Hey guys, not sure if this belongs here but I think this category matched the most. I'm trying to figure out this crypto problem but I think my solution is too simple to be correct. I'm given a PRG $\displaystyle G : \{0,1\}^n \to \{0,1\}^{p(n)}$ where $\displaystyle p(n) > 2n$. Then, I'm asked if $\displaystyle G'(s) = s||G(s)$ is a PRG. My solution is to say no because given a distinguisher that inputs s, all it has to do is check if the output begins with s and if it does, the probability of it coming from the PRG is $\displaystyle \frac{1}{p(n)}$ which is trivial.