Constraint (Logic) Programming

7 November 2005 Foundations of Logic and Constraint Programming 1

Constraint (Logic) Programming

Anoverview

• ExamplesofConstraintDomains

• Constraint(Logic)Programming

• OperationalSemanticsofCLP

• ModellinginCLP

[Dech04]RinaDechter,ConstraintProcessing,


Constraints and Constraint Problems

- Formallyaconstraint(solving)problemcanberegardedasatuple<V,D,C>,

where

- X={X1,...,Xn}isasetofvariables

- D={D1,...,Dn}isasetofdomains(forthecorrespondingvariables)

- C={C1,...,Cm}isasetofconstraints(onthevariables)

- Solving a constraint problem consists of determining values xi Di for each

variableXi,satisfyingalltheconstraintsC.

- Intuitively,aconstraintCiisalimitationonthevaluesofitsvariables.

- More formally, a constraint Ci (with arity k) over variables Xi1, ..., Xik ranging

overdomainsDi1,...,DikisasubsetofthecartesiancartesianDj1...Djk.

CiDj1...Djk


Constraint Problems: Examples

- Examplesoftheseproblemsinclude

ModellingofDigitalCircuits

ProductionPlanning

NetworkManagement

Scheduling(QualitativeandQuantitative)

Assignment(Colouring,Latin/MagicSqaures,Sudoku,Circuits,...)

AssignmentandScheduling(Timetabling,Job-shop)

FillingandContainment

SceneInterpretation


Modeling of Digital Circuits

Goal(Example):Determineatestpatternthatdetectssomefaultygate

Variables: Signalsinthecircuit

Domain: Booleans:0/1(orTrue/False,orHigh/Low)

Constraints: Equality constraints between the output of a gate and its “boolean

operation”(e.g.and,or,not,nand,...)

A

C

D

B

E

F

G

H

I

G1

G2

G3

G4

G5

E = or(A,B) % G1 F = nand(B,C) % G2G = and(B,C) % G3 H = nand(E,F) % G4I = nand(F,G) % G5


Production Planning

Goal(Example):Determineaproductionplan

Variables: Quantitiesofgoodstoproduce

Domain: Rational/RealsorIntegers

Constraints: EqualityandInequality(linear)constraintstomodelresourcelimitations,

minimal quantities toproduce, costsnot to exceed, balance conditions,etc...

Find X, Y and Z such that 4x+ 3y + 6z 1500 % resources used do not exceed 1500 x + y + z >= 300 % production not less than 300 units x z + 20 % x units within z 20 units x z - 20 x, y, z 0 % non negative production


Network Management

Goal(Example):Determineacceptabletrafficonanetwok

Variables: Flowsineachedge

Domain: Rational/Reals(orIntegers)

Constraints: Equalityand Inequality (linear)constraints tomodelcapacity limitations,

flowmaintenance,costs,etc...

Find x,y,z, a,b,c,d,e such that x 6, z 10 % minimum flow a 5, ... , f 6 % capacity% flow maintenance x = a, y = b + c, a + b + d = e, c = d + f, e + f = z x, y ,z, a, b, c, d, e, f 0

5/a

2/d8/e

6/f

3/b

7/c

x

y

z


Schedulling (Quantitative)

Goal(Example):Assigntiming/precedencetotasks

Variables: StartTimingofTasks,DurationofTasks

Domain: Rational/RealsorIntegers

Constraints: PrecedenceConstraints,Non-overlappingconstraints,Deadlines,etc...

Find Sa ,..., Se, such that Sb Sa+Ta, % precedence .... (Sc Sb+Tb) (Sb Sc+Tc) % non overlap ..., 6 Sa+Ta % deadline Sa,..., Se 0

Sa

Sc

Sb

Se

SdTa

Ta

Tb

Tc

Td


Schedulling (Qualitative)

Goal(Example):Completeassignmentoftaskrelations

Variables: Relationsbetweentasks(qualitative;e.gTapreceedesTb)

Domain: 16relations(preceeds,meets,overlap,....)

Constraints: Partialassignments

? What relation between C and D, given that A preceeds B, B preceeds C,

C meets EA preceeds D, D meets E,B overlaps D

A

D

B

E

C


Assignment

Manyconstraintproblemscanbeclassifiedasassignmentproblems. Ingeneral

allthatcanbestatedisthattheseproblemsfollowthegeneralgoalCSPs:Assign

valuestothevariablestosatisfytherelevantconstraints.

– Variables:

Objects/Propertiesofobjects

– Domain:

Integer,enumeratedorRealValues(properties-colours,numbers,..)

Booleansfordecisions

– Constraints:

Compatibility(Equality,Difference,No-attack,ArithmeticRelations)

Someexamplesmayhelptoillustratethisclassofproblems


Assignment (2)

AB

CD

E

F

Graph Colouring (Booleans)

Assign values to A1,A2, .., F1,F2

s.t. A1,A2, .., F1,F2 {0,1}% different colours for A and B

A1 B1 or A2 B2....

E1 F1 or E2 F2

AB

CD

E

F

Graph Colouring (Finite Domains)

Assign values to A, .., F,

s.t. A, B, .., F {red, blue, green}A B, A C, A D,B C, B F, C D, C E, C FD E, E F


Assignment (3)

Q1

Q2

Q3

Q4

N-queens (Finite Domains):

Assign Values to Q1,..., Qn {1,.., n}

s.t. ij noattack (Qi, Qj)

Latin Squares (similarto Sudoku):

Assign Values to X11,..., X33 {1,.., 3}

s.t. i jk Xij Xik % same row

j ik Xij Xkj % same column

X11 X12 X13

X21 X22 X23

X31 X32 X33

Magic Squares:

Assign Values to X11,..., X33 {1,..,9}

s.t. ij k Xki = Xkj % same rows sum

ij Sk Xik = Xjk % same cols sum

ik jl Xij Xkl % all different

X11 X12 X13

X21 X22 X23

X31 X32 X33


Assignment (3)

Travelling Salesperson (Finite Domains)

Find values for A, B, C, D {1,..,4} s.t. A B, ..., C D % a permutation of [A,B,C,D] if A = B+1 then XA = Lab, ... if D = C+1 then XD = Lcd

Xa + Xb + Xc + Xd K

A B

C D

23

18

33

17 1321

Travelling Salesperson (Booleans)

Find decision values for Xab...Xdc {0,1}

s.t. a k Xak = 1

a Sk Xka = 1

no subcycle constraints

ab Xab Lab < k

A B

C D

23

18

33

17 1321


Mixed: Assignment and Scheduling

Goal(Example):Assignvaluestovariables

Variables: StartTimes,Durations,Resourcesused

Domain: Integers(typicaly)orRationals/Reals

Constraints: Compatibility(Disjunctive,Difference,ArithmeticRelations)

Job-Shop

Assign values to Sij {1,..,n} % time slots

and to Mij {1,..,m} % machines available

% precedence within job

j i < k Sij + Dij ≤ Skj

% either no-overlap or different machines

i,j,k,l (Mij ≠ Mkl) (Sij + Dij ≤ Skl) (Skl + Dkl ≤ Sij)

J 1

J 2

J 3

J 4

1 2 3

1 2 3

1 2 3

1 2 3


Filling and Containment

Goal(Example):Assignvaluestovariables

Variables: PointLocations

Domain: Integers(typicaly)orRationals/Reals

Constraints: Non-overlapping(Disjunctive,Inequality)

Fiiling

Assign values to Xi {1,.., Xmax} % X-dimension

Yi {1,.., Ymax} % Y-dimension

% no-overlapping rectangles

i,j (Xi+Lxi Xj) % I to the left of J

(Xj+Lxj Xi) % I to the right of J

(Yi+Lyi Yj) % I in front of J

(Yj+Lxj Xi) % I in back of J

D

G

I

BCA

JH

E

FK


Scene Interpretation

Goal(Example):Assignvaluestojunctions

Variables: Junctions

Domain: JunctionTypes

Constraints: Compatibilityofjunctions


Constraint Programming

ConstraintProgrammingandLanguagesisdrivenbyanumberofgoals

- Expressivity

- Constraint Languages should be able to easily specify the variables,

domainsandconstraints(e.g.conditional,global,etc...);

- DeclarativeNature

- Ideally,programsshouldspecify theconstraints tobesolved,not the

algorithmsusedtosolvethem

- Efficiency

- Solutions should be found as efficiently as possile, i.e. With the

minimumpossibleuseofresources(timeandspace).

These goals are partially conficting goals and have led to the various

developmentsinthisresearchanddevelopmentarea.


Complexity of Constraint Problems

Assuming that a potential

solution can be generated

and tested in 1 μs, the

approximatetimesrequiredto

searchthewholespaceofthe

kn possible solutions isgiven

inthetable

This complexity is in sharp

constrast with the complexity

of deterministic problems

(such as sorting of lists,

matrix multiplication, etc,

which are polinomial on the

sizenofthestructures

10 20 30 40 50 60

2 1.0E-03 1.0E+00 1.1E+03 1.1E+06 1.1E+09 1.2E+12

3 5.9E-02 3.5E+03 2.1E+08 1.2E+13 7.2E+17 4.2E+22

4 1.0E+00 1.1E+06 1.2E+12 1.2E+18 1.3E+24 1.3E+30

5 9.8E+00 9.5E+07 9.3E+14 9.1E+21 8.9E+28 8.7E+35

6 6.0E+01 3.7E+09 2.2E+17 1.3E+25 8.1E+32 4.9E+40

n

k

nk

10 20 30 40 50 60

2 1.0E-04 4.0E-04 9.0E-04 1.6E-03 2.5E-03 3.6E-03

3 1.0E-03 8.0E-03 2.7E-02 6.4E-02 1.3E-01 2.2E-01

4 1.0E-02 1.6E-01 8.1E-01 2.6E+00 6.3E+00 1.3E+01

5 1.0E-01 3.2E+00 2.4E+01 1.0E+02 3.1E+02 7.8E+02

6 1.0E+00 6.4E+01 7.3E+02 4.1E+03 1.6E+04 4.7E+04

n

k

nk

1 hour = 3.6*103 ; 1 day = 8.6*104 ; 1 year = 3.1*107 ; age of universe = 4.7*1023


Complexity of Constraint Problems

Amajorissueinconstraintsolvingproblemsistheircomplexity.Ingeneralthese

arecombinatorialproblemswithanexponentialsearchspace.

- Booleans- NumberofVariables:n- Domaincardinality:2- Numberofpotentialsolutions:2n

- FiniteDomains- NumberofVariables:n- Domaincardinality:d- Numberofpotentialsolutions:dn

- Real/RationalDomains- NumberofVariables:n- Domaincardinality:Theoreticallyinfinite.Inpracticefinite,D,dueto

rounding.- Numberofpotentialsolutions:Dn


Constraint Problems are NP

Notwithstandingtheexponentialnumberofthesearchspace,onecouldexpect

thatinpracticesolutionsarefoundmuchquickerbysome“cleaver”procedure.

Nevertheless,mostdecisionproblemsofinterest,(e.g.thosedescribedbefore)

are NP–complete, i.e. they are of the same complexity of the Satisfiability

problemforBooleanClauses(SAT).

ThisequivalenceofSATandaproblemProb

isproveningeneralbythetransformation(in

polinomialtime)ofSATintoProb,andofProb

intoSAT.

Given this equivalence, if there was a procedure to solve SAT in polinomial

time,allNPproblemswouldbesolvedinpolinomialtimeaswell.

Unfortunately,noonehasever foundsuchprocedure ingeneral.Foragiven

procedure,ithasalwaysbeenpossibletofindinstancesofSATthataresolved

inexponentialtime.

Prob SAT


Constraint Problems are NP

NoticethattheNPnatureofthesedecisionproblemsdoesnotpreventthatfor

many problem instances it is possible to find efficient procedures to solve

them.

Inpractice,aproblem isNP-complete if it ispossible tocheckasolution in

polinomialtime(tofindasolutionmaytakeexponentialtime).

Decisionproblemslieinthiscategory,sincegivensomepotentialsolutionone

hasonlytocheckonebyone,alltheconstraints(anumberusuallyintheorder

ofmagnitudeofthevariables).

Optimisation problems are usually harder than decision ones, hence their

classificationas NP-Hard.

Herecheckingasolutionusuallyrequiresexponentialtime,sinceitrequiresto

compareallsolutions(andhencefindthem,whichrequiresexponentialtime).


Declarative Programming

Programmingacombinatorialproblemrequires

thespecificationoftheconstraintsoftheproblem

Thespecificationofasearchalgorithm

Theseparationof these twoaspectshas fora long timebeenadvocatedby

several programming paradigms, namely functional programming and logic

programming..

Logic programming in particular has a built in mechanism for search

(backtracking)thatmakesitagoodcandidatetoconstraintprogramming.

Maintaining its search mechanism, all that is required is to extend its

underlyingresolutiontoconstraint solving.

This is in factquitenatural,since resolutionmaybe regardedasaparticular

typeofconstraintsolving,namelytheequalityofHerbrandterms(trees),which

isperformedintheunificationofagoalandaclausehead.


Operational Semantics of CLP

TheoperationalsemanticsofConstraintLogicProgrammingcanbedescribed

asatransitionsystemonstates.

Thestateofaconstraintsolvingsystemcanbeabstractedbyatuple

< G, CS > where

Gisasetofconstraintsstilltosolve

CSistheConstraint Store:asetofconstraintsalreadyexamined

InLP,Gisthequery, i.e. thesetofgoalsstill tosolve.SetCSmaintainsthe

compositionofalltheunificationsperformedinthepreviousresolutionsteps.

Aderivationisjustasequenceoftransitions.

Astatethatcannotberewrittenisafinalstate.

An executor of (C)LP aims at finding successful final states of derivations

startingwithaquery.


State Transitions of CLP

Aderivationisfailedifitisfiniteanditsfinalstateisfail.

Aderivationissuccessfulifitisfiniteandprocessesalltheconstraints,i.e.it

endsinastatewithform<Ø, S >

Thestatetransitionsarethefollowing

S(elect): Selectagoal(constraint)

Itisassumedthat

A computation rule selects some goal g, among those still to be

consideredthatrequiressolvingofsomeconstraintc.

The set of unsolved goals is eventually augmented with set G’, with

additionalgoalstobefurtherconsidered.

g requires solving c and other goals G’

< G {g}, CS > s < G G’, CS {c} >



Theothertransitionrulesrequiretheexistenceoftwofunctions,infer(C,S)and

consistent(C),adequatetothedomainsthatareconsidered.

I(nfer): Processaconstraint,inferringitsconsequences

C(heck): ChecktheconsistencyoftheStore

CS’ = infer(CS {c})

< G, CS {c}) > i < G, CS’ >

consistent (CS)

< G, CS > c < G, CS >

¬ consistent (CS)

< G, CS > c fail


State Transitions of LP

InLogic Programming

HandlingagoalgissimplycheckingwhetherthereisaclauseH←Bwhose

head h matches the goal. In this case, the equality constraint between

Herbrandtermsg=hisaddedtothepassivestore.

Functioninfer(C,c)performsunificationofthetermsincludedingandh,after

applyingallsubstitutionsincludedinC

function infer(, g=h) if unify(g,h , ) then infer = (success, )

else infer = (fail,_); end function

Checking consistency is a simple check on whether the previous unification

hasreturnedsuccessorfailure(inpracticethetwofunctionsaremerged).

g is a goal h B < G {g}, CS > s < G B, CS {h = g} >


State Transitions of LP: Example

Forexample,assumingthat

CS = {X / f(Z), Y / g(Z,W)} % a solved form

G includesgoal p(Z,W)

thereisaclause p(a,b) :- B.

thenthefollowingtransitionstakeplace,whengoalp(Z,W) is“called”

< G , {X/f(Z), Y/g(Z,W)} >

→s %goalselection

< G \ {g} B, {X/f(Z), Y/g(Z,W)} {p(Z,W) = p(a,b)}} >

→i,c %unification(inferenceandconsistencycheck)

< G \ {g} B, {X/f(a), Y/g(a,b), Z/a, W/b} >



InConstraint Logic Programming

Handling a goal g is as before. Handling a constraint in a certain domain,usuallybuilt-in,simplyplacestheconstraintinsetCS.

.

Function infer(C,S) propagates the constraint in the Constraint Store, bymethodsthatdependontheconstraintsystemused.Tipically,

Itattemptstoreducethevaluesinthedomainsofthevariables;or

obtainanewsolvedform(likeinunification)

Checkingconsistencyalsodependsontheconstraintsystem.

Itcheckswhetheradomainhasbecomeempty;or

Whetheritwasnotpossibletoobtainasolvedform.

g is a constraint< G {g}, CS > s < G, CS {g} >


State Transitions of CLP: Example (1)

A positive example

Assuming thatvariablesAandBcan takevalues in1..3, then theconstraint

storemayberepresentedas

CS = {A in 1..3, B in 1..3}

Rule S:IftheconstraintselectedbyruleSisA > B,thenthestorebecomes

CS = {A in 1..3, B in 1..3} {A > B}

Rule I:TheInferrulepropagatesthisconstraint intheconstraintstore,which

becomes

C,S = {A in 2..3, B in 1..2, A > B}

Rule C:Thesystemdoesnotfindanyinconsitencyintheactivestore,sothe

systemdoesnotchange.


State Transitions of CLP: Example (2)

A negative example

Assuming that variables A and B can take, respectively values in the sets

{1,3,5}and{2,4,6}.Theconstraintstoremayberepresentedas

CS = { A in {1,3,5} B in {2,4,6} }

Rule S:IftheconstraintselectedbyruleSisA = B,thestorebecomes

CS = { A in {1,3,5} B in {2,4,6} } { A = B }

Rule I:Therulepropagatesthisconstrainttotheactivestore.Sincenovalues

arecommontoAandB,theirdomainsbecomeempty

CS = {A in {}, B in {} }

Rule C: This rule finds the empty domains and makes the transition to fail

(signaling the system to backtrack in order to find alternative successful

derivations).


Solutions in (C)LP

InLPandCLP,solutionsareobtained,byinspectingtheconstraintstoreofthe

finalstateofasuccessfulderivation.

Insystems thatmaintainasolved form, (likeLP,butalsoconstraintsystems

CLP(B) for booleans, and CLP(Q), for rationals), solutions are obtained by

projectionofthestoretotherelevantvariables.

Forexampleiftheinitialgoalwasp(X,Y)andthefinalstoreis

{X/f(Z,a), Y/g(a,b), W/c}

thenthesolutionistheprojectiontovariablesXandY

{X/f(Z,a), Y/g(a,b), W/c}

In systems where a solved form is not maintained, solutions require some

formofenumeration.Forexample,from

C = {X in 2..3, Y in 1..2, X > Y},

the solutions <X=2;Y=1>, <X=3;Y=1> and <X=3;Y=1>, are obtained by

enumeration.


The CLP approach to Constraint Solving

ToprogramaconstraintsolvingprobleminCLPanumberof issuesmustbe

takenintoaccount

• Whatisthepurposeoftheprogram?

• Any(first)solution?

• Allsolutions?(decisionmakingwithuncertainty)

• BestSolution?(optimisation)

• Whatkindofmodeling?

• VariablesandDomains?(FiniteDomains?Booleans?Sets?Mixed?)

• Constraints?(Global?Redundant?)

• Whatsolvershouldbeused?

• CompleteSolver?(Allsolutions,inneficiency?)

• Incompletesolvers?(Singlesolutions,Heuristics?)

Anexample,N-queens,willillustratesomeoftheseissues.


N-Queens

TheN-queensproblemconsistsoffindingpositionsforN

queensinaNNchess-board,sothatno2queensattack

eachother.

Modeling:

BooleanVariables:One0/1variableforeachpositionoftheboard

SATorarithmeticconstraints?

FiniteDomains:One1..Nvariableforeachrowintheboard

Solvers:

Complete(onlyforBooleans):

maintainsallsolutions

Incomplete:Booleans/FiniteDomains:

Onesolutionatatime

Inthiscasethereisnoclearoptimisationpurpose(functiontooptimise)


N-Queens Models

AssumingBooleanvariables,Xij(i,j1..n),oneperposition

• Constraints

• OneperRow(Onepercolumnissimilar)

• SAT: Xij \/ ...\/ Xin foralli1..n

• Xij \/ Xikforalli,jk1..n

• Arithmetic: Xij + ...+ Xin = 1 foralli1..n

• Diagonals

• SAT: Xij \/ Xklforalli+j=k+landalli-j=k-l

• Arithmetic: ij Xij 1 forallij=k-2n-1..2n+1

• AssumingFiniteDomainvariables,Xi(i1..n),),oneperrow

• Constraints(oneperrowisimplicit)

• OneperColumn Xi Xjforallij1..n

• Diagonals: Xi Xj foralli+j=k-2n-1..2n+1


CLP Programs

CLPprogramshaveingeneralthefollowingstructure

Program

• VariableDeclaration%notneededincompletesolvers

• ConstraintDeclaration

• Search % optional in all solutions/complete

solvers

In general, CLP systems require the addition of modules that implement the

intendedsolvers.ForexampleinSICStus

:- use_module(library(clpb)). thecompletesolverforbooleans

:- use_module(library(clpfd)). Thesolverforfinitedomains

Suchsolversidentifyconstraintsandgoalswithadifferentsyntax sat(X+Y =:= 1)implementsan“OR”ofbooleanvariablesXandY X+Y #= 3enforcesthesumofFDvariablesXandYtobe3

Intherest,CLPprogramsadoptthesyntaxofLP(Prolog)


Boolean N-Queens: Complete Solver

Forsimplicity,weshowthe3-queensprogram.Itcanbeeasily

adaptedforlargerboards(3-queenshasnosolutions).

:- clpb.

queens_3(Q):- Q = [Q11, Q12, Q13, Q21, Q22, Q23, Q31, Q32, Q33],

% one and only one queen per row sat(Q11+Q12+Q13 =:= 1),sat(Q11*Q12=:=0),sat(Q11*Q13=:=0),sat(Q12*Q13=:=0), sat(Q21+Q22+Q23 =:= 1),sat(Q21*Q22=:=0),sat(Q21*Q23=:=0),sat(Q22*Q23=:=0), sat(Q31+Q32+Q33 =:= 1),sat(Q31*Q32=:=0),sat(Q31*Q33=:=0),sat(Q32*Q33=:=0),

% one and only one queen per column sat(Q11+Q21+Q31 =:= 1),sat(Q11*Q21=:=0),sat(Q11*Q31=:=0),sat(Q21*Q31=:=0), sat(Q12+Q22+Q32 =:= 1),sat(Q12*Q22=:=0),sat(Q12*Q32=:=0),sat(Q22*Q32=:=0), sat(Q13+Q23+Q33 =:= 1),sat(Q13*Q23=:=0),sat(Q13*Q33=:=0),sat(Q23*Q33=:=0),

% no two queens in \ diagonals sat(Q11*Q22=:=0),sat(Q11*Q33=:=0),sat(Q22*Q32=:=0), sat(Q12*Q23=:=0),sat(Q21*Q32=:=0),

% no two queens in \ diagonals sat(Q13*Q22=:=0),sat(Q13*Q31=:=0),sat(Q22*Q31=:=0), sat(Q12*Q21=:=0),sat(Q23*Q32=:=0).


Boolean N-Queens: SAT Solver

The3-queensprograminSATisshownwithuser-defined

operators(“\/”and“~”),implementedintheFDsolver.

:- op(300,fy,~).:- op(400,xfy,\/).

:- clpfd.

queens_3(Q):-

% variable declaration Q = [Q11, Q12, Q13, Q21, Q22, Q23, Q31, Q32, Q33], domain(Q,0,1),

% one and only one queen per row Q11 \/ Q12\/ Q13, ~Q11 \/ ~Q12, ~Q11 \/ ~Q13, ~Q12 \/ ~Q13, Q21 \/ Q22\/ Q23, ~Q21 \/ ~Q22, ~Q21 \/ ~Q23, ~Q22 \/ ~Q23, Q31 \/ Q32\/ Q33, ~Q31 \/ ~Q32, ~Q31 \/ ~Q33, ~Q32 \/ ~Q33,

% one and only one queen per column Q11 \/ Q21\/ Q31, ~Q11 \/ ~Q21, ~Q11 \/ ~Q31, ~Q21 \/ ~Q31, Q12 \/ Q22\/ Q32, ~Q12 \/ ~Q22, ~Q12 \/ ~Q32, ~Q22 \/ ~Q32, Q13 \/ Q23\/ Q33, ~Q13 \/ ~Q23, ~Q13 \/ ~Q33, ~Q23 \/ ~Q33,

% no two queens in \ diagonals ~Q11 \/ ~Q22, ~Q11 \/ ~Q33, ~Q22 \/ ~Q33, ~Q12 \/ ~Q23, ~Q21 \/ ~Q32,

% no two queens in / diagonals ~Q13 \/ ~Q22, ~Q13 \/ ~Q31, ~Q22 \/ ~Q31, ~Q12 \/ ~Q21, ~Q23 \/ ~Q32,

% enumeration (search) labeling([], Q).


Boolean N-Queens: Arithmetic Solver

The3-queensprogramwithbooleanvariablesmaybemore

easilyexpressedwitharithmeticconstraints.

:- op(300,fy,~).:- op(400,xfy,\/).:- clpfd. queens_3(Q):-

% variable declaration Q = [Q11, Q12, Q13, Q21, Q22, Q23, Q31, Q32, Q33], domain(Q,0,1),

% one and only one queen per row Q11 + Q12 + Q13 #= 1, Q21 + Q22 + Q23 #= 1, Q31 + Q32 + Q33 #= 1,

% one and only one queen per column Q11 + Q21 + Q31 #= 1, Q12 + Q22 + Q32 #= 1, Q13 + Q23 + Q33 #= 1,

% no two queens in \ diagonals Q11+Q21 #=<1, Q11+Q33 #=<1, Q22+Q33 #=<1, Q12+Q23 #=<1, Q21+Q32 #=<1,

% no two queens in / diagonals Q13+Q22 #=<1, Q13+Q31 #=<1, Q22+Q31 #=<1, Q12+Q21 #=<1, Q23+Q32 #=<1,

% enumeration (search) labeling([], Q).


Finite Domains N-Queens

The3-queensprograminFDismuchsimpler,since

withmuchlessvariables.

:- clpfd.

queens_3(Q):-

% declaration of variables Q = [Q1,Q2,Q3], % only one queen in each row domain(Q,1,3),

% one and only queen in each column Q1 #\= Q2, Q2 #\=Q3, Q1 #\= Q3,

% no two queens in \ diagonals Q1+1 #\= Q2+2, Q1+1 #\= Q3+3, Q2+2 #\= Q3+3,

% no two queens in / diagonals Q1-1 #\= Q2-2, Q1-1 #\= Q3-3, Q2-2 #\= Q3-3,

% enumeration (search) labeling([],Q).


Finite Domains N-Queens

Ofcourse,alltheseproblemsmaybegeneralisedforanarbitrary.

sizeoftheboard,usingalltheconstructsofProlog,namelylists.

ThisisthecasewiththeFDmodel,shownbelow.

:-clpfd.

queens_fd(N,Future):- create_vars(N,Future-[]), domainQ(N,Future), compatible([],Future), enumerate(Futuras).

create_vars(0,Z-Z):- !.create_vars(N,L-Z):- M is N-1, create_vars(M,L-[N-_|Z]).

domainQ(_,[]).domainQ(N,[_-C|Variables]):- C in 1 .. N, domainQ(N,Variables).

compatible(_,[]).compatible(Past,[L-C|Future]):- compatible_1(L-C,Future), compatible([L-C|Past], Future).

compatible_1(_,[]).compatible_1(L1-C1,[L2-C2|T]):- no_atack(L1,C1,L2,C2), compatible_1(L1-C1,T).

no_atack(L1,C1,L2,C2):- C1 #\= C2, L1 + C1 #\= L2 + C2, L1 + C2 #\= L2 + C1.

enumerate(Future):- clean(Future,L), labeling([ff],L).

clean([],[]).clean([_-C|T1],[C|T]):- clean(T1,T).

Constraint (Logic) Programming

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