6
votes
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New PRG from old

Is the following statement true: If $G: \{0,1\}^k \to \{0,1\}^n$ is a PRG, then so is $G':\{0,1\}^{k+l} \to \{0,1\}^{n+l}$ defined by $$G'(x||x')=G(x)||x'$$ where $x \in \{0,1\}^k$ and $x' \in ...