# Given a secure PRF $f(k, x)$, is $f(x, k)$ also a secure PRF?

Let $f: \{0,1\}^n \times \{0,1\}^n \rightarrow \{0,1\}^n$ be a secure PRF. Define $F(k, x) = f(x, k)$.
Is $F(k, x)$ also a secure PRF?

• As usual, what have you tried? – fkraiem Aug 21 '18 at 14:43
• Big hint: every secure PRP is also a secure PRF. – Luis Casillas Aug 21 '18 at 18:02

Example based on @LuisCasillas's comment: With a secret random key, you cannot determine the key from input output pairs. If you reverse the role of key and input, then you have the question: Can you learn "input" ($k$) from output ($f(x,k)$) and "key" ($x$)?
Yes. If $F(k, x) = f(x, k) = E_x(k)$, where $E$ is a PRP, then you can recover $k$ from one input/output pair of $F$. If your output is $c = E_x(k)$ then anyone can compute $k$ if they know $c$ and $x$. $$k = E^{-1}_x(c) = E^{-1}_x(E_x(k))$$
(Where $E$ is block encryption and $E^{-1}$ is the inverse, decryption.