computing with oracles · 2017. 5. 21. · computing with oracles from np to beyond np and back...

Computing with Oracles

From NP to Beyond NP and Back Again

Joao Marques-Silva

University of Lisbon, Portugal

Beyond NP Meeting

Paris, France

10 May 2017

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The SAT disruption

• SAT is NP-complete [Cook’71]

– But, CDCL SAT solving is a success story of Computer Science– Hundreds (thousands?) of practical applications

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The SAT disruption


– But, CDCL SAT solving is a success story of Computer Science

– Hundreds (thousands?) of practical applications

2 / 73

The SAT disruption


– But, CDCL SAT solving is a success story of Computer Science– Hundreds (thousands?) of practical applications

2 / 73

SAT solver improvement I[Source: Le Berre 2013]

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SAT solver improvement II[Source: Simon 2015]

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SAT is the engines’ engine

Engines usingSAT engines

Boolean

QBF

MaxSAT

PBO

#SAT

...

FOL SMT

Modelfinding

Theoremproving

...

Other

ASP

LCG

CSP

...

5 / 73

SAT is ubiquitous in problem solving

Problem solvingwith SAT

Embeddings

PBOB&B

Search

Enumeration

OPT SAT

Lazy SMT

LCG

Oracles

MaxSAT

MCS

MUS

Min. Mod-els

Backbones

Enumeration

CEGARSMT

CEGARQBF

MC: ic3

Encodings

MBD

Eager SMT

Planning

BMC

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SAT can make the difference – propositional abduction

10−3 10−2 10−1 100 101 102 103 104

Hyper?

10−3

10−2

10−1

100

101

102

103

104

AbH

S+

1800 sec. timeout

1800

sec.

timeo

ut

• Topic(s): quantified optimization [ECAI’16]

• Instances: KR16 propositional abduction

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SAT can make the difference – axiom pinpointing

10−2 10−1 100 101 102 103 104

EL2MUS

10−2

10−1

100

101

102

103

104

EL

+SA

T

3600 sec. timeout

3600

sec.

timeo

ut

• Topic(s): MUS enumeration; MCSes; implicit hitting sets [SAT’15]

• Instances: EL+ medical ontologies

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Part I

From NP to Beyond NP

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Outline – part A

Background

MaxSAT Solving

2QBF Solving

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Outline

Background

MaxSAT Solving

2QBF Solving

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Beyond decision problems

Answer Problem Type

Yes/No Decision Problems

Some solution Function Problems

All solutions Enumeration Problems

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Answer Problem Type


Some solution

Function Problems


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Answer Problem Type




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Answer Problem Type



All solutions

Enumeration Problems

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Answer Problem Type




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... and beyond NP – decision and function problems

∆p0 = Σp

0 = P = Πp0 = ∆p

1

NP = Σp1 Πp

1 = coNP

PNP = ∆p2

Σp2 Πp

2

∆p3

Σp3 Πp

3

...

F∆p0 = FΣp

0 = FP = FΠp0 = F∆p

1

FNP = FΣp1 FΠp

1 = coFNP

FPNP = F∆p2

FΣp2 FΠp

2

F∆p3

FΣp3 FΠp

3

...

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Oracle-based problem solving – ideal scenario

Decision Procedure

Poly-timeAlgorithm

Yes/No +Witness

SAT, SMT, CSP, ...Solver / Oracle

Bounded # ofcalls / queries

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Oracle-based problem solving – in some settings

Decision Procedure

Poly-timeAlgorithm

Yes/No +Witness

SAT, SMT, CSP, ...Solver / Oracle

Bounded # ofcalls / queries

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Many problems to solve – within FPNP

Answer Problem Type




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Answer Problem Type




Function Problems on Propositional Formulas

MaxSATPBO

MinSAT

Autarkies

Backbones

Prime Implicants

MCSesMUSes Indep. Vars

WBO

MESes

MSSesMNSes

MDSes Implicant Ext.MFSes

MCFSes

Minimal Models

Prime ImplicatesMaximal Models

Implicate Ext.

...

...

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Answer Problem Type




Function Problems on Propositional Formulas

Optimization Problems

Minimal Sets

MaxSATPBO

MinSAT

Autarkies

Backbones

Prime Implicants

MCSesMUSes Indep. Vars

WBO

MESes

MSSesMNSes

MDSes Implicant Ext.MFSes

MCFSes

Minimal Models

Prime ImplicatesMaximal Models

Implicate Ext.

...

...

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Selection of topics

Problem solvingwith SAT

Embeddings

PBOB&B

Search

Enumeration

OPT SAT

Lazy SMT

LCG

Oracles

MaxSAT

MCS

MUS

Min. Mod-els

Backbones

Enumeration

CEGARSMT

CEGARQBF

MC: ic3

Encodings

MBD

Eager SMT

Planning

BMC

MaxSAT solving2QBF solving

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Outline

Background

MaxSAT Solving

2QBF Solving

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Recap MaxSAT

x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1 ¬x1

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4 ¬x4 ∨ x5

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3 ¬x3

• Given unsatisfiable formula, find largest subset of clauses that issatisfiable

• A Minimal Correction Subset (MCS) is an irreducible relaxation ofthe formula

• The MaxSAT solution is one of the smallest cost MCSes

– Note: Clauses can have weights & there can be hard clauses

• Many practical applications

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Recap MaxSAT

x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1 ¬x1

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4 ¬x4 ∨ x5

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3 ¬x3



• The MaxSAT solution is one of the smallest MCSes



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Recap MaxSAT

x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1 ¬x1

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4 ¬x4 ∨ x5

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3 ¬x3



• The MaxSAT solution is one of the smallest cost MCSes



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The MaxSAT (r)evolution – plain industrial instances

0

50

100

150

200

250

300

0 5 10 15 20 25 30 35 40

CP

U tim

e in s

econds

Number of instances

Number x of instances solved in y seconds

Open-WBO-Inpmifumax-13

WPM1-11wbo-1.4a-10

wbo1.6-cnf-12

Source: [MaxSAT 2014 organizers]

48.1% more

instances solved!

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The MaxSAT (r)evolution – plain industrial instances

0

50

100

150

200

250

300

0 5 10 15 20 25 30 35 40

CP

U tim

e in s

econds

Number of instances


Open-WBO-Inpmifumax-13

WPM1-11wbo-1.4a-10

wbo1.6-cnf-12


48.1% more

instances solved!20 / 73

The MaxSAT (r)evolution – partial

0

50

100

150

200

250

300

0 50 100 150 200 250 300 350 400

CP

U tim

e in s

econds

Number of instances


Open-WBO-InQMaxSAT2-mt-13

QMaxSat-g2-12QMaxSat0.4-11

QMaxSat-10


71.5% more

instances solved!

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The MaxSAT (r)evolution – partial

0

50

100

150

200

250

300

0 50 100 150 200 250 300 350 400

CP

U tim

e in s

econds

Number of instances


Open-WBO-InQMaxSAT2-mt-13

QMaxSat-g2-12QMaxSat0.4-11

QMaxSat-10


71.5% more


The MaxSAT (r)evolution – weighted partial

0

50

100

150

200

250

300

0 50 100 150 200 250 300 350

CP

U tim

e in s

econds

Number of instances


Eva500aWPM1-2013

WPM1-11pwbo2.1-12

wbo-1.4a-wcnf-10


51.5% more

instances solved!

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The MaxSAT (r)evolution – weighted partial

0

50

100

150

200

250

300

0 50 100 150 200 250 300 350

CP

U tim

e in s

econds

Number of instances


Eva500aWPM1-2013

WPM1-11pwbo2.1-12

wbo-1.4a-wcnf-10


51.5% more


Many MaxSAT approaches

MaxSATAlgorithms

Branch& Bound

Iterative

CoreGuided

IterativeMHS

ModelGuided

No unit prop;No cl. learning

All cls relaxed

Relax cls givenunsat cores

IterativeMHS & SAT

Relax cls givenmodels

• For practical (industrial) instances: core-guided approaches arethe most effective [MaxSAT14]

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Outline

Background

MaxSAT SolvingIterative SAT SolvingCore-Guided AlgorithmsMinimum Hitting Sets

2QBF Solving

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Basic MaxSAT with iterative SAT solving

x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 12

Example CNF formula

AtMostk/PB constraints

over all relaxation variables

All (possibly many)

soft clauses relaxed

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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 12

Relax all clauses; Set UB = 12 + 1



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 12

Formula is SAT; E.g. all xi = 0 and r1 = r7 = r9 = 1 (i.e. cost = 3)



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 2

Refine UB = 3



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 2

Formula is SAT; E.g. x1 = x2 = 1; x3 = ... = x8 = 0 and r4 = r9 = 1(i.e. cost = 2)



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 1

Refine UB = 2



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 1

Formula is UNSAT; terminate



All (possibly many)


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x6 ∨ x2∨r1 ¬x6 ∨ x2∨r2 ¬x2 ∨ x1∨r3 ¬x1∨r4

¬x6 ∨ x8∨r5 x6 ∨ ¬x8∨r6 x2 ∨ x4∨r7 ¬x4 ∨ x5∨r8

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r11 ¬x3∨r12∑12i=1 ri ≤ 1

MaxSAT solution is last satisfied UB: UB = 2



All (possibly many)


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Outline

Background


2QBF Solving

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MSU3 core-guided algorithm

x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1 ¬x1

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4 ¬x4 ∨ x5

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3 ¬x3

Example CNF formula

AtMostk/PB

constraints used

Relaxed soft clauses

become hard

Some clauses

not relaxed

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x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1 ¬x1

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4 ¬x4 ∨ x5

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3 ¬x3

Formula is UNSAT; OPT ≤ |ϕ| − 1; Get unsat core

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

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x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1∨r1 ¬x1∨r2

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4∨r3 ¬x4 ∨ x5∨r4

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3∨r5 ¬x3∨r6∑6i=1 ri ≤ 1

Add relaxation variables and AtMostk , k = 1, constraint

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

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x6 ∨ x2 ¬x6 ∨ x2 ¬x2 ∨ x1∨r1 ¬x1∨r2

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4∨r3 ¬x4 ∨ x5∨r4

x7 ∨ x5 ¬x7 ∨ x5 ¬x5 ∨ x3∨r5 ¬x3∨r6∑6i=1 ri ≤ 1

Formula is (again) UNSAT; OPT ≤ |ϕ| − 2; Get unsat core

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

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x6 ∨ x2∨r7 ¬x6 ∨ x2∨r8 ¬x2 ∨ x1∨r1 ¬x1∨r2

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4∨r3 ¬x4 ∨ x5∨r4

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r5 ¬x3∨r6∑10i=1 ri ≤ 2

Add new relaxation variables and update AtMostk , k=2, constraint

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

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x6 ∨ x2∨r7 ¬x6 ∨ x2∨r8 ¬x2 ∨ x1∨r1 ¬x1∨r2

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4∨r3 ¬x4 ∨ x5∨r4

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r5 ¬x3∨r6∑10i=1 ri ≤ 2

Instance is now SAT

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

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x6 ∨ x2∨r7 ¬x6 ∨ x2∨r8 ¬x2 ∨ x1∨r1 ¬x1∨r2

¬x6 ∨ x8 x6 ∨ ¬x8 x2 ∨ x4∨r3 ¬x4 ∨ x5∨r4

x7 ∨ x5∨r9 ¬x7 ∨ x5∨r10 ¬x5 ∨ x3∨r5 ¬x3∨r6∑10i=1 ri ≤ 2

MaxSAT solution is |ϕ| − I = 12− 2 = 10

AtMostk/PB

constraints used


become hard

Some clauses

not relaxed

27 / 73

Outline

Background


2QBF Solving

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MHS approach for MaxSAT

c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = ∅

• Find MHS of K:

• SAT(F \ ∅)?

• Core of F : {c1, c2, c3, c4}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = ∅

• Find MHS of K: ∅

• SAT(F \ ∅)?

• Core of F : {c1, c2, c3, c4}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = ∅

• Find MHS of K: ∅• SAT(F \ ∅)?

• Core of F : {c1, c2, c3, c4}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = ∅

• Find MHS of K: ∅• SAT(F \ ∅)? No

• Core of F : {c1, c2, c3, c4}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}}

• Find MHS of K: ∅• SAT(F \ ∅)? No

• Core of F : {c1, c2, c3, c4}. Update K

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}}

• Find MHS of K:

• SAT(F \ {c1})?

• Core of F : {c9, c10, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}}

• Find MHS of K: E.g. {c1}

• SAT(F \ {c1})?

• Core of F : {c9, c10, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}}

• Find MHS of K: E.g. {c1}• SAT(F \ {c1})?

• Core of F : {c9, c10, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}}

• Find MHS of K: E.g. {c1}• SAT(F \ {c1})? No

• Core of F : {c9, c10, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}}

• Find MHS of K: E.g. {c1}• SAT(F \ {c1})? No

• Core of F : {c9, c10, c11, c12}. Update K

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}}

• Find MHS of K:

• SAT(F \ {c1, c9})?

• Core of F : {c3, c4, c7, c8, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}}

• Find MHS of K: E.g. {c1, c9}

• SAT(F \ {c1, c9})?

• Core of F : {c3, c4, c7, c8, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}}

• Find MHS of K: E.g. {c1, c9}• SAT(F \ {c1, c9})?

• Core of F : {c3, c4, c7, c8, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}}

• Find MHS of K: E.g. {c1, c9}• SAT(F \ {c1, c9})? No

• Core of F : {c3, c4, c7, c8, c11, c12}

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3

K = {{c1, c2, c3, c4}, {c9, c10, c11, c12}, {c3, c4, c7, c8, c11, c12}}

• Find MHS of K: E.g. {c1, c9}• SAT(F \ {c1, c9})? No

• Core of F : {c3, c4, c7, c8, c11, c12}. Update K

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3


• Find MHS of K:

• SAT(F \ {c4, c9})?

• Terminate & return 2

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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3


• Find MHS of K: E.g. {c4, c9}

• SAT(F \ {c4, c9})?


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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3


• Find MHS of K: E.g. {c4, c9}• SAT(F \ {c4, c9})?


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c1 = x6 ∨ x2 c2 = ¬x6 ∨ x2 c3 = ¬x2 ∨ x1 c4 = ¬x1

c5 = ¬x6 ∨ x8 c6 = x6 ∨ ¬x8 c7 = x2 ∨ x4 c8 = ¬x4 ∨ x5

c9 = x7 ∨ x5 c10 = ¬x7 ∨ x5 c11 = ¬x5 ∨ x3 c12 = ¬x3


• Find MHS of K: E.g. {c4, c9}• SAT(F \ {c4, c9})? Yes


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MaxSAT solving with SAT oracles

• A sample of recent algorithms:

Algorithm # Oracle Queries Reference

Linear search SU Exponential*** [e.g. LBP10]

Binary search Linear* [e.g. FM06]

FM/WMSU1/WPM1 Exponential** [FM06,MSM08,MMSP09,ABL09a,ABGL12]

WPM2 Exponential** [ABL10,ABGL13]

Bin-Core-Dis Linear [HMMS11,MHMS12]

Iterative MHS Exponential [DB11,DB13a,DB13b]

* O(logm) queries with SAT oracle, for (partial) unweighted MaxSAT

** Weighted case; depends on computed cores*** On # bits of problem instance (due to weights)

• But also additional recent work:– Progression– Soft cardinality constraints (OLL)– MaxSAT resolution– ...

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Outline

Background

MaxSAT Solving

2QBF Solving

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Abstraction refinement for QBF

• Many approaches proposed for solving QBF [...]

• Abstraction-refinement proposed for 2QBF in 2011 [JMS11]

• Extended to QBF in 2012 [JKMSC12]

• Significant impact in QBF competitions

• Influenced research in QBF solvers

– E.g. see conference papers in 2015/2016

• Ack: Slides adapted from M. Janota SAT’11 talk

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Problem definition

Given: ∃X∀Y .φ, where φ is a propositional formula

Question: Is there assignment ν to X variables such that ∀Y .φ[X/ν]?

Example

∃x1, x2 ∀y1, y2. (x1 → y1) ∧ (x2→y2)

solution: x1 = 0, x2 = 0

A simple algorithm

• While true– Pick fresh assignment ν to X variables– Check with SAT solver whether ∀Y .φ[X/ν] holds

I How? Check SAT(¬φ[X/ν]) is unsat

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Problem definition



Example

∃x1, x2 ∀y1, y2. (x1 → y1) ∧ (x2→y2)


A simple algorithm


I How?

Check SAT(¬φ[X/ν]) is unsat

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Problem definition



Example

∃x1, x2 ∀y1, y2. (x1 → y1) ∧ (x2→y2)


A simple algorithm


I How? Check SAT(¬φ[X/ν]) is unsat

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Looking at assignments

Y

X

ξ

µ

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Y

X

ξ

µ

1

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Y

X

ξ

µ

11 0 . . .

. . .0 . . .

. . .1

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Y

X

ξ

µ

11 0 . . .

. . .0 . . .

. . .1

ν 1 1 . . . 1 1 . . . 1

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Y

X

ξ

µ

φ[Y /µ]

11 0 . . .

. . .0 . . .

. . .1

ν 1 1 . . . 1 1 . . . 1

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Expanding into SAT

∃X∀Y . φ =⇒ SAT

∧µ∈B|Y |

φ[Y /µ]

Example

∃x1, x2 ∀y1, y2. (x1 → y1) ∧ (x2→y2)

(x1 → 0) ∧ (x2 → 0)∧ (x1 → 0) ∧ (x2 → 1)∧ (x1 → 1) ∧ (x2 → 0)∧ (x1 → 1) ∧ (x2 → 1)

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Abstraction

• Consider only some set of assignments W ⊆ B|Y |∧µ∈W

φ[Y /µ]

• A solution to the original problem is also a solution to theabstraction ∧

µ∈B|Y |φ[Y /µ] ⇒

∧µ∈W

φ[Y /µ]

• But converse not true

– A solution to an abstraction is not necessarily a solution to theoriginal problem

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CEGAR loop

input : ∃X∀Y .φoutput: (true, ν) if there exists ν s.t. ∀Yφ[X/ν],

(false, –) otherwise

W ← ∅while true do

(outc1, ν)← SAT(∧

µ∈W φ[Y /µ]) // find a candidate

if outc1 = false thenreturn (false,–) // no candidate found

endif ν is a solution // solution checkthen

return (true, ν)else

Grow W // refinement

endend

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Checking for a solution

An assignment ν is a solution to ∃X∀Y .φ iff

∀Y .φ[X/ν] iff UNSAT(¬φ[X/ν])

If SAT(¬φ[X/ν]) for some µ, then µ is a counterexample to ν

Example

∃x1, x2 ∀y1, y2. (x1 → y1) ∧ (x2→y2)

• candidate: x1 = 1, x2 = 1

• ¬φ[X/ν] , ¬y1 ∨ ¬y2

• counterexamples: y1 = 0, y2 = 0y1 = 0, y2 = 1y1 = 1, y2 = 0

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Refinement

YX

ν 1 1 1 1. . . . . .

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Refinement

YX

ν

µ

1 1 1 0. . . . . .

W

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Refinement

YX

ν

µ

1 1 1 0. . . . . .

W

W ′

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2QBF algorithm

input : ∃X∀Y .φoutput: (true, ν) if there exists ν s.t. ∀Yφ[X/ν],

(false, –) otherwise

ω ← 1while true do

(outc1, ν)← SAT(ω) // find a candidate solution

if outc1 = false thenreturn (false,–) // no candidate found

end

(outc2, µ)← SAT (¬φ[X/ν]) // find a counterexample

if outc2 = false thenreturn (true, ν) // candidate is a solution

endω ← ω ∧ φ[Y /µ] // refine

end

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Properties of refinement

Y

X

ν

µ

1 1 1

ν1 1 1 1 0

ν2 1 1 1 0

0. . . . . .

W

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Y

X

νν

µ

1 1 1

ν1 1 1 1 0

ν2 1 1 1 0

0. . . . . .

W

W ′

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Y

X

νν

µ

1 1 1

ν1ν1 1 1 1 00

ν2ν2 1 1 1 00

0. . . . . .

W

W ′

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About refinement step

• No candidate for counterexample appears more than once

• Thus, upper bound on the number of iterations is:

min{

2|X |, 2|Y |}

• Heuristic: look for such counterexamples that are alsocounterexamples to many other candidates, look for µ s.t.

¬φ[X/ν] ∧max(|{ν ′ | ¬φ[X/ν ′,Y /µ]}|

)

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Part II

Back Again (to NP)

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Why back again to NP?

• Fact: There are many hard examples for resolution and CDCL

– One example are pigeonhole formulas (PHP) (more later)

• What we have been looking at?

– Reduce problems to one concrete problem, i.e. Horn MaxSAT– Develop fast algorithms for Horn MaxSAT

I Use IHSes, MHSes, MUSes, etc.

• What we found out?

– Reductions are remarkably effective for PHP in practice

– There exist polynomial time proofs that PHP is unsatisfiable !

I Using core-guided algorithms; andI Using MaxSAT resolution

– But, core-guided algorithms also use CDCL !– Also, MHS MaxSAT algorithms are effective on hard problems

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Why back again to NP?

• Fact: There are many hard examples for resolution and CDCL

– One example are pigeonhole formulas (PHP) (more later)

• What we have been looking at?

– Reduce problems to one concrete problem, i.e. Horn MaxSAT– Develop fast algorithms for Horn MaxSAT

I Use IHSes, MHSes, MUSes, etc.

• What we found out?

– Reductions are remarkably effective for PHP in practice– There exist polynomial time proofs that PHP is unsatisfiable !

I Using core-guided algorithms; andI Using MaxSAT resolution

– But, core-guided algorithms also use CDCL !– Also, MHS MaxSAT algorithms are effective on hard problems

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Plan for part B

1. Recap PHP

2. Reduce SAT to Horn MaxSAT

– Also, what happens to PHP?

3. Develop polynomial time proofs of the unsatisfiability of PHP

– Using an MSU3-like MaxSAT algorithm– Using MaxSAT resolution

4. Experimental results

– PHP, Urquhart, and combinations thereof

5. Detailed description available from:https://arxiv.org/abs/1705.01477

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https://arxiv.org/abs/1705.01477

Outline

Pigeonhole Formulas

Reduction: SAT to Horn MaxSAT

Polynomial Time Proofs

Experimental Results

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Pigeonhole formulas I

• Pigeonhole principle:

– Typical: if m + 1 pigeons are distributed by m holes, then at leastone hole contains more than one pigeon

– Alternative: there exists no injective function mapping from{1, 2, ...,m + 1} to {1, 2, ...,m}, for m ≥ 1

• Propositional formulation:

Does there exist assignment such that the m + 1 pigeons canbe placed into m holes?

• Encoding: xij variables

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Pigeonhole formulas I

• Pigeonhole principle:

– Typical: if m + 1 pigeons are distributed by m holes, then at leastone hole contains more than one pigeon

– Alternative: there exists no injective function mapping from{1, 2, ...,m + 1} to {1, 2, ...,m}, for m ≥ 1

• Propositional formulation:

Does there exist assignment such that the m + 1 pigeons canbe placed into m holes?

• Encoding: xij variables

1

1

pigeons

holes

m + 1i

j m

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Pigeonhole formulas II – propositional encoding PHPm+1m

• Variables:

– xij = 1 iff the i th pigeon is placed in the j th hole, 1 ≤ i ≤ m + 1,1 ≤ j ≤ m

• Constraints:

– Each pigeon must be placed in at least one hole, and each holemust not have more than one pigeon∧m+1

i=1 AtLeast1(xi1, . . . , xim) ∧∧mj=1 AtMost1(x1j , . . . , xm+1 j)

• Example encoding, with pairwise encoding for AtMost1 constraint:

Constraint Clause(s)

∧m+1i=1 AtLeast1(xi1, . . . , xim) (xi1 ∨ . . . ∨ xim)

∧mj=1AtMost1(x1j , . . . , xm+1 j) ∧m+1

r=2 ∧r−1s=1 (¬xrj ∨ ¬xsj)

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Outline

Pigeonhole Formulas




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Reducing SAT to Horn MaxSAT

• Formula F with variables X = {x1, . . . , xt}

• Replace each original variable xi ∈ X by ni and pi , s.t.– ni = 1 iff xi = 0– pi = 1 iff xi = 1– Add (hard Horn) constraint (¬ni ∨ ¬pi ) ⇐ set of clauses P

• Translate each clause cr ∈ F into (hard Horn) clause c ′r ∈ FH :– Literal xi converted to ¬ni– Literal ¬xi converted to ¬pi– Resulting clause is goal clause ⇐ (can do better)

• Soft clauses: S = {(n1), . . . , (nt), (p1), . . . , (pt)}• Horn MaxSAT formula: 〈FH ∪ P,S〉• Claim:

F is SAT iff Horn MaxSAT formula has solution with cost ≤ t– There exists assignment that satisfies hard clauses FH and at least t

soft clauses from S, i.e. cost ≤ t– Due to P clauses, cost ≥ t; thus F is SAT iff cost = t

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• Formula F with variables X = {x1, . . . , xt}• Replace each original variable xi ∈ X by ni and pi , s.t.

– ni = 1 iff xi = 0– pi = 1 iff xi = 1– Add (hard Horn) constraint (¬ni ∨ ¬pi ) ⇐ set of clauses P





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• Soft clauses: S = {(n1), . . . , (nt), (p1), . . . , (pt)}• Horn MaxSAT formula: 〈FH ∪ P,S〉

• Claim:



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An example

• CNF formula:

F = (x1 ∨ ¬x2 ∨ x3) ∧ (x2 ∨ x3) ∧ (¬x1 ∨ ¬x3)

• New variables: {n1, p1, n2, p2, n3, p3}

• Soft clauses: S = {(n1), (p1), (n2), (p2), (n3), (p3)}

• Clauses in P:

P , (¬n1 ∨ ¬p1) ∧ (¬n2 ∨ ¬p2) ∧ (¬n3 ∨ ¬p3)

• Original clauses converted to:

FH , (¬n1 ∨ ¬p2 ∨ ¬n3) ∧ (¬n2 ∨ ¬n3) ∧ (¬p1 ∨ ¬p3)

• Resulting formula: 〈FH ∪ P,S〉

• F is satisfiable iff Horn MaxSAT formula has a solution with cost 3

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PHP as Horn MaxSAT

• New variables nij and pij , for each xij , 1 ≤ i ≤ m + 1, 1 ≤ j ≤ m

• The soft clauses S, with |S| = 2m(m + 1), are given by

{ (n11), . . . , (n1m), . . . , (nm+1 1), . . . , (nm+1m),(p11), . . . , (p1m), . . . , (pm+1 1), . . . , (pm+1m) }

• Clauses in P: P = {(¬nij ∨ ¬pij) | 1 ≤ i ≤ m + 1, 1 ≤ j ≤ m}• AtLeast1 constraints encoded as Li , 1 ≤ i ≤ m + 1

• AtMost1 constraints encoded as Mj , 1 ≤ j ≤ m

• Full reduction of PHP to Horn MaxSAT

〈H,S〉 =⟨∧m+1i=1 Li ∧ ∧mj=1Mj ∧ P,S

⟩• No more than m(m + 1) clauses can be satisfied, due to P• PHPm+1

m is satisfiable iff there exists an assignment that satisfiesthe hard clauses H and m(m + 1) soft clauses from S

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PHP as Horn MaxSAT




• Clauses in P: P = {(¬nij ∨ ¬pij) | 1 ≤ i ≤ m + 1, 1 ≤ j ≤ m}

• AtLeast1 constraints encoded as Li , 1 ≤ i ≤ m + 1



〈H,S〉 =⟨∧m+1i=1 Li ∧ ∧mj=1Mj ∧ P,S



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PHP as Horn MaxSAT







〈H,S〉 =⟨∧m+1i=1 Li ∧ ∧mj=1Mj ∧ P,S



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PHP as Horn MaxSAT







〈H,S〉 =⟨∧m+1i=1 Li ∧ ∧mj=1Mj ∧ P,S

⟩

• No more than m(m + 1) clauses can be satisfied, due to P• PHPm+1


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PHP as Horn MaxSAT







〈H,S〉 =⟨∧m+1i=1 Li ∧ ∧mj=1Mj ∧ P,S



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PHP as Horn MaxSAT II

• Clauses in each Li and in each Mj , with pairwise encoding

Original Constraint Encoded To Clauses

∧m+1i=1 AtLeast1(xi1, . . . , xim) Li (¬ni1 ∨ . . . ∨ ¬nim)

∧mj=1AtMost1(x1j , . . . , xm+1,j) Mj ∧m+1

r=2 ∧r−1s=1 (¬prj ∨ ¬psj)

• Note: constraints with key structural properties:

– Variables in each Li disjoint from any other Lk and Mj , k 6= i– Variables in each Mj disjoint from any other Ml , l 6= j

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PHP as Horn MaxSAT II

• Clauses in each Li and in each Mj , with pairwise encoding

Original Constraint Encoded To Clauses

∧m+1i=1 AtLeast1(xi1, . . . , xim) Li (¬ni1 ∨ . . . ∨ ¬nim)

∧mj=1AtMost1(x1j , . . . , xm+1,j) Mj ∧m+1

r=2 ∧r−1s=1 (¬prj ∨ ¬psj)

• Note: constraints with key structural properties:

Constraint Variables

Li (¬ni1 ∨ . . . ∨ ¬nim)

Lk (¬nk1 ∨ . . . ∨ ¬nkm)

Mj ∧m+1r=2 ∧

r−1s=1 (¬pr j ∨ ¬psj)

Ml ∧m+1r=2 ∧

r−1s=1 (¬pr l ∨ ¬psl)

– Variables in each Li disjoint from any other Lk and Mj , k 6= i– Variables in each Mj disjoint from any other Ml , l 6= j

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Outline

Pigeonhole Formulas




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Main claims

Claim 1

Core-guided MaxSAT produces a lower bound on the number offalsified clauses of ≥m(m + 1) + 1 in polynomial time

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Main claims

Claim 1


Claim 2

MaxSAT resolution produces a lower bound on the number of falsifiedclauses of ≥m(m + 1) + 1 in polynomial time

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Main claims

Claim 1


Claim 2

MaxSAT resolution produces a lower bound on the number of falsifiedclauses of ≥m(m + 1) + 1 in polynomial time

Corollary

Horn MaxSAT encoding enables polynomial time proofs of theunsatisfiability of PHP instances, using CDCL SAT solvers

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Proof of claim 1 – outline

1. Assume MSU3 MaxSAT algorithm– Note: Suffices to analyze disjoint sets separately

2. Relate soft clauses with each Li and each Mj

– Each constraint disjoint from the others (but not from P)

3. Derive large enough lower bound on # of falsified clauses:

4. Each increase in the value of the lower bound obtained by unitpropagation (UP)

– In total: polynomial number of (linear time) UP runs

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1. Assume MSU3 MaxSAT algorithm– Note: Suffices to analyze disjoint sets separately

2. Relate soft clauses with each Li and each Mj

– Each constraint disjoint from the others (but not from P)

3. Derive large enough lower bound on # of falsified clauses:

Constr. type # falsified cls # constr In total

Li 1 i = 1, . . . ,m + 1 m + 1

Mj m j = 1, . . . ,m m ·m

m(m + 1) + 1

4. Each increase in the value of the lower bound obtained by unitpropagation (UP)

– In total: polynomial number of (linear time) UP runs

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Proof of claim 1 – unit propagation steps I

Constr Hard cls Soft cls Relaxed clausesUpdatedAtMostk

constr

LBincr

Li (¬ni1 ∨ . . . ∨ ¬nim) (ni1), . . . , (nim)(sil ∨ ni1),1 ≤ l ≤ m

∑ml=1 sil ≤ 1 1

Mj (¬p1j ∨ ¬p2j) (p1j), (p2j)(r1j ∨ p1j),(r2j ∨ p2j)

∑2l=1 rlj ≤ 1 1

Mj

(¬p1j ∨ ¬p3j),(¬p2j ∨ ¬p3j),

(r1j ∨ p1j),(r2j ∨ p2j),∑2l=1 rlj ≤ 1

(p3j) (r3j ∨ p3j)∑3

l=1 rlj ≤ 2 1

· · ·

Mj

(¬p1j∨¬pm+1j), . . .,(¬pmj ∨ ¬pm+1j),

(r1j ∨ p1j), . . .,(rmj ∨ pmj),∑ml=1 rlj ≤ m − 1

(pm+1j) (rm+1j ∨ pm+1j)∑m+1

l=1 rlj ≤ m 1

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Proof of claim 1 – unit propagation steps II

Clauses Unit Propagation

(pk+1 j) pk+1 j = 1

(¬p1j ∨¬pk+1 j), . . . , (¬pkj ∨¬pk+1 j) p1j = . . . = pkj = 0

(r1j ∨ p1j), . . . , (rkj ∨ pkj) r1j = . . . = rkj = 1∑kl=1 rlj ≤ k − 1

(∑kl=1 rlj ≤ k − 1

)`1⊥

• Key points:

– For each Li , UP raises LB by 1– For each Mj , UP raises LB by m– In total, UP raises LB by m(m + 1) + 1– PHPm+1

m is unsatisfiable

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Proof of claim 2 – recap MaxSAT resolution

• Clauses: (x ∨ A, u) and (¬x ∨ B,w)

• m , min(u,w)

• u w , (u ==>) ?> : u − w , with u ≥ w

• Example MaxSAT resolution steps:

Clause 1 Clause 2 Derived Clauses

(x ∨ A, u) (¬x ∨ B,w)(A ∨ B,m), (x ∨ A, u m), (¬x ∨ B,w m),

(x ∨ A ∨ ¬B,m), (¬x ∨ ¬A ∨ B,m)

(x ∨ A, 1) (¬x ,>) (A, 1), (¬x ,>), (¬x ∨ ¬A, 1)

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• Follow ideas used in previous proof

• Mimic unit propagation steps as MaxSAT resolution steps

• Each increase in LB corresponds to deriving one empty clause

• In total: polynomial number of steps, each running in polynomialtime

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Proof of claim 2 – key steps I

Constraint Clauses Resulting clause(s)

Li(¬ni1 ∨ . . . ∨ ¬nim,>),

(ni1, 1)(¬ni2 ∨ . . . ∨ ¬nim, 1) , . . .

Li(¬ni2 ∨ . . . ∨ ¬nim, 1),

(ni2, 1)(¬ni3 ∨ . . . ∨ ¬nim, 1) , . . .

· · ·

Li(¬nim, 1),(nim, 1)

(⊥, 1) , . . .

Mj(¬p1j ∨ ¬p2j ,>),

(p1j , 1)(¬p2j , 1), (¬p1j ∨ ¬p2j ,>), (p1j ∨ p2j , 1)

Mj(¬p2j , 1),(p2j , 1)

(⊥, 1)

Mj(¬p1j ∨ ¬p3j ,>),

(p1j ∨ p2j , 1)

(p2j ∨ ¬p3j , 1) , (¬p1j ∨ ¬p3j ,>),

(¬p1j ∨¬p3j ∨¬p2j , 1), (p1j ∨ p2j ∨ p3j , 1)

Mj(¬p2j ∨ ¬p3j ,>),

(p2j ∨ ¬p3j , 1)(¬p3j , 1) , (¬p2j ∨ ¬p3j ,>)

Mj(¬p3j , 1),(p3j , 1)

(⊥, 1)

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Proof of claim 2 – key steps II

Constraint Clauses Resulting clause(s)

· · ·

Mj(¬p1j ∨ ¬pm+1j ,>),(p1j ∨ . . . ∨ pmj , 1)

(p2j . . . pmj ∨ ¬pm+1j , 1) , . . .

Mj

(¬p2j ∨ ¬pm+1j ,>),(p2j ∨ . . . ∨ pmj ∨¬pm+1j , 1)

(p3j . . . pmj ∨ ¬pm+1j , 1) , . . .

· · ·

Mj(¬pmj ∨ ¬pm+1j ,>),

(pmj ∨ ¬pm+1j , 1)¬pm+1j , 1) , . . .

Mj(pm+1j , 1),(¬pm+1j , 1)

(⊥, 1)

• Key points:

– For each Li , derive 1 empty clause– For each Mj , derive m empty clauses– In total, derive m(m + 1) + 1 empty clauses– PHPm+1

m is unsatisfiable62 / 73

Outline

Pigeonhole Formulas




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Experimental setup

• Instances:

– PHP-pw (46), PHP-sc (46), Urquhart (84), Comb (96)

• Solvers:

SAT SAT+ IHS MaxSAT CG MaxSAT MRes MIP OPB BDD

minisat glucose lgl crypto maxhs lmhs mscg wbo wpm3 eva lp cc sat4j∗ zres

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Results on PHP instances: pw vs. sc

0 10 20 30 40 50 60instances

10−3

10−2

10−1

100

101

102

103

CPU

time

(s)

lp-cnflp-wcnfmaxhslmhsmscgevalgllmhs-neszresglucoselgl-nocardcc-cnfcc-opb

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lp-cnflp-wcnfmaxhslmhsmscglmhs-neseva

glucoselgl-nocardlglzrescc-cnfcc-opb

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Effect of P clauses

0 20 40 60 80 100instances

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time

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mscg (no P)maxhslmhswbo (no P)mscgeva (no P)evalmhs-nes (no P)lmhs-neswbo

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Effect of P clauses on mscg and wbo

10−3 10−2 10−1 100 101 102 103 104

PHP-nop

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PHP

1800 sec. timeout

1800

sec.

timeo

ut

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1800 sec. timeout

1800

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timeo

ut

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Results on Urquhart & combined instances

0 10 20 30 40 50 60 70 80 90instances

10−2

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time

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zresmaxhslmhslmhs-neslgllp-wcnf

mscglgl-nogaussglucoseevacc-cnf

0 20 40 60 80 100instances

10−1

100

101

102

103

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time

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lmhslmhs-nesmaxhslgl-nocard

zreslgllp-wcnfglucose

mscgevacc-cnf

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More detail in arXiv report

“On Tackling the Limits of Resolution in SAT Solving”

A. Ignatiev, A. Morgado, and J. Marques-Silva


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Part III

Wrap Up

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Conclusions

• Covered some examples of problem solving using SAT oracles– MaxSAT solving– 2QBF solving

• But, many more examples:– MUS & MCS extraction– MUS & MCS enumeration– Prime compilation– Implicit hitting sets– Quantification: decision, QMaxSAT, abduction, ...– Smallest MUSes– Approximate model counting– Also: backbones; autarkies/lean kernels, ...– Also: (many) practical applications

• (Horn) MaxSAT solvers can solve (in polynomial time) hardinstances for resolution

– If equipped with the right reduction

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Some research topics

• Beyond NP:– Query complexity– Enumeration– Quantification– Implicit hitting sets & duality– ...

• Applications:– Diagnosis– Axiom pinpointing– Planning– Reachability– Synthesis– Networking– Configuration– Argumentation– ...

• Also, where to go with Horn MaxSAT?

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Thank You

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computing with oracles · 2017. 5. 21. · computing with oracles from np to beyond np and back...

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