RELATIVIZATION

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RELATIVIZATION

• CSE860
• Vaishali Athale

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Overview

• Introduction
• Idea behind Relativization
• Concept of Oracle
• Review of Diagonalization Proof
• Limits of Diagonalization method
• Proof idea
• Proof
• Implications of proof

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Introduction

• Revisiting question of NPP?
• Diagonalization proof used to show that Halting
  Problem is undecidable
• Can we use it to prove that NPP or
• NP ? P?
• Strong evidence against the possibility of
  solving the P Versus NP problem using
  Diagonalization technique.(BGS theorem, 1975)

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Idea behind Relativization

• Turing Machine provided with some information for
  free
• Concept of Oracle for a language
• Black box that answers membership of a string in
  the given language in one step
• Information affects the outcome of Turing Machine
• TM can solve some problems more easily

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Example of Oracle

• Consider an oracle for SAT
• Ability to solve SAT problem in a single step,
  for any size Boolean formula.
• With the help of an oracle for SAT, Turing
  Machine can solve any NP problem in polynomial
  time
• Regardless of whether NPP, every NP problem is
  polynomial time reducible to SAT
• Such a machine is computing relative to the SAT
  problem Relativization

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Oracle Turing Machine

• Consider an oracle for language A
• Oracle Turing Machine MA gets the result of
  question of whether the given string is in A in a
  single computation step.
• PA
• Class of languages decidable with a polynomial
  time TM MA that uses oracle A.
• NPA
• Class of languages decidable with a
  nondeterministic polynomial time TM MA that uses
  oracle A.

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Example

• TM can solve all NP problems with the help of
  oracle which can solve SAT in single step. Thus,
  NP ? PSAT
• Also, coNP ? PSAT, as deterministic complexity
  class is closed under complementation.

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Review of Diagonalization

• Using Diagonalization to show that Halting
  problem is undecidable
• ATM ltM, wgt M is a TM and M accepts w
• H(ltM,wgt) accept if M accepts w
• rejects if M accepts
• New TM D with H as subroutine
• D accepts when H rejects and vice versa.
• What happens when D uses ltD,wgt as input?
• Concept of Simulator(TM H) and Simulating
  Machine(TM D)

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Limits of Diagonalization

• Goal of BGS theorem(theorem 9.19) - to prove
  that Diagonalization technique is unlikely to
  resolve the P versus NP question.
• Key ideas
• Diagonalization is simulation of one TM by
  another.
• Theorem proved by TMs using the Diagonalization
  method would still hold if both the machines were
  given the same oracle.

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Key ideas(contd.)

• If P? NP is provable using Diagonalization
  method, then even if assistance of an oracle is
  given then they should be different.
• Does not work because BGS theorem proves that
  there exists an oracle B such that PB NPB
• If P NP is provable using Diagonalization
  method, then even if assistance of an oracle is
  given then they should be same.
• Does not work because BGS theorem proves that
  there exists an oracle A such PA ? NPA

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Proof

• Proof Idea
• Oracle B exists whereby PB NPB
• Oracle A exists whereby PA ? NPA
• Proof of existence of oracle B
• Let B be any PSPACE-complete problem, e.g, TQBF
• PB ?NPB as any language solvable by deterministic
  polynomial TM will be solvable by
  non-deterministic polynomial TM.
• To show that NPB ? PB,
• NPB ? NPSPACE ? PSPACE ? PB

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Proof of existence of oracle A

• Goals
• Design A such that certain language LA in NPA
  provably requires brute force search and hence LA
  cannot be in PA.
• LA ? NPA
• LA ? PA
• Construct A such that no polynomial time turing
  machine M1, M2..solves LA

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Goal 1 Identifying Language LA

• Let LA be the following language
• w ? x ? A x w
• i.e., a string is in LA iff there exists some
  string of the same length that is in A.
• Intuition
• There are 2n strings of length n
• For a large enough n (i.e. 2n gt ni) , a
  polynomial time deterministic Turing machine
  cannot check the status of all strings of length
  n.

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Goal 2 LA ? NPA

• Given a string w,
• Guess a string x and verify that
• x w
• String x is in A
• Can be achieved in one step by the oracle for A

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Goal 3 LA ? PA

• Construct A such that no polynomial time turing
  machine M1, M2..solves LA
• Wlog, complexity of Mi is ni
• For each stage i, for a subset of strings of
  increasing length, define membership of those
  strings in A by considering Mi

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Goal 3 LA ? PA (continued)

• For each stage i,
• Choose n such that
• n is larger than all the strings considered in
  stage i 1
• 2n gt n i

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Goal 3 LA ? PA (continued)

• Ensure that 1n ? LA iff Mi rejects 1n
• Run Mi on 1n
• Every time Mi asks the question about membership
  of a string in A
• If the membership for this string was defined
  before then answer consistently
• Otherwise, reject that string, I.e., define that
  it is not in A
• Note that
• Mi has not found even one string of length n
• Mi has not checked all strings of length n

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Goal 3 LA ? PA (continued)

• If Mi accepts 1n then
• Define that all strings of length n are not in A
• If Mi rejects 1n then
• Find one string that was not checked by Mi
• Define it to be in A
• Clearly, Mi cannot accept LA
• Continuing thus, we can show that LA is not
  accepted by any deterministic machine in
  polynomial time

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Key ideas(contd.)

• Any argument which relies on step by step
  simulation, would also apply in presence of an
  oracle.
• BGS theorem(theorem 9.19) shows that oracle can
  relativise both ways.
• Diagonalization method cannot help in solving
  question of P versus NP.
• Instead of simulating, analyzing computations
  might help. Circuit complexity may lead to such
  analysis.

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References

• The history and status of the P versus NP
  question - Annual ACM Symposium on Theory of
  Computing, Author - Michael Sipser
• C. Papadimitriou. Computational Complexity.
  Addison-Wesley, 1994.
• T. P. Baker, J. Gill, R. Solovay. Relativizatons
  of the P ? NP Question. SIAM Journal on
  Computing, 4(4) 431-442 (1975)