Relativization

1
Relativization
Giorgi Japaridze Theory of
Computability
Section 9.2
2
Oracle Turing machines
9.2.a
Giorgi Japaridze Theory of Computability
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?).

3
Theorem 9.20(2)
9.2.b
Giorgi Japaridze Theory of Computability
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 Savitchs theorem.
?3 because TQBF is PSPACE-complete.
Note In this theorem, instead of TQBF, we could
have taken any other PSPACE-complete problem.

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Theorem 9.20(1)
9.2.c
Giorgi Japaridze Theory of Computability
Theorem 9.20(1) An oracle A exists whereby PA
? NPA.
Proof. For any oracle A, let LAw ?x?A
(xw). 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 2ngtni. 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 ys 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.

5
Limits of the simulation method
9.2.d
Giorgi Japaridze Theory of Computability
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
PNP? 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.