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# Gillat Kol joint work with Irit Dinur Covering CSPs.

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Gillat Kol joint work with Irit Dinur Covering CSPs

Constraint Satisfaction Problem CSP = Constraint Satisfaction Problem – Variables: x 1,x 2,…,x n in {-1,1}. – Constraints: ((x 1 =1) v (-x 2 =1) v (x 7 =-1)), (x 2  x 5 = 1), … Goal: – Ideally: Find assignment that satisfies all constraints. – NP-hard, so we approximate.

Optimization Notions Max-CSP: – Restriction: Use only a single asg. – Optimization Goal: Maximize # satisfied constraints. Min-Cover-CSP (this paper): – Restriction: Must satisfy all constraints. – Optimization Goal: Minimize # asgs.

Example: The Dinner Party Problem You want everyone to have at least something to eat. But, would like to cook as few dishes as possible. You invite friends over for dinner. Each has diff dietary constraints:

Covering Number The Covering Number of a CSP instance C, denoted cover(C), is the smallest number of asgs to the variables s.t. every constraint is “covered” (satisfied by at least one asg).

Covering “Extends” Coloring: – [GHS’02]: (Hyper)graph G naturally induces a NAE-CSP instance C G with chromatic(G)  2 cover(C G ). Covering & Coloring x2x2 x2x2 x3x3 x3x3 x4x4 x4x4 x1x1 x1x1 x5x5 x5x5 G x 1 ≠ x 2 x 2 ≠ x 4 x 2 ≠ x 5 x 3 ≠ x 4 x 4 ≠ x 5 CGCG 2 asgs over {+,-} coloring using 4 colors {++,--,+-,-+}

Covering “Extends” Coloring: – [GHS’02]: (Hyper)graph G naturally induces a NAE-CSP instance C G with chromatic(G)  2 cover(C G ). – Covering allows us to “increase the number of colors” in any predicate . Covering & Coloring

Our Results

Covering  Predicate  :{+1,-1} t  {+1,-1} (-1 = true, 1 = false).  -CSP = constraints of the form  (  x 1,…,  x t ). The (c,s)-covering  Problem: Given a  -CSP instance C, decide between (c < s  N): – cover(C) ≤ c. – cover(C)  s. Our Goal: Study  ’s covering behavior. – covering  is hard if  const c s.t.  const s>c: (c,s)-covering  is hard.

Observation: If  is odd (  (x) = -  (-x)), then cover(  )  2. – Proof:  asg A, {A, -A} covers. – covering 3LIN is easy. Observation 2: If  Odd*, then cover(  )  2, where Odd* = {  |  “contains” an odd predicate} = {  |  x:  (x)=true or  (-x)=true}. – covering 3SAT is easy. Easy Predicates

The Characterization of Covering-Hard Predicates? Our Covering Dichotomy Conjecture: covering  is hard iff  Odd*.

Def: 4LIN(x 1,x 2,x 3,x 4 ) = x 1  x 2  x 3  x 4. Result 1: (2,s)-covering 4LIN is NP-hard for any const s>2. – The “first” interesting new predicate. – 4LIN is easy in the max-CSP sense. Challenge: Getting perfect completeness with 2 asgs. We “doable” the label cover, and apply correlated noise. Result 1 NP-Hardness for covering 4LIN

Result 2 Partial Proof for the Dichotomy Conjecture Result 2 [a la Austrin-Mossel 2009]: Under a covering unique games conjecture: If  Odd*, and supports a pairwise independent distribution, then covering  is hard. Challenge: Analyzing soundness for a general predicate. – Observation: Among predicates  Odd*, the predicate  =NAE has the “largest” support.

Result 3 Connecting covering  and NAE Approximate Coloring Problem: Given an O(1)-colorable (hyper)graph, what is the smallest number of colors needed to color it in polynomial time? – lower bound: polylog(n) (hypergraphs) [Khot’02] – upper bound: n 

Result 3 Connecting covering  and NAE Approximate Coloring Problem: Given an O(1)-colorable (hyper)graph, what is the smallest number of colors needed to color it in polynomial time? Result 3 [a la Feige’s R3SAT 2002]: – Hypothesis:   s.t. given a  -CSP instance C, it is hard to tell if C is a random instance, or if cover(C) = 2. – If the hypothesis holds with sufficiently good parameters (density of C), we get polynomial hardness for hypergraph coloring.

Covering Dictatorship Test for 4LIN (part of the proof of Result 1)

Dictatorship Test Hardness results for  are usually obtained through a  -Dictatorship Test. f:{+1,-1} R  {+1,-1} is a dictator if  i s.t. f(x) = x i. A 4LIN-Dict Test for f:{+1,-1} R  {+1,-1} is specified by a distribution over 4-tuples x,y,z,w  {-1,1} R. It draws x,y,z,w and accepts iff f(x)f(y)f(z)f(w) = -1. – Completeness: f is a dictator  Pr[test accepts]  1- . – Soundness: f is “regular”  Pr[test accepts]  ½+ . low influences, “far” from dictator imperfect completeness

Covering Dictatorship Test A 4LIN-Covering Dict Test for f:{+1,-1} R  {+1,-1} is specified by a distribution over x,y,z,w (as before). Let C be the 4LIN-CSP instance induced by the distribution (every 4-tuple x,y,z,w induces a constraint, f is an asg). – Covering Completeness of the test  c:  c dictators that cover C. – Covering Soundness of the test  s: No “regular set” of s functions covers C. every product of functions from the set has low influences.

Covering Dictatorship Test A 4LIN-Covering Dict Test for f:{+1,-1} R  {+1,-1} is specified by a distribution over x,y,z,w (as before). Let C be the 4LIN-CSP instance induced by the distribution (every 4-tuple x,y,z,w induces a constraint, f is an asg). – Covering Completeness of the test  c:  c dictators that cover C. – Covering Soundness of the test  s: No “regular set” of s functions covers C. We want such a test with covering completeness 2 (and super-const covering soundness).

Hastad’s Dictatorship Test Hastad’s Dict Test uses the distribution: – Choose x,y,z  {-1,1} R, independently uniformly at rand. – Choose a noise vector r  {-1,1} R in which each coordinate is independently -1 (noise) w.p. ε. – Set w = -xyzr. Covering Completeness > const: Let f(x) = x 1. – f(x)f(y)f(z)f(w) = x 1 y 1 z 1 w 1 = -r 1. – Thus, f doesn’t cover constraints with noise on r 1 (r 1 =-1). – No const num of dictators cover the test’s constraints!

Getting Perfect Completeness New Dict Test: Same distribution with tweak on noise. – x,y,z random, w = -xyzr. – Partition the noise vector r into pairs (r 1,r 2 ), (r 3,r 4 ),… For each pair, w.p. 2ε have noise one exactly one element of the pair. There is never noise on both! Covering Completeness = 2: Let f(x) = x 1 and g(x) = x 2. – There is never noise on both r 1 and r 2 (noise = -1). – Thus, at least one of the following holds: f(x)f(y)f(z)f(w) = x 1 y 1 z 1 w 1 = -r 1 = -1 g(x)g(y)g(z)g(w) = x 2 y 2 z 2 w 2 = -r 2 = -1 – f and g cover the test’s constraints!

Many Open Problems Covering is a natural notion, pretty much any max-CSP question can be considered in the context of covering. Prove the Covering Dichotomy Conjecture in full. Quantitative results: – We get 4LIN covering soundness  Ω(logloglog n). – Can we get Ω(log n) for some  ? Connecting the covering-UGC to known conjectures – Incomparable to UGC, but implies the UGC with completeness 1/c (instead of 1-ε). Devise ‘direct’ reductions between covering problems.

Thank You!

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