RELATIVIZATION

1. RELATIVIZATION CSE860 Vaishali Athale

2. Overview • Introduction • Idea behind “Relativization” • Concept of “Oracle” • Review of Diagonalization Proof • Limits of Diagonalization method • Proof idea • Proof • Implications of proof

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

4. 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

5. 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 NP=P, every NP problem is polynomial time reducible to SAT Such a machine is computing relative to the SAT problem – “Relativization”

6. 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.

7. 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.

8. Review of “Diagonalization” • Using Diagonalization to show that “Halting problem is undecidable” • ATM = {<M, w> | M is a TM and M accepts w} • H(<M,w>) = 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 <D,w> as input? • Concept of “Simulator”(TM H) and “Simulating Machine(TM D)”

9. 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.

10. 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

11. 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

12. 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

13. 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 > ni) , a polynomial time deterministic Turing machine cannot check the status of all strings of length n.

14. 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

15. 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

16. 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 > n i

17. 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

18. 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

19. 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.

20. 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)