Question

Problem 2 (30 points). Given a true PRG G, show whether or not the following functions are necessarily PRGs. If true, prove it. If not give a counterexample. Note that || denotes concatenation

Answer 1

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Let GG be a pseudorandom generator, and F(s)=G(s)||G(s¯)F(s)=G(s)||G(s¯) (where s¯s¯ denotes the bitwise complement of ss), is FF necessarily a PRG?

My intuition says that it is a PRG, as clearly G(s)G(s) and G(s¯)G(s¯) are PRGs on their own, and concatenating them shouldn't affect the randomness, but rather it will simply increase the expansion factor (length of output). However, I feel like there exists a counterexample where G(s)G(s) and G(s¯)G(s¯)are secure on their own but concatenating them will lose some randomness, i.e. make them predictable (because F(s)F(s) and F(s¯)F(s¯) should look pretty similar). I've thought about some examples where G(s)=H(s)⊕H(s¯)G(s)=H(s)⊕H(s¯), where H(s)H(s) is a PRG (and G(s)G(s) would also still be a PRG here), but I'm not sure if this leads to F(s)F(s) not being a PRG.

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Let GG be a pseudorandom generator, and F(s)=G(s)||G(s¯)F(s)=G(s)||G(s¯) (where s¯s¯ denotes the bitwise complement of ss), is FF necessarily a PRG?

My intuition says that it is a PRG, as clearly G(s)G(s) and G(s¯)G(s¯) are PRGs on their own, and concatenating them shouldn't affect the randomness, but rather it will simply increase the expansion factor (length of output). However, I feel like there exists a counterexample where G(s)G(s) and G(s¯)G(s¯)are secure on their own but concatenating them will lose some randomness, i.e. make them predictable (because F(s)F(s) and F(s¯)F(s¯) should look pretty similar). I've thought about some examples where G(s)=H(s)⊕H(s¯)G(s)=H(s)⊕H(s¯), where H(s)H(s) is a PRG (and G(s)G(s) would also still be a PRG here), but I'm not sure if this leads to F(s)F(s) not being a PRG.