that there are at least $(1-c) \cdot 2^{2n}$ pairs $(s, t)$ with $s, t \in S$ such that $f(s \odot t) = f(s) f(t)$. Show that there exists a function $f'$ such that $f'(s \odot t) = f'(s) f'(t)$ for all $s, t \in S$ and $f(s) = f'(s)$ for at least $(1-10c) \cdot 2^n$ values of $s \in S$.
$$
s = (s_1, s_2, \ldots, s_n),\ t = (t_1, t_2, \ldots, t_n) \in \{-1,1\}^n
$$
define $s \odot t = (s_1 t_1, s_2 t_2, \ldots, s_n t_n)$.
Let $c$ be a positive constant, and let $f: S \rightarrow \{-1,1\}$ be a function such that there are at least $(1-c) \cdot 2^{2n}$ pairs $(s, t)$ with $s, t \in S$ such that $f(s \odot t) = f(s) f(t)$. Show that there exists a function $f'$ such that $f'(s \odot t) = f'(s) f'(t)$ for all $s, t \in S$ and $f(s) = f'(s)$ for at least $(1-10c) \cdot 2^n$ values of $s \in S$.