Download presentation

Presentation is loading. Please wait.

Published byAlden Burbidge Modified over 2 years ago

1
Constraint Satisfaction over a Non-Boolean Domain Approximation Algorithms and Unique Games Hardness Venkatesan Guruswami Prasad Raghavendra University of Washington Seattle, WA

2
Constraint Satisfaction Problem A Classic Example : Max-3-SAT Given a 3-SAT formula, Find an assignment to the variables that satisfies the maximum number of clauses. Equivalently the largest fraction of clauses

3
Constraint Satisfaction Problem General Definition : Domain : {0,1,.. q-1} Predicates :{P 1, P 2, P 3 … P r } P i : [q] k -> {0,1} Arity : Maximum number of variables per constraint (k) Example : Max-3-SAT Domain : {0,1} Predicates : P 1 (x,y,z) = x ѵ y ѵ z Arity = 3 GOAL : Find an assignment satisfying maximum fraction of constraints

4
Approximability Max-3-SAT : 7/8 [Karloff-Zwick],[Hastad] Most Max-CSP problems are NP-hard to solve exactly. Different Max-CSP problems are approximable to varying ratios. 3-XOR : ½ [Hastad] Max-Cut : 0.878 [Goemans-Williamson], [Khot-Kindler-Mossel- O’donnel] Max-2-SAT : 0.94 [Lempel-Livnat- Zwick],[Austrin]

5
Question : Which Max-CSP is the hardest to approximate? Refined Question : Among all Max-CSP problems over domain [q] ={0,..q-1}, and arity k, which is the hardest to approximate? Clearly, the problems become harder as domain size or the arity grows

6
PCP Motivation Probabilistically Checkable Proof (A string over alphabet {0,1,..q-1}) Verifier Random bits ACCEPT/REJECT Completeness(C) : If SAT formula is satisfiable, there is a proof that verifier accepts with probability C Soundness (S): For an unsatisfiable formula, no proof is accepted with probability more than S

7
Among all Max-CSP problems over domain [q] ={0,..q-1}, and arity k, which is the hardest to approximate? PCP Motivation What is the best possible gap between completeness (c) and soundness (s) for a PCP verifier that makes k queries over an alphabet [q] = {0,1,..q-1} ?

8
Boolean CSPs Hardness: For every k, there is a boolean CSP of arity k, which is NP- hard to approximate better than : Algorithm: Every boolean CSP of arity k, can be approximated to a factor : [Samorodnitsky-Trevisan 2000] simplified by [Hastad-Wigderson] [Engebresten-Holmerin] [Samorodnitsky-Trevisan 2006] Assuming Unique Games Conjecture Random Assignment [Trevisan] [Hast] [Charikar-Makarychev-Makarychev]

9
This Work : Non-Boolean CSPs UG Hardness: Assuming Unique Games Conjecture, For every k, and a prime number q, there is a CSP of arity k over the domain [q] ={0,1,2,..q-1}, which is NP-hard to approximate better than Algorithm: The algorithm of [Charikar-Makarychev- Makarychev] can be extended to non-boolean domains. Every CSP of arity k over the domain [q] ={0,1,2,..q-1} can be approximated to a factor

10
Related Work [Raghavendra 08] “Optimal approximation algorithms and hardness results for every CSP, assuming Unique Games Conjecture.” – “Every” so applies to the hardest CSPs too. – Does not give explicit example of hardest CSP, nor the explicit value of the approximation ratio. [Austrin-Mossel 08] “ Assuming Unique Games conjecture, For every prime power q, and k, it is NP-hard to approximate a certain CSP over [q] to a factor > ” – Independent work using entirely different techniques(invariance principle) – Show a more general result, that yields a criteria for Approximation Resistance of a predicate.

11
Techniques We extend the proof techniques of [Samorodnitsky-Trevisan 2006] to non- boolean domains. To this end, we – Define a subspace linearity test. – Show a technical lemma relating the success probability of a function F to the Gower’s norm of F (similar to the standard proof relating the number of multidimensional arithmetic progressions to the Gower’s norm) Along the way, we make some minor simplifications to [Samorodnitsky-Trevisan 2006]. – (Remove the need for common influences)

12
Proof Overview

13
Given a function F : [q] R [q], Make at most k queries to F Based on values of F, Output ACCEPT or REJECT. Distinguish between the following two cases : Dictatorship Testing Problem F is a dictator function F(x 1,… x R ) = x i F is far from every dictator function (No influential coordinate) Pr[ACCEPT ] = Completeness Pr[ACCEPT ] = Soundness Goal : Achieve maximum gap between Completeness and Soundness

14
UG Hardness Proofs UG Hardness Result: Assuming Unique Games Conjecture, it is NP- hard to approximate a CSP over [q] with arity k to ratio better than C/S Using [Khot-Kindler-Mossel-O’Donnell] reduction. Dictatorship Test Over functions F:[q] R -> [q] Completeness = C Soundness = S # of queries = k For the rest of the talk, we shall focus on Dictatorship Testing.

15
Testing Dictatorships by Testing Linearity [Samorodnitsky-Trevisan 2006] Fix {0,1} : field on 2 elements k = 2 d Given a function F : {0,1} R -> {0,1} Pick a random affine subspace A of dimension d. Test if F agrees with some affine linear function on the subspace A. Every dictator F(x 1, x 2,.. x R ) = x i is a linear function over vector space {0,1} R Random Assignment : There are 2 d+1 different affine linear functions on A. There are possible functions on A. So a random function satisfies the test with probability

16
Gower’s Norm For F : {0,1} R -> {0,1}, let f(x) = (-1) F(x). d th Gowers Norm U d (f) = E [ product of f over C ] Expectation over random d- dimensional subcubes C in {0,1} R xx+y 1 x+y 2 x+y 1 +y 2 xx+y 1 x+y 2 x+y 1 +y 2 x x+y 1 x+y 3 + y 2 x+y 1 +y 2 +y 3 x+y 3 x+y 1 +y 3 d-dimensional cube spanned by {x,y 1,y 2,.. y d } is Cubes

17
More Formally, Intuitively, the d th Gower’s norm measures the correlation of the function f with degree d-1 polynomials. Gower’s Norm

18
Testing Dictatorships by Testing Linearity [Samorodnitsky-Trevisan 2006] Lemma : If F : {0,1} R -> {0,1} passes the test with probability then f = (-1) F has high d th Gowers Norm. (k=2 d ) Lemma : If a balanced function f : {0,1} R -> {-1,1} has high d th Gowers Norm, then it has an influential coordinate (k=2 d ) Theorem : If a balanced function F : {0,1} R -> {0,1} passes the test with probability then it has an influential coordinate Using Noise sensitivity, There are only a FEW influential coordinates.

19
Extending to Larger domains Fix [q]: field on q elements(q is a prime). k = q d Given a function F : [q] R -> [q] Pick a random affine subspace A of dimension d. Test if F agrees with some affine linear function on the subspace A. Replace 2 by q in the [Samorodnitsky- Trevisan] dictatorship test.

20
The Difficulty Lemma : If F : {0,1} R -> {0,1} passes the test with probability then f = (-1) F has high d th Gowers Norm. (k=2 d ) Over {0,1} R, Subcube = Affine subspace. Testing linearity over a random affine subspace, can be easily related to expectation over a random cube. Over [q] R, Subcube ≠ Affine subspace. (2 R points) (q R points)

21
Success probability of a function F : {0,1} R -> {0,1}, is related to : let f(x) = (-1) F(x). E [ product of f over A ] Expectation over random d-dimensional affine subspace A in [q] R (Affine subspaces are like multidimensional arithmetic progressions) xx+y 1 x+2y 1 x+(q-2)y 1 x+(q-1)y 1 x+y 2 +y 1 x+y 2 +2y 1 x+y 2 + (q-2)y 1 x+y 2 + (q-1)y 1 x x+y 1 x+2y 1 x+ (q-2)y 1 x+ (q-1)y 1 x+2y 2 +y 1 x+2y 2 +2y 1 x+2y 2 + (q-2)y 1 x+2y 2 + (q-1)y 1 x+(q-2)y 2 +y 1 x+(q-2)y 2 +2y 1 x+(q-2)y 2 + (q-2)y 1 x+(q-2)y 2 + (q-1)y 1 x+(q-2)y 2 x+2y 2 x+y 2 x+(q-1)y 2 x+(q-1)y 2 +y 1 x+(q-1)y 2 +2y 1 x+(q-1)y 2 + (q-2)y 1 x+(q-1)y 2 + (q-1)y 1 Multidimensional Progressions E [ product of f over C ] Expectation over random dq-dimensional subcubes C in [q] R

22
Alternate Lemma Lemma : If F : [q] R -> [q] passes the test with probability then f = (-1) F has high dq th Gower’s Norm. (k=q d ) d-dimensional affine subspace test relates to the dq th Gower’s norm The proof is technical and involves repeated use of the Cauchy-Schwartz inequality. A special case of a more general result by [Green-Tao][Gowers-Wolf], where they define “Cauchy-Schwartz Complexity” of a set of linear forms.

23
Open Questions CSPs with Perfect Completeness: Which CSP is hardest to approximate, under the promise that the input instance is completely satisfiable? Approximation Resistance: Characterize CSPs for which the best approximation achievable is given by a random assignment.

24
Thank You

25
Unique Games A Special Case E2LIN mod p Given a set of linear equations of the form: X i – X j = c ij mod p Find a solution that satisfies the maximum number of equations. x-y = 11 (mod 17) x-z = 13 (mod 17) … …. z-w = 15(mod 17)

26
Unique Games Conjecture [Khot 02] An Equivalent Version [Khot-Kindler-Mossel-O’Donnell] For every ε> 0, the following problem is NP-hard for large enough prime p Given a E2LIN mod p system, distinguish between: There is an assignment satisfying 1-ε fraction of the equations. No assignment satisfies more than ε fraction of equations.

Similar presentations

OK

E3-LIN-2 is hard to approximate Hastad Speaker : Guy Kindler.

E3-LIN-2 is hard to approximate Hastad Speaker : Guy Kindler.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on describing words for grade 1 Ppt on glasgow coma scale Ppt on nature and human quotes Ppt on history of indian constitution Ppt on ssa in india Ppt on standing order template Cell surface display ppt online Chemistry ppt on matter Ppt on eye osmosis Ppt on depth first search java