Scryer Prolog documentation

Introduction

This library provides CLP(B), Constraint Logic Programming over Boolean variables. It can be used to model and solve combinatorial problems such as verification, allocation and covering tasks.

CLP(B) is an instance of the general CLP(X) scheme, extending logic programming with reasoning over specialised domains.

The implementation is based on reduced and ordered Binary Decision Diagrams (BDDs).

Benchmarks and usage examples of this library are available from: https://www.metalevel.at/clpb/

Boolean expressions

A Boolean expression is one of:

where Expr again denotes a Boolean expression.

The Boolean expression card(Is,Exprs) is true iff the number of true expressions in the list Exprs is a member of the list Is of integers and integer ranges of the form From-To. For example, to state that precisely two of the three variables X, Y and Z are true, you can use sat(card([2],[X,Y,Z])).

+(Exprs) and *(Exprs) denote, respectively, the disjunction and conjunction of all elements in the list Exprs of Boolean expressions.

Atoms denote parametric values that are universally quantified. All universal quantifiers appear implicitly in front of the entire expression. In residual goals, universally quantified variables always appear on the right-hand side of equations. Therefore, they can be used to express functional dependencies on input variables.

Interface predicates

The most frequently used CLP(B) predicates are:

• sat(+Expr) True iff the Boolean expression Expr is satisfiable.

• taut(+Expr, -T) If Expr is a tautology with respect to the posted constraints, succeeds with T = 1. If Expr cannot be satisfied, succeeds with T = 0. Otherwise, it fails.

• labeling(+Vs) Assigns truth values to the variables Vs such that all constraints are satisfied.

The unification of a CLP(B) variable X with a term T is equivalent to posting the constraint sat(X=:=T).

Examples

Here is an example session with a few queries and their answers:

?- use_module(library(clpb)).
   true.

?- sat(X*Y).
   X = 1, Y = 1.

?- sat(X * ~X).
   false.

?- taut(X * ~X, T).
   T = 0, clpb:sat(X=:=X).

?- sat(X^Y^(X+Y)).
   clpb:sat(X=:=X), clpb:sat(Y=:=Y).

?- sat(X*Y + X*Z), labeling([X,Y,Z]).
   X = 1, Y = 0, Z = 1
;  X = 1, Y = 1, Z = 0
;  X = 1, Y = 1, Z = 1.

?- sat(X =< Y), sat(Y =< Z), taut(X =< Z, T).
   T = 1, clpb:sat(X=:=X*Y), clpb:sat(Y=:=Y*Z).

?- sat(1#X#a#b).
   clpb:sat(X=:=a#b).

The pending residual goals constrain remaining variables to Boolean expressions and are declaratively equivalent to the original query. The last example illustrates that when applicable, remaining variables are expressed as functions of universally quantified variables.

Obtaining BDDs

By default, CLP(B) residual goals appear in (approximately) algebraic normal form (ANF). This projection is often computationally expensive. We can assert clpb:clpb_residuals(bdd) to see the BDD representation of all constraints. This results in faster projection to residual goals, and is also useful for learning more about BDDs. For example:

?- asserta(clpb:clpb_residuals(bdd)).
   true.

?- sat(X#Y).
node(3)- (v(X, 0)->node(2);node(1)),
node(1)- (v(Y, 1)->true;false),
node(2)- (v(Y, 1)->false;true).

Note that this representation cannot be pasted back on the toplevel, and its details are subject to change. Use copy_term/3 to obtain such answers as Prolog terms.

The variable order of the BDD is determined by the order in which the variables first appear in constraints. To obtain different orders, we can for example use:

?- sat(+[1,Y,X]), sat(X#Y).
node(3)- (v(Y, 0)->node(2);node(1)),
node(1)- (v(X, 1)->true;false),
node(2)- (v(X, 1)->false;true).

Enabling monotonic CLP(B)

In the default execution mode, CLP(B) constraints are not monotonic. This means that adding constraints can yield new solutions. For example:

?-          sat(X=:=1), X = 1+0.
   false.

?- X = 1+0, sat(X=:=1), X = 1+0.
   X = 1+0.

This behaviour is highly problematic from a logical point of view, and it may render declarative debugging techniques inapplicable.

Assert clpb:monotonic to make CLP(B) monotonic. If this mode is enabled, then you must wrap CLP(B) variables with the functor v/1. For example:

?- asserta(clpb:monotonic).
   true.

?- sat(v(X)=:=1#1).
   X = 0.

Example: Pigeons

In this example, we are attempting to place I pigeons into J holes in such a way that each hole contains at most one pigeon. One interesting property of this task is that it can be formulated using only cardinality constraints (card/2). Another interesting aspect is that this task has no short resolution refutations in general.

In the following, we use Prolog DCG notation to describe a list Cs of CLP(B) constraints that must all be satisfied.

:- use_module(library(clpb)).
:- use_module(library(clpz)).
:- use_module(library(lists)).
:- use_module(library(dcgs)).

pigeon(I, J, Rows, Cs) :-
        length(Rows, I), length(Row, J),
        maplist(same_length(Row), Rows),
        transpose(Rows, TRows),
        phrase((all_cards(Rows,[1]),all_cards(TRows,[0,1])), Cs).

all_cards([], _) --> [].
all_cards([Ls|Lss], Cs) --> [card(Cs,Ls)], all_cards(Lss, Cs).

Example queries:

?- pigeon(9, 8, Rows, Cs), sat(*(Cs)).
   false.

?- pigeon(2, 3, Rows, Cs), sat(*(Cs)),
   append(Rows, Vs), labeling(Vs),
   maplist(portray_clause, Rows).
[0,0,1].
[0,1,0].
etc.

Example: Boolean circuit

Consider a Boolean circuit that express the Boolean function =|XOR|= with 4 =|NAND|= gates. We can model such a circuit with CLP(B) constraints as follows:

:- use_module(library(clpb)).

nand_gate(X, Y, Z) :- sat(Z =:= ~(X*Y)).

xor(X, Y, Z) :-
        nand_gate(X, Y, T1),
        nand_gate(X, T1, T2),
        nand_gate(Y, T1, T3),
        nand_gate(T2, T3, Z).

Using universally quantified variables, we can show that the circuit does compute =|XOR|= as intended:

?- xor(x, y, Z).
   clpb:sat(Z=:=x#y).

Acknowledgments

The interface predicates of this library follow the example of SICStus Prolog.

Use SICStus Prolog for higher performance in many cases.