1. 2 March 2006 Constraints - Finite Domains 1 Constraints – Finite Domains  Advanced Constraints Cardinality Constraints Propositional Constraints Constructive Disjunction  Global Constraints Generalised Arc Consistency All-Different, Element, Circuit

2. 2 March 2006 Constraints - Finite Domains 2 Cardinality (Meta-) Constraint ­Many problems, impose various types of precedence betwen tasks, have the requirement that among a list of constraints Lr = [C1, C2,..., Cn] ­The number of constraints to satisfy must lie between a lower bound L and an upper bound U, both inclusive. Syntactically, this meta-constraints may be specified as #(L,U, [C1, C2,..., Cn]) ­In particular, such constraints are quite common in resource management and in scheduling problems, where resources (machines, materials, tasks or even people) must be allocated in a “flexible” way.

3. 2 March 2006 Constraints - Finite Domains 3 Cardinality Constraint: Example Example (Nurse Scheduling): From the 7 nurses available at some hospital ward, at least 4 must be on duty in the first shift. ­This constraint may be modelled by considering boolean variables X ij to represent that nurse i is on duty in shift j. The intended allocation may be modelled by specification, to each shift j, of the cardinality meta-constraint #(4,7, [X 1j =1, X 2j =1,..., X 7j =1]) ­In alternative, 7 variables N ij are considered for each shift j, whose domain are the nurses 0 to 7 (0 meaning no nurse) leading to the formulation #(4, 7, [N 1j  0, N 2j  0,..., N 7j  0])

4. 2 March 2006 Constraints - Finite Domains 4 Cardinality Constraint: Operational Semantics ­The operational semantics of this meta-constraint may be specified by rewriting rules, using the Ask & Tell primitives just presented. The first two rules rewrite the constraint into a simplified form ­Rule 1: If one constraint is satisfied, it may be removed, and the number of constraints to satisfy is decreased by one. #(L,U, [C1, C2,..., Ci,..., Cn]), ask(Ci) #(L-1,U-1, [C1, C2,...,Ci-1,Ci+1,..., Cn]) ­Rule 2: If one constraint is not satisfied, it is simply removed from the list. #(L,U, [C1, C2,..., Ci,..., Cn]), ask(~Ci) #(L,U, [C1, C2,...,Ci-1,Ci+1,..., Cn])

5. 2 March 2006 Constraints - Finite Domains 5 Cardinality Constraint: Operational Semantics ­The two following rules refer to situations where the satisfaction or “dissatisfaction” of the constraints must be imposed. ­Rule 3: If the number of constraints in the list is the same as the lower limit, then all constraints must be satisfied. #(L,U, [C1, C2,..., Ci,..., Cn]), L = n tell(C1), tell(C2),..., tell(Cn) ­Rule 4: If the upper limit becomes 0, all the constraints in the list must be dissatisfied. #(L,U, [C1, C2,..., Ci,..., Cn]), U = 0 tell(~C1), tell(~C2),..., tell(~Cn)

6. 2 March 2006 Constraints - Finite Domains 6 Cardinality Constraint: Operational Semantics ­Finally, the meta-constraint is trivially verified after certain transitions of rules 1 and 2 that decrease the constraints in the list and the lower and upper limits. ­Rule 5: If no more constraints must be satisfied, and the constraints in the list are not more than the upper limit, the meta-constraint succeeds. #(L,U, [C1, C2,..., Cn]), L = 0, U >= n true ­Rule 6: If there are less constraints in the list than the lower bound, the meta- constraint fails #(L,U, [C1, C2,..., Cn]), L > n false

7. 2 March 2006 Constraints - Finite Domains 7 Cardinality Constraint through Reified Constraints ­Although this constraint may be specified directly from the rules presented, the Ask&Tell mechanisms of SICStus based in the reified constraints allow a more efficient form of specification, by counting the satisfied constraints and setting the intended limits, by means of a typical recursive “Prolog-like” definition. cardinal(L,U,Cs):- sats(Cs,N), N #=< U, L #=< N. sats([],0). sats([C1|Cs],N1):- C1 # B, N1 #= N2 + B, sats(Cs,N2).

8. 2 March 2006 Constraints - Finite Domains 8 Cardinality Constraint through Reified Constraints ­The cardinality (meta-)constraint is genericaly applicable to combinations of constraints requiring some form of counting. ­Other combinations of constraints are possible to specify through the use of reified constraints. ­Among others, it is worth mentioning the case of conditional constraints, i.e. those that should only be imposed if others are satisfied. ­In scheduling, for example, it may be the case that only one of tasks T 1 and T 2 must finish before some time limit Z. ­If this is task T 2, then task T 3 must start after some delay interval, I, starting from the end of task T 2.

9. 2 March 2006 Constraints - Finite Domains 9 Conditional Constraints through Reification ­The guarantee that only one of the tasks T 1 or T 2 terminates before Z, may be specified with cardinality card( 1, 1, [S 1 +D 1 #< Z, S 2 +D 2 #< Z] ) ­However, the conditional constraint between T 2 e T 3 is not easily expressed by means of countings, since the number of constraints to be satisfied on tasks T 2 / T 3 may be 0, 1 or 2. ­Conditional constraints may be modelled directly on the variables associated with the reified constraints. In this case, we would have the following reifications S 1 +D 1 # B 1 S 2 +D 2 # B 2 S 3 #> S 2 +D 2 +I # B 3

10. 2 March 2006 Constraints - Finite Domains 10 Conditional Constraints through Reification ­Now, using directly boolean variables from the reified constraints, we may specify both the cardinality constraint on tasks T 1 and T 2 B 1 + B 2 #= 1 as well as the conditional constraint on tasks T 2 and T 3, through the inequality B 3 #>= B 2 ­Such inequality expresses the intended condition. If constraint C 2 is not satisfied (B2=0) the conditional constraint is not effectively posted or “told” (C 3 may be satisfied or not). ­When constraint C 2 is satisfied (B2=1), then B 3 is set to 1, and constraint C 3 is “told”.

11. 2 March 2006 Constraints - Finite Domains 11 Propositional Constraints ­For convenience, rather than working with the boolean variables associated to reified constraints, SICStus allows its direct use in propositional constraints. ­For instance, the constraints on tasks T 1, T 2 and T 3 could be directly specified by means of the following propositional constraints ­Cardinality (one and only one of constraints C 1 and C 2 ) (S 1 +D 1 #< Z) #\ ( S 2 +D 2 #< Z) ­Conditional (if C 2 then C 3 ) (S 3 #> S 2 +D 2 +I ) #<= ( S 2 +D 2 #< Z)

12. 2 March 2006 Constraints - Finite Domains 12 Cardinality Constraint through Reified Constraints The following propositional constraints are defined ­ #\ C1B1 #= 0C1 must be dissatisfied ­C1 #/\ C2B1+B2 #= 2C1 and C2 must both be satisfied ­C1 #\ C2B1+B2 #= 1 One and only one of C1 and C2 must be satisfied ­C1 #\/ C2B1+B2 #>=1At least one of C1 and C2 must be satisfied ­C1 #=> C2B2 #>= B1It C1 (C2) is satisfied, then C2 (C1) must also be C2 #<= C1 ­C1 # C2B1 #= B2 C1 e C2 must both be either satisfied or dissatisfied.

13. 2 March 2006 Constraints - Finite Domains 13 Constructive Disjunction ­The combinations of constraints through propositional constraints follow the general heuristic of least commitment ­However, simply suspending the commitement of constraints might not be sufficiently useful, as the variables concerned have not their domains pruned. ­In some circumstances, one may infer conditions on the domains of the variables, even before commiting to some of the constraints. This is the case with Constructive Disjunction Delay commitement to a constraint until, following the enumeration of variables, make the variables domains to be sufficiently reduced to make such commitment safe.

14. 2 March 2006 Constraints - Finite Domains 14 Constructive Disjunction Example: ­Let us consider two constraints C 1 and C 2 on variable X, whose domain is 1 to 100, of which we want that at least one of them stands (disjunction) C1 : X #>= 50C2 : X #=< 20 ­The previous propositional constraints would suspend until the domain of X be pruned to either above 20 or below 50, in which case, one of the disjuncts would be committed to. ­For example, if X is reduced to 3.. 40, C 2 could be posted safely, further pruning the domain to 3.. 20. ­However, the interval 21.. 49 could have been immediately pruned from the the domain of X, since it does not satisfy either of the disjuncts.

15. 2 March 2006 Constraints - Finite Domains 15 Constructive Disjunction ­Such antecipation of domain pruning is the goal of constructive disjunction. Its modelling may be specified through adequate indexical constraints. disjunction1(X, S1, S2) :- X in S2.. Sup \/ inf.. S1. ?- X in 1..100, disjunction1(X,20,50). X in(1..20)\/(50..100) and compared with the least commitment implementation disjunction2(X, S1, S2) :- X #>= S2 #\/ X #=< S1. ?- X in 1..100, disjunction2(X,20,50). X in 1..100

16. 2 March 2006 Constraints - Finite Domains 16 Global Constraints: Generalised Arc Consistency ­It is often important to define n-ary “global” constraints, for at least two reasons Ease the modelling of a problem Exploitation of specialised algorithms that take the semantics of the constraint for efficient propagation, achieving generalised arc consistency. ­The generalised arc consistency (GAC) criterion sees that no value remains in the domain of a variable with no support in values of each of the other variables participating in the “global” constraint. Example: all_diff ([A 1, A 2,..., A n ] Constrain a set of n variables to be all different among themselves

17. 2 March 2006 Constraints - Finite Domains 17 Global Constraints: all_diff ­The constraint definition based on binary difference constraints (  ) does not pose any problems as much as modelling is concerned. For example, it may be being defined recursively in CLP. ­However, constraint propagation based on binary constraints alone does not provide in general much propagation. ­As seen before, arc consistency is not any better than node consistency, and higher levels of consistency are in general too costly and do not take into account the semantics of the all_different constraint. one_diff(_,[]). one_diff(X,[H|T]):- X #\= H, one_diff(X,T). all_diff([]). all_diff([H|T]):- one_diff(H,T), all_diff(T).

18. 2 March 2006 Constraints - Finite Domains 18 Global Constraints: all_diff Example: X 1 : 1,2,3X 6 : 1,2,3,4,5,6,7,8,9 X 2 : 1,2,3,4,5,6X 7 : 1,2,3,4,5,6,7,8,9 X 3 : 1,2,3,4,5,6,7,8,9X 8 : 1,2,3 X 4 : 1,2,3,4,5,6X 9 : 1,2,3,4,5,6 X 5 : 1,2,3 ­It is clear that constraint propagation based on maintenance of node-, arc- or even path-consistency would not eliminate any redundant label. ­Yet, it is very easy to infer such elimination with a global view of the constraint!

19. 2 March 2006 Constraints - Finite Domains 19 Global Constraints: all_diff ­Variables X 1, X 5 and X 8 may only take values 1, 2 and 3. Since there are 3 values for 3 variables, these must be assigned these values which must then be removed from the domain of the other variables. ­Now, variables X2, X4 and X9 may only take values 4, 5 e 6, that must be removed from the other variables domains. X 1 : 1,2,3 X 2 : 1,2,3,4,5,6 X 3 : 1,2,3,4,5,6,7,8,9 X 4 : 1,2,3,4,5,6 X 5 : 1,2,3 X 6 : 1,2,3,4,5,6,7,8,9 X 7 : 1,2,3,4,5,6,7,8,9 X 8 : 1,2,3X 9 : 1,2,3,4,5,6 X 1 : 1,2,3 X 2 : 1,2,3,4,5,6 X 3 : 1,2,3,4,5,6,7,8,9 X 4 : 1,2,3,4,5,6 X 5 : 1,2,3 X 6 : 1,2,3,4,5,6,7,8,9 X 7 : 1,2,3,4,5,6,7,8,9 X 8 : 1,2,3X 9 : 1,2,3,4,5,6

20. 2 March 2006 Constraints - Finite Domains 20 Global Constraints: all_diff ­In this case, these prunings could be obtained, by maintaining (strong) 4- consistency. ­For example, analysing variables X 1, X 2, X 5 and X 8, it would be “easy” to verify that from the d 4 potential assignments of values to them, no assignment would include X 2 = 1, X 2 = 2, nor X 2 = 3, thus leading to the prunning of X 2 domain. ­However, such maintenance is usually very expensive, computationally. For each combination of 4 variables, d 4 tuples whould be checked, with complexity O(d 4 ). ­In fact, in some cases, n-strong consistency would be required, so its naïf maintenance would be exponential on the number of variables, exactly what one would like to avoid in search!

21. 2 March 2006 Constraints - Finite Domains 21 Global Constraints: all_diff ­However, taking the semantics of this constraint into account, an algorithm based on quite a different approach allows the prunings to be made at a much lesser cost, achieving generalised arc consistency. ­Such algorithm (see [Regi94]), is grounded on graph theory, and uses the notion of graph matching. ­To begin with, a bipartite graph is associated to an all_diff constraints. The nodes of the graphs are the variables and all the values in their domains, and the arcs associate each variable with the values in its domain. ­In polinomial time, it is possible to eliminate, from the graph, all arcs that do not correspond to possible assignments of the variables.

22. 2 March 2006 Constraints - Finite Domains 22 Global Constraints: all_diff ­However, taking the semantics of this constraint into account, an algorithm based on quite a different approach allows the prunings to be made at a much lesser cost, achieving generalised arc consistency. ­Such algorithm (see [Regi94]), is grounded on graph theory, and uses the notion of graph matching. ­To begin with, a bipartite graph is associated to an all_diff constraints. The nodes of the graphs are the variables and all the values in their domains, and the arcs associate each variable with the values in its domain. ­In polinomial time, it is possible to eliminate, from the graph, all arcs that do not correspond to possible assignments of the variables.

23. 2 March 2006 Constraints - Finite Domains 23 Global Constraints: all_diff Key Ideas: ­For each variable-value pair, there is an arc in the bipartite graph. ­A matching, corresponds to a subset of arcs that link some variable nodes to value nodes, different variables being connected to different values. ­A maximal matching is a matching that includes all the variable nodes. ­For any solution of the all_diff constraint there is one and only one maximal matching.

24. 2 March 2006 Constraints - Finite Domains 24 Global Constraints: all_diff Example: A,B:: 1..2, C:: 1..3, D:: 2..5, E:: 3..6, all_diff([A,B,C,D,E]). A B C D E 1 2 3 4 5 6 A = 1 B = 2 C = 3 D = 4 E = 5 Maximal Matching

25. 2 March 2006 Constraints - Finite Domains 25 Global Constraints: all_diff ­The propagation (domain filtering) is done according to the following principles: 1.If an arc does not belong to any maximal matching, then it does not belong to any all_diff solution. 2.Once determined some maximal matching, it is possible to determine whether an arc belongs or not to any maximal matching. 3.This is because, given a maximal matching, an arc belongs to any maximal matching iff it belongs: a)To an alternating cycle; or b)To an even alternating path, starting at a free node.

26. 2 March 2006 Constraints - Finite Domains 26 Global Constraints: all_diff Example: For the maximal matching (MM) shown 6 is a free node; 6-E-5-D-4 is an even alternating path, alternating arcs from the MM (E-5, D-4) with arcs not in the MM (D-5, E-6); A-1-B-2-A is an alternating cycle; E-3 does not belong to any alternating cycle E-3 does not belong to any even alternating path starting in a free node (6) E-3 may be filtered out! A B C D E 1 2 3 4 5 6

27. 2 March 2006 Constraints - Finite Domains 27 Global Constraints: all_diff ­Compaction  Before this analysis, the graph may be “compacted”, aggregating, into a single node, “equivalent nodes”, i.e. those belonging to alternating cycles.  Intuitively, for any solution involving these variables and values, a different solution may be obtained by permutation of the corresponding assignments.  Hence, the filtering analysis may be made based on any of these solutions, hence the set of nodes can be grouped in a single one.

28. 2 March 2006 Constraints - Finite Domains 28 Global Constraints: all_diff ­A-1-B-2-A is an alternating cycle; ­By permutation of variables A and B, the solution = becomes = ­Hence, nodes A e B, as well as nodes 1 and 2 may be grouped together (as may the nodes D/E and 4/5). A/B C 4/5 3 6 D/E 1/2 A B C D E 1 2 3 4 5 6 With these grouping the graph becomes much more compact

29. 2 March 2006 Constraints - Finite Domains 29 Global Constraints: all_diff Analysis of the compacted graph shows that A/B C 4/5 3 6 D/E 1/2 A/B C 4/5 3 6 D/E 1/2 Arc D/E - 3 may be filtered out (notice that despite belonging to cycle D/E - 3 - C - 1/2 - D/E, this cycle is not alternating. Arcs D/E - 1/2 and C - 1/2 may also be filtered. The compact graph may thus be further simplified to

30. 2 March 2006 Constraints - Finite Domains 30 Global Constraints: all_diff By expanding back the simplified compact graph, one gets the graph on the right that A/B C 4/5 3 6 D/E 1/2 Which immediately sets C=3 and, more generaly, filters the initial domains to A,B :: 1..2, C:: 1,2,3, D:: 2,3,4,5, E:: 3,4,5,6 A B C D E 1 2 3 4 5 6

31. 2 March 2006 Constraints - Finite Domains 31 Global Constraints: all_diff ­Upon elimination of some labels (arcs), possibly due to other constraints, the all_diff constraint propagates such prunings, incrementally. ­There are 3 situations to consider: 1.Elimination of a vital arc (the only arc connecting a variable node with a value node): The constraint cannot be satisfied. A B C D E 1 2 3 4 5 6 A B C D E 1 2 3 4 5 6 ?

32. 2 March 2006 Constraints - Finite Domains 32 Global Constraints: all_diff 2.Elimination of a non-vital arc which is a member to the maximal matching  Determine a new maximal matching and restart from there. A B C D E 1 2 3 4 5 6 A B C D E 1 2 3 4 5 6 Arc A-4 does not belong to the even alternating path started in node 6. D-5 also leaves this path, but it still belongs to an alternating cycle.

33. 2 March 2006 Constraints - Finite Domains 33 Global Constraints: all_diff 3.Elimination of a non-vital arc which is not a member to the maximal matching  Eliminate the arcs that do not belong any more to an alternating cycle or path. A new maximal matching includes arcs D-5 and E-6. In this matching, arcs E-4 and E-5 do not belong to even alternating paths or alternating cycles. A B C D E 1 2 3 4 5 6 A B C D E 1 2 3 4 5 6

34. 2 March 2006 Constraints - Finite Domains 34 Global Constraints: all_diff Time Complexity: Assuming n variables, each of which with d values, and where D is the cardinality of the union of all domains, 1.It is possible to obtain a maximal matching with an algorithm of time complexity O(dn  n). 2.Arcs that do not belong to any maximal matching may be removed with time complexity O( dn+n+D). 3.Taking into account these results, we obtain complexity of O(dn+n+D+dn  n). Since D < dn, the total time complexity of the algorithm is dominated by the last term, thus becoming O(dn  n). which is much better than the poor result with a naïf analysis.

35. 2 March 2006 Constraints - Finite Domains 35 Global Constraints: all_diff Availability: 1.The all_diff constraint first appeared in the CHIP system (algorithm?). 2.The described implementation is incorporated into the ILOG system, and avalable as primitive IlcAllDiff. 3.This algorithm is also implemented in SICStus, through buit-in constraint all_distinct/1. 4.Other versions of the constraint, namely all _different/2, are also available, possibly using a faster algorithm but with less pruning, where the 2 nd argument controls the available pruning options.

36. 2 March 2006 Constraints - Finite Domains 36 Global Constraints: all_diff Example: Sudoku Cuts in green are all that are found by maïve all_diff. Global all_diff finds all values without backtracking. The first cuts are illustrated in the figure. (where the indices show a possible order in which the cuts are made) sudoku_cp.pl Files vh_1 e vh_2

37. 2 March 2006 Constraints - Finite Domains 37 Global Constraints: circuit ­The previous global constraints may be regarded as imposing a certain “permutation” on the variables. ­In many problems, such permutation is not a sufficient constraint. It is necessary to impose a certain “ordering” of the variables. ­A typical situation occurs when there is a sequencing of tasks, with precedences between tasks, possibly with non-adjacency constraints between some of them. ­In these situations, in addition to the permutation of the variables, one must ensure that the ordering of the tasks makes a single cycle, i.e. there must be no sub-cycles.

38. 2 March 2006 Constraints - Finite Domains 38 Global Constraints: circuit ­These problems may be described by means of directed graphs, whose nodes represent tasks and the directed arcs represent precedences. ­The arcs may even be labelled by “features” of the precedences, namely transition times. ­This is a situation typical of several problems of the travelling salesman type. AB CD AB CD 2 3 4 5 6 9 8 7 2 2 2

39. 2 March 2006 Constraints - Finite Domains 39 Global Constraints: circuit ­Filtering: For these type of problems, the arcs that do not belong to any hamiltonian circuit should be eliminated. ­In the graph, it is easy to check that the only possible circuits are A -> B -> D -> C -> A and A -> C-> D-> B-> A. ­Certain arcs (e.g. B -> C, B -> B,...), may not belong to any hamiltonian circuit and can be safely pruned. AB CD AB CD

40. 2 March 2006 Constraints - Finite Domains 40 Global Constraints: circuit ­The pruning of the arcs that do not belong to any circuit is the goal of the global constraint circuit/1, available in SICStus. ­This constraint is applicable to a list of domain variables, where the domain of each corresponds to the arcs connecting that variable to other variables, denoted by the order in which they appear in the list. For example: A in 2..3, B in 1..4, C in 1..4, D in 2..3, circuit([A,B,C,D]). A/1B/2 C/3D/4 A=2 A=3

41. 2 March 2006 Constraints - Finite Domains 41 Global Constraints: circuit ­Global constraint circuit/1 incrementally achieves the pruning of the arcs not in any hamiltonian circuit. For example, posing A, D in 2..3, B in 1,2,3,4, C in 1,2,3,4, circuit([A,B,C,D]). The following prunning is achieved A in 2..3, B in 1,2,3,4, C in 1,2,3,4, D in 2..3, since the possible solutions are [A,B,C,D] = [2,4,3,1] and [A,B,C,D] = [3,1,4,2] AB CD AB CD

42. 2 March 2006 Constraints - Finite Domains 42 Global Constraints: element For example, the value X from the arc that leaves node A, depends on the arc chosen: if A = 2 then X = 3, if A = 3 then X = 4; otherwise X = undefined A/1B/2 C/3D/4 2 3 4 5 6 9 8 7 2 2 2 The disjunction implicit in this definition raises, as well known, problems of efficiency to constraint propagation. Often a variable not only has its values constrained by the values of other variables, but it is actually defined conditionally in function of these values.

43. 2 March 2006 Constraints - Finite Domains 43 Global Constraints: element In fact, the value of X may only be known upon labelling of variable A. Until then, a naïf handling of this type of conditional constraint would infer very little from it. A/1B/2 C/3D/4 2 3 4 5 6 9 8 7 2 2 2 However, if other problem constraints impose, for example, X < 4, an efficient handling of this constraint would impose not only X = 3 but also A = 2. if A = 2 then X = 3, if A = 3 then X = 4; otherwise X = undefined

44. 2 March 2006 Constraints - Finite Domains 44 Global Constraints: element ­The efficient handling of this type of disjunctions is the goal of global constraint element/3, available in SICStus and CHIP. element(X, [V 1,V 2,...,V n ], V) ­In this constraint, X is a variable with domain 1..n, and both V and the V i s are either finite domain constraints or constants. The semantics of the constraint can be expressed as the equivalence X = i  V = V i ­From a propagation viewpoint, this constraint imposes arc consistency in X and bounds consistency in V. It is particularly optimised for when all V i s are ground.

45. 2 March 2006 Constraints - Finite Domains 45 Global Constraints: element ­The previous global constraints may be regarded as imposing a certain “permutation” on the variables. ­In many problems, such permutation is not a sufficient constraint. It is necessary to impose a certain “ordering” of the variables. ­A typical situation occurs when there is a sequencing of tasks, with precedences between tasks, possibly with non-adjacency constraints between some of them. ­In these situations, in addition to the permutation of the variables, one must ensure that the ordering of the tasks makes a single cycle, i.e. there must be no sub-cycles.

46. 2 March 2006 Constraints - Finite Domains 46 Global Constraints: element ­Global constraints may be used together. In particular, constraints element and circuit may implement the travelling salesman (satisfaction problem (travelling.pl): For some graph, determine an hamiltonian circuits whose length does not exceed Max (say 20). circ([A,B,C,D], Max, Cost):- A in 2..3, B in 1..4, C in {1}\/{3,4}, D in 2..3, circuit([A,B,C,D]), element(A,[_,3,4,_],Ca), element(B,[2,2,5,6],Cb), element(C,[2,_,2,9],Cc), element(D,[_,8,7,_],Cd), Cost #= Ca+Cb+Cc+Cd, Cost #=< Max, labeling([],[A,B,C,D]). A/1B/2 C/3D/4 2 3 4 5 6 9 8 7 2 2 2

47. 2 March 2006 Constraints - Finite Domains 47 Global Constraints: cycle ­Sometimes, the goal is to form more than one (1) circuit connecting all the nodes. ­This situation corresponds to the classical problem of the Multiple Traveling Salesman, that is highly suitable to model several practical applications. ­For example, the case where a number of petrol stations has to be serviced by a number N of tankers. ­For such type of applications, CHIP extended the global constraint circuit/1, to another cycle/2, where the first argument is a domain variable of an integer to account for the number of cycles. Note: Constraint circuit(L) is implemented in CHIP as cycle(1,L). Challenge: Implement element/3 with reified constraints.

48. 2 March 2006 Constraints - Finite Domains 48 Global Constraints: cycle Example: X1::[1,3,4], X2::[1,3], X3::[2,5,6] X4::[2,5], X5::[6], X6::[4,5] ?- N::[1,2,3], X = [X1,X2,X3,X4,X5,X6], cycle(N,X). N = 1 N = 2 N = 3 X =[3,1,5,2,6,4]X=[3,2,1,5,6,4] X=[3,1,5,2,6,4] 2 4 6 1 3 5 2 4 6 1 3 5 2 4 6 1 3 5 2 4 6 1 3 5