21 november 2006 foundations of logic and constraint programming 1 constraint (logic) programming an...

21 November 2006 Foundations of Logic and Constraint Programming 1

Constraint (Logic) Programming

Anoverview

• BooleanConstraintPropagation

• FiniteDomainsandConstraintNetworks

• ConsistencyCriteria

• MaintaningNodeConsistency

• MaintainingArcConsistency

[Dech04]RinaDechter,ConstraintProcessing,


Boolean Satisfiability

InmostBooleanproblems, theuseofa completesolver isquite innefficient,

sinceitmaintainsthesetofallsolutionsinasolvedform.

Alternatively, a Boolean satisfiability problem may be solved, usually much

moreefficiently,bysearchingforasolution.

In this case, two alternatives are available, with respect to the role of

constraints:

Constraints are used passively. Once the variable appearing in the

constraintareinstantiated,theconstraintischecked.Ifitisnotsatisfied,

thesearchbacktracks.

Constraintsareusedactively,toreducethedomainofthevariablesstill

toinstantiate.

Thesealternativesareillustratedbymodellingallconstraintsasclauses.


Boolean Clauses

Syntatically,

• A literal:variableoritsnegation

Variablesaredenotedbyuppercaseidentifiers,andnegationwith.

• Aclause:setofliterals.

Clausesarerepresentedwiththedisjunctionoperator,e.g.A B C

• ABoolean Satisfiabilty Problem:setofclauses.

Semantically,

• BooleanvariablesmaytakevaluesTrue / False(orT / For1 / 0).

• Anegation““reverses,ofcourse,thisvalue.

• A clause is interpreted as adisjunction of literals. Hence, a clause issatisfiedifat least oneofitsliteralstakesvalueTrue.

• Solvingaproblemcorresponds to findvalues to thevariablessuch thatallclausesaresatisfied.


Search Tree

• Everyproblemmayberepresentedasa(search) tree,whereeachnodehas

twochildren,correspondingtoavariableanditsnegationnotappearinginthe

pathfromtheroottothenode.

• Asolutionoftheproblem,acompleteasignmentofT/Fvaluestoallvariables,

correspondstoapathfromtheroottoaleaf,thatsatisfiesallconstraints.

• Thesearchprocessmaybeimplementedbytraversingthesearchtree.

• Acompletesearchwillvisit(ifnecessary)allleafs.

CC CC

B B

BB BB

C C

A A


Search Tree

• Agenerate and testapproachtoconstraintsolvingwillvisitallleafs(generate

potential solutions) and test the satisfiability of all clauses for the

correspondingassignmentsofvaluestovariables.

• Searchmay be improved bybacktracking - at each node (not only in the

leafs)theclauseswithvariablesdefinedaretestedforsatisfiability.

• In this form, by testing incomplete solutions, backtracking avoids the

“completion”ofmanyimpossiblepartialsolutions.

c1: A B

c2: A C

c3: B C

CC CC

B B

BB BB

C C

A A

c3 c2

c1

c3


Search Tree

• Still,backtrackinghandlesconstraintspassively.First,valuesareassignedto

variables;then,constraints(clauses)aretested.

• Alternatively,constraintsarehandledactivelywith“constraintpropagation”.

• IntheBooleanConstraintPropagation(BCP):

whenall literalsbut oneofaclauseareassignedFalse,thentheclause

becomesactiveandpropagatestheremaining literalisassignedTrue.

c1: A B

c2: A C

c3: B C

CC CC

B B

BB BB

C C

A AA = 1 B 1 (c1) C 1(c2) ? (c3)

C = 1 B 1 (c3)


• This propagation can be noticed in the 4-queens problem. Here are some

examples:

1. Propagate a 0-assignment

2. Propagate a 1

Boolean Constraint Propagation

1 1 0

X11 X12 X12 0

X21 X22 X23 X24 X24 1

1 0 000 0 0

000

1 0 000 0 0

000

1


• Hereisthecompletesearchprocesswithconstraintpropagation

Assign X11 Propagate X23 & Propagate

X24 & Propagate & Propagate

X12 & Propagate & Propagate & Propagate & Propagate

Boolean Constraint Propagation

1 1 0 0 0

00

0

0

0

0

1 0 0 0

00

0

0

0

0

100

0

0

1 0 0 0

0

0

0

0

0

0

1

0

0

0

1 0 0 0

0

0

0

0

0

0

1

0

0

0

1

0

1 000

0 0 0

00

0

1 000

0 0 0

00

0

1

0

0

1 000

0 0 0

00

0

1

0

01

0

1 000

0 0 0

00

0

1

0

01

0 1

Fail !X31 X32 X33 X34

Backtrack X23

Fail !X41 X42 X43 X44

Backtrack X24


Boolean Clauses in SICStus

• SICStusdoesnotprovideanyconstraintsolverforthesatisfiabilityofBoolean

clauses(SAT).

• Nevertheless,itcanbeeasilyimplementedasaninstanceofafinitedomains

solver,wherethedomainofthevariablesare{0,1}.

• A clause is thus converted into an inequality constraint over an arithmetic

expression.Anexampleillustratesthismodelling

• ClauseX Y Z

X + (1-Y) + Z 1 X – Y + Z 0

• Propagation 1:X = 0, Y = 1 0-1+Z 0 Z 1 Z 1

• Propagation 2:X = 0, Z = 0 0-Y+0 0 Y 0 Y 0

(seedetailsinqueens4_sat_fd.pl)


Why Finite Domains ?

Inmostproblems,andnotwithstandingthetheoreticalandpracticalinterestin

SAT,problemsaremorenaturallyspecifiedwithvariableswhosedomainisnot

restrictedto0/1values,butmaytakeafinitesetofvalues.

In particular, the N queens problem is more “easily” programmed with n

variables,oneforeachrow,whosevaluedenotesthecolumnthatthequeenof

that row must be in. The values allowed are, of course, the number of the

column,i.e.integersfrom1 to n.

Such specification, not only is more natural, but it also may lead to a

substantialincreaseintheefficiencyofthesolvingprocess.

FiniteDomain encodings of problemswheren variables can takek different

values implement a search spacewith a substantially different size from the

correspondingBooleancodingwhereeachvalue is representedbyadistinct

Booleanvariable.


Why Finite Domains ?

To compare the search spacesofBooleanandFiniteDomain encodings forproblemswherenvariablescantakekdifferentvalueswenotethat:

BooleanEncodings-k*nbooleanvariables(0/1).

SearchSpace=2kn.

FiniteDomains-nvariableseachwithkpossiblevalues

SearchSpace=kn

RatiooftheSearchSpaces

r=2kn/2log2k*n=2(k-log2(k))*n

Forexample,intheN-queensproblem,wehave

n=8k=8,andr=2(8-3)*8=2401012

n=16k=16,andr=2(16-4)*16=21921058


Finite Domains

Ofcourse, theratioof thesearchspacesthatareactuallysearchedisnotso

large,namelyduetoconstraintpropagation.Still,thedifferencesmaybequite

significant.

TheefficiencyoftheCLPprogramsoverfinitedomains(includingtheBoolean

case)maystillbeimprovedbymakinguseof

AdequatePropagation Algorithms

Appropriate Heuristics for selection of the Variable to label and the

Valuetoassign

Beforeinvestigatingtheseissues,anumberofconceptsareintroduced

ConstraintsandConstraint Network

Partial andTotal Solutions

SatisfactionandConsistency


Basic Concepts: Domains

Definition(Domain of a Variable):

Thedomainofavariableisthe(finite)setofvaluesthatcanbeassignedtothatvariable.

Given some variableX, its domainwill be usually referred to asdom(X) or,simply,Dx.

Example:TheNqueensproblemmaybemodelledbymeansofNvariables,X1 to Xn,allwiththedomainfrom1ton.

Dom(Xi) = {1,2, ..., n} or Xi :: 1..n.


Basic Concepts: Labels

Toformalisethenotionofthestateofavariable(i.e.itsassignmentwithoneof

thevaluesinitsdomain)wehavethefollowing

Definition(Label):

AlabelisaVariable-Valuepair,wheretheValueisoneoftheelements

ofthedomainoftheVariable.

Thenotionofapartialsolution,inwhichsomeofthevariablesoftheproblem

havealreadyassignedvalues,iscapturedbythefollowing

Definition (Compound Label):

Acompound labelisasetoflabelswithdistinctvariables.


Basic Concepts: Constraints

Wecomenowtotheformaldefinitionofaconstraint

Definition(Constraint):

Given a set of variables, aconstraint is a set of compound labels on

thesevariables.

Alternatively,aconstraintmaybedefinedsimplyasarelation,i.e.asubsetof

the cartesian product of the domains of the variables involved in that

constraint.

Forexample,givenaconstraintRijkinvolvingvariablesXi, XjandXk,then

Rijk dom(Xi) x dom(Xj) x dom(Xk)


Basic Concepts: Constraints

GivenaconstraintC,thesetofvariablesinvolvedinthatconstraintisdenoted

byvars(C).

Simetrically, thesetofconstraints inwhichvariableXparticipates isdenoted

bycons(X).

Noticethataconstraintisarelation,notafunction,sothatitisalwaysCij = Cji.

Inpractice,constraintsmaybespecifiedby

• Extension: through an explicit enumeration of the allowed compound

labels;

• Intension: through some predicate (or procedure) that determines the

allowedcompoundlabels.


Basic Concepts: Constraint Definition

Forexample, theconstraintC13 involvingX1and X3 in the4-queensproblem,maybespecified

By extension(labelform):

C13 = {{X1-1,X3-2},{X1-1,X3-4},{X1-2,X3-1},{X1-2,X3-3},

{X1-3,X3-2},{X1-3,X3-4},{X1-4,X3-1},{X1-4,X3-3}}.

or,intuple(relational)form,omittingthevariables

C13 = {<1,2>,<1,4>,<2,1>,<2,3>,<3,2>,<3,4>,<4,1>,<4,3>}.

By intension:

C13 = (X1 X3) (1+X1 3+X3) (3+X1 1+X3).


Basic Concepts: Constraint Arity

Definition(Constraint Arity):

The constraint arity of some constraint C is the number of variables

overwhichtheconstraintisdefined,i.e.thecardinalityofthesetVars(C).

Despite the fact that constraints may have an arbitrary arity, an important

subsetoftheconstraintsisthesetofbinary constraints.

Theimportanceofsuchconstraintsistwofold

Allconstraintsmaybeconvertedintobinaryconstraints

A number of concepts and algorithms are appropriate for these

constraints.


Basic Concepts: Binary Constraints

Conversion to Binary Constraints

Ann-aryconstraint,C,definedbyk compoundlabelsonitsvariablesX1 to Xn,

isequivalenttonbinaryconstraints,C0i,throughtheadditionofanewvariable,

X0,whosedomainistheset1 to k.

Inpractice,

The kn-arylabelsmaybesortedinsomearbitraryorder.

Each of the n binary constraints C0i relates the new variable X0 with the

variableXi.

Thecompoundlabel{Xi-vi, X0-j}belongstoconstraintC0iiffXi-vibelongstothe

j-thcompoundlabelthatdefinesC.


Basic Concepts: Binary Constraints

Example:

GivenvariablesX1 to X3,withdomains1..3,thefollowingternaryconstraintCiscomposedof6labels..

C(X1, X2, X3) = {<1,2,3>,<1,3,2>,<2,1,3>,<2,3,1> ,<3,1,2>,<3,2,1>}

Eachofthelabelsmaybeassociatedtoavaluefrom1to6

1 : <1,2,3>, 2 : <1,3,2>, ... , 6: <3,2,1>

Now, the following binaryconstraints,C01 toC03, areequivalent to the initialternaryconstraintC

C01(X0, X1) = {<1,1>, <2,1>, <3,2>, <4,2>, <5,3>, <6,3>}

C02(X0, X2) = {<1,2>, <2,3>, <3,1>, <4,3>, <5,1>, <6,2>}

C03(X0, X3) = {<1,3>, <2,2>, <3,3>, <4,1>, <5,2>, <6,1>}


Basic Concepts: Constraint Satisfaction

Definition(Constraint Satisfaction 1):

A compound label satisfies a constraint if their variables are the same

andifthecompoundlabelisamemberoftheconstraint.

Inpractice, it is convenient togeneraliseconstraint satisfaction tocompound

labelsthatstrictlycontaintheconstraintvariables.

Definition(Constraint Satisfaction 2):

A compound label satisfies a constraint if its variables contain the

constraint variables and the projection of the compound label to these

variablesisamemberoftheconstraint.


Basic Concepts: Constraint Satisfaction Problem

Definition(Constraint Satisfaction Problem):

Aconstraint satisfaction problemisatriple<V, D, C>where

Visthesetofvariablesoftheproblem

Disthedomain(s)ofitsvariables

Cisthesetofconstraintsoftheproblem

Definition(Problem Solution):

AsolutiontoaConstraintSatisfactionProblemP:<V, D, C>, isacompound

labeloverthevariablesVoftheproblem,whichsatisfiesallconstraintsinC.


Basic Concepts: Constraint Graph

Forconvenience,theconstraintsofaproblemmaybeconsideredasforminga

specialgraph.

Definition(Constraint Graph or Constraint Network):

The Constraint Graph or Constraint Network of a binary constraint

satisfactionproblemisdefinedasfollows

Thereisanodeforeachofthevariablesoftheproblem.

For each non-trivial constraint of the problem, involving one or two

variables,thegraphcontainsanarclinkingthecorrespondingnodes.

When the problems include constraints with arbitrary arity, the Constraint

Network may be formed after converting these constraints on its binary

equivalent.


Basic Concepts: Constraint Hyper-Graph

Whenitisnotconvenienttoconvertn-aryconstraintstotheirbinaryequivalent,

theproblemmayberepresentedbyanhiper-graph.

Definition(Constraint Hyper-Graph):

Any constraint satisfaction problem with arbitrary n-ary constraints may be

representedbyaConstraints Hiper-Graphformedasfollows:

Thereisanodeforeachofthevariablesoftheproblem.

For each constraint of the problem, the graph contains an hiper-arc

linkingthecorrespondingnodes.

Of course, when the problem has only binary constraints, the hiper-graph

degeneratesintothesimplergraph.


Example: 4-Queens

Example:

The4queensproblemmaybespecifiedbythefollowingconstraint network:

X1in1..4

X4in1..4

X3in1..4X2in1..4

C12

R23

R14

C24

C34

C13

C13:<1,2>, <1,4>, <2,1>, <2,3>,<3,2>, <3,4>, <4,1>, <4,3>


Basic Definitions – Graph Density

Definition(Complete Constraint Network):

A constraint netwok is complete, when there is an arc connecting any two

nodesofthegraph(i.e.whenthereisaconstraintoveranypairofvariables).

The4queensproblem(infact,anynqueensproblem)hasacompletegraph.

However,thisisoftennotthecase:mostgraphsaresparse.Inthiscase,itis

importanttomeasurethedensityoftheconstraintnetwork.

Definition(Density of a Constraint Graph):

Thedensityofaconstraintgraph is the ratiobetween thenumberof itsarcs

andthenumberofarcsofacompletegraphonthesamenumberofnodes


Basic Definitions – Problem Difficulty

Inprinciple,the“difficulty”ofaconstraintproblemisrelatedtothedensityofits

graph.

Intuitively, the denser the graph is, themore difficult the problem becomes,

sincetherearemoreopportunitiestoinvalidateacompoundlabel.

It is important to distinguish between the difficulty of a problem and the

difficultyofsolvingtheproblem.

Inparticular,adifficultproblemmaybesodifficultthatitistrivialtofindit

tobeimpossible!

The“difficulty”ofaproblemisalsorelatedtothedifficultyofsatisfyingeachof

itsconstraints.Suchdifficultymaybemeasuredthroughits“tightness”.


Basic Definitions – Constraint Tightness

Definition(Tightness of a Constraint):

GivenaconstraintC onvariables X1 ... Xn,withdomainsD1 to Dn, the

tightnessofC isdefinedas theratiobetweenthenumberof labels that

define the constraint and the size (i.e. the cardinality) of the cartesian

productD1 x D2 x .. Dn.

For example, the tightness of constraint C13 of the 4 queens problem, on

variablesX1andX3withdomains1..4

C13 = {<1,2>, <1,4>, <2,1>, <2,3>, <3,2>, <3,4>, <4,1>, <4,3>}

is½,sincethereare8tuplesintherelationoutofthepossible16(4*4).

Thenotionoftightnessmaybegeneralisedforthewholeproblem.


Basic Definitions – Problem Tightness

Definition(Tightness of a Problem):

ThetightnessofaconstraintsatisfactionproblemwithvariablesX1 ... Xn,

istheratiobetweenthenumberofitssolutionsandthecardinalityofthe

cartesianproduct

D1 x D2 x .. Dn.

Forexample,the4queensproblem,thathasonlytwosolutions<2,4,1,3>and

<3,1,4,2>

hastightness2/(4*4*4*4) = 1/128.


Basic Definitions – Problem Solving Difficulty

Thedifficultyinsolvingaconstraintsatisfactionproblemisrelatedbothwiththe

Thedensityofitsgraphand

Itstightness.

Assuch,whentestingalgorithmstosolvethis typeofproblems, it isusual to

generaterandom problem instances,parameterisednotonlybythenumber

ofvariablesandthesizeoftheirdomains,butalsobythedensityofitsgraph

andthetightnessofitsconstraints.

Thestudyoftheseissueshasledtotheconclusionthatconstraintsatisfaction

problemsoftenexhibitaphase transition,whichshouldbetakenintoaccount

inthestudyofthealgorithms.

This phase transition typically contains the most difficult instances of the

problem,andseparatestheinstancesthataretriviallysatisfiedfromthosethat

aretriviallyinsatisfiable.


Basic Definitions – Problem Solving Difficulty

Forexample,inSAT,ithasbeenfoundthatthephasetransitionoccurswhen

therationofclausestovariablesisaround4.3.

0 5 10 15

# clauses / # variables

difficulty

4.3


Basic Definitions – Redundancy

Another issue to consider in the difficulty of solving a constraint satisfaction

problem is the potential existence of redundant values and labels in its

constraints.

Definition(Redundant Value):

A value inthedomainofavariableisredundant,ifitdoesnotappearin

anysolutionoftheproblem.

Definition(Redundant Label):

Acompound labelofaconstraintisredundantifitisnottheprojection

totheconstraintvariablesofasolutiontothewholeproblem.

Redundantvaluesand labels increase thesearchspaceuseless,andshould

thusbeavoided.There isnopoint in testingavalue thatdoesnotappear in

anysolution!



Example:

The4queensproblemonlyadmitstwosolutions:

<2,4,1,3> and <3,1,4,2>.

Hence,values1and4areredundantinthedomainofvariablesX1andX4,and

values2and3areredundantinthedomainofvariablesX2andX3.

Regardingredundantlabels, labels<2,3>(showninthefigure)and<3,2>are

redundantinconstraintC13.



Example:

The 4 queens problem, which only admits the two solutions <2,4,1,3> and

<3,1,4,2> may be “simplified” by elimination of the redundant values and

labels.

C13:<1,2>, <1,4>, <2,1>, <2,3>,<3,2>, <3,4>, <4,1>, <4,3>

X1in1,2,3,4

X4in1,2,3,4

X3in1,2,3,4X2in1,2,3,4

C12

R23

R14

C24

C34

C13


Basic Definitions – Equivalent Problems

Ofcourse,anyproblemshouldbeequivalenttoanyofitssimplifiedversions.

Formally,

Definition(Equivalent Problems):

TwoproblemsP1 = <V1, D1, C1>andP2 = <V2, D2, C2>areequivalent iff

theyhavethesamevariables(i.e.V1 = V2)andthesamesetofsolutions.

The“simplification”ofaproblemmayalsobeformalised

Definition(Reduced Problem):

AproblemP=<V,D,C>isreducedtoP’=<V’, D’, C’>if

PandP’areequivalent;

ThedomainsD’areincludedinD;and

TheconstraintsC’areatleastasrestrictiveasthosein C.


Constraint Solving Methods

As shown before, and independently of any problem reduction, a generate

and testproceduretosolveaConstraintSatisfactionProblemisusuallyvery

inneficient.

Nevertheless, this is the approach taken in local searchmethods (simulated

annealingorgeneticalgorithms)–althoughmostlyinanoptimisationcontext!

It is thus often preferable to use a solvingmethod that isconstructive and

incremental,whereby a compound label is being completed (constructive),

onevariableatatime(incremental),untilasolutionisreached.

However,onemustcheck thatateverystep in theconstructionofasolution

theresultinglabelhasstillthepotentialtoreachacompletesolution.


Constraint Solving Methods

Ideally,acompoundlabelshouldbetheprojectionofsomeproblemsolution.

Unfortunately,intheprocessofsolvingaproblem,itssolutionsarenotknown

yet!

Hence, one must use a notion weaker than that of a (complete) solution,

namely

Definition(k-Partial Solution):

Ak-partial solution of a constraint solvingproblemP = <V,D,C>, is a

compoundlabelonasubsetofk ofitsvariables,Vk,thatsatisfiesallthe

constraintsinCwhosevariablesareincludedinVk.


Constructive Solving Methods

This is of course, the basis of the solving methods that use some form of

backtracking. If conveniently performed, backtrackingmay be regarded as a

treesearch,wherethepartialsolutionscorrespondtotheinternalnodesofthe

treeandcompletesolutionstoitsleaves.

2

4

1

3

31 4

1

4

2


Problem Search Space

Clearly,themoreaproblemisreduced,theeasieritis,inprinciple,tosolveit.

Given a problem P=<V,D,C> with n variables X1,..,Xn the potential search

space where solutions can be found (i.e. the leaves of the search tree with

compoundlabels{<X1-v1>, ..., <Xn-vn>})hascardinality

#S = #D1 * #D2 * ... * #Dn

Assuming identicalcardinality (orsomekindofaverageof thedomainssize)

forallthevariabledomains,(#Di = d)thesearchspacehascardinality

#S = dn

whichisexponentialonthe“size”noftheproblem.


If instead of the cardinality d of the initial problem, one solves a reduced

problemwhosedomainshavelowercardinalityd’(<d)thesizeofthepotential

searchspacealsodecreasesexponentially!

S’/S=d’n/dn=(d’/d)n

Such exponential decrease may be very significant for “reasonably” large

valuesofn,asshowninthetable.

10 20 30 40 50 60 70 80 90 100

7 6 4.6716 21.824 101.95 476.29 2225 10395 48560 226852 1E+06 5E+06

6 5 6.1917 38.338 237.38 1469.8 9100.4 56348 348889 2E+06 1E+07 8E+07

5 4 9.3132 86.736 807.79 7523.2 70065 652530 6E+06 6E+07 5E+08 5E+09

4 3 17.758 315.34 5599.7 99437 2E+06 3E+07 6E+08 1E+10 2E+11 3E+12

3 2 57.665 3325.3 191751 1E+07 6E+08 4E+10 2E+12 1E+14 7E+15 4E+17

d d'

S/S'

n

Problem Search Space


Reduction of the Search Space

Inpractice,thispotentialnarrowingofthesearchspacehasacostinvolvedin

findingtheredundantvalues(andlabels).

Adetailedanalysisof thecostsandbenefits inthegeneralcaseisextremely

complex,sincetheprocessdependshighlyontheinstancesoftheproblemto

besolved.

However, it is reasonable to assume that the computational effort spent on

problemreductionisnotproportionaltothereductionachieved,becomingless

andlessefficient.

Aftersomepoint,thegainobtainedbythereductionofthesearchspacedoes

notcompensatetheextraeffortrequiredtoachievesuchreduction.



• Qualitatively,thisprocessmayberepresentedbymeansofthefollowinggraph

Com

p uta

ti ona

l Co s

t

R - Reduction Cost

S- Search Cost

R+S Combined Cost

Effort spent in solving the problem

Amount of Reduction Achieved



The effort in reducing the domains must be considered within the general

schemetosolvetheproblem.

InConstraint Logic Programming, the specification of the constraints usually

precedestheenumerationofthevariables.

Problem(Vars):-

Declaration of Variables and Domains,

Specification of Constraints,

Labelling of the Variables.

Ingeneral,searchisperformedexclusivelyonthelabellingofthevariables.

The execution model alternates enumeration with propagation, making it

possibletoreducetheproblematvariousstagesofthesolvingprocess.



Hence,givenaproblemwithnvariablesX1toXntheexecuctionmodelfollows

thefollowingpattern:

Declaration of Variables and Domains,

Specification of Constraints,

indomain(X1), % value selection with backtraking

propagation, % reduction of problem X2 to Xn

indomain(X2),

propagation, % reduction of problem X3 to Xn

...

indomain(Xn-1)

propagation, % reduction of problem Xn

indomain(Xn)


Domain Reduction – Consistency Criteria

Once formallydefined thenotionofproblemreduction,onemustdiscuss the

actualprocedurethatmaybeusedtoachieveit.

First of all, onemust ensure thatwhatever procedure is used, the reduction

keepstheproblemequivalenttotheinitialone.

Herewehaveasmallproblem,sincethedefinitionofequivalencerequiresthe

solutionstobethesameandwedonotknowingeneralwhatthesolutionsare!

Nevertheless, thesolutionswillbethesameif in theprocessofreductionwe

have theguarantee that “nosolutionsare lost”.Suchguaranteesaremetby

severalcriteria.


Domain Reduction – Consistency Criteria

Consistency criteria enable to establish redundant values in the variables

domains in an indirect form, i.e. requiring no prior knowledge on the set of

problemsolutions.

Hence, procedures that maintain these criteria during the “propagation”

phases,willeliminateredundantvaluesandsodecreasethesearchspaceon

thevariablesyettobeenumerated.

For constraint satisfaction problems with binary constraints, the most usual

criteriaare,inincreasinglycomplexityorder,

Node Consistency

Arc Consistency

Path Consistency


Domain Reduction – Node Consistency

Definition(Node Consistency):

Aconstraint satisfactionproblem isnode-consistent ifnovalueon the

domainofitsvariablesviolatestheunaryconstraints.

Thiscriterionmayseembothobviousanduseless.Afterall,whowouldspecify

adomainthatviolatestheunaryconstraints?!

However, this criterionmust be regardedwithin the context of the execution

modelthatincrementallycompletespartialsolutions.Constraintsthatwerenot

unary in the initial problem become so when one (or more) variables are

enumerated.



Example:

Aftertheinitialpostingofthe

constraints, the constraint

network model at the right

represents the 4-queens

problem.

Afterenumerationofvariable

X1, i.e.X1=1,constraintsC12,

C13andC14becomeunary!!

X1in1..4

X4in1..4

X3in1..4X2in1..4

C12

C23

R14

C24

C34

R13

X4in1..4

X3in1..4X2in1..4C23

C24

C34

X21,2

X31,3

X41,4



Analgoriththatmaintainsnodeconsistencyshouldremovefromthedomains

ofthe“future”variablestheappropriatevalues.

Maintainingnodeconsistencyachievesadomainreductionsimilartowhatwas

achievedinpropagationwiththebooleanformulation.

X4 in 2,3

X3 in 2,4X2 in 3,4C23

C24

C34

X2 1,2 X3 1,3

X4 1,4

X4 1,4

1 1

1 1

1 1

X2 1,2X3 1,3


Maintaining Node Consistency

Giventhesimplicityofthenodeconsistencycriterion,analgorithmtomaintainitisverysimpleandwithlowcomplexity.

ApossiblealgorithmisNC-1,below.

procedure NC-1(V, D, R); for X in V for v in Dx do for Cx in {Cons(X): Vars(Cx) = {X}} do if not satisfy(X-v, Cx) then Dx <- Dx \ {v} end for end for end forend procedure



Space Complexity of NC-1: O(nd)

Assumingn variables in the problem, eachwithd values in its domain, and

assuming that thevariable’sdomainsare representedbyextension,aspace

ndisrequiredtokeepexplicitelythedomainsofthevariables.

Forexample,theinitialdomain1..4ofvariablesXiinthe4queensproblemis

representedbyabooleanvector(where1meansthevalueisinthedomain)

Xi = [1,1,1,1] or 1111.

After enumeration X1=1, node consistency prunes the domain of other

variablestoX1 = 1000,X2 = 0011,X3 = 0101andX4 = 0110

AlgorithmNC-1doesnotrequireadditionalspace,soitsspacecomplexityis

O(nd).



Time Complexity of NC-1: O(nd)

Assumingnvariables in theproblem,eachwithdvalues in itsdomain,

andtakingintoaccountthateachvalueisevaluatedonesingletime,itis

easytoconcludethatalgorithmNC-1hastimecomplexityO(nd).

The low complexity, both temporal and spatial, of algorithm NC-1, makes it

suitabletobeusedinvirtualallsituationsbyasolver.

However,nodeconsistencyisratherincomplete,notbeingabletodetectmany

possiblereductions.


Domain Reduction – Arc Consistency

A more demanding and complex criterion of consistency is that of arc-

consistency

Definition(Arc Consistency):

Aconstraintsatisfactionproblemisarc-consistentif,

• Itisnode-consistent;and

• ForeverylabelXi-viofeveryvariableXi,andforallconstraintsCij,defined

overvariablesXiandXj, theremustexistavaluevj thatsupportsvi, i.e.

suchthatthecompoundlabel{Xi-vi,Xj-vj}satisfiesconstraintCij.



Example:

AfterenumerationofvariableX1=1,andmaking thenetworknode-consistent,

the4queensproblemhasthefollowingconstraintnetwork:

However, label X2-3 has no support in variable X3, since neither compound

label{X2-3,X3-2}nor{X2-3,X3-4}satisfyconstraintC23.

Therefore,value3canbesafelyremovedfromthedomainofX2.

X4 in 2,3

X3 in 2,4X2 in 3,4C23

C24

C34

X2 1,2 X3 1,3

X4 1,4X4 1,4

1 1

1 1

1 1

X2 1,2X3 1,3



Example(cont.):

In fact, none (!) of the values ofX3 has support in variables X2 andX4., as

shownbelow:

label X3-4 has no support in variable X2, since none of the compound

labels{X2-3,X3-4}and{X2-4,X3-4}satisfyconstraintC23.

label X3-2 has no support in variable X4, since none of the compound

labels{X3-2,X4-2}and{X3-2,X4-3}satisfyconstraintC34.

X4 1,4

1 1

1 1

1 1

X2 1,2X3 1,3



Example(cont.):

Sincenoneof thevalues from thedomainofX3hassupport invariables X2

andX4,maintenanceofarc-consistencyemptiesthedomainofX3!

Hence, maintenance of arc-consistency not only prunes the domain of the

variablesbutalsoantecipatesthedetectionofunsatisfiabilityinvariableX3!In

thiscase,backtrackingofX1=1maybestartedevenbeforetheenumerationof

variableX2.

Given the good trade-of between pruning power and simplicity of arc-

consistency,anumberofalgorithmshavebeenproposedtomaintainit.

X4 1,4

1 1

1 1

1 1

X2 1,2X3 1,3


Maintaining Arc Consistency: AC-1

AC-1,shownbelow,isaverysimplealgorithmtomaintainArcConsistency´

procedure AC-1(V, D, C);

NC-1(V,D,C); % node consistency

Q = {aij | Cij C Cji C }; % see note repeat

changed <- false;

for aij in Q do

changed <- changed or revise_dom(aij,V,D,C)

end for

until not change

end procedure

Note:forconstraintCij twodirectedarcs,aijandaji,areconsidered



Predicate rev_dom(aij,V,D,R) succeeds iff it prunes values from thedomainofXi

predicate revise_dom(aij,V,D,C): Boolean;

success <- false;

for vi in dom(Xi) do

if there is no vj in dom(Xj) such that

satisfies({Xi-vi,Xj-vj},Cij) then

dom(Xi) <- dom(Xi) \ {vi};

success <- true;

end if

end for

revise_dom <- success;

end predicate


Time Complexity of AC-1

Time Complexity of AC-1 : O(nad3)

Assuming nvariablesintheproblem,eachwithdvaluesinitsdomain,anda

total of a arcs, in the worst case, predicate revise_dom, checks d2 pairs of

values.

The number of arcs aij in queue Q is 2a (2 directed arcs aij and aji are

consideredforeachconstraintCij).Foreachvalueremovedfromonedomain,

revise_domiscalled2atimes.

Intheworstcase,onlyonevaluefromonevariableisremovedineachcycle,

andthecycleisexecutedndtimes.

Therefore,theworst-casetimecomplexityofAC-1isO(d22a*nd),i.e.

O(nad3)


Space Complexity of AC-1

Space Complexity of AC-1 : O(ad2) = O(n2d2)

AC-1mustmaintain a queueQ,withmaximum size2a. Hence the inherent

spacialcomplexityofAC-1isO(a).

To this space, one has to add the space required to represent the domains

O(nd)andtheconstraintsoftheproblem.Assumingaconstraintsanddvalues

in each variable domain the space required is O(ad2), and a total space

requirementof

O(nd + ad2)

whichdominatesO(a).

For“dense”constraintnetworks”,a n2/2.Thisisthenthedominantterm,and

thespacecomplexitybecomes

O(ad2) = O(n2d2)


Maintaining Arc Consistency: From AC-1 to AC-3

InefficiencyofAC-1

Every time a value vi is removed from the domain of variable Xi by

predicaterevise_dom(aij,V,D,R), allarcsarereexamined.

However,onlythearcsaki(fork iandk j)shouldbereexamined.

This isbecause theremovalofvimayeliminate thesupport fromsome

valuevkofsomevariableXkforwhichthereisaconstraintCki(orCik).

SuchinefficiencyiseliminatedinalgorithmAC-3.



AC-3onlyrevisitsarcsforwhichthedomainofthe“leading”variablehasbeen

“revised”(narrowed).

procedure AC-3(V, D, C);

NC-1(V,D,R); % node consistency

Q = {aij | Rij C Cji C }; while Q do Q = Q \ {aij} % removes an element from Q

if revise_dom(aij,V,D,C) then % revised Xi

Q = Q {aki | Cki C k i k j} end if

end while

end procedure

Intuitively,AC-3musthavenotonlybetterworst-case,butalsoabettertypical-

casecomplexitythanAC-1.


Time and Space Complexity of AC-3

Time Complexity of AC-3: O(ad3)

Each arc aki is only added to Q when some value vi is removed from the

domainofXi.

Intotal,eachofthe2aarcsmaybeaddedtoQ(andremovedfromQ)d times.

Everytimethatanarcisremoved,predicaterevise_domiscalled,tocheckat

mostd2pairsofvalues.

All thingsconsidered,andincontrastwithAC-1,withcomplexityO(nad3), the

timecomplexityofAC-3,intheworstcase,isO(2ad * d2),i.e.O(ad3).

Space Complexity of AC-3: O(ad2)

As tospacecomplexityAC-3hasthesamerequirements thanAC-1,andthe

same worst-case space complexity of O(ad2) O(n2d2), due to the

representationofconstraintsbyextension.

21 november 2006 foundations of logic and constraint programming 1 constraint (logic) programming an...

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