lec13 - Alternating quantifiers Pleasing viewpoint: CS151...

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1 CS151 Complexity Theory Lecture 13 May 10, 2011 May 10, 2011 2 Alternating quantifiers Pleasing viewpoint: P NP coNP Δ 3 Σ 2 Π 2 Δ 2 9 8 98 89 Σ 3 Π 3 PSPACE PH 989 898 9898989 …” Σ i Π i 989 …” 898 …” const. # of alternations poly(n) alternations May 10, 2011 3 PH collapse Theorem : if Σ i = Π i then for all j > i Σ j = Π j = Δ j = Σ i “the polynomial hierarchy collapses to the i-th level” • Proof: – sufficient to show Σ i = Σ i+1 – then Σ i+1 = Σ i = Π i = Π i+1 ; apply theorem again P NP coNP Σ 3 Π 3 Δ 3 PH Σ 2 Π 2 Δ 2 May 10, 2011 4 PH collapse – recall: L Σ i+1 iff expressible as L = { x | 9 y (x, y) R } where R Π i – since Π i = Σ i , R expressible as R = { (x,y) | 9 z ( (x, y), z) R’ } where R’ Π i-1 – together: L = { x | 9 (y, z) (x, (y, z)) R’} – conclude L Σ i May 10, 2011 5 Karp-Lipton • we know that P = NP implies SAT has polynomial-size circuits. • suppose SAT has poly-size circuits – any consequences? – might hope: SAT P/poly PH collapses to P , same as if SAT P Theorem (KL): if SAT has poly-size circuits then PH collapses to the second level. May 10, 2011 6 BPP PH • Recall: don’t know BPP different from EXP Theorem (S,L,GZ): BPP ( Π 2 Σ 2 ) • don’t know Π 2 Σ 2 different from EXP but believe much weaker
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2 May 10, 2011 7 BPP PH • Proof: BPP language L: p.p.t. TM M: x L Pr y [M(x,y) accepts] ≥ 2/3 x L Pr y [M(x,y) rejects] ≥ 2/3 – strong error reduction : p.p.t. TM M’ • use n random bits (|y’| = n) • # strings y’ for which M’(x, y’) incorrect is at most 2 n/3 • (can’t achieve with naïve amplification) May 10, 2011 8 BPP PH • view y’ = ( w , z ) , each of length n/2 • consider output of M’(x, ( w , z )) : 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 w = 000…00 000…01 000…10 111…11 x L x L so few ones, not enough for whole disk May 10, 2011 9 BPP PH • proof (continued): – strong error reduction: # bad y’ < 2 n/3 – y’ = (w, z) with |w| = |z| = n/2 – Claim: L = {x : 9 w 8 z M’(x, (w, z)) = 1 } – x L: suppose 8 w 9 z M’(x, (w, z)) = 0 • implies 2 n/2 0’s; contradiction – x L: suppose 9 w 8 z M’(x, (w, z)) = 1 • implies 2 n/2 1’s; contradiction May 10, 2011 10 BPP PH – given BPP language L: p.p.t. TM M: x L Pr y [M(x,y) accepts] ≥ 2/3 x L Pr y [M(x,y) rejects] ≥ 2/3 – showed L = {x : 9 8 z M’(x, (w, z)) = 1} – thus BPP Σ 2 BPP closed under complement BPP Π 2 – conclude: BPP ( Π 2 Σ 2 ) May 10, 2011 11 New Topic The complexity of counting May 10, 2011 12 Counting problems • So far, we have ignored function problems
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lec13 - Alternating quantifiers Pleasing viewpoint: CS151...

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