CS276 Lecture 13 (draft)


Today we complete the proof that it is possible to construct a pseudorandom generator from a one-way permutation

1. Pseudorandom Generators from One-Way Permutations

Last time we proved the Goldreich-Levin theorem.

Theorem 1 (Goldreich and Levin) Let {f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a {(t,\epsilon)}-one way permutation computable in time {r\leq t}. Then the predicate {x,r \rightarrow \langle x,r \rangle} is {(\Omega( t \cdot \epsilon^2 \cdot n^{-O(1)} , 3\epsilon)} hard core for the permutation {x,r \rightarrow f(x),r}.

A way to look at this result is the following: suppose {f} is {(2^{\Omega(n)},2^{-\Omega(n)})} one way and computable in {n^{O(1)}} time. Then {\langle x,r \rangle} is a {(2^{\Omega(n)},2^{-\Omega(n)})} hard-core predicate for the permutation {x,r \rightarrow f(x),r}.

From now on, we shall assume that we have a one-way permutation {f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} and a predicate {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} that is {(t,\epsilon)} hard core for {f}.

This already gives us a pseudorandom generator with one-bit expansion.

Theorem 2 (Yao) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} is {(t,\epsilon)}-hard core for {f}. Then the mapping

\displaystyle  x \rightarrow f(x),P(x)

is {(t-O(1), \epsilon)}-pseudorandom generator mapping {n} bits into {n+1} bits.

We will amplify the expansion of the generator by the following idea: from an {n}-bit input, we run the generator to obtain {n+1} pseudorandom bits. We output one of those {n+1} bits and feed the other {n} back into the generator, and so on. Specialized to above construction, and repeated {k} times we get the mapping

\displaystyle   G_k (x) := P(x), P(f(x)), P(f(f(x)), \ldots, P(f^{(k-1)} (x), f^{(k)} (x) \ \ \ \ \ (1)

Theorem 3 (Blum-Micali) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n} be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}} is {(t,\epsilon)}-hard core for {f} and that {f,P} are computable with complexity {r}.

Then {G_k : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{n+k}} as defined in (1) is {(t-O(rk), \epsilon k)}-pseudorandom.

Thinking about the following problem is a good preparation for the proof the main result of the next lecture.

Exercise 1 (Tree Composition of Generators) Let {G:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^{2n}} be a {(t,\epsilon)} pseudorandom generator computable in time {r}, let {G_0(x)} be the first {n} bits of the output of {G(x)}, and let {G_1(x)} be the last {n} bits of the output of {G(x)}.

Define {G' : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{4n}} as

\displaystyle  G' (x) = G(G_0(x)), G( G_1 (x))

Prove that {G'} is a {(t-O(r),3\epsilon)} pseudorandom generator.

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