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 M A gets the result of question of whether the given string is in A in a single computation step. P A –Class of languages decidable with a polynomial time TM M A that uses oracle A. NP A –Class of languages decidable with a nondeterministic polynomial time TM M A 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  P SAT Also, coNP  P SAT, as deterministic complexity class is closed under complementation.

8. Review of “Diagonalization” Using Diagonalization to show that “Halting problem is undecidable” A TM = { | M is a TM and M accepts w} H( ) = 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 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 P B =NP B 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 P A  NP A

11. Proof Proof Idea –Oracle B exists whereby P B =NP B –Oracle A exists whereby P A  NP A Proof of existence of oracle B –Let B be any PSPACE-complete problem, e.g, TQBF –P B  NP B as any language solvable by deterministic polynomial TM will be solvable by non-deterministic polynomial TM. –To show that NP B  P B, NP B  NPSPACE  PSPACE  P B

12. Proof of existence of oracle A Goals 1.Design A such that certain language L A in NP A provably requires brute force search and hence L A cannot be in P A. 2.L A  NP A 3.L A  P A Construct A such that no polynomial time turing machine M 1, M 2 ……..solves L A

13. Goal 1: Identifying Language L A Let L A be the following language –{ w |  x  A [|x| = |w| } –i.e., a string is in L A iff there exists some string of the same length that is in A. –Intuition: There are 2 n strings of length n For a large enough n (i.e. 2 n > n i ), a polynomial time deterministic Turing machine cannot check the status of all strings of length n.

14. Goal 2: L A  NP A 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: L A  P A Construct A such that no polynomial time turing machine M 1, M 2 ……..solves L A –Wlog, complexity of M i is n i –For each stage i, for a subset of strings of increasing length, define membership of those strings in A by considering M i

16. Goal 3: L A  P A (continued) For each stage i, –Choose n such that n is larger than all the strings considered in stage i – 1 2 n > n i

17. Goal 3: L A  P A (continued) Ensure that 1 n  L A iff M i rejects 1 n Run M i on 1 n –Every time M i 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 M i has not found even one string of length n M i has not checked all strings of length n

18. Goal 3: L A  P A (continued) If M i accepts 1 n then –Define that all strings of length n are not in A If M i rejects 1 n then –Find one string that was not checked by M i –Define it to be in A Clearly, M i cannot accept L A Continuing thus, we can show that L A 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)