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This is not homework, but a random thought.

Let $G(x)$ be a PRG. We define the following function: $$F_k(x) = G(k\oplus x)$$

My intuition is that it shouldn't be a PRF but I couldn't come with an example yet.

I tried to build a reduction from identifying G to identifying F but I'm not certain how you can simulate an oracle call when all you have is some string.

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  • $\begingroup$ This allows you to construct related PRG inputs which probably is exploitable for a suitable PRG definition. $\endgroup$ – SEJPM Feb 20 at 15:41
  • $\begingroup$ Consider the counter-example in your recent question. $\endgroup$ – Maeher Feb 20 at 17:10
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    $\begingroup$ One possible proof technique here might be to assume that $G$ is not some arbitrary PRG, but one that you opportunistically construct to make it easily exploitable in the $F$ construction. I.e., don't try to write a proof that attacks any choice of $G$, but rather figure out what properties $G$ could have that don't disqualify it as a PRG but are exploitable when used as a component in $F$. And one way to construct a suitable $G$ might be to assume you're given an arbitrary PRG $G'$ and define some construction that delegates to it but makes some changes to its input or output. $\endgroup$ – Luis Casillas Feb 20 at 17:34
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Given some PRG $G : \{0,1\}^n \to \{0,1\}^m$ with $m > n+1$, define $G' : \{0,1\}^{n+1} \to \{0,1\}^m$ like this:

$$ \begin{align} G'(0\|x) & = G(x) \\ G'(1\|x) & = G(x) \end{align} $$


Lemma: If $G$ is a PRG, so is $G'$.

Proof: Let $s \in \{0,1\}^{n+1}$ be a randomly chosen seed, and let $s' \in \{0,1\}^{n}$ be its suffix (with the first bit of $s$ removed), which is also distributed uniformly at random. Since $G$ is a PRG, then $G(s')$ has a pseudorandom distribution. And since $G'(s) = G(s')$, so does $G'(s)$ have a pseudorandom distribution. So then, since $G'(s)$ is pseudorandom for random $s$, $G'$ is a PRG.


Theorem: $F_k(x) = G'(k \oplus x)$ is not a PRF.

Proof: Because of the definitions of $F$ and $G'$, we see that for all $k$ and $s$, $F_k(0\|s) = F_k(1\|s)$. Given oracle access to an $f$ that's either $F_k$ or a random function, we have the following distinguisher:

  1. Pick an arbitrary $s$;
  2. Query the oracle for $x_0 = f(0\|s)$;
  3. Query the oracle for $x_1 = f(1\|s)$;
  4. If $x_0 = x_1$, then output $1$, otherwise output $0$.

If $f$ is a random function, this outputs $1$ with probability $2^{-m}$. But if $f$ is $F_k$, then it outputs $1$ with probability $1$.


So, there exists a choice of PRG for which your proposed construction is not a PRF.

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