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\paragraph{Problem 2 (18 points)}
Let $G:=\{ G^n: \{0,1\}^n \to \{0,1\}^{2n} \}_{n\in\mathbb{N}}$ be a family of PRGs. For all $n\in\mathbb{N}$, let $h^n: \{0,1\}^n \to \{0,1\}$ be a hardcore bit of $G^n$.
Let $L:=\{ L^n: \{0,1\}^{n} \to \{0,1\}^{2n} \}_{n\in\mathbb{N}}$ be defined by: for any $x\in\{0,1\}^n$
\[ L^n(x) := G^n(x) \oplus (0^{2n-1} | h^n(x)). \]
That is, $L^n(x)$ outputs the first $2n-1$ bits of $G^n(x)$, and also the last bit of $G^n(x)$ XORed with $h^n(x)$.
Prove that $L$ is a family of PRGs. |