- Documentation
- Reference manual
- The SWI-Prolog library
- library(clpfd): CLP(FD): Constraint Logic Programming over Finite Domains
- Introduction
- Arithmetic constraints
- Declarative integer arithmetic
- Example: Factorial relation
- Combinatorial constraints
- Domains
- Example: Sudoku
- Residual goals
- Core relations and search
- Example: Eight queens puzzle
- Optimisation
- Reification
- Enabling monotonic CLP(FD)
- Custom constraints
- Applications
- Acknowledgments
- CLP(FD) predicate index
- Closing and opening words about CLP(FD)

- library(clpfd): CLP(FD): Constraint Logic Programming over Finite Domains

- The SWI-Prolog library
- Packages

- Reference manual

Using CLP(FD) constraints to solve combinatorial tasks typically consists of two phases:

**Modeling**. In this phase, all relevant constraints are stated.**Search**. In this phase,*enumeration predicates*are used to search for concrete solutions.

It is good practice to keep the modeling part, via a dedicated
predicate called the **core relation**, separate from the actual
search for solutions. This lets us observe termination and determinism
properties of the core relation in isolation from the search, and more
easily try different search strategies.

As an example of a constraint satisfaction problem, consider the cryptoarithmetic puzzle SEND + MORE = MONEY, where different letters denote distinct integers between 0 and 9. It can be modeled in CLP(FD) as follows:

puzzle([S,E,N,D] + [M,O,R,E] = [M,O,N,E,Y]) :- Vars = [S,E,N,D,M,O,R,Y], Vars ins 0..9, all_different(Vars), S*1000 + E*100 + N*10 + D + M*1000 + O*100 + R*10 + E #= M*10000 + O*1000 + N*100 + E*10 + Y, M #\= 0, S #\= 0.

Notice that we are *not* using labeling/2
in this predicate, so that we can first execute and observe the modeling
part in isolation. Sample query and its result (actual variables
replaced for readability):

?- puzzle(As+Bs=Cs). As = [9, A2, A3, A4], Bs = [1, 0, B3, A2], Cs = [1, 0, A3, A2, C5], A2 in 4..7, all_different([9, A2, A3, A4, 1, 0, B3, C5]), 91*A2+A4+10*B3#=90*A3+C5, A3 in 5..8, A4 in 2..8, B3 in 2..8, C5 in 2..8.

From this answer, we see that this core relation *terminates*
and is in fact *deterministic*. Moreover, we see from the residual
goals that the constraint solver has deduced more stringent bounds for
all variables. Such observations are only possible if modeling and
search parts are cleanly separated.

Labeling can then be used to search for solutions in a separate predicate or goal:

?- puzzle(As+Bs=Cs), label(As). As = [9, 5, 6, 7], Bs = [1, 0, 8, 5], Cs = [1, 0, 6, 5, 2] ; false.

In this case, it suffices to label a subset of variables to find the puzzle's unique solution, since the constraint solver is strong enough to reduce the domains of remaining variables to singleton sets. In general though, it is necessary to label all variables to obtain ground solutions.

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