Lecture 9: Connecting PH, P/poly and BPP

Transcription

1 Comutational Comlexity Theory, Fall 010 Setember Lecture 9: Connecting PH, P/oly and BPP Lecturer: Kristoffer Arnsfelt Hansen Scribe: Martin Sergio Hedevang Faester Although we do not know how to searate P from NP, It is a standard hyosthesis of comutational comlexity theory that the two classes differ. Hyothesis 1. P NP Another common hyothesis is that NP and conp differ. We note P = NP NP = conp (since P is closed under comlement of languages). This could also be written as NP conp P NP. Thus the hyothesis that NP conp is a stronger hyothesis than P NP. In general we may hyothesize that the entire olynomial time hiearchy is a strict hierarchy. Hyothesis (generalized). For all k: 1. Σ k+1 Σ k. Σ k Π k NP P conp Figure 1: Polynomial-time hierarchy The second hyothesis at least as strong as the first: Proosition 3. Σ k+1 = Σ k Σ k = Π k (equivalently Σ k Π k Σ k+1 Σ k ) Proof. We will show that Π k Σ k. Consider the following Π k Σ k+1 by asm. = Σ k By comlements, we also have Σ k Π k. We next show that if the hyothesis Σ k Π k collases to the kth level. fails, then the entire olynomial time hierarchy 1

2 Theorem 4. If Σ k = Π k for some k then P H = Σ k Proof. By induction, we will rove that Σ k+i+1 = Σ k (= Π k ). The base case is done in the same way as the inductive ste. Let L Σ k+i+1. Then there exists some olynomial and R P such that Now, define L Π k+i by x L y 1 y y 3 Q y k+i+1 x, y 1, y,..., y k+i+1 R x, y L y y 3 Q y k+i+1 x, y 1, y,..., y k+i+1 R By induction L Π k = Σ k so there is a olynomial ˆ and ˆR P such that Then x, y 1 L ˆ z 1 ˆ z Qˆ z k x, y 1, z 1, z,..., z k ˆR x L This concludes the roof that L Σ k. merge +ˆ y 1,z 1 {}}{ y 1 ˆ z 1 ˆ z Qˆ z k x, y 1, z 1, z,..., z k ˆR Corollary 5. If P H has a comlete roblem then P H collases (i.e. P H = Σ k for some k) Proof. Assume L is comlete for P H. Then L Σ k for some k. Σ k since any L P H reduces to L, P H Σ k. is closed under reduction and Corollary 6. If P H = P SP ACE then P H collases Proof. P SP ACE has comlete roblems. Theorem 7 (Kar-Liton). If NP P/oly then P H = Σ = Π Proof. Assume NP P/P oly. We will show Π Σ. Let L Π then there is some olynomial and R P such that x L y z x, y, z R Define L NP by x, y L z x, y, z R and define L re NP by x, y, z re L re z such that x, y, z R and z re is a refix of z. By assumtion we have olynomial-sized circuits (L re P/oly). Using circuits for L re we can find a witness z, given x, y, in olynomial time, if it exists. Converting this to a multi-outut circuit C, we obtain that C( x, y ) oututs a witness z if it exists. Now, we can write x L C y x, y, C( x, y ) R. Thus L Σ. Theorem 8 (Meyer). If EXP P/oly then EXP = Σ = Π.

3 nk z i, t at time t, store content of cell i is the head scanning the cell? if so, internal state Figure : We can comute the content of the cell by looking at the three neighbors below. Proof. Let L EXP and let M be a 1-tae Turing machine deciding L in time nk. Consider the comutation tableau for M, and assume each osition (i, t) encodes as a string z i,t the content of cell i at time t, whether the tae head is scanning cell i at time t, and if so, stores the internal state of the machine. Now consider the following tableau language for M. L M = { x, i, t, z on inut x we have z i,t = z for M} By simulating M we have L M EXP P/oly. Using olynomial size circuits for L M we can construct a olynomial-size multi-outut circuit C such that C( x, i, t ) = z. Now we write x L C i, t : C( x, i, t ) follow from C( x, i t, t 1 ), C( x, i, t 1 ) and C( x, i+1, t t1 ) and C( x, 1, nk ) is acceting. Theorem 9. BP P P/oly Proof. The roof technique here is known as the robabilistic method, which is used for showing existence of an object by deriving that the robability of finding the object is strictly greater than 0 (which yields the existence). We shall assume our alhabet is Σ = {0, 1}. Let L BP P. By success amlification there is a robabilistic olynomial time Turing machine M that comutes L with error < 1. n Fix x {0, 1} n. Taking the robability over the for internal coin flis, Pr[M errs on x] < 1. n By a union bound we have the following Pr[M errs on some x {0, 1} n ] Σ x {0,1} n Pr[M errs on x] < n 1 n = 1 Thus, there are some internal coin-flis where M makes no error on any x {0, 1} n. Convert M into a circuit and hardwire the lucky coin-flis (mentioned right above). Theorem 10. BP P Σ Π 3

4 Proof. We will rove BP P Σ. Suose L BP P and let us recall the definition of this: L BP P olynomial, L P x L : Pr[ x, y L ] 3 x L : Pr[ x, y L ] 3 for y {0, 1} ( x ). Recall the details of success amlification: If we use m coin-flis to get error < 1 3 then by running O(k) indeendent trials, i.e., by using O(km) coin-flis, we can get error < 1. k Then there exists a Turing machine M using m = m(n) random bits to get error < 1 m. That is, we have L P such that x L : Pr[ x, y L ] 1 1 m x L : Pr[ x, y L ] < 1 m where y {0, 1} m. For a given x, let S x = {y {0, 1} m x, y L }: S x S x Figure 3: (*) [on the left] and (**) [on the right] x L : S x (1 1 m )m (*) x L : S x < 1 m m (**) The idea now is that by a few random shifts one can cover the entire robability sace in (*) but not in (**): Notation: S {0, 1} m, y {0, 1} m ; Shift S by y S := {y x x S x }. Claim 1. x L u 1, u,..., u m {0, 1} m r {0, 1} m : m [ x, r u i L ] Note: The above exression is a Σ statement as needed for the conclusion. 4 i

5 This may also be rewritten as u 1, u,..., u m {0, 1} m r {0, 1} m i : r u i S x and further (note r u i S x r u i S x ) u 1, u,..., u m {0, 1} m : u i S x = {0, 1} m Proof of claim. x L: Let u 1, u,..., u m be arbitrary. u i S x i m u i S x m m m < m Hence for all choices for u 1,..., u m we have that m u i S x {0, 1} m. x L: Pick u 1, u,..., u m at random. Fix r {0, 1} m and consider the rewrites Pr[r u i S x ] = Pr[ i : r u i S x ] = Pr[ i : u i r S x ] = m Pr[u i r S x ] ( 1 m )m < 1 m (1) Taking a union bound over all r, we get that Pr[r m u i S x for some x] < 1. Hence concluding by the robabilistic method there exists some choice for u 1,..., u m such that m u i S x = {0, 1} m. 5