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Giorgi Japaridze Theory of Computability. Relativization. Section 9.2. 9.2.a. Giorgi Japaridze Theory of Computability. Oracle Turing machines. Definition 9.17 An oracle for a language A is device that is capable of reporting

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Giorgi Japaridze

Theory of Computability

Relativization

Section 9.2

9.2.a

Giorgi JaparidzeTheory of Computability

Oracle Turing machines

Definition 9.17 An oracle for a language A is device that is capable of reporting

whether any given string w is a member of A. An oracle Turing machine (OTM)

MA is a modified Turing machine that has the additional capability of querying an

oracle. Whenever MA writes a string on a special oracle tape, it is informed whether

that string is a member of A, in a single computation step.

Let PA be the class of languages decidable with a polynomial time OTM that

uses oracle A. DefineNPA similarly.

Example 9.18 NPPSAT(why?).

9.2.b

Giorgi JaparidzeTheory of Computability

Theorem 9.20(2)

Theorem 9.20(2)PTQBF= NPTQBF.

Proof. The containment PTQBF NPTQBF is trivial. And the containment

NPTQBF PTQBF follows from the following chain of containments:

NPTQBF 1 NPSPACE 2 PSPACE 3PTQBF

1because we can convert the nondeterministic polynomial time OTM to a

nondeterministic polynomial space TM that computes the answers to queries

regarding TQBF instead of using the oracle.

2follows from Savitch’s theorem.

3because TQBF is PSPACE-complete.

Note: In this theorem, instead of TQBF, we could have taken any other

PSPACE-complete problem.

9.2.c

Giorgi JaparidzeTheory of Computability

Theorem 9.20(1)

Theorem 9.20(1) An oracle Aexists whereby PA ≠ NPA.

Proof. For any oracle A, let LA={w | xA (|x|=|w|)}. Obviously LA is in NPA

(why?). To show that, on the other hand, LA is not in PA , we design A as follows.

Let M1,M2, … be a list of all polynomial time OTMs. For simplicity, we may

assume that each Miruns in time ni. Construction proceeds in stages, each stage

declaring certain finitely many strings to be in or out of A. Initially we have no

information about A. We begin with stage 1.

Stage i: We choose n greater than the length of any string whose membership (in A) status

has already been determined, also making sure that n is large enough to satisfy 2n>ni. Then we

run Mion input 1nand respond to its oracle queries as follows. If Mi queries a string y whose

status has already been determined, we respond consistently. If y’s status is undetermined, we

respond NO to the query and declare y to be out of A.

We continue simulation until Mi halts. If it accepts 1n, we declare all the remaining strings of

length n to be out of A. If Mirejects 1n, we find a string of length n that Mi has not queried and

declare that string to be in A. Such a string must exist because, within the ni steps available to

Mi, it could not have queried all of the 2n strings of length n.

It can be seen that Mi accepts 1n iff 1nLA. Hence Midoes not decide LA.

9.2.d

Giorgi JaparidzeTheory of Computability

Limits of the simulation method

We have proved so many theorems using the method of simulation (of one machine

by another). An import of Theorem 9.20 is that the same method is unlikely to be

successfully used for solving the P=NP? problem.

Indeed, if a machine M can simulate a machine N, then, for any oracle Q, MQcan

also simulate NQ, because whenever NQ queries the oracle, so can MQ, and

therefore the simulation can proceed as before.

Consequently, if we could prove by simulating that P and NP are the same, we could

conclude that they are the same relative to any oracle as well. But Theorem 9.20(1)

shows that they are not the same relative to the oracle A.

Similarly, if we could prove by simulating that P and NP are different, we could

conclude that they are different relative to any oracle as well. But Theorem 9.20(2)

shows that they are not the same relative to the oracle TQBF.