# Is it possible to build a secure PRG from two functions one of them being a PRG?

Having two deterministic functions $$G_1, G_2 : \left\{0,1\right\}^\lambda \rightarrow \left\{0,1\right\}^{\lambda+l}$$, at least one of which is a secure PRG.

Being $$\alpha$$ a constant, is it possible to build a secure PRG $$G' : \left\{0,1\right\}^{\alpha\lambda} \rightarrow \left\{0,1\right\}^{\alpha(\lambda+l)}$$ using $$G_1$$ and $$G_2$$?

My intuition is that it can't be done, but I don't know how to prove it.

• In order to show that something can’t be done, you need to explain the rules of the game - what would count as a solution. – Yuval Filmus Nov 14 '18 at 22:04
• It can be done. Suppose $G_1$ is the PRG. Ignore $G_2$ and use a lengthening method on $G_1$. If this approach isn't admissible, please clarify the question. – Reinstate Monica Nov 15 '18 at 13:54
• @user2712414 Just to be sure, are $\lambda$, $l$ and $\alpha$ positive integers? – John L. Nov 16 '18 at 16:56