going after the -sat threshold
DESCRIPTION
Going after the -SAT Threshold. Konstantinos Panagiotou ( with Amin Coja-Oghlan ). -SAT Formulas. Given: boolean variables a Boolean formula in - conjunctive normal form ( -CNF) where is a variable or the negation of a variable - PowerPoint PPT PresentationTRANSCRIPT
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Going after the -SAT Threshold
Konstantinos Panagiotou(with Amin Coja-Oghlan)
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-SAT Formulas
• Given:– boolean variables – a Boolean formula in -conjunctive normal form (-CNF)
where is a variable or the negation of a variable
• An assignment is called satisfying (for ), if it satisfies all clauses
• A clause is satisfied (by ) if at least one literal in it is satisfied
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Example ()
• Assignment is satisfying• Assignment is not
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The -SAT Problem• Question: given , is there any satisfying
assignment?• This is a central problem in computer science.• If , then it is easy:– is satisfiable iff no variable appears both negated
and not negated• If , then there is a linear time algorithm [Aspvall,
Plass & Tarjan (1979)]
• If , then the problem is -complete [Cook & Levin (1971)]
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Random Formulas
• Setup:– variables
– is a -CNF with clauses, where each clause is drawn uniformly at random from the set of all possible clauses
• We call the density of the formula
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A Generative Procedure
• Generate as follows:– for // Generate – for // Generate th literal in • , where is uar (uniformly at random) from • With probability set (i.e. negate the occurrence
of the variable)
• All random decisions are independent– Particularly, the choice of the variable and of its
„sign“ are distinct processes
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Many Questions…
• For which densities (# clauses ) is satisfiable whp (with high probability)?– Initial motivation for studying random -SAT: the
„most difficult“ instances seem to be around a specific
• Other properties that hold whp?• Algorithms?
• We will consider only the case here.
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Picture - SatisfiabilityPr[ is satisfiable] as
c (density)
1
0
?
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A First Bound• Consider the obvious random variable
= # of satisfying assignments of • If for the fixed value of we can show
as , then and is not satisfiable whp. • Let , where the sum is over all possible
assignments in and
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A First Bound (cont.)
𝔼 [ 𝑋 ]=∑𝜎Pr [𝜎 satisfies 𝐹𝑛 ,𝑐𝑛 ]¿∑
𝜎Pr [∀1≤ 𝑖≤𝑐𝑛:𝜎 satisfies𝐶𝑖]
¿∑𝜎
∏1≤ 𝑖≤ 𝑐𝑛
Pr [𝜎 satisfies𝐶𝑖 ]
𝐶𝑖=…∨…∨…∨…𝐶𝑖=𝑥𝑖 1∨𝑥 𝑖2∨𝑥𝑖 3∨𝑥 𝑖4
In
¿∑𝜎
(1−2−𝑘 )𝑐𝑛
¿2𝑛 (1−2−𝑘 )𝑐𝑛
≈ exp (𝑛 (ln 2−2−𝑘𝑐))
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PicturePr[ is satisfiable] as
c (density)
1
0
2𝑘 ln 2
?
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(Some) Previous Work• Friedgut ’05: There is a sharp threshold
sequence :– If , then is satisfiable whp– If , then it is not whp
• Kirousis et al. ’98:
• Achlioptas and Peres ’04:
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Before our WorkPr[ is satisfiable] as
c (density)
1
0
2𝑘 ln 2−(1+ln 2)/22𝑘 ln 2−𝑘 ln 2
Gap:
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Our Result
Coja-Oghlan, P. ‘13:
1
0
Gap:
2𝑘 ln 2−(1+ln 2)/2
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Random CSP‘s• Many examples
– Variations of random -SAT (NAESAT, XORSAT, …)– -coloring random graphs– 2-coloring random -uniform hypergraphs
• For no random version of these problems (in the NP-hard cases) the threshold is known
• Statistical physicists have developed sophisticated but non-rigorous techniques– detailed picture about the structural properties– several conjectures, algorithms– many papers: Krzakala, Montanari, Parisi, Ricci-Tersenghi,
Semerjian, Zdeborova, Zecchina, …• Mathematical treatment: Talagrand
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One Conjecture for -SATPr[ is satisfiable] as
c (density)
1
0
2𝑘 ln 2−(1+ln 2)/2
Gap:
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The Second Moment Problem
• If is a non-negative random variable
• We can apply this to , the number of satisfying assignments of
• If for the given , then we are done!• Problem: for all we have that is exponentially
larger than !
Pr [𝑍>0 ] ≥ 𝔼 [𝑍 ]2
𝔼 [𝑍2]Paley-Zygmund InequalitySecond Moment Method
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Bound for 2nd Moment𝔼 [ 𝑋 2 ]=∑
𝜎 , 𝜏Pr [𝜎 ,𝜏 satis fy 𝐹 𝑛 ,𝑐𝑛]
¿∑𝜎 ,𝜏
∏1≤ 𝑖≤ 𝑐𝑛
Pr [𝜎 ,𝜏 satisfy𝐶𝑖]
𝜎𝜏
𝛼𝑛
≥ max0<𝛼<1
∑𝜎 , 𝜏
𝑜𝑣𝑒𝑟𝑙𝑎𝑝 (𝜎 ,𝜏 )=𝛼𝑛
∏1≤ 𝑖≤𝑐𝑛
Pr [𝜎 ,𝜏 satisfy𝐶𝑖 ]
≥ max0<𝛼<1
2𝑛( 𝑛𝛼𝑛)¿ ¿
𝛼=12 :≈2
𝑛 ∙2𝑛 ∙ (1−2−𝑘 )2𝑐𝑛=𝔼 [ 𝑋 ]2𝛼=12+
12−𝑘
:≫𝔼 [ 𝑋 ]2
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Why?
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An Asymmetry• Consider a thought experiment• Suppose that somebody makes the promise
„ appears in exactly times …… and all these appearances are positive“
• What value do we assign to ? • Other promise:
„ appears in exactly times …… and 51% of the appearances are positive“
• We (should) set again to
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The Majority
• Our „best guess“ for a satisfying assignment is the majority vote:– Somebody tells us how often each variable appears positively
and negatively, and nothing else– If appears more often positively, assign it to , and otherwise
to • This assignment maximizes the probability that is
satisfied• Even more: assignments that are „close“ to the majority
vote have a larger probability of being satisfying
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Picture of the Situation
• Majority assignment• Largest probability of
being satisfiable
• Distance 1• Less probability of
being satisfiable
• Distance 2• Even smaller probability
of being satisfiable
→Thesatisfying assignments𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑒 !
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Symmetry vs. Asymmetry
• There is an asymmetry in the meaning of the assignment „true“ and „false“
• In many other problems this isn‘t so.– In graph coloring all colors play the same role– In not-all-equal SAT the roles of true and false can
be interchanged– …
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Getting a Grip on the Majority
• Generate in two steps as follows:1. For each variable choose randomly the number of
positive occurences and the number of negative occurences.
2. Choose randomly a formula where each variable appears times positively and times negatively.
• Want: distributions of are the same.• Step 1– It is easy to see in that and are distributed like Po,
and they are almost independent
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Step 2• How do we choose a formula where each
variable appears times positively and times negatively?
• Configuration model:
𝑑1
𝑥1
𝑑2
𝑥2
𝑑𝑛
𝑥𝑛…
𝐹=𝐶1∧𝐶2∧…∧𝐶𝑚
Random Matching: variable occurences
to positions in clauses
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A Weighting Scheme• Let
• measures how „distinct“ the majority vote is• Lemma. . Proof. is a sum of (almost) independent random variables.• Lemma. Let . Then
Proof. The number of satisfied variable occurences in the majority assignment increases linearly with w.• Conclusion: beats the main contribution to is from highly
atypical .
𝑤 (𝐹 𝑛 ,𝑚)= 1𝑛 ∑1≤ 𝑖≤𝑛
¿𝑑𝑖−𝑑𝑖∨¿¿
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Ingredient Nr. 1
• This is the first ingredient in our proof: fix w!– Actually, we fix the whole sequence such that it
enjoys many typical properties of independent Po random variables.
• Generate in two steps as follows:1. For each variable choose randomly the number of
positive occurences and the number of negative occurences.
2. Choose randomly a formula where each variable appears times positively and times negatively.
instead
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More Things go Wrong…
• From now on let be the number of satisfying assignments of , where the degree sequence is typical.
• It turns out: second moment fails again.• Other parameters start to fluctuate– Number of unsatisfied clauses under the majority
assignment– …
• Need to control everything at once.
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Recall the Situation
• Majority assignment• Largest probability of
being satisfiable
• Distance 1• Less probability of
being satisfiable
• Distance 2• Even smaller probability
of being satisfiable
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Our Variable – 2nd Ingredient
• We do not count all satisfying assignments!• Intuition: if a variable appears times positively
and times negatively, then assign it to true with some probability that depends on only.
• Map • Set also , • Meaning: a -fraction of the literals is satisfied
under the assignments that we consider.
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More formally• Set • This is the set of different „types“ of variable
occurences (equivalent )• We say that has -marginals if for all
• That is, a t-fraction of the variable occurences is set to true, for all
• Question: how do we choose ?
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Pictorially
{𝑥 𝑖 :𝑑𝑖−𝑑𝑖=0 }{𝑥 𝑖 :𝑑𝑖−𝑑𝑖=1 }
{𝑥 𝑖 :𝑑𝑖−𝑑𝑖=−1}{𝑥 𝑖 :𝑑𝑖−𝑑𝑖=−𝑠 } {𝑥 𝑖 :𝑑𝑖−𝑑𝑖=𝑠 }
𝑝 (0)𝑝 (1)
𝑝 (𝑠)
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Detour: Physics
• For let be the fraction of satisfying assignments in which is set to true in
• It is NP-hard to compute • According to physicists: can be computed by a
message passing algorithm called Belief Propagation [Montanari et al ‘07]
• So-called Replica Symmetric Phase: unique fixed point solution exists
• Condition: density
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Conjecture
• Belief Propagation leads to a stronger prediction – Conjecture for up to an error of as – This stronger conjecture is not explicit form– it does depend on many parameters
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Our Choice
• This matches the conjecture on the „bulk“ of the variables– Recall that – Except of a very small fraction, all other variables
have the property
𝑝 (𝑧 )={12 +𝑧2𝑘+1 ,if |𝑧|<10√𝑘2𝑘 ln𝑘
12, otherwise
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Summary & Outlook
• We can determine the replica symmetric -SAT threshold with high accuracy
• We manage for the first time to get a grip on an asymmetric problem
• On the way we use algorithmic insights gained by physicists
• Catching the -SAT threshold?
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Thank you!
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