# If $G'(s)=G(s0^{|s|})$ and $G$ is a PRNG, is $G'$ necessarily a PRNG? [closed]

I argue it is because there should be no more correlation between the outputs $G(s0^{|s|})$ than between the outputs $G(s)$. Thus we can say the $G'(s)$ is a PRNG that satisfies $|G'(s)| \geq 4|s|$. If we let $s' = s0^{|s|}$ and denote the set of all $s'$ by $S'$, then $s'$ should be chosen uniformly at random from $S'$. Since, as I stated there should be no correlation between $G'(s')$ for any $s'$, then effectively we are only increasing the expansion factor $l$.

Disagree? Please tell me why.

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## closed as unclear what you're asking by D.W., figlesquidge, e-sushi♦, rath, archieMar 15 '14 at 0:23

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

I attempted the same editing, but wondered if $s0$ is meant to be $s_0$, or something else; and in either case those come from nowhere and I fail to parse the question. –  fgrieu Mar 2 '14 at 18:10

Suppose g is a PRNG. $\:$ Define G by the following pseudocode:

def G(w):
m = length(w)
n = ceiling(m/2)
set x equal to the n leftmost bits of w
set y equal to the m-n rightmost bits of w
z = concatenate(g(x),y)
output z


In this case, G is a PRNG and G' is not a PRNG.

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The way you wrote this was not clear. If I understand you correctly you mean to say: "Let G(s) be a PRNG. Then where $s = s_1, \dots ,s_n$ define $G'(s) = G(s_1, \dots ,s_{n/2})||(s_{(n/2)+1, \dots, s_n)$, where $||$ denotes the concatenation. In this case G'(s) is not a PRNG." –  fowlslegs Mar 2 '14 at 17:36