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Unformatted text preview: ICS 180: Introduction to Cryptography 5/13/2004 Homework 4 Due Tuesday , 5/20/2004 1 One-bit-streching PRG implies polynomially-streching PRG Assume that G is a PRG which stretches input by only one bit, i.e. for all inputs x , the length | G ( x ) | , of the output of G on x is equal to | x | + 1. 1.1 For any polynomial p ( ), use the 1-bit stretching PRG G to construct a PRG G which stretches the (random) k-bit input into a (pseudorandom) output of length p ( k ). Prove that your construction G is indeed a PRG if G is a PRG. Hint(s) : First try to construct a two-bit stretching G , i.e. do it for p ( k ) = k + 2. (Note that in the subsection below you have some wrong ways of making the 2-bit stretching PRG. I think that all ways where you try to use G just once will fail, and to get (2+ k )-bit output you need to use G twice.) If you do get it for 2-bit stretching PRG, chances are that your construction generalizes to any polynomial number of extra bits, and that you can prove this generalized construction using the proof you did for the 2-bit case and induction....
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- Spring '04