The solver interface for both Q and R consists of the following
predicates which are exported from module(linear).
{+Constraint}
Constraint is a term accepted by the the grammar below. The corresponding constraint is added to the current constraint store and checked for satisfiability. Use the module prefix to distinguish the solvers if both clp(Q) and clp(R) were loaded
| ?- clpr:{Ar+Br=10}, Ar=Br, clpq:{Aq+Bq=10}, Aq=Bq.
Aq = 5,
Ar = 5.0,
Bq = 5,
Br = 5.0
Although 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 in argument 2 of clpq:=/2
! a rational number expected, but 5.0 found
! goal: _118=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 constants
Conjunctive 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 Monash Examples.
entailed(+Constraint)
Succeeds iff the linear Constraint is entailed by the current constraint store. This predicate does not change the state of the constraint store.
clp(q) ?- {A =< 4}, entailed(A=\=5).
{A=<4}
yes
clp(q) ?- {A =< 4}, entailed(A=\=3).
no
inf(+Expr, -Inf)
inf(+Expr, -Inf, +Vector, -Vertex)
Computes the infimum of the linear expression Expr and unifies it
with Inf. If given, Vector should be a list of variables
relevant to Expr, and Vertex will be unified a list of the
same length as Vector containing the values for Vector, such
that the infimum is produced when assigned.
Failure indicates unboundedness.
sup(+Expr, -Sup)
sup(+Expr, -Sup, +Vector, -Vertex)
Computes the supremum of the linear expression Expr and unifies it with Sup. If given, Vector should be a list of variables relevant to Expr, and Vertex will be unified a list of the same length as Vector containing the values for Vector, such that the supremum is produced when assigned. Failure indicates unboundedness.
clp(q) ?- { 2*X+Y =< 16, X+2*Y =< 11,
X+3*Y =< 15, Z = 30*X+50*Y
}, sup(Z, Sup, [X,Y], Vertex).
Sup = 310,
Vertex = [7,2],
{Z=30*X+50*Y},
{X+1/2*Y=<8},
{X+3*Y=<15},
{X+2*Y=<11}
minimize(+Expr)
Computes the infimum of the linear expression Expr and equates it with the expression, i.e. as if defined as:
minimize(Expr) :- inf(Expr, Expr).
maximize(+Expr)
Computes the supremum of the linear expression Expr and equates it with the expression.
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)
Computes the infimum of the linear expression Expr under the additional constraint that all of variables in the list Ints assume integral values at the infimum. This allows for the solution of mixed integer linear optimization problems; see MIP.
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)
Computes the infimum of the linear expression Expr under the
additional constraint that all of variables in the list Ints
assume integral values at the infimum. Eps is a positive number
between 0 and 0.5 that specifies how close a number X must be to the next
integer to be considered integral: 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.
bb_inf/5 works properly for non-strict inequalities only!
Disequations (=\=) and higher dimensional strict inequalities
(>,<) are beyond its scope. Strict bounds on the decision
variables are honored however:
clp(q) ?- {X >= Y+Z, Y > 1, Z > 1}, bb_inf([Y,Z],X,Inf,Vertex,0).
Inf = 4,
Vertex = [2,2],
{Y>1},
{Z>1},
{X-Y-Z>=0}
The limitation(s) can be addressed by:
{X + Y > 0} becomes {X
+ Y >= 1} for integral X and Y;
ordering(+Spec)
Provides a means to control one aspect of the presentation of the
answer constraints; see Variable Ordering.
dump(+Target, -NewVars, -CodedAnswer)
Reflects the constraints on the target variables into a term, where Target and NewVars are lists of variables of equal length and CodedAnswer is the term representation of the projection of constraints onto the target variables where the target variables are replaced by the corresponding variables from NewVars (see Turning Answers into Terms).
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}
projecting_assert/1(:Clause)
If you use the dynamic data base, the clauses you assert might have
constraints associated with their variables. Use
this predicate instead of assert/1 in order to ensure that only
the relevant and projected constraints get stored in the data base.
It will transform the clause into one with plain variables and extra
body goals which set up the relevant constraint when called.