How can I prove that if $G$ is a Pseudo Random Generator, then other generator is a PRG (while the range's size is different)?

For example: for the PRG $G$ (from $n$ to $t(n)$ where $n < t(n)$), the generator $G'$ (from $n$ to $n+1$), where $G'$ returns the first $n+1$ bits of $G$) is PRG too?

Is the fact that the ranges are different affect the indistinguishability between the PRGs?

  • 1
    $\begingroup$ For questions like these we require that you give us a better indication on where you are stuck in solving it yourself. Please edit to provide those details. $\endgroup$
    – mikeazo
    Jan 23, 2017 at 18:48
  • $\begingroup$ It is IMHO often the case that with a "practical" PRNG, that delivers n bits on each invocation, one doesn't have the n bits of the same statistical quality. $\endgroup$ Jan 24, 2017 at 11:17

1 Answer 1


Well I think you can prove it by reduction:

If there exist an efficient predictor $P$ that given the first $n$ bits of $G'$ is able to predict the $n+1$ bit then we can construct a distinguisher $D$ for $G$ defined as an elementary wrapper of $P$, that can distinguish $G$ from a truly random generator. Since by assumption $G$ is a (secure?) pseudo-random generator there is not such a $D$.

  • $\begingroup$ I see, is it possible to prove this by the hybrid argument? $\endgroup$
    – Adi Ml
    Jan 26, 2017 at 11:12
  • $\begingroup$ I am not completely sure on how to prove it by the Hybrid argument but from the definition of computational indistinguishability you have that $G_{t(n)} \approx_p R_{t(n)}$ then for every $m < t(n)$ the truncation of $G$ to the first m bits is indistinguishable from the truncation of $R$ to the first m bits, since you have that $G'_m \equiv G_m$ for $m=n+1$ then by transitivity $G'_m \approx_p R_m$. $\endgroup$
    – th3n3rd
    Jan 26, 2017 at 13:18

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