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Unformatted text preview: 6.841 Advanced Complexity Theory March 18, 2009 Lecture 13 Lecturer: Madhu Sudan Scribe: Alex Cornejo 1 Overview of todays lecture Todas Theorem: PH := S k N P k P #P , steps: Prove some properties concerning C , C , C , BP C Do some operator calculus to prove PH BP P . Prove that BP P P #P 2 Review: Operator definitions Let L be a language, C a complexity class and q ( n ) some growing function of n . BP q ( n ) L = n q ( n ) yes ( L ) , q ( n ) no ( L ) o q ( n ) yes ( L ) = n x  Pr[( x,y ) L ] 1 2 q ( n ) o q ( n ) no ( L ) = n x  Pr[( x,y ) L ] 2 q ( n ) o When we omit the q ( n ) subscript we assume that q ( n ) P . BP L = { yes ( L ) , no ( L ) } BP C = { BP L  L C } L = { x  #( y ) s.t. ( x,y ) L is even } C = { L  L C } L = { x  #( y ) s.t. ( x,y ) L is odd } C = { L  L C } L = { x  y s.t. ( x,y ) L } C = { L  L C } L = { x  y s.t. ( x,y ) L } C = { L  L C } For this lecture, the correct way to think about an expression involving Todas complexity operators is visualizing the execution tree that represents the expression. For example consider a language L BP P , then for x L we have the following tree: BP + M ( x,y,z ) ... M ( x,y,z ) ... + M ( x,y,z ) ... M ( x,y,z ) y z z 131 3 Operator properties To prove Todas theorem we need to prove the following properties: Property 1. C = C Property 2. BP BP C = BP C Property 3. BP C = BP C Property 4. C , C BP C Observe that for our purposes it would suffice to prove them for C { P , P , BP P } . Lets warm up....
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 Spring '09
 MadhuSudan

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