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ThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHThe Polynomial HierarchyA.Antonopoulos18/1/2010

ThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PH1 Optimization ProblemsIntroductionThe Class DPOracle Classes2 The Polynomial HierarchyDefinitionBasic TheoremsBPP and PH

DP Class DefinitionThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHDefinitionA language L is in the class DP if and only if there are twolanguages L 1 ∈ NP and L 2 ∈ coNP such that L = L 1 ∩ L 2 .DP is not NP ∩ coNP!Also, DP is a syntactic class, and so it has completeproblems.

Complete Problems for DPThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PH

Complete Problems for DPThePolynomialHierarchyA.AntonopoulosTheoremSAT-UNSAT is DP-complete.OutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PH

Complete Problems for DPThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremSAT-UNSAT is DP-complete.ProofFirstly, we have to show it is in DP.So, let:L 1 ={(φ, φ ′ ): φ is satisfiable}.L 2 ={(φ, φ ′ ): φ ′ is unsatisfiable}.It is easy to see, L 1 ∈ NP and L 2 ∈ coNP, thusL ≡ L 1 ∩ L 2 ∈ DP.For completeness, let L ∈ DP. We have to show thatL ≤ P SAT-UNSAT. L ∈ DP ⇒ L = L 1 ∩ L 2 , L 1 ∈ NP andL 2 ∈ coNP.SAT NP-complete⇒ ∃R 1 :L 1 ≤ P SAT and R 2 :L 2 ≤ P SAT.Hence, L ≤ P SAT-UNSAT,by R(x) = (R 1 (x), R 2 (x))

Complete Problems for DPThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremEXACT TSP is DP-complete.ProofEXACT TSP ∈ DP, by L 1 ≡TSP ∈ NP and L 2 ≡TSPCOMPLEMENT ∈ coNPCompleteness: we’ll show that SAT-UNSAT≤ P EXACTTSP.3SAT≤ P HP: (φ, φ ′ ) → (G, G ′ )Broken Hamilton Path (2 node-disjoint paths that coverall nodes)Almost Satisfying Truth Assignement (satisfies all clausesexcept for one)

Complete Problems for DPThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofWe define distances:1 If (i, j) ∈ E(G) or E(G’): d(i, j) ≡ 12 If (i, j) /∈ E(G), but i and j ∈ V(G): d(i, j) ≡ 23 Otherwise: d(i, j) ≡ 4Let n be the size of the graph.1 If φ and φ ′ satisfiable, then optCost = n2 If φ and φ ′ unsatisfiable, then optCost = n + 33 If φ satisfiable and φ ′ not, then optCost = n + 24 If φ ′ satisfiable and φ not, then optCost = n + 1“yes” instance of SAT-UNSAT ⇔ optCost = n + 2Let B ≡ n + 2!

Other DP-complete problemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHAlso:CRITICAL SAT: Given a Boolean expression φ, is it truethat it’s unsatisfiable, but deleting any clause makes itsatisfiable?CRITICAL HAMILTON PATH: Given a graph, is it truethat it has no Hamilton path, but addition of any edgecreates a Hamilton path?CRITICAL 3-COLORABILITY: Given a graph, is it truethat it is not 3-colorable, but deletion of any node makesit 3-colorable?are DP-complete!

The Classes P NP and FP NPThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHAlternative DP DefinitionDP is the class of languages that can be decided by an oraclemachine which makes 2 queries to a SAT oracle, and accepts iffthe 1st answer is yes, and the 2nd is no.P SAT is the class of languages decided in pol time with aSAT oracle.Polynomial number of queriesQueries computed adaptivelySAT NP-complete ⇒ P SAT =P NPFP NP is the class of functions that can be computed by apol-time TM with a SAT oracle.Goal: MAX OUTPUT≤ P MAX-WEIGHT SAT≤ P SAT

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHMAX OUTPUT DefinitionGiven NTM N, with input 1 n , which halts after O(n),withoutput a string of length n. Which is the largest output,of anycomputation of N on 1 n ?

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHMAX OUTPUT DefinitionGiven NTM N, with input 1 n , which halts after O(n),withoutput a string of length n. Which is the largest output,of anycomputation of N on 1 n ?TheoremMAX OUTPUT is FP NP -complete.ProofMAX OUTPUT ∈ FP NP .Let F : Σ ∗ → Σ ∗ ∈ FP NP ⇒ ∃ pol-time TM M ? , s.t.M SAT (x) = F (x)We’ll show: F ≤MAX OUTPUT!Reductions R and S (log space computable) s.t.:∀x, R(x) is a instance of MAX OUTPUTS(max output of R(x)) → F (x)

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofNTM N:Let n = p 2 (|x|), p(· ), is the pol bound of SAT.N(1 n ) generates x on a string.M SAT query state (φ 1 ):If z 1 = 0 (φ 1 unsat), then continue from q NO .If z 1 = 1 (φ 1 sat), then guess assignment T 1 :If test succeeds, continue from q YES .If test fails, output=0 n and halt. (Unsuccessfulcomputation)Continue to all guesses (z i ), and halt, with output= z 1 z 2 ....00} {{ }n(Successful computation)

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofWe claim that the successful computation that outputs thelargest integer, correspond to a correct simulation:Let j the smallest integer,s.t.: z j = 0, while φ j was satisfiable.Then, ∃ another successful computation of N, s.t.: z j = 1.The computations agree to the first j − 1 digits,⇒ the 2 ndrepresents a larger number.The S part: F (x) can be read off the end of the largest outputof N.

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHMAX-WEIGHT SAT DefinitionGiven a set of clauses, each with an integer weight, find thetruth assignment that satisfies a set of clauses with the mosttotal weight.

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHMAX-WEIGHT SAT DefinitionGiven a set of clauses, each with an integer weight, find thetruth assignment that satisfies a set of clauses with the mosttotal weight.TheoremMAX-WEIGHT SAT is FP NP -complete.ProofMAX-WEIGHT SAT is in FP NP : By binary search, and a SAToracle, we can find the largest possible total weight of satisfiedclauses, and then, by setting the variables 1-1, the truthassignment that achieves it.MAX OUTPUT≤MAX-WEIGHT SAT:

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofNTMN(1 n ) → φ(N, m):Any satisfying truth assignment of φ(N, m) → legal comp.of N(1 n )Clauses are given a huge weight (2 n ), so that any t.a. thataspires to be optimum satisfy all clauses of φ(N, m).Add more clauses: (y i ): i = 1, ..n with weight 2 n−i .Now, optimum t.a. must not represent any legalcomputation, but this which produces the largest possibleoutput value.S part: From optimum t.a. of the resulting expression (orthe weight), we can recover the optimum output of N(1 n ).

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHAnd the main result:TheoremTSP is FP NP -complete.

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHAnd the main result:TheoremTSP is FP NP -complete.CorollaryTSP COST is FP NP -complete.

FP NP -complete ProblemsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHFigure: The overall construction (17-2)

NP[log n]The Class PThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHDefinitionP NP[logn] is the class of all languages decided by a polynomialtime oracle machine, which on input x asks a total ofO(log |x|) SAT queries.FP NP[logn] is the corresponding class of functions.

NP[log n]The Class PThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHDefinitionP NP[logn] is the class of all languages decided by a polynomialtime oracle machine, which on input x asks a total ofO(log |x|) SAT queries.FP NP[logn] is the corresponding class of functions.CLIQUE SIZE DefinitionGiven a graph, determine the size of his largest clique.

NP[log n]The Class PThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHDefinitionP NP[logn] is the class of all languages decided by a polynomialtime oracle machine, which on input x asks a total ofO(log |x|) SAT queries.FP NP[logn] is the corresponding class of functions.CLIQUE SIZE DefinitionGiven a graph, determine the size of his largest clique.TheoremCLIQUE SIZE is FP NP[logn] -complete.

ConclusionThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PH1 TSP (D) is NP-complete.2 EXACT TSP is DP-complete.3 TSP COST is FP NP -complete.4 TSP is FP NP -complete.And now,P NP → NP NP ?Oracles for NP NP ?

OutlineThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PH1 Optimization ProblemsIntroductionThe Class DPOracle Classes2 The Polynomial HierarchyDefinitionBasic TheoremsBPP and PH

The Polynomial HierarchyThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHPolynomial Hierarchy Definition∆ 0 P = Σ 0 P = Π 0 P = P∆ i+1 P = P Σ i PΣ i+1 P = NP Σ i PΠ i+1 P = coNP Σ i PPH ≡ ⋃ i0Σ i P

The Polynomial HierarchyThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHPolynomial Hierarchy Definition∆ 0 P = Σ 0 P = Π 0 P = P∆ i+1 P = P Σ i PΣ i+1 P = NP Σ i PΠ i+1 P = coNP Σ i PPH ≡ ⋃ i0Σ i P

The Polynomial HierarchyThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHPolynomial Hierarchy Definition∆ 0 P = Σ 0 P = Π 0 P = P∆ i+1 P = P Σ i PΣ i+1 P = NP Σ i PΠ i+1 P = coNP Σ i PPH ≡ ⋃ i0Σ i PΣ 0 P = P∆ 1 P = P , Σ 1 P = NP , Π 1 P = coNP∆ 2 P = P NP , Σ 2 P = NP NP , Π 2 P = coNP NP

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremLet L be a language , and i ≥ 1. L ∈ Σ i P iff there is apolynomially balanced relation R such that the language{x; y : (x, y) ∈ R} is in Π i−1 P andProof (by Induction)L = {x : ∃y, s.t. : (x, y) ∈ R}For i = 1{x; y : (x, y) ∈ R} ∈ P,so L = {x|∃y : (x, y) ∈ R} ∈ NPFor i > 1If ∃R ∈ Π i−1 P, we must show that L ∈ Σ i P ⇒∃ NTM with Σ i−1 P oracle: NTM(x) guesses a y and asksΣ i−1 P oracle whether (x, y) /∈ R.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofIf L ∈ Σ i P, we must show the existence or R.L ∈ Σ i P ⇒ ∃ NTM M K , K ∈ Σ i−1 P, which decides L.K ∈ Σ i−1 P ⇒ ∃S ∈ Π i−2 P : (z ∈ K ⇔ ∃w : (z, w) ∈ S)We must describe a relation R (we know: x ∈ L ⇔accepting comp of M K (x))Query Steps: “yes”→ z i has a cerfificate w i st (z i , w i ) ∈ S.So, R(x) =“(x, y) ∈ R iff yrecords an acceptingcomputation of M ? on x , together with a certificate w i foreach yes query z i in the computation.”We must show {x; y : (x, y) ∈ R} ∈ Π i−1 P.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHCorollaryLet L be a language , and i ≥ 1. L ∈ Π i P iff there is apolynomially balanced relation R such that the language{x; y : (x, y) ∈ R} is in Σ i−1 P andCorollaryL = {x : ∀y, |y| ≤ |x| k , s.t. : (x, y) ∈ R}Let L be a language , and i ≥ 1. L ∈ Σ i P iff there is apolynomially balanced, polynomially-time decicable (i + 1)-aryrelation R such that:L = {x : ∃y 1 ∀y 2 ∃y 3 ...Qy i , s.t. : (x, y 1 , ..., y i ) ∈ R}where the i th quantifier Q is ∀, if i is even, and ∃, if i is odd.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremIf for some i ≥ 1, Σ i P = Π i P, then for all j > i:Σ j P = Π j P = ∆ j P = Σ i POr, the polynomial hierarchy collapses to the i th level.ProofIt suffices to show that: Σ i P = Π i P ⇒ Σ i+1 P = Σ i PLet L ∈ Σ i+1 P ⇒ ∃R ∈ Π i P: L = {x|∃y : (x, y) ∈ R}Since Π i P = Σ i P ⇒ R ∈ Σ i P(x, y) ∈ R ⇔ ∃z : (x, y, z) ∈ S, S ∈ Π i−1 P.Thus, x ∈ L ⇔ ∃y; z : (x, y, z) ∈ S, S ∈ Π i−1 P, which meansL ∈ Σ i P.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHCorollaryIf P=NP, or even NP=coNP, the Polynomial Hierarchycollapses to the first level.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHCorollaryIf P=NP, or even NP=coNP, the Polynomial Hierarchycollapses to the first level.MINIMUM CIRCUIT DefinitionGiven a Boolean Circuit C, is it true that there is no circuitwith fewer gates that computes the same Boolean function

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHCorollaryIf P=NP, or even NP=coNP, the Polynomial Hierarchycollapses to the first level.MINIMUM CIRCUIT DefinitionGiven a Boolean Circuit C, is it true that there is no circuitwith fewer gates that computes the same Boolean function

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHQSAT i DefinitionGiven expression φ, with Boolean variables partitioned into isets X i ,is φ satisfied by the overall truth assignment of theexpression:∃X 1 ∀X 2 ∃X 3 .....QX i φ, where Q is ∃ if i is odd, and ∀ if i is even.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHQSAT i DefinitionGiven expression φ, with Boolean variables partitioned into isets X i ,is φ satisfied by the overall truth assignment of theexpression:∃X 1 ∀X 2 ∃X 3 .....QX i φ, where Q is ∃ if i is odd, and ∀ if i is even.TheoremFor all i ≥ 1 QSAT i is Σ i P-complete.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremIf there is a PH-complete problem, then the polynomialhierarchy collapses to some finite level.ProofLet L is PH-complete.Since L ∈ PH, ∃i ≥ 0 : L ∈ Σ i P.But any L ′ ∈ Σ i+1 P reduces to L. Since PH is closed underreductions, we imply that L ′ ∈ Σ i P, so Σ i P = Σ i+1 P.

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremIf there is a PH-complete problem, then the polynomialhierarchy collapses to some finite level.ProofLet L is PH-complete.Since L ∈ PH, ∃i ≥ 0 : L ∈ Σ i P.But any L ′ ∈ Σ i+1 P reduces to L. Since PH is closed underreductions, we imply that L ′ ∈ Σ i P, so Σ i P = Σ i+1 P.TheoremPH ⊆ PSPACE

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremIf there is a PH-complete problem, then the polynomialhierarchy collapses to some finite level.ProofLet L is PH-complete.Since L ∈ PH, ∃i ≥ 0 : L ∈ Σ i P.But any L ′ ∈ Σ i+1 P reduces to L. Since PH is closed underreductions, we imply that L ′ ∈ Σ i P, so Σ i P = Σ i+1 P.TheoremPH ⊆ PSPACE

Basic TheoremsThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremIf there is a PH-complete problem, then the polynomialhierarchy collapses to some finite level.ProofLet L is PH-complete.Since L ∈ PH, ∃i ≥ 0 : L ∈ Σ i P.But any L ′ ∈ Σ i+1 P reduces to L. Since PH is closed underreductions, we imply that L ′ ∈ Σ i P, so Σ i P = Σ i+1 P.TheoremPH ⊆ PSPACEPH ? = PSPACE (Open). If it was, then PH has completeproblems, so it collapses to some finite level.

BPP and PHThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHTheoremBPP ⊆ Σ 2 P ∩ Π 2 PProofBecause coBPP = BPP,we prove only BPP ⊆ Σ 2 P.Let L ∈ BPP (L is accepted by “clear majority”).For |x| = n, let A(x) ⊆ {0, 1} p(n) be the set of acceptingcomputations.We have:x ∈ L ⇒ |A(x)| ≥ 2 p(n) ( 1 − 1 2 n )x /∈ L ⇒ |A(x)| ≤ 2 p(n) ( 12 n )Let U be the set of all bit strings of length p(n).For a, b ∈ U, let a ⊕ b be the XOR:a ⊕ b = c ⇔ c ⊕ b = a, so “⊕b” is 1-1.

BPP and PHThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofFor t ∈ U, A(x) ⊕ t = {a ⊕ t : a ∈ A(x)} (translation of A(x)by t). We imply that: |A(x) ⊕ t| = |A(x)|If x ∈ L, consider a random (drawing p 2 (n) bits) sequence oftranslations: t 1 , t 2 , .., t p(n) ∈ U.For b ∈ U, these translations cover b, if b ∈ A(x) ⊕ t j ,j ≤ p(n).b ∈ A(x) ⊕ t j ⇔ b ⊕ t j ∈ A(x) ⇒ Pr[b /∈ A(x) ⊕ t j ]= 1 2 nPr[b is not covered by any t j ]=2 −np(n)Pr[∃ point that is not covered]≤ 2 −np(n) |U| = 2 −(n−1)p(n)

BPP and PHThePolynomialHierarchyA.AntonopoulosOutlineOptimizationProblemsIntroductionThe Class DPOracle ClassesThePolynomialHierarchyDefinitionBasic TheoremsBPP and PHProofSo,T = (t 1 , .., t p(n) ) has a positive probability that it covers allof U.If x /∈ L,|A(x)| is exp small,and (for large n) there’s not T thatcover all U.(x ∈ L) ⇔ (∃T that cover all U)So,L = {x|∃(T ∈ {0, 1} p2 (n) )∀(b ∈ U)∃(j ≤ p(n)) : b ⊕ t j ∈ A(x)}which is precisely the form of languages in Σ 2 P.The last existential quantifier (∃(j ≤ p(n))...) affects onlypolynomially many possibilities,so it doesn’t “count” (can bytested in polynomial time by trying all t j ’s).