constraint (logic) programming

25 November 2005 Foundations of Logic and Constraint Programming 1

Constraint (Logic) Programming

Anoverview

• BooleanConstraints

• CompleteSolversforSAT

• BooleanUnification

• CompleteSolversforLinearConstraints

• Applications

[Dech04]RinaDechter,ConstraintProcessing,


Boolean Constraints

Booleandomains(oftenreferredtoas0/1variables)hasmanyapplications,namely

InDigitalCircuits

o Exemple:Half-adder

Inproblemsmodellingbinarychoices

o Exemple:N-Queens

Inproblemsdealingwithsets

B

G1

G2

AC

S


Approaches to Boolean Constraint Solving

The satisfaction of Boolean constraints may be addresses in various and alternativeways

Simbolically• BooleanUnification• CompleteSolver

• SAT• AllconstraintsareconvertedintoClauses(CNF)• Solutionconstruction(backtracking)orrepairing(localsearch)

• Finite Domains• BooleansarejustaninstanceofaFinitedomain(0/1)• TechniquescommontootherFiniteDomains


Boolean Constraints

To handle Boolean satisfaction simbolically, it is convenient to convert all

BooleanconstraintsusingexclusivelyequalityconstraintonBooleanFormulas.

BooleanFormulasarecomposedof

TwoBooleanoperators

#(exclusiveor)and·(conjunction),

Booleanconstants

0e1(andpossiblyotherconstants,domaindependent)

variábles

denotedbyuppercaseletters

Thisconversionisalwayspossible,sincetheset{0, 1, #, ·}iscomplete.

Givenarbitrarytermsaandb,all theusualoperatorsandconstantsmaybe

expressedusingexclusivelythe“#”and“·”operators.

a = 1 # a ; a b = a · b ; a b= a # b # a · b

a b = 1 # a # a · b ; a b = 1 # a # b


Boolean Rings

More formally, the tuple< A, 0, 1, #, · >, where A any domain that

includestwoelements0and1,isabooleanring,ifthefollowingpropertiesare

satisfied

Associativity

a#(b#c) = (a#b)#c a·(b·c) = (a·b)·c Comutativity

a # b = b # a a · b = b · a Distribution

a #(b·c) = (a#b)·(a#c) a·(b#c) = a·b#a·c NeutralElement

a#0 = a a·1 = a ExclusivityandIdempotence

a#a = 0 a·a = a AbsorbingElement

a·0 = 0


Boolean Constraints on Sets

AlthoughBooleanconstraintsareusuallyconsideredondomainswithonlytwo

elements,anotherinterestingdomainofapplicationisSets.

Moreprecisely,consideringadomainPofallallsetscomposedofelements

a1,a2,a3,...,an

Constant1correspondstotheUniversalset{a1,a2,a3,...,an} Constant1correspondstotheEmptyset{} Operator“#”correspondstotheSetExclusiveUnion Operator“·” correspondstoSetIntersection

Other constraintsonsetsmaybeconverted intoequality.Forexample,given

theequivalence

a b a b = a a b = b

theincludesconstraint““canberewrittenas

a # b # a·b = b a # b # b # a·b = b # b

a # a·b = 0 a·b = a


Constraint Solving through Boolean Unification

The implementation of a symbolical and complete constraint solver fo

Booleans is the core of CLP(B), the instance of the CLP(X) scheme to the

BooleanDomain.

In this solver,Booleanconstraintsaremaintained ina solved form,obtained

throughBooleanunification.

ABooleanconstraintt1 = t2 (where termst1 et2 areexclusivelywritten

has the formwithoperators “#” and “·” .Such constraint canbe satisfied iff

thereisaBooleanUnifierfortermst1andt2.

ABooleanunifierisasetofpairsx / twherexandtarerespectivelyBoolean

variablesandterms(muchinthesamewayofunifiersinlogicprogramming),

andwherenoneofthevariablesmayappearintheterms.

As in logic programming, Boolean unifiers will be denoted by greek letters.


Boolean Unification: Examples

Example 1:Termst1=1 # A and t2 = A·BadmitthefollowingBooleanunifier

= {A/1, B/0}

Infact,denotingbyt(ort)theapplicationofsubstitution to term t,

t1 = (1#A){A/1, B/0}= 1 # 1 = 0

t2 = (A·B){A/1, B/0}= 1 · 0 = 0

whichguaranteesthatconstraintt1 = t2issatisfiable.

Example 2:Termst1 = 1 # A·Bandt2= C # Dadmitunifiers

= {C/1#A·B#D} and ={C/1#D, A/0}

since

t1 = (1#A·B) {C/1#A·B#D} = 1#A·B

t2 = (C#D) {C/1#A·B#D} = 1#A·B#D#D = 1#A·B

and

t1 = (1#A·B) {C/1#D, A/0} = 1#0·B = 1

t2 = (C#D) {C/1#D, A/0} = (1#D#D) = 1


Multiple Boolean Unifiers

Like in LP it is convenient to consider a hierarchy of unifiers, related by the

“moregeneral”relationship.

Unifierismore generalthen

thereisasubstitutionsuchthat

Example:Asseenabove,termst1 = 1 # A·Bandt2= C # D,admitunifiers

= {C/1#A·B#D} and ={C/1#D, A/0}

Inthiscase,ismoregeneralthan,sincefor{A/0}, itis.

= {C/1#A·B#D } o {A / 0} = {C/1#D, A/0} =


Most General Boolean Unifiers

Ingeneral,giventwoBooleanterms,t1et2,theremightbenotonlymorethan

oneunifiers,butalsomorethanonemostgeneralunifier.

Example: Booleantermst1 = 1 # A·Bandt2= C # Dmaybeunifiedbythe

mostgeneralunifiers

1 = { C / 1 # A·B # D}

2 = { D / 1 # A·B # C}

3 = { A / 1 # C # D, B / 1}

4 = { A / 1, B / 1 # C # D}

Noneof theunifiersmaybeobtainedby thecompositionofanyof theothers

andasubstitution.


Boolean Unification Algorithm

Taking into account that constraint t1 = t2 is equivalent to t1#t2= 0, the

unification of terms t1 and t2 is equivalent to constrain term t1#t2 to 0. The

conditionsnecessarytozeroatermt,maybeanalysedfollows:

1. Zeroingaconstanttermistriviallycheckable

a. Ift = 0thetermisalreadyzero;

b. Ift = 1termcannotbezeroed

• Given distribution, a non constant term t, can always be written in the form

a·U#b(whereUisoneofthevariablesoccurringint)factoringtinordertoU.

• A term t = a·U # b may onlty be zeroed if either a = 1 or b = 0, or both.

Otherwise, i.e.a = 0andb = 1, itwouldbe t = 1 0,whatever thevalue

chosenforU.

4. Suchcondition(a = 1or b = 0) isguaranteedifftheterm(1 # a)·b is

zeroed.


Boolean Unification Algorithm

5. Onceconditiona = 1or b = 0isguaranteed,

5. ifa = 0 (andb = 0), then variableU may take an arbitraru value,since0 = 0·U # 0)

6. ifa = 1,thenvariableUmusttakevalurb(inthiscase,0 = 1·U # b)

6. The assignment to U of these terms can be implemented by means of a

fresh variable, not occurring in t. Denoting by _U this new variable, the

previousconditionisequivalentto

U = (1#a)·_U # b

I. Infact,

I. Ifa = 0(andb = 0)thenU = _U,i.e.Umaytakeanyarbitraryvalue,sincevariable_U,beingafreshvariable,isnotconstrainedanywhere.

II. Ifa = 1thenU = (1#1)·_U # b,i.e.U = b,asintendend.


Boolean Unification Procedure

• The reasoning above can be used in the implementation of predicate

unif_bool,shownbelow.

predicate unif_bool(in: t1, t2; out: );

t t1 # t2;

unif_bool zero(t,);end predicate.

• Thepredicateinputstermst1andt2,thataretobeunified,andsucceedsif

predicatezero succeedes,returningunifierobtainedfromthispredicate.


Boolean Unification Procedure

Predicatezerocannowbeimplementedasshpownbelow

predicate zero (in: t; out: );case t = 0: zero = True, = {};case t = 1: zero = False;

otherwise:

A·u # B t; s (1#A)·B;

if zero(s,σ) then

zero True; {U/(1#A)·_U#B} o σ

else

zero = False

end if

end caseend predicate


Boolean Unification Procedure: Example

{U/(1#A)·_U#B} o σ

Noticethatsubstitutionreturnedbypredicatezero isobtainedthroughthe

composition of substitutions {U/(1#A)·U#B} and σ obtained from the

recursivecallofthepredicate.

Example:SatisfyconstraintX # X·Z = Y·Z # 1

Unify X#X·Z and Y·Z#1

zero X#X·Z#Y·Z#1

= (1#Z)·X#Y·Z#1 % Ax·X#Bx

zero (1#(1#Z))·(Y·Z#1) % (1#Ax)·Bx

= Z·(Y·Z#1)

= Z·Y#Z % Ay·Y#By

zero (1#Z)·Z %(1#Ay)·By

= 0


Boolean Unification Algorithm: Example

Nowletusanalysetheunifierthatisreturnedbythesequenceofcalls

Unify X#X·Z and Y·Z#1

zero X#X·Z#Y·Z#1 = (1#Z)·X#Y·Z#1

zero (1#(1#Z))·(Y·Z#1)=Z·Y#Z %(1#Ax)·Bx

zero (1#Z)·Z = 0 %(1#Ay)·By

σz = {}

σy = {Y/(1#Ay)·_Y#By} o σz

= {Y/(1#Z) ·_Y#Z} o {}

= {Y/(1#Z) ·_Y#Z}

σx = {X/(1#Ax)·_Y#Bx} o σy

= {X/(1#1#Z)·_X # Y·Z#1} o {Y/(1#Z)·_Y#Z}

= {X/ Z·_X #Y·Z#1} o {Y/(1#Z)·_Y#Z}

= {X/ Z·_X #((1#Z)·_Y#Z)·Z#1, Y/(1#Z)·_Y#Z}

= {X/ Z·_X #Z#1, Y/(1#Z)·_Y#Z}

= {X/ Z·_X #Z#1, Y/(1#Z)·_Y#Z}


Interpreting Unifiers

Hence, constraint X#X·Z = Y·Z#1 is satisfiable, since the unification of

termsX#X·ZandY·Z#1succeeds,returningthemostgeneralunifier

= {X/Z·_X #Z#1, Y/(1#Z)·_Y#Z}

Thisresultcanbeconfirmed:

(X#X·Z) = (Z·_X#Z#1)#(Z·_X#Z#1)·Z

= (Z·_X#Z#1)#Z·_X

= Z#1

(Y·Z#1) = ((1#Z)·_Y#Z)·Z#1)

= Z#1

Analysingtheunifierwenotethatthetwotermsunifywhenever Z=0andX = 1andYisarbitrary(Y =_Y) Z=1andY = 1 andXisarbitrary(X =_X)

Theunifierhasthereforethegroundinstances

{X/1, Y/0, Z/0}, {X/1, Y/1, Z/0},

{X/0, Y/1, Z/1}, {X/1, Y/1, Z/1}.


Digital Circuits Application

Asimpleexample:1. ModelcircuitbelowwithBooleanconstraints2. Howdoestheoutputrelates(symbolically)withtheinputs.

R1:Unifier1isobtainedbysolvingR1

C = 1 # A·B 1 ={C / 1 # A·B}

R2:Then1isappliedtoR2

R2’: R2 1 : D = (1#A·C){C/1#A·B}

D = (1#A·(1#A·B))

D = 1 # A # A·B

R1: C = 1 # A·B

R2: D = 1 # A·C

R3: E = 1 # B·C

R4: F = 1 # D·E

CR1 FR4

DR2

ER3

B

A



SolvingR2’: D = 1 # A # A·B unifier’2isobtained

’2 = {D / 1 # A # A·B}

2isobtainedbycompositionof2’with1

2 = 1

o 2’

= {C/1#A·B} o {D/1#A#A·B} = {C/1#A·B, D/1#A#A·B}

R3:Applying2toR3,then

R3’: R3 2 : E = (1#B·C) {D/1#A#A·B, C/1#A·B}: E = 1#B·(1#A·B): E = 1#B·(1#A·B)

SolvingR3’,unifier3’ isobtained

3’ = {E/1#B#A·B}

Now,3’iscomposedwith2

3 = 2

o 3’

= {E/1#B#A·B} o {D/1#A#A·B} o {C/1#A·B}= {E/1#B#A·B, D/1#A#A·B, C/1#A·B}



R4:Finally,beforesolvingR4 ,3isapplied

R4’: R4 3

: F = (1#D·E) {E/1#B#A·B, D/1#A#A·B, C/1#A·B}

: F = 1#(1#A#A·B)·(1#B#A·B)

: F = 1#1#B#A·B#A#A·B#A·B#A·B#A·B#A·B

: F = A # B

Now,R4’issolvedandunifier4’ isobtained

4’ = {F/A#B}

Composing4’with3

,itisobtained4,aunifierforalltheconstraints

4 = 3

o 4’

= {E/1#B#A·B, D/1#A#A·B, C/1#A·B}o{F/A#B}

= {E/1#B#A·B, D/1#A#A·B, C/1#A·B, F/A#B}

Byinterpreting4wemayconcludethatthiscircuitwith4nandgatesimplements

theexclusive-or,since

F / A # B.


CLP(B) in SICStus

OnceloadedSICStusPrologthemoduleonBooleanConstraintsisloaded

withdirective

:- use_module(library(clpb)).

BooleanConstraintsmaybespecifiedwithbuilt-inpredicatesat(E),where

equalityisspecifiedas‘=:=’andseveralBooleanoperatorscanbeused,

namely

C = or(A,B) specified as C =:= A+B

C = nor(A,B) specified as C =\= A+B

C = xor(A,B) specified as C =\= A#B

C = and(A,B) specified as C =:= A*B

C = nand(A,B) specified as C =\= A*B

C = not(A) specified as C =\= A


CLP(B) in SICStus

SomeExamples:

1. ?- sat(A#B=:=F).

sat(A =:= B#F)?

%ConstraintA#B=FissatisfiablewithunifierA/B+F

2. ?- sat(A*B=:=1#C*D).

sat(A=\=C*D#B)?

%ConstraintA·B=1+C·DissatisfiablewithunifierA/1+C·D+B

3. ?- sat(A#B=:=1#C*D), sat(C#D=:=B).

sat(B=:=C#D),

sat(A=\=C*D#C#D) ?

%ConstraintsA+B=1+C·DandB=C+Daresatisfiablewithunifier

{A/1+C·D+C+D, B/C+D}


CLP(B) in SICStus

Amorecompleteexample,relatedtothepreviouscircuit

| ?- circuit(A,B,[C,D,E],1).C = 1,E = A,sat(B=\=A),sat(D=\=A) ?

Note: Comparethisanswerwithunifier 5 = {E/1+B, D/B, C/1, F/1, A/1+B } foundbefore.

CG1

FG4

DG2

EG3

B

A:- use_module(library(clpb)).nand_gate(X,Y,Z):- sat(X*Y =:= 1#Z).

circuit(A,B,[C,D,E],F):- nand_gate(A,B,C), nand_gate(A,C,D), nand_gate(B,C,E), nand_gate(D,E,F).


CLP(Q) Application

Anotherdomainforwhichacompletesolverexistsisthatoflinearconstraints

overtherationalsorreals.TheintegrationofsuchasolverwithLPforms

theCLP(Q)orCLP(R)instanceoftheCLP(X)scheme.

CLP(Q)andCLP(R)differintheirapproachhandlearithmetics:

Ifthecoeficientsarerational(divisionbetweenintegers)thesolutiona

arealsorational.HenceCLP(Q)avoidsroundingerrorsbyhandling

properlywitharithmetics(“+”,”–”,“*”,“/”)overrationals(witharithmetics

withoutrounding

CLP(R)roundsrationalsasfloatingpointnumbers,andismore

“efficient”ifless“precise”

Theseconstraintsarequitecommoninapplications.Evenwhenconstraints

arenotlinear,problemsareoften“linearised”(i.e.modelledwithlinear

constraintssuchtahttheerrorsareacceptable),giventheveryefficientway

ofsolvingtheseconstraints,basedontheSIMPLEXalgorithm.


CLP(Q) : Example

Contraints:% Minimum required flow x 6, z 10 % minimum flow% flow capacity of arcs a 5, b 3, c 7, d 2, e 8, f 6% processing capacity of nodes x 7, y 9, a + b 9, c 8, a + b + d 9, e + f 13% flow maintenance x = a, y = b + c, a + b + d = e, c = d + f, e + f = z% non-negative constraints x, y ,z, a, b, c, d, e, f 0

7

8

9 9

13

5/a

2/d 8/e

6/f

3/b

7/c

x

y

z

The following problem illustrates a typical use

oflinearconstraints:

Problem (Network Management):

Findthevalueoftrafficfrompointsx,ytoz

such that the traffic does not exceed the

flow capacity of each arc, nor the

processingcapacityofeachnode.


CLP(Q) Solver: SIMPLEX

TheSIMPLEXalgorithmaimsatconvertingalllinearconstraintsintoasolved

form, composed of equality constraints on linear expressions. This

conversionisdoneintwosteps.

1. All inequality constraints on decision variables are converted into equality

constraintswiththeadditionof“slack”(non-negative)variables

a X1 +...+ an Xn ≤ b a X1 +...+ an Xn + S = b

a X1 +...+ an Xn ≥ b a X1 +...+ an Xn - S = b

2. In general m constraints on n decision variables are converted into m

constraints on k+n non-negative variables, where k m ( k = m if all

constraintsareinequalityconstraints).

3. Asolvedformisobtainedifapartitioncanbefoundonthek+nvariablesinm

basicvariablesXi(i1..n)andp = k+n-mnonbasicvariablesYj(j1..p),

suchthatthemconstraintscanbewrittenas

Xi = di + ci1 Y1+ ... + cip Yp (foralli1..n)

whereallfreecoefficientsdiarenon-negative.


SIMPLEX: Geometric Interpretation

X2 = 0

X2

X1

X3 = 0

X5 = 0

X1 = 0

X4 = 0

X3 = 8 - 2 X1 - X2X4 = -3 + X1 + X2X5 = 5 + X1 - X2

x1 = 1 - x3/3 + x5/3x2 = 6 - x3/3 - 2 x5/3x4 = 4 - 2 x3/3 - X5/3

X1 = 5 - X3 - X4X2 = -2 + X3 + 2 X4X5 = 12 - 2 X3 - 3 X4

X2 = 5 + X1 - X5X3 = 3 - 3 X1 + X5X4 = 2 + 2 X1 - X5

2 X1 + X2 ≤ 8 X1 + X2 ≥ 3 X1 - X2 ≥ -5 X1 , X2 ≥ 0

2 X1 + X2 + X3 = 8 X1 + X2 - X4 = 3 X1 - X2 - X5 = -5


SICStus CLP(Q)/CLP(R) Solvers

SICStusPrologsupportsconstraintsolvingovertheRational/Realnumbers.As

for Booleans, to handle these constraints, a module for CLP(Q) or CLP(R)

mustbeloaded,withdirectives:

:- use_module(user:library(clpq)).

:- use_module(user:library(clpr)).

ContraintsmaynowbespecifiedwiththeusualPrologsyntaxbutinside{}.Forexamplethepreviousproblemmaybedirectlyspecifiedas

p1([X1,X2]) :-

{2*X1 + X2 =< 8, X1 + X2 >= 3, X1 - X2 >= -5 },

{ X1 >= 0, X2 >= 0 }.

orspecifiedwithexplicitslackvariables

p2([X1,X2,X3,X4,X5]) :-

{2*X1+X2+X3 = 8, X1+X2–X4 = 3, X1-X2–X5 = -5 },

{ X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0, X5 >= 0} .



Once defined these programs can be run. Answers projected to the called

variables,withsomeextravariablesincludedwhennecessary

p1([X1,X2]) :-

{2*X1 + X2 =< 8, X1 + X2 >= 3, X1 - X2 >= -5 },

{ X1 >= 0, X2 >= 0 }.

| ?- p1([X1,X2]).

{X1+1/2*X2=<4},

{X1>=0},

{X2>=0},

{X1+X2>=3},

{X1-X2>= -5} ?

X2 >= 0

X2

2 X1 + X2 = 8

X1 - X2 = -5X1 >= 0

X1 + X2 = 3

X1



Thesolvedformismoreeasily(?)identifiedwiththeexplicitslackvariables.

p2([X1,X2,X3,X4,X5]) :-

{2*X1+X2+X3 = 8, X1+X2–X4 = 3, X1-X2–X5 = -5 },

{ X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0, X5 >= 0} .

| ?- p2([X1,X2,X3,X4,X5]).

{X5=9-3/2*X2-1/2*X3},

{X4=1+1/2*X2-1/2*X3},

{X1=4-1/2*X2-1/2*X3},

{X2+X3=<8},

{X2+1/3*X3=<6},

{X2>=0},

{X3>=0},

{X2-X3>= -2} ?

SolvedformidentifiesX1 = 4, X2 = 0

X2 >= 0

X2

2 X1 + X2 = 8

X1 - X2 = -5X1 >= 0

X1 + X2 = 3

X1



Ofcourse,iftheproblemhasnosolutionstheanswerisno!

p3([X1,X2]) :-

{2*X1 + X2 >= 8, X1 + X2 =< 3, X1-X2 =< -5 },

{ X1 >= 0, X2 >= 0} .

| ?- p3([X1,X2]).

no ?

X2 >= 0

X2

2 X1 + X2 = 8

X1 - X2 = -5X1 >= 0

X1 + X2 = 3

X1


SICStus CLP(Q)/CLP(R): Example

Example:

Acertainamountofmoneyislentatsomeinterestrateforanumberoftime

periods,tobepayedwithconstantinstalments.

Model

Assumingthat Iisthenumberoftimeperiodsremainingtopaythedebt Di representsthedebtatthebeginingoftimeperiodi.

Ristheinterestrate(inpercentagepoints) Pisthe(constant)instalmenttobepayedattheendofeachtimeperiod

thenwehavethefollowingrelations:

Thedebtatthebeginingofaperiodisthedebtatthebeginingofthepreviousperiod,increasedbytheinterest,lessthepaymentmade.

Di = Di+1 * (1 +R/100) – P for i > 1

Nodebtshouldexistattheendofthelasttimeperiodlasttimeperiod

0 = D1 * (1 +R/100) – P



TheseconstraintscanbeeasilyexpressedinCLP(Q)/CLP(R)asshowninthe

programsbelow.

Notice that such program could not be easily converted to pure Prolog, in

statementslikeV is Exp,sinceonedoesnotknowwhatvaluesaregivenand

whatareknow.

% 0 = D1 * (1 +R/100) – P debt(Debt, Rate, TimeLeft, Payment):- { TimeLeft = 1, Payment = Debt * (1 + Rate/100) }.

% Di = Di+1 * (1 +R/100) – P for i > 1debt(Debt, Rate, TimeLeft, Payment):- { TimeLeft > 1, NextTimeLeft = TimeLeft - 1, NextDebt = Debt * (1 + Rate/100) – Payment }, debt(NextDebt, Rate, NextTimeLeft, Payment).



Someinteractionwiththeprogram

Whatyearlypaymentisduewhen100unitsarelentfor7yearsata5%

interestrate.Whatisthetotalpayment?

| ?- debt(100,5,7,P), {T = 7* P}.

P = 17.28198184461708,

T = 120.97387291231955 ?

Whatamountofmoneycanbeborrowedbysomeonethatiswillingtopay

afixedinstalmentof10unitsfor7years,whnetheinterestrateis5%?

| ?- debt(D,5,7,10).

D = 57.86373397397567 ?

Whatistherelationshipbetweentheyearlypaymentandtheinitialdebt,

when7yearsareusedandtheinterestrateis5%?

| ?- debt(D,5,7,P).

{P=0.17281981844617078*D} ?



ItisimportanttonoticethatCLP(Q)/CLP(R)isonlycompletewhenthe

constraintsarelinear.Whentheyarenottheyarefrozenintheconstraintstore,

pendingfurtherinstantiation

Whatistherelationbetweentheyearlypaymentandtheinitialdebt,when

7yearsareusedandtheinterestrateisR%

| ?- debt(D,R,7,10).

clpr:{10.0-D-0.01*(R*D)+_A=0.0}, % 7 years to go

clpr:{10.0-0.01*(_B*R)-_B=0.0}, % 6 years to go

clpr:{10.0+_C-0.01*(_A*R)-_A=0.0}, % 5 years to go

clpr:{10.0+_D-0.01*(_C*R)-_C=0.0}, % 4 years to go

clpr:{10.0+_E-0.01*(_D*R)-_D=0.0}, % 3 years to go

clpr:{10.0+_F-0.01*(_E*R)-_E=0.0}, % 2 years to go

clpr:{10.0+_B-0.01*(_F*R)-_F=0.0} ? % 1 years to go

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