0
$\begingroup$

This expression appears on my cryptography homework:

$G'(k) = G(k \oplus 1^s)$

It is a description of a Pseudorandom Number Generator where $G:\{0,1\}^s \to \{0,1\}^n$

I just cant see what the "prime" adds, to me it would read the same as:

$G(k) = k \oplus 1^s$

Am I missing something?

1 Answers 1

1

Given one generator $G \colon \{0,1\}^s \to \{0,1\}^n$ we define another generator as follows: Given $k$, first swap all bits, that is xor with $1^s$, then apply $G$, as a formula: We define $G'\colon \{0,1\}^s \to \{0,1\}^n$ by $G'(k) = G(k \oplus 1^s)$.

For example: $G'(100) = G(001)$, $G'(111) = G(000)$ and so on.

  • 0
    Ahh so G prime means something like "a specific instance of G"2012-11-08