# Is this construction a OWF?

Given the OWF function $$f: \{0,1\}^{2\lambda} \rightarrow \{0,1\}^{2\lambda}$$ and the PRG $$G: \{0,1\}^{\lambda} \rightarrow \{0,1\}^{2\lambda}$$, is the following function $$f^*$$ a OWF?

\begin{align} f^*: \{0,1\}^{\lambda} &\to \{0,1\}^{2\lambda}\\ x &\to f^*(x) = f(G(x)\oplus(0^{\lambda}||x)) \end{align}

My idea is that it is secure, mainly because the function $$f$$ is a OWF itself, but I haven't been able to prove it. Moreover, I thought that the collision probability of its input might be involved, but it's nothing more than intuition.

• Hint: Let $G'$ be a PRG. Then $G$ defined as $G(x_1\|x_2) = G(x_1)||x_2$ for long enough $x_1$ is also a PRG. Jan 11 at 12:41
• @Maeher long enough =? Jan 11 at 12:50
• @kelalaka Any constant fraction will do. Say 1/2 of the input length for a concrete construction. Jan 11 at 15:18
• @Maeher should I build a counterexample with your hint? Jan 11 at 16:03
• Hint 2: A OWF is only guaranteed to be hard to invert if the inputs are sampled according to the uniform distribution. A distribution that only contains inputs that end in many zeroes is very far (and easily distinguishable from) uniform. Jan 12 at 14:08