Constraint Logic Programming (CLP)

1. Constraint Logic Programming (CLP) Luis Tari March 10, 2005

2. Why CLP? • “Generate-and-test” approach is a common methodology for logic programming. • Generate possible solutions • Test and eliminate non-solutions • Disadvantages of “generate-and-test” approach: • Passive use of constraints to test potential values • Inefficient for combinatorial search problems • CLP languages use the global search paradigm. • Actively pruning the search space • Recursively dividing a problem into subproblems until its subproblems are simple enough to be solved

3. Why CLP and AnsProlog? • AnsProlog is inefficient in dealing with numerical values, due to • Implementation of the current solvers • “generate-and-test” paradigm • Goal of CLP is to pick numerical values from pre-defined domains for certain variables so that the given constraints on the variables are all satisfied. • Idea: use CLP to define and reason with numerical constraints and assignments

4. List Notation in Prolog • Example of list notation • [eggs, tea, milk, steak, spinach, toothpaste] • Head of a list • First element of a list • eggs • Tail of a list • The remaining list apart from the head • [tea, milk, steak, spinach, toothpaste] • [H|T] • H = eggs • T = [tea, milk, steak, spinach, toothpaste]

5. An example of List • Use the predicate append to define appending two lists into one. append([X|XL],YL,[X|ZL]) :- append(XL,YL,ZL). append([],YL,YL). ?- append([1],[2],X). X = [1,2]

6. CLP with Sicstus Prolog • To invoke the CLP library: | ?- :- use_module(library(clpfd)). • A constraint is called as if it is a Prolog predicate. | ?- X in 1..5, Y in 2..8, X+Y #= T. X in 1..5, Y in 2..8, T in 3..13 • The above constraint says that given X,Y, assign values for T so that the constraint is satisfiable. | ?- X in 1..5, T in 3..13, X+Y #= T. X in 1..5, T in 3..13, Y in -2..12

7. Typical Steps of a CLP Program • Step 1. Declare the domains of the variables • Step 2. Post the problem constraints • Step 3. Look for a feasible solution via backtrack search, or look for an optimal solution via branch-and-bound search

8. Constraint Satisfaction Problem • Let’s consider the Send More Money puzzle. • The variables are the letters S, E, N, D, M, O, R and Y. • Each letter represents a digit between 0 and 9. • Assign a value to each digit, such that SEND + MORE equals MONEY.

9. Send More Money Exampleusing CLP | ?- mm([S,E,N,D,M,O,R,Y], []). D = 7, E = 5, M = 1, N = 6, O = 0, R = 8, S = 9, Y = 2

10. Reified Constraints • To reflect its truth value into a 0/1 variable B, so that: • The constraint is posted if B is set to 1 • The negation of the constraint is posted if B is set to 0 • B is set to 1 if the constraint becomes entailed • B is set to 0 if the constraint becomes disentailed. • A reified constraint is written as: | ?- Constraint #<=> B.

11. An example using reified constraint • Define exactly(X,L,N) so that it is true if X occurs exactly N times in the list L. • If the constraint “element X is the same as element Y” becomes entailed, then assign B to be 1 else 0.

12. Other Constraints • Arithmetic constraints • | ?- X in 1..2, Y in 3..5, X#=<Y #<=> B. B = 1, X in 1..2, Y in 3..5 • If the constraint “value in X is <= value in Y” becomes entailed, then assign B to be 1 else 0. • Propositional constraints • X #= 4 #\/ Y #= 6 • Disjunction of the two equality constraints • Many more….

13. Defining Search Variants • labeling(:Options, +Variables) • Where Variables is a list of domain variables or integers and Options is a list of search options. • | ?- constraints(Variables), labeling([], Variables). %same as [leftmost,step,up,all] • leftmost: the leftmost variable is selected • step: makes a binary choice between X#=B and X#\=B, where B is the lower or upper bound of X. • up: the domain is explored in ascending order • all: all solutions are enumerated by backtracking

14. An Example - Cumulative Scheduling • There are 7 tasks where each task has a fixed duration and a fixed amount of used resource. • Goal: find a schedule that minimizes the completion time for the schedule while not exceeding the capacity 13 of the resource.

15. Constraining the tasks such that for each task j has a start time Ssj, a duration Dsj, and a resource Rsj, so that the total resource consumption does not exceed 13 at any time. Find an assignment that minimizes the domain variable of Vars Cumulative Scheduling (2) :- use_module(library(clpfd)). :- use_module(library(lists), [append/3]). schedule(Ss, End) :- length(Ss, 7), Ds=[16,6,13,7,5,18,4], Rs=[2,9,3,7,10,1,11], domain(Ss, 1, 30), domain([End], 1, 50), after(Ss, Ds, End), cumulative(Ss, Ds, Rs, 13), append(Ss, [End], Vars), labeling([minimize(End)], Vars). after([], [], _). after([S|Ss], [D|Ds], E) :- E #>= S+D, after(Ss, Ds, E).

16. Cumulative Scheduling (3) • Ds = [16,6,13,7, 5,18,4] • Rs = [2 ,9, 3,7,10, 1,11] • Capacity = 13 | ?- schedule(Ss, End). Ss = [1,17,10,10,5,5,1], End = 23

17. References • Sicstus Prolog Manual Version 3.12, Chapter 34 on Constraint Logic Programming over Finite Domains • P. Van Hentenryck, H. Simonis, M. Dincbas, Constraint satisfaction using constraint logic programming, Journal of Artif. Intell., 58(1-3), pages 113--159, 1992.