practical sat solving - kit · 2016. 5. 31. · sat solver resolution refutations based on clause...

INSTITUTE FOR THEORETICAL COMPUTER SCIENCE

Practical SAT SolvingLecture 6

Carsten Sinz, Tomáš Balyo, Markus Iser | May 30, 2016

KIT – University of the State of Baden-Wuerttemberg and

National Laboratory of the Helmholtz Association

www.kit.edu

http://www.kit.edu

Review: Last Week

Details on implementing DPLLDecision HeuristicsRestartsImplementing Unit Propagation

Carsten Sinz, Tomáš Balyo, Markus Iser – SAT Solving May 30, 2016 2/23

Lecture Outline: Today

More Details on implementing DPLLPhase Saving

Towards CDCLReview: ResolutionNon-Chronological Backtracking (Backjumping)Implication GraphClause Learing


Phase Saving

First implemented in RSAT (2006)http://reasoning.cs.ucla.edu/rsat/

Observeration: Frequent Restartsdecrease performance on some SATinstances

Goal: Cache partial solutions to subsets ofthe formula and reuse them after a restart

Idea: Remember last assignment of eachvariable and use it first in branching

Result: Phase Saving stabilizes positiveeffect of restarts; best results incombination with non-chronologicalbacktracking (later in this lecture)

A B

C�

F

Example: A and B aresatisfied, search works oncomponent C


http://reasoning.cs.ucla.edu/rsat/

Review: Resolution

General deduction rule for clauses: (A ∨ l), (B ∨ l) =⇒ (A ∨ B)

Can be used to extend DPLL Search by adding resolved clauses tothe input formula

Special Case: Unit Resolution (central part of DPLL Search)

Resolution Operator: C ◦l D denotes the resolvent of clauses C andD on literal l


Terminology

TrailThe partial assignment that represents the current path in the search tree

Decision LevelStarting with decision level 0 every branching step increases the decisionlevel by 1. Note: Unit-propagation does not change the decision level.

Conflicting ClauseClause with all literals assigned to 0 under the current trail (empty underunit resolution)


DPLL: Chronological Backtracking

A

B

C

X

[Y=1]

[Z]�

[Z]�

X

� �

[C=1]

X

� �

[B=1]

1 0

10

1 0

1 0 1 0 1 0

F = {{A,B},{B,C},{¬A,¬X ,Y},{¬A,X ,Z},{¬A,¬Y ,Z},{¬A,X ,¬Z},{¬A,¬Y ,¬Z}}

Trail: A,B,C,X ,Y ,Z

Conflicting Clause:{¬A,¬Y ,¬Z}


DPLL: Chronological Backtracking

A

B

C

X

[Y=1]

[Z]�

[Z]�

X

� �

[C=1]

X

� �

[B=1]

1 0

10

1 0

1 0 1 0 1 0

F = {{A,B},{B,C},{¬A,¬X ,Y},{¬A,X ,Z},{¬A,¬Y ,Z},{¬A,X ,¬Z},{¬A,¬Y ,¬Z}}

Trail: A,B,C,¬X ,Z

Conflicting Clause:{¬A,X ,¬Z}


Non-Chronological Backtracking

A

B

C

X

[Y=1]

[Z=1, Z=0]�

[Z=1, Z=0]�

[ B = 1 ]

1 0

1

1

1 0

The first two conflicting clauses{¬A,¬Y ,¬Z}, {¬A,X ,¬Z} containonly a fraction of the assignments onthe trail

Assignments to B and C obviously playno role in the present conflicting stateand we could immediately backtrack toflip the assignment to A

How to find out which assignments onthe trail are relevant for the actualconflict?


Implication Graph

Implication Graph

The Implication Graph GI = (V ∪ {�},E) (for a given formula F and atrail T ending in a conflict) is a directed acyclic graph (DAG) where eachvertex v ∈ V is a tuple (x = b, d) consisting of an assignment x = b onthe trail and its decision level d . There is an additional vertex�(with noassignment and decision level) indicating a conflicting assignment. Foreach clause C = (l1, . . . , lk , u) that became unit and caused a unitpropagation of u there are edges e ∈ E connecting the assignment ofliteral li to the assignment of unit u. Similarly, for a clause C, in which allliterals are assigned false, there are edged from each literal in C to�.


Example: Implication GraphThe graph shows the implication graph for the conflicting state under thetrail A,B,C,X ,Y ,Z . The edge labels denote clauses, node labelsindicate an assignment and its decision level.

0A = 1

1B = 1

2C = 1

3X = 1

3Y = 1

3Z = 1 �

3 57

57

37

F = {{A,B}, =: 1

{B,C}, =: 2

{¬A,¬X ,Y}, =: 3

{¬A,X ,Z}, =: 4

{¬A,¬Y ,Z}, =: 5

{¬A,X ,¬Z}, =: 6

{¬A,¬Y ,¬Z}} =: 7


Implication GraphIn the implication graph

the sink is always the conflicting assignment

the sources are the desicion literals that take part in the conflict

we can use it to detect the reasons for a conflict

0A = 1

1B = 1

2C = 1

3X = 1

3Y = 1

3Z = 1 �

3 57

57

37


Implication Graph: Implementation

Trail with conflicting clause {¬A,¬Y ,¬Z}:

A

B

C

X

Y

Z

null

null

null

null

{¬A,¬X ,Y}

{¬A,¬Y ,Z}for each assignment store a pointer tothe reason clause, the one that hasbecome unit

decision literals store a null pointer

with the clause pointers and the trail wecan trace all the implications back totheir sources


Clause Learing

Back-jumping does not fully avoid the occurrence of the same conflict, thesame (partial) assignments may generate the same conflict in anotherpart of the search tree.

Idea: store no-good constraint like in CSP (No-Good-Learning)

Great in SAT: no-good is clause!

generate conflict clauses and add them to CNF

the literals contributing to a conflict form a partial assignment and itsnegation is a clause (implied by the original CNF)

adding this clause avoids the conflicting partial assignment


Conflict Analysis: ExampleIn the first conflict of the previous example we can learn the clause{¬A,¬X} in order to prevent the solver to pick the same partialassignment again. This can also be expressed as the following sequenceof resolution steps along the edges in the implication graph:C = (7 ◦Z 5) ◦Y 3

0A = 1

1B = 1

2C = 1

3X = 1

3Y = 1

3Z = 1 �

3 57

57

37

F = {{A,B}, =: 1

{B,C}, =: 2

{¬A,¬X ,Y}, =: 3

{¬A,X ,Z}, =: 4

{¬A,¬Y ,Z}, =: 5

{¬A,X ,¬Z}, =: 6

{¬A,¬Y ,¬Z}} =: 7


Confict Clause

Trail with conflicting clause{¬A,¬Y ,¬Z}:

A

B

C

X

Y

Z

null

null

null

null

{¬A,¬X ,Y}

{¬A,¬Y ,Z}

Given the trail the conflict clause (learnedclause) can be generated by subsequentresolution, starting with the conflict clause{¬A,¬Y ,¬Z} and the first reason clause onthe trail {¬A,¬Y ,Z} we can resolve the in-termediate clause {¬A,¬Y}. An additionalresolution step results in the conflict clauseC = {¬A,¬X}.

Properties of conflict clause C:

F |= C

F ∪ ¬C `UP ⊥D /∈ F , ∀D ⊆ C


Unique Implication Points (UIP)In general: several possibilities to learn a clause from an implicationgraph. A learned clause can be resolved for every cut in the theimplication graph.

Graphic from Beame et al.: “Understandingthe Power of Clause Learning”

UIP is a dominator in theimplication graph

A node v is a dominator(for�), if all paths fromthe last decision to�contain v

FirstUIP: “first”dominator (seen fromconflict side)


Common Implementation (1UIP LearningScheme)

FirstUIP-clause: resolve conflicting clause and reason clauses untilonly a single literal of the current decision level remains

Advantage: Stopping at a UIP always leads to an asserting clause.Algorithm becomes easier: backtrack until clause becomes asserting

Undo all decisions until that level where the learned clause becomesasserting. The assertion level is the second highest level in a conflictclause.


Backtracking with Clause LearningIn our previous example the conflict clause (1UIP) with backtracking toassertion level changes the decision tree like this:

A

B

C

X

[Y=1]

[Z=1, Z=0]�

[Y=0]

[X=0]

[Z=1, Z=0]�

[ B = 1 ]

1 1 0

1

1

1

F = {{A,B}, {B,C},{¬A,¬X ,Y},{¬A,X ,Z},{¬A,¬Y ,Z},{¬A,X ,¬Z},{¬A,¬Y ,¬Z}}

Trail: A,B,C,X ,Y ,ZConflicting Clause: {¬A,¬Y ,¬Z}Conflict Clause (1UIP): {¬A,¬Y}


Backtracking with Clause LearningIn our previous example the conflict clause (1UIP) with backtracking toassertion level changes the decision tree like this:

A

B

C

X

[Y=1]

[Z=1, Z=0]�

[Y=0]

[X=0]

[Z=1, Z=0]�

[ B = 1 ]

1 1 0

1

1

1

F = {{A,B}, {B,C},{¬A,¬X ,Y},{¬A,X ,Z},{¬A,¬Y ,Z},{¬A,X ,¬Z},{¬A,¬Y ,¬Z}{¬A,¬Y}}

Trail: A,¬Y ,¬X ,ZConflicting Clause: {¬A,X ,¬Z}


Clause Database Maintenance

Without restrictions the clause database grows exponentially. Besides therisk of running out of memory this also slows down unit propagation.Clause database is regularly cleaned up. Quality measures for clauses(e.g. clause size, clause activity, LBD value)


CDCL Algorithm

Data: Formula FResult: SAT / UNSAT

1 dl ← 02 V ← ∅3 if unit propagation (F, V) == CONFLICT then4 return UNSAT

5 while not all variables assigned do6 (x , l)← pick branching literal7 dl ← dl + 18 V ← ∪(x , l)9 if unit propagation (F, V) == CONFLICT then

10 bl← analyze conflict11 if bl < 0 then12 return UNSAT

13 else14 backtrack to bl15 dl← bl

16 return SAT


Resolution Proof

With Clause Learning a complete resolution refutation can be derived withCDCL

proof can serve as a certificate for validating the correctness of theSAT solver

resolution refutations based on clause learning find key practicalapplications (e.g. model checking)

can help to determine minimally unsatisfiable subsets in anunsatisfiable formula


NEWS: Boolean Pythagorean triplesproblem was recently cracked using SATtechnology

The problem asks whether it is possible to colour each positive integereither red or blue, so that no trio of integers a, b and c that satisfyPythagoras famous equation a2 + b2 = c2 are all the same colour.

Solutions exist for numbers up to 7824

Proof that from 7825 onwards there is no such coloring

"Largest math proof" ever built (200 TB)

Paper will be presented at the SAT conference 2016.


References

Phase Saving: “A Lightweight Component Caching Scheme forSatisfiability Solvers” (Pipatsrisawat and Darwiche)

Clause Learning: “Understanding the Power of Clause Learning”(Beame et al.)

Two-Coloring of Pythagorean Triplets:http://www.nature.com/news/

two-hundred-terabyte-maths-proof-is-largest-ever-1.

19990


http://www.nature.com/news/two-hundred-terabyte-maths-proof-is-largest-ever-1.19990



practical sat solving - kit · 2016. 5. 31. · sat solver resolution refutations based on clause...

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