My Slides for a Course on Satisfiability

Home , Negation normal form, Resolution (logic), Satisfiability, Tautology (logic)

Satisﬁability

Victor W. Marek

Computer Science

University of Kentucky

Spring Semester 2005

Satisﬁability – p.1/97 Satisﬁability story

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Á Satisﬁability - invented in the of XX century by philosophers and mathematicians (Wittgenstein, Tarski)

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Á Shannon (late ) applications to what was then known as electrical engineering

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Á Fundamental developments: and – both mathematics of it, and fundamental algorithms

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Á - progress in computing speed and solving moderately large problems

Á Emergence of “killer applications” in Computer Engineering, Bounded Model Checking

Satisﬁability – p.2/97 Current situation

Á SAT solvers as a class of software

Á Solving large cases generated by industrial applications

Á Vibrant area of research both in Computer Science and in Computer Engineering

¯ Various CS meetings (SAT, AAAI, CP)

¯ Various CE meetings (CAV, FMCAD, DATE)

¯ CE meetings Stressing applications

Satisﬁability – p.3/97 This Course

Á Providing mathematical and computer science foundations of SAT

¯ General mathematical foundations

¯ Two- and Three- valued logics

¯ Complete sets of functors

¯ Normal forms (including ite)

¯ Compactness of propositional logic

¯ Resolution rule, completeness of resolution, semantical resolution

¯ Fundamental algorithms for SAT

¯ Craig Lemma

Satisﬁability – p.4/97 And if there is enough of time and will...

Á Easy cases of SAT

¯ Horn

¯ 2SAT

¯ Linear formulas

Á And if there is time (but notes will be provided anyway)

¯ Expressing runs of polynomial-time NDTM as SAT, NP completeness

¯ “Mix and match”

¯ Learning in SAT, partial closure under resolution

¯ Bounded Model Checking

Satisﬁability – p.5/97 Various remarks

Á Expected involvement of Dr. Truszczynski

Á An extensive set of notes (> 200 pages), covering most of topics will be provided f.o.c. to registered students (in installments)

Á No claim to absolute correctness made

Á Syllabus: http://www.cs.uky.edu/ marek/htmldid.dir/686.html

Á Homeworks every other week

Á Midterm

Á Individually-negotiated project a ¯ Write your own SAT solver

¯ Modify Chaff

¯ Use SAT solver for some reasonable task

Á Questions?

aHic Rhodes, Hic Salta

Satisﬁability – p.6/97 Basic concepts

Á Sets and operations on sets

Á Relations

Á Partial orderings (posets)

Á Elements’ classiﬁcation

Á Lattices

Á Boolean Algebras

Á Chains in posets, Zorn Lemma

Á Well-orderings, ordinals, induction

Á Inductive proofs

Á Inductive deﬁnability

Satisﬁability – p.7/97 Fixpoint theorem

Á Complete Lattices

Á Monotone operators (functions) in lattices

Á (Knaster-Tarski Fixpoint Theorem) If is a complete lattice and ! is

monotone operator in then possesses a ﬁxpoint. In fact ﬁxpoints of form a

complete lattice under the ordering of , and thus there is a least and largest

ﬁxpoint of

Á Continuous monotone functions

Á The least ﬁxpoint (but not the largest ﬁxpoint) of a continuous operator reached

in or less steps

Á Generalizations of ﬁxpoint theorem

Satisﬁability – p.8/97 Syntax of propositional logic

Á Variables of some (problem-dependent) set Var

Á For each set Var a separate propositional logic

Inductive deﬁnition of the set of formulas Î aÖ

Á Thinking about formulas as binary trees, the rank of formula

Satisﬁability – p.9/97 Semantics

Á Valuations of variables

Á Partial valuations of variables

Á Two-valued logic and valuations

Á Type Bool

Á Tables for operations in Bool

Á Valuations acting on formulas

Á Valuations uniquely extend from variables to formulas

Á Characterizing valuations as complete sets of literals

Á Characterizing valuations as sets of variables

Two-valued truth function ¾

Á Satisfaction relation j

Á Consistent theories

Satisﬁability – p.10/97 Localization and joint-consistency

Á Interactions between sets of variables

Á Restrictions of valuations

Á ¾ Ä

Localization theorem: If ½ ¾ and Î aÖ , is a valuation of ¾ ,

¼ ¼

j j j

is the restriction of to ½ then if and only if

Á Complete sets of formulas

Á Lemma: Complete sets of formulas determine valuations and conversely, valuations determine completes sets of formulas

(Robinson joint-consistency) Let ½ , ¾ are two sets of formulas in ½ , ¾

resp. Let us assume that ½ ¾ and that both ½ , ¾ are complete

[

for and coincide. Then ½ ¾ is consistent

Satisﬁability – p.11/97 Partial valuations, 3-valued logic

Á Need for partial valuations?

Á Partial valuation as complete valuations but in f g

Á Post ordering and Kleene ordering in the set f g

Á Product ordering

Á f g

Post ordering and Kleene ordering in the multiple-copies product of ,

and

Á Getting Kleene and Post orderings of partial valuations

Á Valuations as maximal partial valuations

Satisﬁability – p.12/97 Tables for Kleene 3-valued logic

Three-valued truth function ¿

¾ and ¿ coincide if is a (two-valued) valuation

If then for every formula , ¿ ¿

Á Autarkies

Á Fundamental effect of this result in SAT

Á Restriction result for 3-valued valuations

Satisﬁability – p.13/97 Tautologies and Satisﬁability

Á A tautology - formula true under all valuation of its variables

Á Satisﬁable formulas

Á A formula is satisﬁable if and only if : is not a tautology

Á Consequences of this fact: satisﬁability checkers as tautology checkers

Á Common tautologies

Á How many are there tautologies?

Satisﬁability – p.14/97 Substitutions to formulas

Á Substitution

Á Valuations acting on substitutions

Á Substitution Lemma ¾ : Let f g be a formula in propositional

½ Ñ

h i

variables ½ and let ½ be a sequence of propositional

formulas. Let be a valuation of all variables occurring in ½ . Finally,

¼ ¼

let be a valuation of variables ½ deﬁned by ,

. Then

Satisﬁability – p.15/97 Substitutions to formulas, cont’d

Let be a tautology, with variables of among ½ . Then for every

h i

choice of formulas ½ , the formula is a tautology.

Á There are inﬁnitely many tautologies

Satisﬁability – p.16/97 Lindenbaum Algebra

Á Relation : if for all valuations (of all variables occurring in , )

Á is an equivalence relation

Á iff the formula is a tautology

Á We can form the cosets

Á Operations in

Á Independence from the choice of representatives

Á Lindenbaum Algebra

Á Lindenbaum Algebra is a Boolean Algebra

Satisﬁability – p.17/97 Permutations of atoms and literals

Á Permutations of atoms

Á Permutations of literals, consistent permutations of literals

Á Shifts

Á Consistent permutations form a group

Á Decomposing consistent permutations of literals into shifts and permutation of variables

Á There is ¡ of consistent permutations of literals over a set of size

Á Consistent permutations of literals (and thus permutations of variables) preserve completeness of sets of literals

Á Every consistent and complete set of literals can be mapped onto any other consistent and complete set of literals by a suitably chosen consistent permutation of literals

Satisﬁability – p.18/97 Permutations and formulas

Á Permutations act on formulas

Á (Permutation Theorem) If is a formula, is a valuation and a consistent

permutation of literals then j if and only if j

Á Set-representation of valuations and permutation

Satisﬁability – p.19/97 Semantical consequence

Á Operation Cn, entailment of the (sets of) formulas

f 8 j µ j g

Á is an operator in the complete Boolean algebra of sets of formulas

Á For all sets ,

Á is monotone and idempotent

Á is continuous (but we have no means to prove it, yet)

Á ; consists of tautologies and nothing else

Satisﬁability – p.20/97 Deduction Theorem

Á Implication functor and consequence operation

Á The following are equivalent

µ ¾

¾ [ f g

Satisﬁability – p.21/97 Operations Mod and Th

Á From sets of formulas to valuations: f j g

Á From sets of valuations to formulas f for all ¾ , j g

Á Connection: Let be a valuation and a set of valuations. Then

¾ if and only if for every ﬁnite set of variables there is a

¾ j j

valuation such that .

Á Let be a ﬁnite set of propositional variables. Let be a collections of

valuations of set and be a valuation of . Then j if and

only if ¾ .

Satisﬁability – p.22/97 Functors and formulas

Á An -ary functor is a -ary function in

Á There are n-ary functors

Á Table - Boolean function of ﬁnite number of variables

Á Altogether there is inﬁnitely many tables

Á Given a set C of names for functors we can form - the set of formulas

based on C (obvious syntactic restrictions)

Satisﬁability – p.23/97 Completeness of sets of functors

Á Valuations as before, functions on into Bool

Á Assigning tables to well-formed formulas (i.e. trees, but no longer binary trees)

table associated with

Á C ¾

is complete if every table is equal to for some

Satisﬁability – p.24/97 Completeness of sets of functors, cont’d

Á C C C C C C

If ½ are two sets of functors, is complete and ½ then ½ is also complete.

Á The set f: ^ _g is a complete set of functors.

Á Thus f: ^ _ µ g is a complete set of functors.

Á C C C

Let ½ and ¾ be two sets of functors. Let us assume that ¾ is a complete set of

¾ C ¾

functors, and that for every functor ¾ there is a formula such

that the table is identical with the table . Then ½ is also a complete set of functors.

Á C C C

Let ½ and ¾ be two sets of functors. Let us assume that ¾ is a complete set of

¾ C ¾

functors, and that for every functor ¾ there is a formula such

that the formula ½ ½ is a tautology. Then ½ is a complete set of functors.

Satisﬁability – p.25/97 Completeness of sets of functors, cont’d

Á f: ^g is a complete set of functors

Á f: _g is a complete set of functors

Á fµg is a complete set of functors.

Á But in this last case we allowed use of ?

Á f g is complete, but use of constants needed

Á Beautiful property of ite:

Let be a propositional formula and let be a variable. Then for all variables

and formulas and the equivalence

is a tautology.

Satisﬁability – p.26/97 More on completeness

Á The set f^ _g is not a complete set of functors, even in the weak sense.

Á The set fg is not complete

Satisﬁability – p.27/97 Normal Forms, starting with negation

Á Negation normal form, pushing negation downwards

Á Double negation rewrite rule

Á De Morgan laws as rewrite rules

Á Canonical negation normal form

Satisﬁability – p.28/97 Occurrences of variables

Á Inductive deﬁnition of positive and negative occurrences

Á A variable may occur both positively and negatively in a formula

Á Canonical negative normal form preserves both positive and negative occurrences of variables

Satisﬁability – p.29/97 Alternative way of getting the occurrences result

Á Treating formulas as trees

Á Counting negations on the unique path from the leaf to the root

Á Positive occurrrence if this number is even, odd occurrence o/w

Á Rewrite rules, when interpreted as operations on trees preserve even and odd number of occurrences on paths

Á In number of occurrences or

Satisﬁability – p.30/97 cDNF, cCNF

Á Canonical disjunctive and conjunctive normal forms:

Á Four distributive rules

^ _ ^ _ ^

_ ^ ^ _ ^

_ ^ _ ^ _

^ _ _ ^ _

Satisﬁability – p.31/97 Handling cDNF

Á Formula is already in cNN form

¯ if is a variable

¯ : : if is a variable

½ ½ ½ ¾ ¾ ¾

_ _ _ _ _ _

Assuming ½ and

½ ¾ ½ ¾

½ ¾

¾ , deﬁne

½ ¾

^ ^

1. ½ ¾

½ ¾

_ _

2. ½ ¾ ½ ¾

Á For every formula in negation normal form, the formula is in

disjunctive normal form. Moreover, .

Á (Preparing for the complete DNF).

For every formula and a variable there exist two formulas ,

such that

1. ,

^ _ : ^

2. The formula ½ ¾ is a tautology.

Satisﬁability – p.32/97 Conjunctive Normal Form

Á Assuming in cNN form

Á if is a variable

Á : : if is a variable

½ ½ ½ ¾ ¾ ¾

Á ^ ^ ^ ^ ^ ^

Assuming ½ and

½ ¾ ½ ¾

½ ¾

¾ , deﬁne

½ ¾

_ _

1. ½ ¾

½ ¾

^ ^

2. ½ ¾ ½ ¾

Á For every formula in negation normal form, the formula is in

conjunctive normal form. Moreover, .

Satisﬁability – p.33/97 What if only positive (negative) occurrences?

Á If a variable has only positive occurences in , f g does not have to be an autarky, but...

Á Let be a set of formulas in which a variable has only positive (resp. negative)

occurrences. Then is satisﬁable if and only if [ f g (resp. [ f: g ) is

satisfiable. In other words, is satisfiable if and only if it is satisfiable by a

valuation such that (resp. ).

Á Look at the proof: it shows equivalent formulas may not have same autarkies

Satisﬁability – p.34/97 Reduced Normal Forms

Á _ _ _ : _ _ :

A clause ½ ½ is a tautology if and only if for

some , and , ,

Á j

Given two clauses ½ and ¾ , ½ ¾ if and only if all literals occurring in ½

occur in ¾ , that is ½ subsumes ¾ .

Á Reduced CNF: no subsumption between clauses

Á Subsumption can be eliminated in polynomial time (but it still may be too much work)

Á Similar result for DNF

Á CNF machine a device to convert to

Á CNF machine as a DNF machine

Satisﬁability – p.35/97 Complete Normal Forms

Á A minterm is an elementary conjunction containing one literal from each set

f : g

Á For every formula and every minterm , either the formula ^ is a

tautology, or ^ is false.

Given a valuation of ﬁnite set of variables, is conjunction of literals such

that

¾ ÅÓ d ´ µ

Each entails

For every formula , is a tautology. Up to the order of variables, and the

listing of minterms the representation is unique.

Satisﬁability – p.36/97 More on canonical forms

Á Minterms: just rows in the table where the value is

Á Maxclauses: clauses non-tautological clauses mentioning all variables

Formula - conjunction of maxclauses entailed by

For every formula , is a tautology. Up to the order of variables, and the

listing of maxclauses the representation is unique.

Satisﬁability – p.37/97 Consequences for Lindenbaum Algebra

Á Equivalence classes of minterms are atoms in the Lindenbaum Algebra

Á Since there are minterms, Lindedenbaum algebra has atoms ( - number of variables)

Á Hence for ﬁnite set the Lindenbaum algebra of is a ﬁnite Boolean algebra

Á Not every ﬁnite Boolean algebra is Lindenbaum algebra

Á But if a ﬁnite Boolean algebra has atoms then it is a Lindenbaum algebra

Satisﬁability – p.38/97 Other normal forms

Á Implication normal form (only implication, ? and variables)

Á Hence by adding additional variables we can reduce every formula to collection of implications

Á if-then-else normal form

Satisﬁability – p.39/97 Compactness of propositional logic

Á König Lemma: an infinite, finitely splitting tree possesses an infinite branch

Á (Compactness Theorem) If a set of formulas is unsatisﬁable then there is a

ﬁnite subset ¼ such that is unsatisﬁable

Á Equivalently: when all finite subsets of are satisfiable then is satisfiable

Á Constructing a tree associated with a family of clauses

Á Corollary: The operator is continuous

Satisﬁability – p.40/97 Resolution

Á Clausal logic: logic where only formulas are clauses

Á Alternative approach: negation and inﬁnitely many functors, -ary disjunctions, each with its own table

Á Such logic expresses the same theories as the full propositional logic

Á Since such disjunctions are commutative w.r.t. all permutations we treat clauses as sets of literals, no repetition

Á Fundamental operation on clauses: resolution

_ _

½ ¾

Á Resolution rule is sound

Satisﬁability – p.41/97 Closure under Resolution

Á Given a set of clauses there is a least set of clauses such that

2. Whenever ½ and ¾ are two clauses in and the operation ½ ¾ is

executable and results in a non-tautological clause then ½ ¾

belongs to .

Operator (in the complete lattice of subsets of the set of clauses)

[ f ¾ [

½ ¾ ½ ¾

and ½ ¾ is deﬁned and non-tautological

has a least ﬁxpoint. This is the closure under resolution

Satisﬁability – p.42/97 Derivations, proof-theoretic approach

Á Derivation of a clause: a labeled binary tree: leaves labeled with elements of , internal nodes labeled with resolvents

Á Proof-theorists invert those trees. Root is at the bottom

Á As every ordered tree, such trees can be represented as strings (with depth-ﬁrst traversal)

Á In this way we can deﬁne proofs as sequences of clauses: 1. Conclusion of the rule application application occurs always later than premises

2. If a clause occurs in the sequence then it belongs to or is a conclusion of an application of resolution to earlier elements of the clause 3. The proven clause is last element of the sequence

Á Both approaches (trees, proofs) are equivalent

set of clauses that have proof

Satisﬁability – p.43/97 Is resolution complete?

Á Resolution alone is not complete. That is there are clauses semantically entailed, but not provable by resolution

Á Reason: resolution does not add variables

Á Example: f _ g , _ _ . Clearly j , but the only thing we

can prove (using resolution) is the only clause in

Satisﬁability – p.44/97 Minimal Resolution Consequence

Á Elements of - resolution consequences

Á Inclusion-minimal elements of - minimal resolution consequences

Á If is a resolution consequence of , then there must be minimal resolution

consequence of included in

Satisﬁability – p.45/97 Subsumption rule

Á Subsumption rule:

Á Subsumption rule is sound

Á We can deﬁne the notion of derivation (tree) using subsumption and resolution

Á We can deﬁne the the notion of proof (sequence of clauses)

Á Whatever can be proved with one can be proved with the other

Satisﬁability – p.46/97 Quine theorem, completeness

Á A non-tautological clause is a consequence of a set of clauses if and only if

there is a Resolution consequence of , , such that .

Á A non-tautological clause is a consequence of a CNF if and only if for some

minimal Resolution consequence of , .

Á A CNF is satisﬁable if and only if ; ¾ .

Á (Quine Theorem Proof system consisting of Resolution and Subsumption rules is

complete for clausal logic. That is, given a set of clauses and a clause ,

j if and only if there exists a derivation (using Resolution and

Subsumption rules) that proves

Á We can limit the proof to admit a single application of Subsumption.

Satisﬁability – p.47/97 Resolution refutation

Á Given a clause ,

: f ¾ g

Á If is a clause then : is a set of unit clauses

Á We say that follows from by Resolution refutation if the closure under

Resolution of [ : contains an empty clause (i.e. is inconsistent).

Á Let be a CNF, and let be a clause. Then j if and only if follows

from by Resolution refutation.

Satisﬁability – p.48/97 The basis of resolution consequences

Á Basis of

3. forms an antichain, i.e. for ½ ¾ , if ½ ¾ then ½ ¾

4. Every element of is subsumed by some element of

Á Let be a set of clauses. Then possesses a unique basis

Á How to compute a basis w/o computing the entire ?

Satisﬁability – p.49/97 Basis, cont’d

Á Start with an CNF . Due to the discussion above we can assume that is subsumption-free.

Non-deterministically, select from a pair of resolvable clauses ½ and ¾ which has not been previously resolved.

If ½ ¾ is subsumed by some other clause in or is a tautology, do nothing and select next pair.

If ½ ¾ is not subsumed by some other clause in , do two things:

(a) Compute f ¾ g . Eliminate from all clauses subsumed

by , that is Ò

(b) Set [ f g .

Á Do this until does not change.

Satisﬁability – p.50/97 Basis, cont’d

Á Two invariants:

¯ Antichain

¯ Consequences

Á This implies termination

Satisﬁability – p.51/97 Another preprocessing for query ans.

Á Pruning , given :

f ¾ and can not be resolved with g

Á j if and only if j

Satisﬁability – p.52/97 Davis-Putnam reduct

The reduct is

f Ò f g ¾ ^ ¾ g

Á The set splits into three sets: of clauses mentioning , of clauses mentioning

and those not mentioning j j at all

arises from the second and the third

Á Let be a set of non-tautological clauses, and let be a literal. Then is

unsatisﬁable if and only if both and are unsatisﬁable.

Á Dual form: Let be a set of non-tautological clauses, and let be a literal. Then

is satisﬁable if and only if at least one of , is satisﬁable.

Satisﬁability – p.53/97 Free literal

Á is free for if no clause of contain

Á Let be a set of clauses. If is a free literal in then is satisﬁable if and only

if is satisﬁable.

Moreover, if is a valuation satisfying , then a valuation deﬁned by

¾ f g

satisﬁes .

Satisﬁability – p.54/97 Semantic Resolution

given. Let us ﬁx . splits in ¼ – clauses of unsatisﬁed by , ½ clauses of

satisﬁed by

If unsatisﬁable then ¼ , ½ nonempty

Á Ordering variables induces order of literals in each non-tautological clauses (via ordering of underlying variables)

Á Semantic resolution rule (remember ﬁxed) applied to an ordered pair of

clauses of

(a)

(b)

Satisﬁability – p.55/97 Semantic Resolution, cont’d

Á ¡

Operator

f ¾ _ 9 ¾ [ ^

¾ [ ^ ^ ^ g

The operator is monotone. Thus it possesses a least ﬁxpoint.

Á Completeness of semantic resolution; given , , , is unsatisﬁable if and

;

only if contains

Satisﬁability – p.56/97 Testing satisﬁability

Á Table method

Á Tableaux

Á Clausal Logic: Davis-Putnam two-phase algorithm

Á Clausal Logic: Davis-Putnam-Logemann-Loveland algorithm

Satisﬁability – p.57/97 Hintikka Sets

Á No immediate contradiction (? , no contradictory atoms)

Á Standard decomposition principles for : ^ _

Á Hintikka set is always consistent

Satisﬁability – p.58/97 Tableaux

Á Binary trees with nodes decorated with signed formulas

Á Tableau for a theory

Á Open and closed branches

Á Managing branches via expansion rules

Á Finished tableaux

Satisﬁability – p.59/97 Fundamental tableaux theorem

Á Open branch in a ﬁnished tableau is a Hintikka set

Á Thus if the a tableau for has an open branch then is satisﬁable

Á Canonical tableau for

Á is satisﬁable if and only if the canonical ﬁnished tableau for has an open branch

Á is unsatisﬁable if and only if the canonical ﬁnished tableau for has no open branches

Á (Completeness for tableaux). j if and only if canonical ﬁnished tableau for

[ f: g is closed

Satisﬁability – p.60/97 Variable elimination

Á Operation , handling free literals

Variable elimination resolution

¯ Select variable

¯ If , or if is free in ,

¯ O/w clauses without occurrence of , and clauses obtained by resolving w.r.t

, tautologies eliminated

Thus has no occurrence of

Á Clauses of that have maintained separately

Satisﬁability – p.61/97 Variable Elimination

If satisﬁes then satisﬁes

Á We will assume that we have some heuristic function selecting a

variable from an input set of variables

Á This heuristic funtion guides variable elimination resolution

Á Decomposition of w.r.t :

¯ f ¾ : ¾ g ,

¯ f ¾ ¾ g ,

f ¾ : ¾ g

Satisﬁability – p.62/97 Introducing DP algorithm, part 1

Á Selection function assumed

Á h i h i h i

Three sequences: , ,

¼ ½

[

¼ , ¼ Î aÖ , ¼ , ½ ¼

F ¼ ¼ ¼

;

If , halt.

¼ ½

[

O/w Î aÖ , ,

·½

Á ; f;g

Possible outcomes: ½ or ½

In the latter case is unsatisﬁable because is included in

Á We can stop at any moment once ; computed

Satisﬁability – p.63/97 Going back, part II of DP algorithm

Á We assume part I did not return the string “input theory unsatisﬁable"

is empty. Why?

Á Two possible rasons:

¯ The last variable selected occurred either positively in all clauses or negatively in all clauses

¯ All we produced during the resolution w.r.t last variables were

tautologies, so ½ is empty

Ò ½

Satisﬁability – p.64/97 Going back, part II of DP algorithm, cont’d

Á The idea: construct a sequence of partial valuations, going backwards

This partial valuation deﬁned on all variables occurring in ½ and nothing else

Base Case 1. ½ free in ½ . We set ½ , all other variables still

occurring in ½ zeroed (but we do not have to do so, b.t.w.)

Á :

Base Case 2. ½ free in ½ . We set ½ , all other variables still

occurring in ½ zeroed (but we do not have to do so, b.t.w.)

Base case 3. Some occurrences of ½ in ½ positive, some negative. Tricky!

Satisﬁability – p.65/97 Going back, part II of DP algorithm, cont’d

Á Inductive case

Á Reduct w.r.t. a partial valuation - eliminating clauses that are already satisﬁed, eliminating literals that can not be used for satisfaction

Inductive assumption is that we satisﬁed all layers , with a

valuation

Goal: Satisfy ½

We reduce ½ by current . Variables of current no longer occur in the reduced CNF

Á We deﬁne an extension of current to new so that is deﬁned on all

variables of ½ , and those occurring in later and nothing else. How?

Satisﬁability – p.66/97 Going back, part II of DP algorithm, cont’d

Á Four, not three cases possible

Reduct is empty. That is all clauses in ½ already satisﬁed. We zero

any variable occurring in ½ but not in current

Literal ½ is free in the reduct. We set the value of ½ to , zero all other variable

½ occurs both positively and negatively in ½ . We proceed similarly to case 3 of base

Á Proof of correctness of this procedure

Satisﬁability – p.67/97 Wrapping up DP

Á The DP algorithm is complete. That is, if is a set of clauses, and

¡ is a selection function, then:

1. If is satisﬁable then the DP algorithm returns on the input and

the selection function a valuation satisfying

2. If is unsatisﬁable, then the DP returns the string ‘input formula unsatisﬁable’.

Á In the second phase we have plenty of leverage, we do not have to use anything already computed, but can plug in other partial valuations, as long as they satisfy the rest

Satisﬁability – p.68/97 The tree of partial valuations

Á We assume a ﬁnite set of clauses

Á We arrange partial valuation in a search tree (i.e. label full binary tree - but see below - with partial valuations)

Á We assume that, like in DP algorithm we have a heuristic function ¡ that

assigns to a partial valuation a variable outside the domain of (providing

)

Á [ f g

Then node has two children: ¼ , ½ . The ﬁrst one labeled by ,

the other labeled by [ f: g

Satisﬁability – p.69/97 Pruning, BCP

Á Given a set of literals

f _ _ _ ¾ ¾ g

There is ½ ½

Á ¡ is a monotone operator in the complete lattice of sets of literals

Á is the least ﬁxpoint of the operator ¡

Á Proof-theoretic representation of BCP, unit resolution.

Á Both approaches equivalent

Satisﬁability – p.70/97 DPLL algorithm

Á Let be a CNF. Then is satisﬁable if and only if consistent and the

reduct of w.r.t. is satisﬁable

Á An obvious observation: Given a CNF , is satisﬁable if and only if [ f g

is satisﬁable or [ f: g is satisﬁable

Á If is the reduct of by means of then ;

Á But often it may happen that ; but [ f g 6 ;

Á DPLL: systematic backtracking search of the tree of partial valuations, with BCP as a pruning algorithm and user-provide selection function

Á DPLL is complete

Á Improvements to DPLL

Satisﬁability – p.71/97 Craig Lemma

Á Given a valid implication µ , an interpolant is any formula such that both

µ and µ are valid

Á Craig Lemma: Given a valid implication µ , there is an interpolant so that

variables in occur both in and in

Á Stronger form will be shown

Á Recent reports of the use of Craig Lemma in Bounded Model Checking

Satisﬁability – p.72/97 Few lemmas etc.

Á _ _ µ

If is DNF, ½ then is a tautology if and only if for all

, , is a tautology

Á j

If is an elementary conjunction and set of variables, then is the result

of eliminating from literals based on variables not in

Á Let be a noncontradictory elementary conjunction, let be an arbitrary

µ j µ

formula. Then is a tautology if and only if Î aÖ is a tautology

Á µ _ _

Thus, assuming is a tautology, ½ a DNF, all non-contradictory. Then

j _ j µ

½ Î aÖ Î aÖ

³ ³

is a tatutolgy

Satisﬁability – p.73/97 Craig Lemma

Á For every formula , the formula

_ _ µ j _ _ j

½ ½

Î aÖ ´ µ Î aÖ ´ µ

is a tautology.

Á If µ is a tautology, then there exists a formula such that

\ and such that both µ and µ are tautologies

Satisﬁability – p.74/97 Improving interpolant

Á If is a formula with only positive occurrences of variable then there are two

formulas: ½ and ¾ with no occurrence of so that

¸ ^ _

½ ¾

is a tautology

Á If is a formula with only negative occurrences of variable then there are two

formulas: ½ and ¾ with no occurrence of so that

¸ : ^ _

½ ¾

is a tautology

Satisﬁability – p.75/97 Eliminating literals

Á ^ ^ µ

Assume that ½ is a tautology, and ½ does not occur in . Then

^ ^ µ

¾ is a tautology

Á Let be an noncontradictory elementary conjunction, and let us assume that

µ is a tautology. Let be the conjunction of those literals in which

occur in . Then µ is a tautology

Á (Craig Lemma, strong form) If µ is a tautology then there is an interpolant

with the property that whenever a variable occurs in then it occurs in (and

also in ) with the same polarities.

Satisﬁability – p.76/97 Is DNF in predecessor needed?

Á Of course not, if we have CNF of the consequent we can do the same

Á We can either transform µ into : µ : and use the previos technique, or prove analogous lemmata for CNF in the consequent

Á We can also compute a "canonical candidate for interpolant”

Satisﬁability – p.77/97 Satisfaction for positive and negative formulas

Á Formula is positive if it has no negative occurrences of variables

Á Formula is negative if it has no positive occurrences of variables

Á Valuation ½ satisﬁes positive formulas and so sets of positive formulas are satisﬁable.

Á Valuation ¼ satisﬁes negative formulas and so sets of negative formulas are satisﬁable.

Satisﬁability – p.78/97 Satisfying Horn formulas

Á A Horn clause: at most one positive literal

Á Basic classiﬁcation of Horn clauses

¯ facts

¯ program clauses

¯ constraints

Satisﬁability – p.79/97 Horn theories and families closed under intersection

Á A family F of sets is closed under intersections if for every nonempty Y , Y X ,

Y belongs to X

Á Families of sets closed under intersections appear all over CS and Mathematics

Á We will use representation of valuations as sets of variables

Á Let be a ﬁnite set of atoms and let Ë È be a nonempty family of sets.

Then Ë is of the form for some collection of Horn clauses over if

and only if Ë is closed under intersections.

Satisﬁability – p.80/97 Program clauses and constraints

Á Program clause: Horn clause with exactly one positive literal

Á Constraint clause: Horn clause with no positive literals

Á CNFs consisting of program clauses are satisﬁable

Á CNFs consisting of program clauses has a least model

Á CNFs consisting of constraint clauses are satisﬁable

Satisﬁability – p.81/97 Model existence for Horn theories

Each Horn theory decomposes into program part ½ and constraint part ¾

If is Horn, then is satisﬁable if and only if least model of ½ satisﬁes ¾

Associating an operator with the Horn program

is monotone and continuous (regardless of the size of )

Least model of the Horn program coincides with the least ﬁxpoint of

Satisﬁability – p.82/97 Representability of monotone operators

Á If is ﬁnite then for every monotone operator in È there is program

such that

Á Dowling-Gallier algorithm for computation of least model of for programs

Á Testing satisﬁability of Horn in linear time

Á Completeness of unit resolution for querying the least model of Horn theory

Satisﬁability – p.83/97 Dual Horn formulas

Á Dual Horn clause – at most one negative literal

Á Permutation

Á Permutation is consistent and transforms Horn formulas to dual Horn and conversely

Á j if and only if Ò j

Á is complement operation

Á acts on families of sets, and transforrms families closed under intersections to families closed under unions

Satisﬁability – p.84/97 Dual Horn formulas

Á Various results on Horn theories are lifted to dual-Horn case

Á BUT: operator associated with dual-Horn theories is monotone

Á There is largest ﬁxpoint, and this is the largest model of

Á Unit resolution complete for dual-Horn theories, but now one of inputs can be required to be negative literal c

Satisﬁability – p.85/97 Krom formulas, 2SAT

Á 2-clause, or Krom clause, a clause with at most two literals

Á The set of 2-clauses is closed under resolution

Á There is 2-clauses

Á Thus in DP the space used by the algorithm (and thus time) is bounded by polynomial

Á DP solves the satisﬁability problem for sets of 2-clauses in polynomial time (but we will see better performance)

Satisﬁability – p.86/97 A closure property of sets of 2-clauses

Let be a 2-CNF, and let be a literal such that ¾ . Let

[ f g

, and let ½ . Then:

1. ½

2. If is consistent then: is consistent if and only if ½ is consistent

3. If is consistent then a satisfying valuation for can be computed from and a

satisfying valuation for ½ in linear time.

Satisﬁability – p.87/97 BCP and 2-SAT

Á \ 6 ;

A partial valuation touches clause if

Á If consists of 2-clauses then for every literal , if [ f g is

consistent then satisﬁes every clause in which touches

Á If consists of 2-clauses then is satisﬁable if and only if for every literal , at

least one of [ f g , [ f g is satisﬁable

Á Autarkies for Krom formulas

Satisﬁability – p.88/97 G

The graph F

Á After initialization the resulting formula is a subset of and each clause has exactly 2 literals

Vertices of : Literals of

Á _

Edges of : whenever belongs to , generate two edges: ,

is a directed graph

Satisﬁability – p.89/97 Strong components and 2-CNF

Á A strong component of a directed graph: for every ¾ , 6 , there is a

directed cycle containing both and

If are in same strongly connected component of then

[ f g [ f g

Á a 2-CNF, all clauses of length . Then is satisﬁable if and only if no strong

connected component of contains a pair of dual literals

Á Complexity of testing satisﬁability of 2-CNF

Satisﬁability – p.90/97 Renameable variants of classes of formulas

Á Given C , a class of CNFs, renameable C , all formulas which can be obtained from

formulas of by consistent permutations of literals

Á Shift permutation: does not change underlying variable, may only change sign

Á is renameable Horn if for some renaming , is Horn, thus if for some

shift , is Horn

Satisﬁability – p.91/97 Describing shift permutations

Á There is a one-to-one correspondence between shift permutations and sets of

atoms of the form

Á Representing shifting into Horn clauses Given a clause

_ _ _ : _ _ :

½ ½

Deﬁne a collection of 2-clauses consisting of three groups:

Group 1. ,

: _ :

Group 2. ,

: _

Group 3. , , .

Satisﬁability – p.92/97 Carrying on shifting

is 2-CNF

There is a one-to-one correspondence between valuations satisfying and shifts of F into a Horn CNF

Á Testing if is renameable Horn is polynomial, in fact quadratic in

Á is renameable Horn if and only if it is renameable dual Horn

Satisﬁability – p.93/97 Linear formulas

Á Operation ¨ , a.k.a. XOR

Á Ä Ç Ç hB ¨ ^ ? >i Z

The structure is a ﬁeld, called ¾

Á >

Linear equation over that ﬁeld: formula in ½ and possibly

Á Z

Special form of Gauss-Jordan elimination over ¾ – add it!

Á Issues with the number of equations

Á Triangular form of a set of equations (ﬁrst pass, Gauss)

Á Substitute (second pass, Jordan)

Satisﬁability – p.94/97 All you need is 3-CNF

Á 3-clauses

Á With additional variables (linear number of variables) we can reduce to 3-clauses.

Speciﬁcally: for every ﬁnite set of formulas there is a set of 3-clauses such

that , contains a special variable and there is a

one-to-one correspondence between satisfying valuations for and satisfying

valuations for that evaluate as

Satisﬁability – p.95/97 Combinatorial circuits

Á Treating a set of formulas as an input-output devise

Á Acyclic digraphs with at most two parents per node

Á Inputs and outputs

Á Circuit representing a collection of formulas (one output per formula)

Á 3-CNF representation of a circuit

Satisﬁability – p.96/97 Some complexity issues

Á The problem SAT (i.e. the language consisting of satisﬁable formuals) is NP-complete (proof in the notes, via coding of accepting computations of Turing machines where the length of operation is bounded by a ﬁxed polynomial in the length of the input)

Á Thus, the problem UNSAT is co-NP-complete

Á By our reduction, the problem 3-SAT (satisﬁability of the sets of clauses consisting of 3-clauses) is also NP-complete

Á Thus, the problem 3-UNSAT is also co-NP-complete

Á All nontrivial combinations of “easy” classes considered in this lecture also result in NP-complete classes (for instance: unions of Horn and dual-Horn are NP-complete)

Á But there is plenty of other “easy classes”

Satisﬁability – p.97/97