4 Oracles and the polynomial hierarchy

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Complexity Theory WS 2009/10

Prof. Dr. Erich Grädel

Mathematische Grundlagen der Informatik RWTH Aachen

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Dieses Werk ist lizensiert uter:

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http://www.logic.rwth-aachen.de

4 Oracles and the polynomial hierarchy

4.1 Oracle Turing machines

Definition 4.1. A deterministic (respectively nondeterministic)oracle Turing machineis a Turing machine with a designated oracle tape and three special states ? (query), Y (yes) and N (no).

A configurationCof an Oracle Turing machine withkworking tapes and a distinguished oracle tape is a tuple

C= (q,p₀,· · ·,p_k,w₀,· · ·,w_k), where

•qis the state of the Turing machine,

•p₀, . . . ,p_k are the head positions (p₀ is the head position of the oracle tape), and

•w0. . . ,wk are the head inscriptions (w0 is the inscription of the oracle tape).

The computation (respectively the computation tree) of an oracle Turing machine depends on a previously defined oracle set A ⊆Σ^∗ (where Σ is the alphabet of M). The successor configurations of a configurationC are defined as usual for q ̸= ? while the successor configurationC^′forq=? is defined as :

C^′=





(Y, 0,p1, . . . ,p_k,ε,w1, . . .w_k) ifw0∈A (N, 0,p₁, . . . ,p_k,ε,w₁, . . . ,w_k) ifw₀̸∈A

whereεis the empty word. The oracle therefore determines whether or not w₀ (the inscription of the oracle tape) is in A. The machine consequently enters the corresponding state (Y or N) and the inscription of the oracle tape is erased.

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4.1 Oracle Turing machines

Definition 4.2. Let Mbe an Oracle Turing machine and A ⊆Σ^∗ be some oracle set. Then the accepted language is

L(M^A):={x:Maccepts the inputxwith oracleA}.

Based on the oracle set A, we define the following complexity classes:

(i) P^A:={L: there is a deterministic Oracle TMM

that decidesLusing oracleAin polynomial time}. (ii) NP^A:={L: there is a nondeterministic Oracle TMM

that decidesLusing oracleAin polynomial time}. LetCbe some class of languages, e.g., a complexity class. Then P^C= ^[

A∈C

P^A and NP^C= ^[

A∈C

NP^A.

Example4.3.

(a) LetB ∈ NP. Then B ∈ P^sat. Since satis NP-hard, there is a polynomially computable function f withx∈B⇐⇒ f(x)∈sat.

The following oracle algorithm then decidesB:

Input:x Compute f(x)

Query the oracle whether f(x)∈sat ifYthenaccept

ifNthenreject

(b) It is likely that NP(P^satas everyB∈coNP is in P^sat. One can use the preceding algorithm and interchange the behaviour for the answers Yes and No.

(c) LetB:={(G,k): Ga graph,ω(G) =k}whereω(G)is the max- imal number of nodes of cliques inG. Reminder: The problem clique:={(G,k) : k≤ ω(G)}is NP-complete. It is straightforward to see thatB∈P^clique:

4 Oracles and the polynomial hierarchy

Input: G,k

Query the oracle whether(G,k)∈clique ifNthenrejectelse

Query the oracle whether(G,k+1)∈clique ifNthenaccept

ifYthenreject endif

4.2 The polynomial hierarchy

Definition 4.4. We define the complexity classesΣ_k^p,Π^p_k, and∆_k^pfor allk∈N:

•Σ^p₀:=Π^p₀:=∆₀^p:=P

•Σ^p_k+1:=NP^Σ^p^k

•Π_k^p:=coΣ_k^p={A: A∈Σ_k^p}

•∆_k+1^p :=P^Σ^k^p

Theorem 4.5. The classesΣ_k^p,Π_k^p, and∆_k^phave the following elemen- tary properties:

(i)∆₁^p=P.

(ii)Σ^p₁=NP,Π₁^p=coNP.

(iii)Σ^p_k+1=NP^Π^p^k =NP^∆^p^k+1. (iv) P^∆^p^k =∆_k^p.

Proof. (i)∆₁^p=P^P=P.

(ii)Σ^p₁=NP^P=NP,Π^p₁=coΣ^p₁=coNP.

(iii) LetB ∈ Σ_k+1^p , B = L(M^A) and A ∈ Σ_k^p. Further, letM^′ be the machine obtained from Mby interchanging the states Y and N.

Obviously, B = L(M^′^A) and therefore B ∈ NP^Π^k^p. In addition, NP^∆^k+1^p =NP^P^Σ^p^k =NP^Σ^p^k =Σ^p_k+1holds.

(iv) k=0: P^∆^p⁰ =P^P=P=∆^p₀.

k>0: P^∆^p^k =P^P^Σ^k−1^p =P^Σ^k−1^p =∆_k^p. q.e.d.

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4.2 The polynomial hierarchy

Theorem 4.6. For allk,Σ_k^p∪Π^p_k⊆∆^p_k+1⊆Σ_k+1^p ∩Π^p_k+1.

Proof. For k = 0, the theorem states P ⊆ P ⊆ NP∩coNP. This is obviously true.

Fork>0:

•Σ_k^p⊆P^Σ^k^pand therefore alsoΠ^p_k⊆P^Σ^p^k because P^Σ^p^k is closed under complement. Hence,Σ_k^p∪Π_k^p⊆∆_k+1^p .

•∆_k+1^p =P^Σ^k^p=coP^Σ^k^p⊆coNP^Σ^p^k =Π_k+1^p .

∆_k+1^p = P^Σ^k^p ⊆ NP^Σⁿ^p = Σ_k+1^p . Therefore, ∆^p_k+1 ⊆ Σ_k+1^p ∩Π_k+1^p . q.e.d.

Theorem 4.7. If there is aksuch thatΣ_k+1^p =Σ_k^p, thenΣ^p_k+i=Π^p_k+i= Σ^p_kfor alli>0.

Proof. For i = 1, Σ^p_k+i = Σ_k+1^p = Σ_k^p by assumption. By induction hypothesis, assumeΣ_k+i^p =Σ^p_k. Then,

Σ_k+i+1^p =NP^Σ^k+i^p =NP^Σ^p^k =Σ_k+1^p =Σ^p_k, and therefore also

Π^p_k+i⊆Σ_k+i+1^p =Σ^p_k for alli.

In particular, Π_k^p ⊆ Σ^p_k holds. It remains to show thatΣ^p_k ⊆ Π_k^p. If B∈Σ^p_k, thenB∈Π^p_k⊆Σ^p_kand, hence,B∈Π^p_k. q.e.d.

Corollary 4.8. If there is ak>0 withΣ^p_k̸=P, then P̸=NP.

Definition 4.9. PH :=^S_k∈NΣ^p_kis called thepolynomial hierarchy.

In caseΣ_k+i^p =Σ_k^p, we say that the polynomial hierarchy collapses at levelk.

Theorem 4.10. PH⊆Pspace.

Proof. By induction overkwe show thatΣ_k^p⊆Pspacefor allk∈N:

Σ₀^p=P⊆Pspace

Σ_k+1^p =NP^Σ^k^p⊆NP^Pspace⊆Pspace^Pspace=Pspace.

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4 Oracles and the polynomial hierarchy Here, Pspace^Pspaceis the class of languages that can be decided by a deterministic polynomially-space bounded Oracle Turing machine with some oracle in Pspace. When we speak about space complexity, we also count the space used on the oracle tape. q.e.d.

If PH=Pspace, the polynomial hierarchy collapses:

Theorem 4.11. If PH=Pspace, there is somekwith PH=Σ^p_k. Proof. If PH=Pspace, thenqbf∈ PH holds. Consequently, there is someksuch thatqbf∈Σ_k^p. For eachA∈Pspace, we haveA≤^p_mqbf, i.e.,A∈Σ_k^p. Thus, we obtain PH=Σ^p_k. q.e.d.

It is assumed thatΣ^p_k(Σ_k+1^p for allk, the polynomial hierarchy is strict and therefore, PH(Pspace.

4.2.1 Additions

There are two natural complete problems forΣ_k^pandΠ_k^p:

Σ_k-QBF={ψ= (∃X₁)(∀X₂). . .(Q_kX_k)ϕ : ϕquantifier free, ψtrue} Here,Q_kis the universal quantifier ifkis even and the existential quantifier otherwise.Πk-QBF is defined analogously but the formulae begin with universal quantifiers. The problemΣ_k-QBF isΣ_k^p-complete and, analogously,Π_k-QBF is Π_k^p-complete. This generalises the NP- completeness ofsat.

Recall the definition of NP. A problem A ∈ NP if, and only if, there is someB∈ P and some polynomial p(n) such thatA ={x :

∃^py(x#y∈B)}where∃^pyis an abbreviation for∃y : |y| ≤p(|x|). We can generalise this definition to obtain a characterisation ofΣ_k^pandΠ_k^p as follows:

•A∈Σ^p_k if, and only if, there is someB∈P and some polynomial p(n)such that

A={x : (∃^py1)(∀^py2)(∃^py3). . .(Q_k^py_k)x#y1#y2# . . . #y_k∈B}.

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4.3 Relativisations

Here,Q_kis the existential quantifier ifkis odd and the universal quantifier otherwise.

•Π^p_k can be characterised analogously, using formulae that begin with universal quantifiers.

4.3 Relativisations

To approach the P=NP question, it is interesting to see whether there are oraclesA,Bsuch that

• P^A=NP^A

• P^B ̸=NP^B.

The first question is easy to answer since P^qbf=NP^qbf=Pspace.

The answer to the second question is not as straightforward.

Theorem 4.12. There is an oracleBsuch that P^B ̸=NP^B.

Proof. Just as Turing machines, polynomially time-bounded oracle Tur- ing machines can also be enumerated recursively (exercise). We choose one such recursive enumeration{Mi :i ∈N}of deterministic polynomial oracle Turing machines such that the following holds for all oraclesA:

(1) P^A={L(M_i^A):i∈N}.

(2) There is a sequence{pi(n):i∈N}of polynomials with:

(i)M_iisp_i-time bounded, (ii)pi(n)≤pi+1(n)for alli,n.

Any given sequence{q_i(n):i∈N}of time bounds for{M_i:i∈N} can be modified to such a sequencepi+1(n)by setting:

• p0(n):=q0(n),

• pi+1(n):=max(pi(n),qi+1(n)).

For everyC⊆ {0, 1}^∗, letS(C) ={0ⁿ: there is anx∈Cwith|x|=n}. We obviously have:

Lemma 4.13. S(B)∈NP^Bfor allB.

4 Oracles and the polynomial hierarchy The goal now is to find some Bsuch thatS(B) ̸∈P^B. This Bis constructed as follows: At the beginning, initialiseB0:=∅andk0:=0.

Forn>0, constructBn,knas follows:

• Setkn so it is the smallest integer with 2^kⁿ > pn(kn) andkn >

p_n−1(k_n−1).

• If 0^kⁿ ∈ L(M^Bnⁿ⁻¹), then setBn := Bn−1. Otherwise, letw(n)be the lexicographically first word in{0, 1}^k(n)for which the oracle is not queried during the computation ofM_n^Bⁿ⁻¹on input 0^kⁿ. Such a word exists sincepn(kn)<2^kⁿ. SetBn=Bn−1∪ {w(n)}.

SetB:=^S_n∈NBn.

Lemma 4.14. 0^kⁿ∈L(M_n^B) ⇐⇒ 0^kⁿ ∈L(Mn^Bⁿ⁻¹)for alln.

The oracle is never queried forw∈B\Bn−1during the computation ofM^B_non 0^kⁿ:

• forw=w(n)by construction and

• forw=w(m)withm>nbecause|w(m)|=k(m)>pn(kn). Lemma 4.15. S(B)̸∈P^B.

Otherwise, there would be some n ∈ N with S(B) = L(M_n^B). However, this cannot be the case since, by definition, 0^kⁿ ∈S(B)if, and only if, there is somewsuch that|w|=knandw∈B. By construction, this is the case if, and only if, 0^kⁿ̸∈L(M^B_nⁿ⁻¹). This follows from the fact that a word of lengthknis added toBif, and only if, 0^kⁿ ̸∈L(Mn^Bⁿ⁻¹). By Lemma 4.14, 0^kⁿ ̸∈ L(Mn^Bⁿ⁻¹) is equivalent to 0^kⁿ ̸∈ L(M^B_n), and

therefore,S(B)̸=L(M^B_n). q.e.d.

The problem whether C₁ = C₂ remains open for many pairs of complexity classesC₁andC₂. In most cases, there are oraclesAandB such that:

C₁^A=C₂^A and C₁^B̸=C₂^B.

It has further been shown that for almost all oracles D: C₁^D ̸= C₂^D. Contradictory relativisations of this kind show that, with respect to the

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4.3 Relativisations

problemC₁̸=C₂, proof techniques that ‘relativise’ (i.e., techniques that are independent of oracles) fail.

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4 Oracles and the polynomial hierarchy

Complexity Theory WS 2009/10

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4 Oracles and the polynomial hierarchy