Axomatic approach to barriers in complexity

Axiomatic Approach to Barriers in ComplexityRussell Impagliazzo (IAS & UCSD)Valentine Kabanets (IAS & SFU)Antonina Kolokolova (MUN)

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P ? NPBPP =? PIP=PSPACEP EXPAlgebrization ridgeRelativization gorge2

Lets compute everything!Hilbert

You cant.Gdel, Turing

Lets prove we cant in efficient realm, too!Cook, Karp,... The way you are trying, you cant.BGS, RR,AW...

3Proving unprovability1900: Hilberts programLets axiomatize all of mathematics!1930: Godels incompleteness theoremThere are true mathematical statements that cannot be proven in a (reasonable) theory.1937: Turings uncomputabilityUses Cantors diagonalization as a major tool.If both L and its complement are semi-decidable (verifiable), then L is decidable.

____Halt

Halt

4Diagonalization Idea: list all possible Turing machines For each of them, look at the inputs it accepts (its language). Show that there is a TM that is not on the list by constructing one different from any TM in the list on some input. Therefore, there is a language D on which any Turing machine will make an error. 010001.....M1YNNY...M2NNYY...M3NYNN...M4YYYY.........DNYYN...5Scaling down to complexity1964-66 Cobham, Edmonds, RabinPolynomial time = efficient. 1965: Hartmanis/StearnsThere are problems solvable in exponential time, but not polynomial time (use diagonalization). Same kind of hierarchy theorems can be proven for space and non-deterministic time and space.

Can this be extended to solve P vs. NP? 6P vs. NP1956: Yablonky (in Russia)Thought he solved the perebor (brute-force search) problem.Showed that for some class of algorithms it is impossible to eliminate brute-force search. 1956: Godels lost letterDo tautologies have proofs of size not too much larger than the size of formulas?1971: Cook,Levin, then KarpConcept of NP-completeness, showing many problems are NP-complete. 7The Plot Summary c. 1971: Quest for P NP

1975: Relativization

1990s: IP=PSPACE, NEXP=MIP, NP=PCP[log n, 1], 2008: Algebrization

PA = NPA PB NPB

PA = NPA PB NPB ~~8Natural Proof Monster Razborov 95, bounded arithmetic framework

NPP/poly[RR97]9Oracles1939: Turing, already, considered allowing his machines access to a source of intuitionSuch a machine could ask queries to some source of knowledge (call it an oracle) by writing a query on a tape and immediately getting an answer

But the diagonalization argument for undecidability works for such machines!

010001.....M1OYNNY...M2ONNYY...M3ONYNN...M4OYYYY.........DONYYN...

10Relativization 1975: Baker/Gill/SolovayThere is an oracle A such that some polynomial-time machine with access to it is as powerful as any NP machine with access to that oracle. Take A to be any PSPACE-complete language. There is another oracle B such that NP with access to it is provably stronger than P with access to it.Construct B using diagonalization, or take a random oracle. Diagonalization alone cannot be used to resolve P vs NP.Same is true for many other complexity questions.

PA = NPA PB NPB 11Logically speaking What does it mean to have contradictory relativizations ?

Intuitively, P vs. NP should be independentof the Relativizing Complexity Theory.Oracle worlds models of a theory

PA = NPA PB NPB [BGS75]12Oracle worlds and models

13AIV92 approachTake a theory of arithmetic (unbounded) and add function symbols.Give limited amount of information about properties of these functions.

What can be proven about these functions with the full power of arithmetic, but knowing only a few facts about them?

14Relativizing Complexity Theory (RCT) [Arora, Impagliazzo, Vazirani 92] Axiomatization of PolyTime Computation s.t.{ PA | any oracle A } = { stand. models of RCT }.

Interpretation: Complexity statements provable in RCT are relativizing statements. Consequence: Non-relativizing statements (such as P NP ) are independent of RCT.

15Cobhams characterization of P[Cob64]: P is the minimal class of functions thatContain basic functions+, |x|,x*y, bit(x,i),projection, 2|x|*|x|Closed under function compositionIf f and g are in P, then so is f o g.And under limited recursion on notationf(x,0)=g(x), f(x,k)=h(x,f(x,k/2)), |f(x,k)|