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
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. Define NPA similarly.
Example 9.18
NPPSAT (why?).
9.2.b
Giorgi Japaridze
Theory 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 3 PTQBF
1 because 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.
2 follows from Savitch’s theorem.
3 because TQBF is PSPACE-complete.
Note: In this theorem, instead of TQBF, we could have taken any other
PSPACE-complete problem.
9.2.c
Giorgi Japaridze
Theory of Computability
Theorem 9.20(1)
Theorem 9.20(1) An oracle A exists 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 Mi runs 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 Mi on input 1n and 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 Mi rejects 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 Mi does not decide LA.
9.2.d
Giorgi Japaridze
Theory 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, MQ can
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.