Presentation is loading. Please wait.

Presentation is loading. Please wait.

Introduction to PCP and Hardness of Approximation

Similar presentations


Presentation on theme: "Introduction to PCP and Hardness of Approximation"— Presentation transcript:

1 Introduction to PCP and Hardness of Approximation
Dana Moshkovitz Princeton University and The Institute for Advanced Study

2 Hardness of Approximation
This Talk A Groundbreaking Discovery! (From ) I offer a time travel, including: The 1991 discovery by [FGLSS91] of the PCP<-->hardness of approximation equivalence, the proof of the PCP Theorem by [AS92,ALMSS92], building upon substantial work from the 1980’s, the [BGS95] scheme for hardness of approximation, the Unique Games Conjecture by [Khot02], etc. What allowed researchers to finally prove hardness of approximation results? And how the substantial work that was done in the 1980’s on generalizations of the notion of a “proof” [interactive proofs, randomized verification, program checking etc] fit in? What turned out to be “the right” [?] problem to study for hardness of approximation? Is it the maximization version of 3SAT? Is it something else? Where are we stuck? What are the barriers? What are the possibilities? What turned out to be? The PCP Theorem and Hardness of Approximation

3 A Canonical Optimization Problem
MAX-3SAT: Given a 3CNF Á, what fraction of the clauses can be satisfied simultaneously? Á = (x7 : x12  x1) Æ … Æ (:x5  : x9  x28) x1 x2 x3 x4 x5 x6 x7 x8 xn-3 xn-2 xn-1 xn

4 Good Assignment Exists
Claim: There must exist an assignment that satisfies at least 7/8 fraction of clauses. Proof: Consider a random assignment. x1 x2 x3 xn

5 1. Find the Expectation Let Yi be the random variable indicating whether the i-th clause is satisfied. For any 1im,  F T

6 1. Find the Expectation The number of clauses satisfied is a random variable Y=Yi. By the linearity of the expectation: E[Y] = E[ Yi] =  E[Yi] = 7/8m

7 2. Conclude Existence Thus, there exists an assignment which satisfies at least the expected fraction (7/8) of clauses. 

8 ®-Approximation (Max Version)
For every input x, computed value C(x): ® ¢ OPT(x) · C(x) · OPT(x) OPT OPT(x) Corollary: There is an efficient ⅞-approximation algorithm for MAX-3SAT.

9 Better Approximation? Fact: An efficient tighter than ⅞-approximation algorithm is not known. Our Question: Can we prove that if P≠NP such algorithm does not exist? ?

10 Computation  Decision
Hardness of distinguishing far off instances  Hardness of approximation OPT A B OPT(x) gap

11 Gap Problems (Max Version)
Instance: … Problem: to distinguish between the following two cases: The maximal solution ≥ B The maximal solution < A YES NO

12 Gap NP-Hard  Approximation NP-hard
Claim: If the [A,B]-gap version of a problem is NP-hard, then that problem is NP-hard to approximate to within factor A/B.

13 Gap NP-Hard  Approximation NP-hard
Proof (for maximization): Suppose there is an approximation algorithm that, for every x, outputs C(x) ≤ OPT so that C(x) ≥ A/B¢OPT. Distinguisher(x): * If C(x) ≥ A, return ‘YES’ * Otherwise return ‘NO’ A B

14 Gap NP-Hard  Approximation NP-hard
(1) If OPT(x) ≥ B (the correct answer is ‘YES’), then necessarily, C(x) ≥ A/B¢OPT(x) ≥ A/B·B = A (we answer ‘YES’) (2) If OPT(x)<A (the correct answer is ‘NO’), then necessarily, C(x) ≤ OPT(x) < A (we answer ‘NO’).

15 New Focus: Gap Problems
Can we prove that gap-MAX-3SAT is NP-hard?

16 Connection to Probabilistic Checking of Proofs [FGLSS91,AS92,ALMSS92]
Claim: If [A,1]-gap-MAX-3SAT is NP-hard, then every NP language L has a probabilistically checkable proof (PCP): There is an efficient randomized verifier that queries 3 proof symbols: xL: There exists a proof that is always accepted. xL: For any proof, the probability to err and accept is ≤A. Note: Can get error probability ² by making O(log1/²) queries.

17 Probabilistic Checking of xL?
If yes, all of Á clauses are satisfied. If no, fraction ≤A of Á clauses can be satisfied. Prove xL! This assignment satisfies Á! Enough to check a random clause! x1 x2 x3 x4 x5 x6 x7 x8 xn-3 xn-2 xn-1 xn

18 Other Direction: PCP  Gap-MAX-3SAT NP-Hard
Note: Every predicate on O(1) Boolean variables can be written as a conjunction of O(1) 3-clauses on the same variables, as well as, perhaps, O(1) more variables. If the predicate is satisfied, then there exists an assignment for the additional variables, so that all 3-clauses are satisfied. If the predicate is not satisfied, then for any assignment to the additional variables, at least one 3-clause is not satisfied.

19 The PCP Theorem Theorem […,AS92,ALMSS92]: Every NP language L has a probabilistically checkable proof (PCP): There is an efficient randomized verifier that queries O(1) proof symbols: xL: There exists a proof that is always accepted. xL: For any proof, the probability to accept is ≤½. Remark: Elegant combinatorial proof by Dinur, 05.

20 Hardness of Approximation
Conclusion Probabilistic Checking of Proofs (PCP) Hardness of Approximation

21 Tight Inapproximability?
Corollary: NP-hard to approximate MAX-3SAT to within some constant factor. Question: Can we get tight ⅞-hardness?

22 The Bellare-Goldreich-Sudan Paradigm, 1995
Projection Games Theorem (aka Hardness of Label-Cover, or low error two-query projection PCP) Long-code based reduction Tight Hardness of Approximating 3SAT [Håstad97]

23 The Bellare-Goldreich-Sudan Paradigm, 1995
Projection Games Theorem (aka Hardness of Label-Cover, or low error two-query projection PCP) Long-code based reduction e.g., Set-Cover [Feige96] Tight Hardness of Approximation for Many Problems

24 Projection Games & Label-Cover
Bipartite graph G=(A,B,E) Two sets of labels §A, §B Projections ¼e:§A§B Players A & B label vertices Verifier picks random e=(a,b)2E Verifier checks ¼e(A(a)) = B(b) Value = maxA,BP(verifier accepts) A ? B ? ¼e Label-Cover: given projection game, compute value.

25 Equivalent Formulation of PCP Thm
Verifier randomness Theorem […,AS92,ALMSS92]: NP-hard to approximate Label-Cover within some constant. Proof: by reduction to Label-Cover (see picture). Proof entries Verifier queries… Projection = consistency check symbol Accepting verifier view

26 Projection Games Theorem: Low Error PCP Theorem
Claim: There is an efficient 1/k-approximation algorithm for projection games on k labels (i.e., |§A|,|§B|·k). Projection Games Theorem For every ²>0, there is k=k(²), such that it is NP-hard to decide for a given projection game on k labels whether its value = 1 or < ². Claim is exercise.

27 The Bellare-Goldreich-Sudan Paradigm
Projection Games Theorem (aka Hardness of Label-Cover, or low error two-query projection PCP) Tight Hardness of Approximation for Many Problems

28 How To Prove The Projection Games Theorem?
[AS92,ALMSS92] PCP Theorem Parallel repetition Theorem [Raz94] [M-Raz08] Construction ?? Projection Games Theorem Disadvantage of parallel repetition: the size of instance is raised to power O(log1/eps). Advantage of [MR08]: the size of the instance in the reduction is almost-linear. This is both much more efficient and allows for eps to be sub-constant (i.e., tend to 0 as n tends to infinity). Hardness of Approximation

29 The Khot Paradigm, 2002 Unique Games Conjecture
Long-code based reduction Constraint Satisfaction Problems [Raghavendra08] e.g., Max-Cut [KKMO05] e.g., Vertex-Cover [DS02,KR03] Tight Hardness of Approximation for More Problems

30 Thank You!


Download ppt "Introduction to PCP and Hardness of Approximation"

Similar presentations


Ads by Google