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

• 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

Example ()

• Assignment is satisfying• Assignment is not

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)]

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

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

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.

Picture - SatisfiabilityPr[ is satisfiable] as

c (density)

1

0

?

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

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))

PicturePr[ is satisfiable] as

c (density)

1

0

2𝑘 ln 2

?

(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:

Before our WorkPr[ is satisfiable] as

c (density)

1

0

2𝑘 ln 2−(1+ln 2)/22𝑘 ln 2−𝑘 ln 2

Gap:

Our Result

Coja-Oghlan, P. ‘13:

1

0

Gap:

2𝑘 ln 2−(1+ln 2)/2

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

One Conjecture for -SATPr[ is satisfiable] as

c (density)

1

0

2𝑘 ln 2−(1+ln 2)/2

Gap:

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

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

Why?

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

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

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 !

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– …

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

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

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∨¿¿

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

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.

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

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.

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 ?

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)

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

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

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

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?

Thank you!

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