Do there exists permutations $\pi_1,\pi_2$ and polynomial size CFG that describe the finite language $\{w \pi_1(w) \pi_2(w)\}$ over alphabet {0,1}? Do there exists permutations $\pi_1,\pi_2$ and polynomial size CFG that describe the finite language {$w \pi_1(w) \pi_2(w)$} over alphabet {0,1}?
Polynomial size in $|w|=n$
 A: Without loss of generality, we can assume that a grammar for $L_n$ (such a language with specific $\pi_1,\pi_2 \in S_n$) is in Chomsky Normal Form. The language $L_n$ consists of the words $w(x) = x\pi_1(x)\pi_2(x)$ for all $x \in \{0,1\}^n$.
Using the Subword Lemma (see my previous answers), for each $w(x)$ we can find a substring $s(x)$ of length $$\frac{n}{2} \leq |s(x)| < n$$ generated by some symbol $A(x)$ and occurring at position $p(x)$.
Suppose that $p(x) = p(y)$ and $A(x) = A(y)$. Since $|s(x)|<n$, the subword $s(x)$ cannot intersect both the $x$ part and the $\pi_2(x)$ part of $w(x)$; we can assume it is disjoint from the $x$ part. Thus $w(x)$ is of the form $x \alpha s(x) \beta$. This implies that $A(x)$ generates exactly one string, namely $s(x)$. Therefore $s(x) = s(y)$.
Now $s(y)$ intersects either $\pi_1(y)$ or $\pi_2(y)$ in at least $n/4$ places, and thus determines at least $n/4$ bits of $y$. Therefore at most $2^{3n/4}$ strings $y \in \{0,1\}^n$ can have $p(x) = p(y)$ and $A(x) = A(y)$. Since there are at most $3n$ possibilities for $p(y)$, we get that there are at least $$\frac{2^{n/4}}{3n}$$ different non-terminals in the grammar.
Comment: The same proof works if $\pi_1,\pi_2 \in S_{\{0,1\}^n}$, i.e. are arbitrary permutations on the set of all $n$-bit words. Given $n/4$ bits of $\pi_i(y)$, there are exactly $2^{3n/4}$ preimages $y$.
