The clp(Q,R) system described in this document is an instance of the general Constraint Logic Programming scheme introduced by [Jaffar & Michaylov 87].
The implementation is at least as complete as other existing clp(R) implementations: It solves linear equations over rational or real valued variables, covers the lazy treatment of nonlinear equations, features a decision algorithm for linear inequalities that detects implied equations, removes redundancies, performs projections (quantifier elimination), allows for linear dis-equations, and provides for linear optimization.
The full clp(Q,R) distribution, including a stand-alone manual
and an examples directory that is possibly more up to date than
the version in the SICStus Prolog distribution, is available from:
http://www.ai.univie.ac.at/clpqr/
.
When referring to this implementation of clp(Q,R) in publications, you should use the following reference:
Holzbaur C.: OFAI clp(q,r) Manual, Edition 1.3.3, Austrian Research Institute for Artificial Intelligence, Vienna, TR-95-09, 1995.
The development of this software was supported by the Austrian Fonds zur Foerderung der Wissenschaftlichen Forschung under grant P9426-PHY. Financial support for the Austrian Research Institute for Artificial Intelligence is provided by the Austrian Federal Ministry for Science and Research.
We include a collection of examples that has been distributed with the Monash University version of clp(R) [Heintze et al. 87], and its inclusion into this distribution was kindly permitted by Roland Yap.
Until rational numbers become first class citizens in SICStus Prolog, rational arithmetics has to be emulated. Because of the emulation it is too expensive to support arithmetics with automatic coercion between all sorts of numbers, like you find it in CommonLisp, for example.
You must choose whether you want to operate in the field of Q (Rationals) or R (Reals):
| ?- use_module(library(clpq)). or | ?- use_module(library(clpr)).
You can also load both modules, but the exported predicates listed below
will name-clash (see section Importation).
You can avoid the interactive resolution dialog if the importation is skipped,
e.g. via: use_module(library(clpq),[]),use_module(library(clpr),[])
.
Throughout this chapter, the prompts clp(q) ?-
and clp(r) ?-
are used to differentiate between clp(Q) and clp(R) in exemplary interactions.
In general there are many ways to express the same linear relationship. This degree of freedom is manifest in the fact that the printed manual and an actual interaction with the current version of clp(Q,R) may show syntactically different answer constraints, despite the fact the same semantic relationship is being expressed. There are means to control the presentation; see section Variable Ordering. The approximative nature of floating point numbers may also produce numerical differences between the text in this manual and the actual results of clp(R), for a given edition of the software.
The solver interface for both Q and R consists of the following
predicates which are exported from module(linear)
.
{+Constraint}
| ?- clpr:{Ar+Br=10}, Ar=Br, clpq:{Aq+Bq=10}, Aq=Bq. Aq = 5, Ar = 5.0, Bq = 5, Br = 5.0Although clp(Q) and clp(R) are independent modules, you are asking for trouble if you (accidently) share variables between them:
| ?- clpr:{A+B=10}, clpq:{A=B}. {TYPE ERROR: _118=5.0 - arg 2: expected 'a rational number', found 5.0}This is because both solvers eventually compute values for the variables and Reals are incompatible with Rationals. Here is the constraint grammar:
Constraint --> C | C , C conjunction C --> Expr =:= Expr equation | Expr = Expr equation | Expr < Expr strict inequation | Expr > Expr strict inequation | Expr =< Expr nonstrict inequation | Expr >= Expr nonstrict inequation | Expr =\= Expr disequation Expr --> variable Prolog variable | number floating point or integer | + Expr unary plus | - Expr unary minus | Expr + Expr addition | Expr - Expr subtraction | Expr * Expr multiplication | Expr / Expr division | abs(Expr) absolute value | sin(Expr) trigonometric sine | cos(Expr) trigonometric cosine | tan(Expr) trigonometric tangent | pow(Expr,Expr) raise to the power | exp(Expr,Expr) raise to the power | min(Expr,Expr) minimum of the two arguments | max(Expr,Expr) maximum of the two arguments | #(Const) symbolic numerical constantsConjunctive constraints
{C,C}
have been made part of the syntax
to control the granularity of constraint submission, which will be exploited by
future versions of this software.
Symbolic numerical constants are provided for compatibility only;
see section Monash Examples.
entailed(+Constraint)
clp(q) ?- {A =< 4}, entailed(A=\=5). {A=<4} yes clp(q) ?- {A =< 4}, entailed(A=\=3). no
inf(+Expr, -Inf)
sup(+Expr, -Sup)
clp(q) ?- { 2*X+Y =< 16, X+2*Y =< 11, X+3*Y =< 15, Z = 30*X+50*Y }, sup(Z, Sup). Sup = 310, {Z=30*X+50*Y}, {X+1/2*Y=<8}, {X+3*Y=<15}, {X+2*Y=<11}
minimize(+Expr)
minimize(Expr) :- inf(Expr, Expr).
maximize(+Expr)
clp(q) ?- { 2*X+Y =< 16, X+2*Y =< 11, X+3*Y =< 15, Z = 30*X+50*Y }, maximize(Z). X = 7, Y = 2, Z = 310
bb_inf(+Ints, +Expr, -Inf)
clp(q) ?- {X >= Y+Z, Y > 1, Z > 1}, bb_inf([Y,Z],X,Inf). Inf = 4, {Y>1}, {Z>1}, {X-Y-Z>=0}
bb_inf(+Ints, +Expr, -Inf, -Vertex, +Eps)
abs(round(X)-X) < Eps
.
The predicate bb_inf/3
uses Eps = 0.001
.
With clp(Q), Eps = 0
makes sense.
Vertex is a list of the same length as Ints and contains
the (integral) values for Ints, such that the infimum is produced
when assigned. Note that this will only generate one particular solution,
which is different from the situation with minimize/1
, where the
general solution is exhibited.
ordering(+Spec)
dump(+Target, -NewVars, -CodedAnswer)
clp(q) ?- {A+B =< 10, A>=4}, dump([A,B],Vs,Cs), dump([B],Bp,Cb). Cb = [_A=<6], Bp = [_A], Cs = [_B>=4,_C+_B=<10], Vs = [_C,_B], {A>=4}, {A+B=<10}The current version of
dump/3
is incomplete with respect to
nonlinear constraints. It only reports nonlinear constraints that
are connected to the target variables. The following example has
no solution. From the top level's report we have a chance to deduce
this fact, but dump/3
currently has no means to collect
global constraints ...
q(X) :- {X>=10}, {sin(Z)>3}. clp(r) ?- q(X), dump([X],V,C). C = [_A>=10.0], V = [_A], clpr:{3.0-sin(_B)<0.0}, {X>=10.0}
Equality constraints are added to the store implicitly each time variables that have been mentioned in explicit constraints are bound - either to another such variable or to a number.
clp(r) ?- {2*A+3*B=C/2}, C=10.0, A=B. A = 1.0, B = 1.0, C = 10.0
Is equivalent modulo rounding errors to
clp(r) ?- {2*A+3*B=C/2, C=10, A=B}. A = 1.0, B = 0.9999999999999999, C = 10.0
The shortcut bypassing the use of {}/1
is allowed and makes sense because the interpretation of
this equality in Prolog and clp(R) coincides.
In general, equations involving interpreted functors, +/2
in this case,
must be fed to the solver explicitly:
clp(r) ?- X=3.0+1.0, X=4.0. no
Further, variables known by clp(R) may be bound directly to floats only. Likewise, variables known by clp(Q) may be bound directly to rational numbers only; see section Rationals. Failing to do so is rewarded with an exception:
clp(q) ?- {2*A+3*B=C/2}, C=10.0, A=B. {TYPE ERROR: _165=10.0 - arg 2: expected 'a rational number', found 10.0}
This is because 10.0
is not a rational constant. To make clp(Q) happy
you have to say:
clp(q) ?- {2*A+3*B=C/2}, C=rat(10,1), A=B. A = 1, B = 1, C = 10
If you use {}/1
, you don't have to worry about such details.
Alternatively, you may use the automatic expansion facility, check section Syntactic Sugar.
What was covered so far was how the user populates the constraint store. The other direction of the information flow consists of the success and failure of the above predicates and the binding of variables to numerical values and the aliasing of variables. Example:
clp(r) ?- {A-B+C=10, C=5+5}. B = A, C = 10.0
The linear constraints imply A=B
and the solver consequently
exports this binding to the Prolog world, which is manifest in the fact
that the test A==B
will succeed.
More about answer presentation in section Projection and Redundancy Elimination.
The clp(Q,R) system is restricted to deal with linear constraints because the decision algorithms for general nonlinear constraints are prohibitively expensive to run. If you need this functionality badly, you should look into symbolic algebra packages. Although the clp(Q,R) system cannot solve nonlinear constraints, it will collect them faithfully in the hope that through the addition of further (linear) constraints they might get simple enough to solve eventually. If an answer contains nonlinear constraints, you have to be aware of the fact that success is qualified modulo the existence of a solution to the system of residual (nonlinear) constraints:
clp(r) ?- {sin(X) = cos(X)}. clpr:{sin(X)-cos(X)=0.0}
There are indeed infinitely many solutions to this constraint
(X = 0.785398 + n*Pi
), but clp(Q,R) has no direct means to find
and represent them.
The systems goes through some lengths to recognize linear expressions as such. The method is based on a normal form for multivariate polynomials. In addition, some simple isolation axioms, that can be used in equality constraints, have been added. The current major limitation of the method is that full polynomial division has not been implemented. Examples:
This is an example where the isolation axioms are sufficient to determine the value of X.
clp(r) ?- {sin(cos(X)) = 1/2}. X = 1.0197267436954502
If we change the equation into an inequation, clp(Q,R) gives up:
clp(r) ?- {sin(cos(X)) < 1/2}. clpr:{sin(cos(X))-0.5<0.0}
The following is easy again:
clp(r) ?- {sin(X+2+2)/sin(4+X) = Y}. Y = 1.0
And so is this:
clp(r) ?- {(X+Y)*(Y+X)/X = Y*Y/X+99}. {Y=49.5-0.5*X}
An ancient symbol manipulation benchmark consists in rising the expression
X+Y+Z+1
to the 15th power:
clp(q) ?- {exp(X+Y+Z+1,15)=0}. clpq:{Z^15+Z^14*15+Z^13*105+Z^12*455+Z^11*1365+Z^10*3003+... ... polynomial continues for a few pages ... =0}
Computing its roots is another story.
Binding variables that appear in nonlinear residues will reduce the complexity of the nonlinear expressions and eventually results in linear expressions:
clp(q) ?- {exp(X+Y+1,2) = 3*X*X+Y*Y}. clpq:{Y*2-X^2*2+Y*X*2+X*2+1=0}
Equating X and Y collapses the expression completely and even determines the values of the two variables:
clp(q) ?- {exp(X+Y+1,2) = 3*X*X+Y*Y}, X=Y. X = -1/4, Y = -1/4
These axioms are used to rewrite equations such that the variable to be solved for is moved to the left hand side and the result of the evaluation of the right hand side can be assigned to the variable. This allows, for example, to use the exponentiation operator for the computation of roots and logarithms, see below.
A = B * C
A = B / C
X = min(Y,Z)
X = max(Y,Z)
X = abs(Y)
X = pow(Y,Z), X = exp(Y,Z)
clp(r) ?- { 12=pow(2,X) }. X = 3.5849625007211565 clp(r) ?- { 12=pow(X,3.585) }. X = 1.9999854993443926 clp(r) ?- { X=pow(2,3.585) }. X = 12.000311914286545
X = sin(Y)
clp(r) ?- { 1/2 = sin(X) }. X = 0.5235987755982989
X = cos(Y)
X = tan(Y)
The fact that you can switch between clp(R) and clp(Q) should solve most of your numerical problems regarding precision. Within clp(Q), floating point constants will be coerced into rational numbers automatically. Transcendental functions will be approximated with rationals. The precision of the approximation is limited by the floating point precision. These two provisions allow you to switch between clp(R) and clp(Q) without having to change your programs.
What is to be kept in mind however is the fact that it may take quite big rationals to accommodate the required precision. High levels of precision are for example required if your linear program is ill-conditioned, i.e., in a full rank system the determinant of the coefficient matrix is close to zero. Another situation that may call for elevated levels of precision is when a linear optimization problem requires exceedingly many pivot steps before the optimum is reached.
If your application approximates irrational numbers, you may be out of space particularly soon. The following program implements N steps of Newton's approximation for the square root function at point 2.
% % from file: library('clpqr/examples/root') % root(N, R) :- root(N, 1, R). root(0, S, R) :- !, S=R. root(N, S, R) :- N1 is N-1, { S1 = S/2 + 1/S }, root(N1, S1, R).
It is known that this approximation converges quadratically, which means that the number of correct digits in the decimal expansion roughly doubles with each iteration. Therefore the numerator and denominator of the rational approximation have to grow likewise:
clp(q) ?- use_module(library('clpqr/examples/root')). clp(q) ?- root(3,R),print_decimal(R,70). 1.4142156862 7450980392 1568627450 9803921568 6274509803 9215686274 5098039215 R = 577/408 clp(q) ?- root(4,R),print_decimal(R,70). 1.4142135623 7468991062 6295578890 1349101165 5962211574 4044584905 0192000543 R = 665857/470832 clp(q) ?- root(5,R),print_decimal(R,70). 1.4142135623 7309504880 1689623502 5302436149 8192577619 7428498289 4986231958 R = 886731088897/627013566048 clp(q) ?- root(6,R),print_decimal(R,70). 1.4142135623 7309504880 1688724209 6980785696 7187537723 4001561013 1331132652 R = 1572584048032918633353217/1111984844349868137938112 clp(q) ?- root(7,R),print_decimal(R,70). 1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 R = 4946041176255201878775086487573351061418968498177 / 3497379255757941172020851852070562919437964212608
Iterating for 8 steps produces no further change in the first 70 decimal digits
of sqrt(2)
. After 15 steps the approximating rational number has a
numerator and a denominator with 12543 digits each, and the next step runs
out of memory.
Another irrational number that is easily computed is e.
The following program implements an alternating series
for 1/e
, where the absolute value of last term is an
upper bound on the error.
% % from file: library('clpqr/examples/root') % e(N, E) :- { Err =:= exp(10,-(N+2)), Half =:= 1/2 }, inv_e_series(Half, Half, 3, Err, Inv_E), { E =:= 1/Inv_E }. inv_e_series(Term, S0, _, Err, Sum) :- { abs(Term) =< Err }, !, S0 = Sum. inv_e_series(Term, S0, N, Err, Sum) :- N1 is N+1, { Term1 =:= -Term/N, S1 =:= Term1+S0 }, inv_e_series(Term1, S1, N1, Err, Sum).
The computation of the rational number E that approximates e up
to at least
1000 digits in its decimal expansion requires the evaluation of 450 terms of
the series, i.e. 450 calls of inv_e_series/5
.
clp(q) ?- e(1000,E). E = 7149056228932760213666809592072842334290744221392610955845565494 3708750229467761730471738895197792271346693089326102132000338192 0131874187833985420922688804220167840319199699494193852403223700 5853832741544191628747052136402176941963825543565900589161585723 4023097417605004829991929283045372355639145644588174733401360176 9953973706537274133283614740902771561159913069917833820285608440 3104966899999651928637634656418969027076699082888742481392304807 9484725489080844360397606199771786024695620205344042765860581379 3538290451208322129898069978107971226873160872046731879753034549 3130492167474809196348846916421782850086985668680640425192038155 4902863298351349469211627292865440876581064873866786120098602898 8799130098877372097360065934827751120659213470528793143805903554 7928682131082164366007016698761961066948371407368962539467994627 1374858249110795976398595034606994740186040425117101588480000000 0000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000 / 2629990810403002651095959155503002285441272170673105334466808931 6863103901346024240326549035084528682487048064823380723787110941 6809235187356318780972302796570251102928552003708556939314795678 1978390674393498540663747334079841518303636625888963910391440709 0887345797303470959207883316838346973393937778363411195624313553 8835644822353659840936818391050630360633734935381528275392050975 7271468992840907541350345459011192466892177866882264242860412188 0652112744642450404625763019639086944558899249788084559753723892 1643188991444945360726899532023542969572584363761073528841147012 2634218045463494055807073778490814692996517359952229262198396182 1838930043528583109973872348193806830382584040536394640895148751 0766256738740729894909630785260101721285704616818889741995949666 6303289703199393801976334974240815397920213059799071915067856758 6716458821062645562512745336709063396510021681900076680696945309 3660590933279867736747926648678738515702777431353845466199680991 73361873421152165477774911660108200059
The decimal expansion itself looks like this:
clp(q) ?- e(1000, E), print_decimal(E, 1000). 2. 7182818284 5904523536 0287471352 6624977572 4709369995 9574966967 6277240766 3035354759 4571382178 5251664274 2746639193 2003059921 8174135966 2904357290 0334295260 5956307381 3232862794 3490763233 8298807531 9525101901 1573834187 9307021540 8914993488 4167509244 7614606680 8226480016 8477411853 7423454424 3710753907 7744992069 5517027618 3860626133 1384583000 7520449338 2656029760 6737113200 7093287091 2744374704 7230696977 2093101416 9283681902 5515108657 4637721112 5238978442 5056953696 7707854499 6996794686 4454905987 9316368892 3009879312 7736178215 4249992295 7635148220 8269895193 6680331825 2886939849 6465105820 9392398294 8879332036 2509443117 3012381970 6841614039 7019837679 3206832823 7646480429 5311802328 7825098194 5581530175 6717361332 0698112509 9618188159 3041690351 5988885193 4580727386 6738589422 8792284998 9208680582 5749279610 4841984443 6346324496 8487560233 6248270419 7862320900 2160990235 3043699418 4914631409 3431738143 6405462531 5209618369 0888707016 7683964243 7814059271 4563549061 3031072085 1038375051 0115747704 1718986106 8739696552 1267154688 9570350354
Once a derivation succeeds, the Prolog system presents the bindings for the
variables in the query. In a CLP system, the set of answer constraints is
presented in analogy. A complication in the CLP context are variables and
associated
constraints that were not mentioned in the query. A motivating example is
the familiar mortgage
relation:
% % from file: library('clpqr/examples/mg') % mg(P,T,I,B,MP):- { T = 1, B + MP = P * (1 + I) }. mg(P,T,I,B,MP):- { T > 1, P1 = P * (1 + I) - MP, T1 = T - 1 }, mg(P1, T1, I, B, MP).
A sample query yields:
clp(r) ?- use_module(library('clpqr/examples/mg')). clp(r) ?- mg(P,12,0.01,B,Mp). {B=1.1268250301319698*P-12.682503013196973*Mp}
Without projection of the answer constraints onto the query variables we would observe the following interaction:
clp(r) ?- mg(P,12,0.01,B,Mp). {B=12.682503013196973*_A-11.682503013196971*P}, {Mp= -(_A)+1.01*P}, {_B=2.01*_A-1.01*P}, {_C=3.0301*_A-2.0301*P}, {_D=4.060401000000001*_A-3.0604009999999997*P}, {_E=5.101005010000001*_A-4.10100501*P}, {_F=6.152015060100001*_A-5.152015060099999*P}, {_G=7.213535210701001*_A-6.213535210700999*P}, {_H=8.285670562808011*_A-7.285670562808009*P}, {_I=9.368527268436091*_A-8.36852726843609*P}, {_J=10.462212541120453*_A-9.46221254112045*P}, {_K=11.566834666531657*_A-10.566834666531655*P}
The variables _A ... _K are not part of the query, they originate from the mortgage program proper. Although the latter answer is equivalent to the former in terms of linear algebra, most users would prefer the former.
In general, there are many ways to express the same linear relationship
between variables. clp(Q,R) does not care to distinguish between them,
but the user might. The predicate ordering(+Spec)
gives
you some control over the variable ordering. Suppose that instead of
B, you want Mp to be the defined variable:
clp(r) ?- mg(P,12,0.01,B,Mp). {B=1.1268250301319698*P-12.682503013196973*Mp}
This is achieved with:
clp(r) ?- mg(P,12,0.01,B,Mp), ordering([Mp]). {Mp= -0.0788487886783417*B+0.08884878867834171*P}
One could go one step further and require P to appear before (to the left of) B in a addition:
clp(r) ?- mg(P,12,0.01,B,Mp), ordering([Mp,P]). {Mp=0.08884878867834171*P-0.0788487886783417*B}
Spec in ordering(+Spec)
is either a list of variables with
the intended ordering, or of the form A<B
.
The latter form means that A goes to the left of B.
In fact, ordering([A,B,C,D])
is shorthand for:
ordering(A < B), ordering(A < C), ordering(A < D), ordering(B < C), ordering(B < D), ordering(C < D)
The ordering specification only affects the final presentation of the
constraints. For all other operations of clp(Q,R), the ordering is immaterial.
Note that ordering/1
acts like a constraint: you can put it anywhere
in the computation, and you can submit multiple specifications.
clp(r) ?- ordering(B < Mp), mg(P,12,0.01,B,Mp). {B= -12.682503013196973*Mp+1.1268250301319698*P} yes clp(r) ?- ordering(B < Mp), mg(P,12,0.01,B,Mp), ordering(P < Mp). {P=0.8874492252651537*B+11.255077473484631*Mp}
In meta-programming applications one needs to get a grip on the results
computed by the clp(Q,R) solver. You can use the predicates dump/3
and/or call_residue/2
for that purpose:
clp(r) ?- {2*A+B+C=10,C-D=E,A<10}, dump([A,B,C,D,E],[a,b,c,d,e],Constraints). Constraints = [e<10.0,a=10.0-c-d-2.0*e,b=c+d], {C=10.0-2.0*A-B}, {E=10.0-2.0*A-B-D}, {A<10.0}
clp(r) ?- call_residue({2*A+B+C=10,C-D=E,A<10}, Constraints). Constraints = [ [A]-{A<10.0}, [B]-{B=10.0-2.0*A-C}, [D]-{D=C-E} ]
As soon as linear inequations are involved, projection gets more
demanding complexity wise. The current clp(Q,R) version uses a
Fourier-Motzkin algorithm for the projection of linear inequalities.
The choice of a suitable algorithm is somewhat dependent on the number
of variables to be eliminated, the total number of variables, and other
factors. It is quite easy to produce problems of moderate size where the
elimination step takes some time. For example, when the dimension
of the projection is 1, you might be better off computing the supremum
and the infimum of the remaining variable instead of eliminating
n-1
variables via implicit projection.
In order to make answers as concise as possible, redundant constraints are removed by the system as well. In the following set of inequalities, half of them are redundant.
% % from file: library('clpqr/examples/eliminat') % example(2, [X0,X1,X2,X3,X4]) :- { +87*X0 +52*X1 +27*X2 -54*X3 +56*X4 =< -93, +33*X0 -10*X1 +61*X2 -28*X3 -29*X4 =< 63, -68*X0 +8*X1 +35*X2 +68*X3 +35*X4 =< -85, +90*X0 +60*X1 -76*X2 -53*X3 +24*X4 =< -68, -95*X0 -10*X1 +64*X2 +76*X3 -24*X4 =< 33, +43*X0 -22*X1 +67*X2 -68*X3 -92*X4 =< -97, +39*X0 +7*X1 +62*X2 +54*X3 -26*X4 =< -27, +48*X0 -13*X1 +7*X2 -61*X3 -59*X4 =< -2, +49*X0 -23*X1 -31*X2 -76*X3 +27*X4 =< 3, -50*X0 +58*X1 -1*X2 +57*X3 +20*X4 =< 6, -13*X0 -63*X1 +81*X2 -3*X3 +70*X4 =< 64, +20*X0 +67*X1 -23*X2 -41*X3 -66*X4 =< 52, -81*X0 -44*X1 +19*X2 -22*X3 -73*X4 =< -17, -43*X0 -9*X1 +14*X2 +27*X3 +40*X4 =< 39, +16*X0 +83*X1 +89*X2 +25*X3 +55*X4 =< 36, +2*X0 +40*X1 +65*X2 +59*X3 -32*X4 =< 13, -65*X0 -11*X1 +10*X2 -13*X3 +91*X4 =< 49, +93*X0 -73*X1 +91*X2 -1*X3 +23*X4 =< -87 }.
Consequently, the answer consists of the system of nine non-redundant inequalities only:
clp(q) ?- use_module(library('clpqr/examples/elimination')). clp(q) ?- example(2, [X0,X1,X2,X3,X4]). {X0-2/17*X1-35/68*X2-X3-35/68*X4>=5/4}, {X0-73/93*X1+91/93*X2-1/93*X3+23/93*X4=<-29/31}, {X0-29/25*X1+1/50*X2-57/50*X3-2/5*X4>=-3/25}, {X0+7/39*X1+62/39*X2+18/13*X3-2/3*X4=<-9/13}, {X0+2/19*X1-64/95*X2-4/5*X3+24/95*X4>=-33/95}, {X0+2/3*X1-38/45*X2-53/90*X3+4/15*X4=<-34/45}, {X0-23/49*X1-31/49*X2-76/49*X3+27/49*X4=<3/49}, {X0+44/81*X1-19/81*X2+22/81*X3+73/81*X4>=17/81}, {X0+9/43*X1-14/43*X2-27/43*X3-40/43*X4>=-39/43}
The projection (the shadow) of this polyhedral set into the X0,X1
space can be computed via the implicit elimination of non-query variables:
clp(q) ?- example(2, [X0,X1|_]). {X0+2619277/17854273*X1>=-851123/17854273}, {X0+6429953/16575801*X1=<-12749681/16575801}, {X0+19130/1213083*X1>=795400/404361}, {X0-1251619/3956679*X1>=21101146/3956679}, {X0+601502/4257189*X1>=220850/473021}
Projection is quite a powerful concept that leads to surprisingly terse executable specifications of nontrivial problems like the computation of the convex hull from a set of points in an n-dimensional space: Given the program
% % from file: library('clpqr/examples/elimination') % conv_hull(Points, Xs) :- lin_comb(Points, Lambdas, Zero, Xs), zero(Zero), polytope(Lambdas). polytope(Xs) :- positive_sum(Xs, 1). positive_sum([], Z) :- {Z=0}. positive_sum([X|Xs], SumX) :- { X >= 0, SumX = X+Sum }, positive_sum(Xs, Sum). zero([]). zero([Z|Zs]) :- {Z=0}, zero(Zs). lin_comb([], [], S1, S1). lin_comb([Ps|Rest], [K|Ks], S1, S3) :- lin_comb_r(Ps, K, S1, S2), lin_comb(Rest, Ks, S2, S3). lin_comb_r([], _, [], []). lin_comb_r([P|Ps], K, [S|Ss], [Kps|Ss1]) :- { Kps = K*P+S }, lin_comb_r(Ps, K, Ss, Ss1).
we can post the following query:
clp(q) ?- conv_hull([ [1,1], [2,0], [3,0], [1,2], [2,2] ], [X,Y]). {Y=<2}, {X+1/2*Y=<3}, {X>=1}, {Y>=0}, {X+Y>=2}
This answer is easily verified graphically:
| 2 - * * | | 1 - * | | 0 -----|----*----*---- 1 2 3
The convex hull program directly corresponds to the mathematical definition of the convex hull. What does the trick in operational terms is the implicit elimination of the Lambdas from the program formulation. Please note that this program does not limit the number of points or the dimension of the space they are from. Please note further that quantifier elimination is a computationally expensive operation and therefore this program is only useful as a benchmark for the projector and not so for the intended purpose.
A beautiful example of disequations at work is due
to [Colmerauer 90]. It addresses the task of tiling a rectangle
with squares of all-different, a priori unknown sizes. Here is a
translation of the original Prolog-III
program to clp(Q,R):
% % from file: library('clpqr/examples/squares') % filled_rectangle(A, C) :- { A >= 1 }, distinct_squares(C), filled_zone([-1,A,1], _, C, []). distinct_squares([]). distinct_squares([B|C]) :- { B > 0 }, outof(C, B), distinct_squares(C). outof([], _). outof([B1|C], B) :- { B =\= B1 }, % *** note disequation *** outof(C, B). filled_zone([V|L], [W|L], C0, C0) :- { V=W,V >= 0 }. filled_zone([V|L], L3, [B|C], C2) :- { V < 0 }, placed_square(B, L, L1), filled_zone(L1, L2, C, C1), { Vb=V+B }, filled_zone([Vb,B|L2], L3, C1, C2). placed_square(B, [H,H0,H1|L], L1) :- { B > H, H0=0, H2=H+H1 }, placed_square(B, [H2|L], L1). placed_square(B, [B,V|L], [X|L]) :- { X=V-B }. placed_square(B, [H|L], [X,Y|L]) :- { B < H, X= -B, Y=H-B }.
There are no tilings with less than nine squares except the trivial one where the rectangle equals the only square. There are eight solutions for nine squares. Six further solutions are rotations of the first two.
clp(q) ?- use_module(library('clpqr/examples/squares')). clp(q) ?- filled_rectangle(A, Squares). A = 1, Squares = [1] ? ; A = 33/32, Squares = [15/32,9/16,1/4,7/32,1/8,7/16,1/32,5/16,9/32] ? ; A = 69/61, Squares = [33/61,36/61,28/61,5/61,2/61,9/61,25/61,7/61,16/61]
Depending on your hardware, the above query may take a few minutes. Supplying the knowledge about the minimal number of squares beforehand cuts the computation time by a factor of roughly four:
clp(q) ?- length(Squares, 9), filled_rectangle(A, Squares). A = 33/32, Squares = [15/32,9/16,1/4,7/32,1/8,7/16,1/32,5/16,9/32] ? ; A = 69/61, Squares = [33/61,36/61,28/61,5/61,2/61,9/61,25/61,7/61,16/61]
There is a package that
transforms programs and queries from a eval-quote variant of clp(Q,R)
into corresponding programs and queries in a quote-eval variant.
Before you use it, you need to know that
in an eval-quote language, all symbols are interpreted unless explicitly quoted.
This means that interpreted terms cannot be manipulated syntactically directly.
Meta-programming in a CLP context by definition manipulates interpreted terms, therefore
you need quote/1
(just as in LISP) and some means to put syntactical terms
back to their interpreted life: {}/1
.
In a quote-eval language, meta-programming is (pragmatically) simpler because everything is implicitly quoted until explicitly evaluated. On the other hand, now object programming suffers from the dual inconvenience.
We chose to make our version of clp(Q,R) of the quote-eval type because this matches the intended use of the already existing boolean solver of SICStus. In order to keep the users of the eval-quote variant happy, we provide a source transformation package. It is activated via:
| ?- use_module(library('clpqr/expand')).
Loading the package puts you in a mode where the arithmetic functors
like +/2
, */2
and all numbers (functors of arity 0) are
interpreted semantically.
clp(r) ?- 2+2=X. X = 4.0
The package works by purifying programs and queries in the sense that all references to interpreted terms are made explicit. The above query is expanded prior to evaluation into:
{2.0+2.0=X}
The same mechanism applies when interpreted terms are nested deeper:
some_predicate(10, f(A+B/2), 2*cos(A))
Expands into:
{Xc=2.0*cos(A)}, {Xb=A+B/2}, {Xa=10.0}, some_predicate(Xa, f(Xb), Xc)
This process also applies when files are consulted or compiled. In fact, this is the only situation where expansion can be applied with relative safety. To see this, consider what happens when the top level evaluates the expansion, namely some calls to the clp(Q,R) solver, followed by the call of the purified query. As we learned in section Feedback and Bindings, the solver may bind variables, which produces a goal with interpreted functors in it (numbers), which leads to another stage of expansion, and so on.
We recommend that you only turn on expansion temporarily while
consulting or compiling files needing expansion with expand/0
and noexpand/0
.
This collection of examples has been distributed with the Monash University Version of clp(R) [Heintze et al. 87], and its inclusion into this distribution was kindly permitted by Roland Yap.
In order to execute the examples, a small compatibility package has to be loaded first:
clp(r) ?- use_module(library('clpqr/monash')).
Then, assuming you are using clp(R):
clp(r) ?- expand, [library('clpqr/examples/monash/rkf45')], noexpand. clp(r) ?- go. Point 0.00000 : 0.75000 0.00000 Point 0.50000 : 0.61969 0.47793 Point 1.00000 : 0.29417 0.81233 Point 1.50000 : -0.10556 0.95809 Point 2.00000 : -0.49076 0.93977 Point 2.50000 : -0.81440 0.79929 Point 3.00000 : -1.05440 0.57522 Iteration finished ------------------ 439 derivative evaluations
The Monash examples have been written for clp(R). Nevertheless,
all but rkf45
complete nicely in clp(Q). With rkf45
,
clp(Q) runs out of memory. This is an instance of the problem
discussed in section Numerical Precision and Rationals.
The Monash University clp(R) interpreter features a dump/n
predicate. It is used to print the target variables
according to the given ordering.
Within this version of clp(Q,R), the corresponding functionality
is provided via ordering/1
. The difference is that ordering/1
does only specify the ordering of the variables and no printing
is performed. We think Prolog has enough predicates to perform output already.
You can still run the examples referring to dump/n
from
the Prolog top level:
clp(r) ?- expand, [library('clpqr/examples/monash/mortgage')], noexpand. % go2 % clp(r) ?- mg(P,120,0.01,0,MP), dump([P,MP]). {P=69.7005220313972*MP} % go3 % clp(r) ?- mg(P,120,0.01,B,MP), dump([P,B,MP]). {P=0.30299477968602706*B+69.7005220313972*MP} % go4 % clp(r) ?- mg(999, 3, Int, 0, 400), dump. clpr:{_B-_B*Int+_A+400.0=0.0}, clpr:{_A-_A*Int+400.0=0.0}, {_B=599.0+999.0*Int}
The predicates bb_inf/3, bb_inf/5
implement a simple
Branch and Bound search algorithm for Mixed Integer Linear (MIP) Optimization
examples. Serious MIP is not trivial. The implementation
library('clpqr/bb.pl')
is to be understood as a starting point
for more ambitious users who need control over branching, or who
want to add cutting planes, for example.
Anyway, here is a small problem from miplib, a collection of MIP models, housed at Rice University:
NAME: flugpl ROWS: 18 COLUMNS: 18 INTEGER: 11 NONZERO: 46 BEST SOLN: 1201500 (opt) LP SOLN: 1167185.73 SOURCE: Harvey M. Wagner John W. Gregory (Cray Research) E. Andrew Boyd (Rice University) APPLICATION: airline model COMMENTS: no integer variables are binary
% % from file: library('clpqr/examples/mip') % example(flugpl, Obj, Vs, Ints, []) :- Vs = [ Anm1,Anm2,Anm3,Anm4,Anm5,Anm6, Stm1,Stm2,Stm3,Stm4,Stm5,Stm6, UE1,UE2,UE3,UE4,UE5,UE6], Ints = [Stm6, Stm5, Stm4, Stm3, Stm2, Anm6, Anm5, Anm4, Anm3, Anm2, Anm1], Obj = 2700*Stm1 + 1500*Anm1 + 30*UE1 + 2700*Stm2 + 1500*Anm2 + 30*UE2 + 2700*Stm3 + 1500*Anm3 + 30*UE3 + 2700*Stm4 + 1500*Anm4 + 30*UE4 + 2700*Stm5 + 1500*Anm5 + 30*UE5 + 2700*Stm6 + 1500*Anm6 + 30*UE6, allpos(Vs), { Stm1 = 60, 0.9*Stm1 +1*Anm1 -1*Stm2 = 0, 0.9*Stm2 +1*Anm2 -1*Stm3 = 0, 0.9*Stm3 +1*Anm3 -1*Stm4 = 0, 0.9*Stm4 +1*Anm4 -1*Stm5 = 0, 0.9*Stm5 +1*Anm5 -1*Stm6 = 0, 150*Stm1 -100*Anm1 +1*UE1 >= 8000, 150*Stm2 -100*Anm2 +1*UE2 >= 9000, 150*Stm3 -100*Anm3 +1*UE3 >= 8000, 150*Stm4 -100*Anm4 +1*UE4 >= 10000, 150*Stm5 -100*Anm5 +1*UE5 >= 9000, 150*Stm6 -100*Anm6 +1*UE6 >= 12000, -20*Stm1 +1*UE1 =< 0, -20*Stm2 +1*UE2 =< 0, -20*Stm3 +1*UE3 =< 0, -20*Stm4 +1*UE4 =< 0, -20*Stm5 +1*UE5 =< 0, -20*Stm6 +1*UE6 =< 0, Anm1 =< 18, 57 =< Stm2, Stm2 =< 75, Anm2 =< 18, 57 =< Stm3, Stm3 =< 75, Anm3 =< 18, 57 =< Stm4, Stm4 =< 75, Anm4 =< 18, 57 =< Stm5, Stm5 =< 75, Anm5 =< 18, 57 =< Stm6, Stm6 =< 75, Anm6 =< 18 }. allpos([]). allpos([X|Xs]) :- {X >= 0}, allpos(Xs).
We can first check whether the relaxed problem has indeed the quoted infimum:
clp(r) ?- example(flugpl, Obj, _, _, _), inf(Obj, Inf). Inf = 1167185.7255923203
Computing the infimum under the additional constraints that Stm6
,
Stm5
, Stm4
, Stm3
, Stm2
, Anm6
,
Anm5
, Anm4
, Anm3
, Anm2
, Anm1
assume
integer values at the infimum is computationally harder, but the query
does not change much:
clp(r) ?- example(flugpl, Obj, _, Ints, _), bb_inf(Ints, Obj, Inf, Vertex, 0.001). Inf = 1201500.0000000005, Vertex = [75.0,70.0,70.0,60.0,60.0,0.0,12.0,7.0,16.0,6.0,6.0]
The system consists roughly of the following components:
The internal data structure for rational numbers is rat(Num,Den)
.
Den is always positive, i.e. the sign of the rational number is the
sign of Num. Further, Num and Den are relative prime.
Note that integer N looks like rat(N,1)
in this representation.
You can control printing of terms with portray/1
.
Once one has a working solver, it is obvious and attractive to run the
constraints in a clause definition at read time or compile time and
proceed with the answer constraints in place of the original
constraints. This gets you constant folding and in fact the full
algebraic power of the solver applied to the avoidance of computations
at runtime. The mechanism to realize this idea is to use
dump/3, call_residue/2
for the expansion of {}/1
,
via hook predicate user:goal_expansion/3
).
If you use the dynamic data base, the clauses you assert might have constraints on the variables occurring in the clause. This works as expected:
clp(r) ?- {A < 10}, assert(p(A)). {A<10.0} yes clp(r) ?- p(X). {X<10.0}
Please send bug reports to <christian@ai.univie.ac.at>
.
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