covering csps
DESCRIPTION
Covering CSPs. Gillat Kol joint work with Irit Dinur. Constraint Satisfaction Problem. CSP = Constraint S atisfaction P roblem 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 : - PowerPoint PPT PresentationTRANSCRIPT
Gillat Koljoint work with Irit Dinur
Covering CSPs
Constraint Satisfaction Problem• CSP = Constraint Satisfaction Problem
– Variables: x1,x2,…,xn in {-1,1}.– Constraints: ((x1=1) v (-x2=1) v (x7=-1)), (x2x5 = 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 CG with chromatic(G) 2cover(CG).
Covering & Coloring
x2
x3
x4
x1
x5
G x1 ≠ x2
x2 ≠ x4
x2 ≠ x5
x3 ≠ x4
x4 ≠ x5
CG
2 asgs over }-,+{
coloring using 4 colors }+-,-+,--,++{
• Covering “Extends” Coloring: – [GHS’02]: (Hyper)graph G naturally induces a NAE-CSP
instance CG with chromatic(G) 2cover(CG). – 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 (x1,…,xt).
• 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, whereOdd* = { | “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(x1,x2,x3,x4) = x1x2x3x4.
• 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 1NP-Hardness for covering 4LIN
Result 2Partial 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 3Connecting 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 3Connecting 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) = xi.
• 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) = x1.– f(x)f(y)f(z)f(w) = x1 y1 z1 w1 = -r1.– Thus, f doesn’t cover constraints with noise on r1 (r1=-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 (r1,r2), (r3,r4),… 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) = x1 and g(x) = x2.
– There is never noise on both r1 and r2 (noise = -1).– Thus, at least one of the following holds:
• f(x)f(y)f(z)f(w) = x1 y1 z1 w1 = -r1 = -1• g(x)g(y)g(z)g(w) = x2 y2 z2 w2 = -r2 = -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!