andrew-yao &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/andrew-yao/ Feed of posts on WordPress.com tagged "andrew-yao" Tue, 21 May 2013 10:14:53 +0000 http://en.wordpress.com/tags/ en <![CDATA[Alan Turing Centenary Conference, 22nd-25th June 2012]]> http://duncan.hull.name/2012/06/15/turing100/ Fri, 15 Jun 2012 14:38:47 +0000 Duncan Hull http://duncan.hull.name/2012/06/15/turing100/ Alan Turing by Michael Dales

The Alan Turing statue at Bletchley Park. Creative commons licensed picture via Michael Dales on Flickr

Next weekend, a bunch of very distinguished computer scientists will rock up at the magnificent Manchester Town Hall for the Turing Centenary Conference in order to analyse the development of Computer Science, Artificial Intelligence and Alan Turing’s legacy [1].

There’s an impressive and stellar speaker line-up including:

Tickets are not cheap at £450 for four days, but you can sign up for free public lectures by Jack Copeland on Turing: Pioneer of the Information Age and Roger Penrose on the problem of modelling a mathematical mind. Alternatively, if you can lend some time, the conference organisers are looking for volunteers to help out in return for a free conference pass. Contact Vicki Chamberlin for details if you’re interested.

References

  1. Chouard, T. (2012). Turing at 100: Legacy of a universal mind Nature, 482 (7386), 455-455 DOI: 10.1038/482455a see also nature.com/turing
]]> <![CDATA[CS276 Lecture 13 (draft)]]> http://lucatrevisan.wordpress.com/2009/03/02/cs276-lecture-13-draft/ Tue, 03 Mar 2009 06:27:10 +0000 luca http://lucatrevisan.wordpress.com/2009/03/02/cs276-lecture-13-draft/

Summary

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}&fg=000000 be a {(t,\epsilon)}&fg=000000-one way permutation computable in time {r\leq t}&fg=000000. Then the predicate {x,r \rightarrow \langle x,r \rangle}&fg=000000 is {(\Omega( t \cdot \epsilon^2 \cdot n^{-O(1)} , 3\epsilon)}&fg=000000 hard core for the permutation {x,r \rightarrow f(x),r}&fg=000000.

A way to look at this result is the following: suppose {f}&fg=000000 is {(2^{\Omega(n)},2^{-\Omega(n)})}&fg=000000 one way and computable in {n^{O(1)}}&fg=000000 time. Then {\langle x,r \rangle}&fg=000000 is a {(2^{\Omega(n)},2^{-\Omega(n)})}&fg=000000 hard-core predicate for the permutation {x,r \rightarrow f(x),r}&fg=000000.

From now on, we shall assume that we have a one-way permutation {f: \{ 0,1 \}^n \rightarrow \{ 0,1 \}^n}&fg=000000 and a predicate {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}}&fg=000000 that is {(t,\epsilon)}&fg=000000 hard core for {f}&fg=000000.

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

Theorem 2 (Yao) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n}&fg=000000 be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}}&fg=000000 is {(t,\epsilon)}&fg=000000-hard core for {f}&fg=000000. Then the mapping

\displaystyle x \rightarrow f(x),P(x) &fg=000000

is {(t-O(1), \epsilon)}&fg=000000-pseudorandom generator mapping {n}&fg=000000 bits into {n+1}&fg=000000 bits.

We will amplify the expansion of the generator by the following idea: from an {n}&fg=000000-bit input, we run the generator to obtain {n+1}&fg=000000 pseudorandom bits. We output one of those {n+1}&fg=000000 bits and feed the other {n}&fg=000000 back into the generator, and so on. Specialized to above construction, and repeated {k}&fg=000000 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)&fg=000000

Theorem 3 (Blum-Micali) Let {f:\{ 0,1 \}^n \rightarrow \{ 0,1 \}^n}&fg=000000 be a permutation, and suppose {P:\{ 0,1 \}^n \rightarrow \{ 0,1 \}}&fg=000000 is {(t,\epsilon)}&fg=000000-hard core for {f}&fg=000000 and that {f,P}&fg=000000 are computable with complexity {r}&fg=000000.

Then {G_k : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{n+k}}&fg=000000 as defined in (1) is {(t-O(rk), \epsilon k)}&fg=000000-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}}&fg=000000 be a {(t,\epsilon)}&fg=000000 pseudorandom generator computable in time {r}&fg=000000, let {G_0(x)}&fg=000000 be the first {n}&fg=000000 bits of the output of {G(x)}&fg=000000, and let {G_1(x)}&fg=000000 be the last {n}&fg=000000 bits of the output of {G(x)}&fg=000000.

Define {G' : \{ 0,1 \}^n \rightarrow \{ 0,1 \}^{4n}}&fg=000000 as

\displaystyle G' (x) = G(G_0(x)), G( G_1 (x)) &fg=000000

Prove that {G'}&fg=000000 is a {(t-O(r),3\epsilon)}&fg=000000 pseudorandom generator.

]]>