Finding irreducible infeasible sets in inconsistent constraint satisfaction problems. Alain Hertz

Transcription

1 Finding irreducible infeasible sets in inconsistent constraint satisfaction problems Alain Hertz GERAD École Polytechnique Montréal, Canada This is a joint work with Philippe Galinier and Christian Desrosiers Ecole Polytechnique, Montréal

2 I II III IV V VI VII VIII IX The main idea An example : the k-coloring problem Problem formulation A crazy assumption Solution techniques A more realistic assumption A related problem What to do in practice Computational results

3 I The main idea Let P be a problem for which we are not able to determine any feasible solution. How can we prove that P has no feasible solution?. Determine a subproblem P for which we are also not able to determine any feasible solution. Prove that P has no feasible solution this proves that P also has no feasible solution

4 II An example : the k-coloring problem Input : Problem : a graph G, a number k is it possible to color the vertices of G with at most k colors such that no two adjacent vertices get the same color? If the answer is yes, then G is k-colorable Example This graph is 3-colorable but not -colorable The smallest k such that G is k-colorable is called the chromatic number of G. How can we prove that G is not -colorable? Idea : and are also not -colorable

5 First formulation A solution is any assignment of a color to each vertex. An edge having its endpoints of the same color is called a conflicting edge. Problem : find a solution with as few conflicting edges as possible. No-conclicting edge G is k-colorable a -coloring with one conflicting edge (optimal solution for k=) 5

6 Second formulation A solution is any legal partial coloring (without conflicting edge). Problem : find a solution with as few non-colored vertices as possible. All vertices colored G is k-colorable a -coloring with one non-colored vertex (the grey one) (optimal solution for k=)

7 Up to a permutation there are different (non necessarily legal) -colorings of G G is not -colorable because all these -colorings have at least one conflicting edge 7

8 Let E i denote the set of -colorings in which edge i is conflicting Example G is not -colorable because the sets E i cover the set E of possible -colorings 8

9 These 3 sets E i also cover E They also explain why G is not -colorable G is not -colorable because of This is an irreducible infeasible set 9

10 The graph has a second irreducible infeasible set (for k=). Indeed, it is not -colorable because of the existence of 0

11 For the second formulation Let E = set of legal partial k-colorings (i.e., there are no conflicting edges, but there may exist non-colored vertices) Question : does E contain a solution where all vertices are colored? Let E i denote the set of solutions in E where vertex i is non-colored. The sets Ei cover E G is not -colorable

12 Example is not -colorable This is the set of legal partial -colorings (up to a permutation of the colors)

13 This is the subset E i where the middle vertex is non-colored 3

14 These three subsets already cover E They explain why G is not -colorable G is not -colorable because of This is an irreducible infeasible set

15 The constraint satisfaction problem (CSP) - a finite set of variables - each variable is associated with a domain of values - a set of constraints. A constraint specifies, for a particular subset of variables, a set of incompatible combinations of values for these variables. A solution of a CSP is an assignment of a value to each variable from its domain such that all constraints are satisfied. A CSP is consistent if it has at least one solution; otherwise it is inconsistent (or overconstrained, or infeasible). 5

16 Example : the k-coloring problem - the vertices of the graph are the variables - the set of possible colors {,,k} is the domain of each variable - the inequality constraints between adjacent vertices are the constraints

17 A partial assignment satisfies a constraint if it can be extended to a complete assignment that satisfies this constraint. A partial assignment is legal if it satisfies all constraints. A subset S of constraints is infeasible if no complete assignment satisfies all constraints in S simultaneously A subset of variables is infeasible if there is no legal partial assignment of these variables 7

18 A subset of constraints is called an irreducible infeasible set (IIS) of constraints if it is infeasible, but becomes feasible when any one constraint is removed. A subset of variables is called an irreducible infeasible set (IIS) of variables if it is infeasible, but becomes feasible when any one variable is removed. 8

19 Illustration with k= has IISs of constraints and and IIS of variables 9

20 Why is it interesting to detect IISs? 9 An IIS represents a part of the problem that gives a partial explanation to its infeasibility. 9 Determining IISs of constraints or variables can be very helpful to prove the inconsistency of a CSP. Indeed, IISs contain typically a small amount of constraints and variables when compared to the original problem a proof of inconsistency is possibly easier to obtain on an IIS rather than on the original problem. 0

21 A typical example Question : how can we prove the optimality of a solution delivered by a heuristic method?. Determine an IIS of constraints or variables. Prove that this IIS is not feasible Example : which is the chromatic number of? A heuristic method was able to color this graph with 3-colors, but not with. How can we prove that 3 is the chromatic number of this graph?. Determine an IIS of constraints or variables : Ÿ or. Prove that this IIS is not -colorable

22 The General Procedure. use a heuristic algorithm do determine a k-coloring of a given graph G with the smallest possible number k of colors. consider the problem P of determining a (k-)-coloring of G. This problem has probably no solution 3. generate a subgraph of G that has no solution for P. prove that this subgraph is indeed not (k-)-colorable Ÿ this proves that P has no solution Ÿ this proves that the chromatic number is k Illustration for G=. determine a 3-coloring. consider the -coloring problem on G 3. generate the subgraph G = or. prove that G is not -colorable Ÿ this proves that G is not -colorable Ÿ this proves that the chromatic number of G is equal to 3.

23 III Problem formulation A more general problem : the Large Set Covering Problem (LSCP) The Unicost set covering problem (USCP). Given a finite set E and a family F={E,,E m } of subsets of E such that F covers E, the USCP is to determine the smallest possible subset of F that also covers E. The LSCP is a variant of the USCP which differs from the USCP in that E and the subsets E i are not given in extension because they may be very large, and are possibly infinite sets. 3

24 Definition A subset I of {,,m} such that i I E i = E is called a cover. The LCSP is to determine a minimal (inclusion wise) cover. We also consider the problem, called minimum LSCP which is to determine a minimum cover.

25 The problems of finding IISs of constraints and variables in an inconsistent CSP are two special cases of the LSCP. Given an inconsistent CSP,. Define as an element of E as any complete assignment.. To each constraint c i (i=,,m) of the CSP, associate the subset E i of E containing all assignments that violate c i. A set of constraints is infeasible if and only if it covers E. finding an IIS of constraints is equivalent to solving the LSCP. 5

26 Given an inconsistent CSP,. Define as an element of E any legal partial assignment.. To each variable v i of the CSP, associate the subset E i of E containing all legal partial assignments in which v i has no value. A subset of variables is infeasible if and only if it covers E. finding an IIS of variables is equivalent to solving the LSCP. Conclusion IISs in an inconsistent CSP are minimal covers for the LSCP.

27 IV A crazy assumption 7

28 We are given a procedure Is-Element(e,i) that returns value true if and only if e E i. First model : given any k-coloring, the procedure tells you whether a given edge is conflicting or not Second model : given any partial legal k-coloring, the procedure tells you whether a given vertex is colored or not 8

29 We are given (continued) given any weighting function ω that assigns a weight ω(i) to each set E i, we can use a procedure Min_Weight(Z) that returns an element e E such that ω(i) is minimum. e First model : if all edges have a weight ω(i)=, then Min_Weight(ω) returns a k-coloring with as few conflicting edges as possible E i Second model : if all vertices have a weight ω(i)=, then Min_Weight(ω) returns a legal partial k-coloring with as few non-colored vertices as possible 9

30 V Solution techniques Notations F I (e) = subset of elements i I such that e E i. Such a subset can easily be generated by means of Is-Element. (it is - the subset of constraints violated by solution e, or - the subset of non-instantiated variables in solution e) 30

31 Given a subset I {,,m}, procedure Cover(I) returns the value true if E, and false otherwise. E i i I = Consider the weighting function ω that assigns a weight ω(i)= to each index i I ω(i)=0 to the other indices, Let e be the output of Min_Weight(ω). Then Cover(I) returns the value true if and only if F I (e). 3

32 We denote Min(I,J) the procedure that produces an element e E that minimizes M F I (e ) + F J where M is any number larger than J. (e ), 3

33 Algorithm Removal Input: a cover {,,m} Output: a minimal cover I {,,m}. Set I:={,,m};. For i= to m do If Cover(I-{-i})= true then set I:=I-{-i}. Property Algorithm Removal produces a minimal cover. 33

34

35 35

36 Algorithm Insertion Input: a cover {,,m} Output: a minimal cover I i. Set I 0 :=, J 0 :={,,m} and i:=0;. Set e i :=Min(I i,j i ); 3. If F Ii (e i ) then STOP: I i is a minimal cover;. Choose h i in F Ji (e i ), set I i+ :=I i {h i }, J i+ :=J i - F Ji (e i ), i:=i+ and go to Step. Property Algorithm Insertion produces a minimal cover. 3

37 3 37

38 38

39 Algorithm Hitting-Set Input: Output: a cover I={,,m} a minimum cover. Set I 0 := and i:=0;. Set e i :=Min(I i,i-i i ); 3. If F I i (ei) then STOP: I i is a minimum cover;. Set i:=i+, I i := Best_HS(F I (e 0 ),, F I (e i- )), and go to Step. Property Algorithm Hitting-Set produces a minimum cover. 39

40 0

41

42 Comparisons Algorithm Removal and Insertion always produce a minimal cover Algorithm Hitting-Set always produces a minimum cover Removal finds a minimal cover in m steps Insertion finds a minimal cover I in I steps Hitting-Set finds a minimum cover in an exponential number of steps But Hitting-Set can be stopped at any time to produce a lower bound on the size of a minimum cover

43 Procedure Lower_Bound_ Input: a cover {,,m} Output: a lower bound on the size of a minimum cover. Set I 0 :=, J 0 :={,,m} and i:=0;. Set e i :=Min(I i,j i ); 3. If F Ii (e i ) then STOP: i is a lower bound on the size of a minimum cover;. set I i+ :=I i F Ji (e i ), J i+ :=J i - F Ji (e i ), i:=i+ and go to Step. Property Procedure Lower_Bound_ produces a lower bound on the size of a minimum cover 3

44

45 5

46 Given m+ integers a,,a m and b we compute f(a,,a m ; b) computed as follows:. the integers a,,a m are first ordered in a non-increasing order, say a j a jm. f(a,,a m ; b) is then set equal to the smallest index r such that r i = a ji b. For example, f(,,3,,5; ) is equal to since the integers,,3,,5 are first ordered so that 5 3 and 5+3+< while

47 Procedure Lower_Bound_ Input: Output: a cover {,,m}; a given number q a lower bound on the size of a minimum cover. Set i:=0 and L:=0; set n(r):=0 for all r=,,m;. Set e i :=Min-Weight(ω) with ω(r)=m n(r) for all r=,,m (where M is a number >m); 3. Set n(r):= n(r)+ for all r F {,,m} (e i );. If n(r)=q for some element r, then STOP: L is a lower bound on the size of a minimum cover; 5. Set i:=i+, L:=max{L, f(n(),,n(m); i)} and go to Step. Property Procedure Lower_Bound_ produces a lower bound on the size of a minimum cover 7

48 8

49 , +, ++ 3,3 +++, ,5 +++, +++,7 +++, , , , , ,3 9

50 VI A more realistic assumption Procedure Min_Weight(Z) solves an NP-hard problem For the k-coloring with Z(i)= for all i, Min_Weight(Z) determines a k-coloring with as few conflicting edges as possible the output of Min_Weight(Z) has no conflicting edge if and only if the graph is k-colorable one can determine the chromatic number by using Min_Weight(Z) 50

51 What happens if one replaces Min_Weight(Z) by a heuristic algorithm? Property If the output of Removal is a cover, then it is a minimal cover If the output of Insertion is a cover, then it is a minimal cover If the output of Hitting-Set is a cover, then it is a minimum cover 5

52 Let P be a problem for which we are not able to determine any feasible solution. How can we prove that P has no feasible solution?. Determine a subproblem P for which we are also not able to determine any feasible solution. Prove that P has no feasible solution this proves that P also has no feasible solution In other words,. Detect a possible cover (IIS) by means of the heuristic versions of Removal, Insertion or Hitting-Set. Prove that this possible cover is indeed a cover (IIS) 5

53 VII A related problem Let E be a set, and let E,,E m be m subsets of E such that m E E i i = = Suppose that E and the subsets E i are very large sets that are possibly infinite. We consider the problem of finding the largest possible subset IŽ^,,m` such that i I E i E. Such a subset is called a maximum non-cover. 53

54 Example given a graph that is not k-colorable, find the largest possible subgraph that is k-colorable Input, k= Ouput: for the constraint formulation for the variable formulation 5

55 This problem can easily be solved using procedure Min-Weight(Z). Indeed, let e be the output of Min-Weight(Z) where Z(i)= for all i=,,m. A maximum non-cover is obtained by setting I equal to the set of elements I such that e E i. If we have to use a heuristic version of Min-Weight(Z), then we can use the following algorithm 55

56 Algorithm Max-non non-cover Input: a cover I={,,m} Output: a subset of ^,,m` that is not a cover or a message of error. Set e:=min-(i, ) and LB:=m- F I (e) ;. Set i:= and H 0 := ; 3. Call Removal, Insertion or Hitting-Set with input I-H i- ; If the output is a message of error then STOP: an error occurred; Else let I i denote this output;. If Cover(I i )= false then STOP: an error occurred; 5. Set H i := Best_HS(I,,I i ); UB:= m- H i ;. If LB=UB then STOP: I-F I (e) is a maximum non-cover; 7. If Cover(I-H i )= false then STOP: I-H i is a maximum non-cover; 8. Set i:=i+ and go to Step 3. Property This algorithm is finite and it either stops with a message of error, or it produces a maximum non-cover 5

57 57

58 5 58

59 VIII What to do in practice Heuristic algorithms for Min-Weight(Z). Minimize the number the total weight of the conflicting edges Tabu search algorithm with - search space = set of k-colorings - objective function = total weight of the conflicting edges - neighborhood = move a vertex from one color class to another 59

60 Heuristic algorithms for Min-Weight(Z) (continued). Minimize the the total weight of the non-colored vertices Tabu search algorithm with - search space = set of partial k-colorings - objective function = total weight of the non-colored vertices - neighborhood = assign a color i to a non-colored vertex v and remove the color from each vertex adjacent to v having color i. (i-swaps defined by Morgenstern, 99) 0

61 5 5

62 What else can we do in practice? Procedure Pre-filtering Input: a cover {,,m} Output: a lower bound on the size of a minimum cover. Set I 0 :=, J 0 :={,,m} and i:=0;. Set e i :=Min-(I i,j i ); 3. If F I i (e i ) then STOP: Set G:=G[I i ];. set I i+ :=I i F (e J i i ), J i+ :=J i - F Ji (e i ), i:=i+ and go to Step.

63 5 3

64 Speed up procedure When the procedure that solves Min (I,J) detects a solution e with F I (e) = then insert the unique element of F I (e) into I.. Min (I,J) Min (I,J)

65 Heuristic for choosing h i in the Insertion algorithm. Partial k-colorings Choose the vertex v with maximum total weight in its neighborhood (minimise u N(v) ω(u) ). Non-legal k-colorings Choose the edge e with maximum total weight in its neighborhood e N(e) ω(e ) (minimise ) 5

66 +9+9= = = m = =5

67 = = n = 8+8= =7 7

68 IX Computational results Example Original graph G 80,0. : 80 vertices, 03 edges Exact method : more than backtracks IIS : - 78 vertices - 7 edges - Exact method : 037 backtracks 8

69 Typical behavior. Number of backtracks for the original graph Number of backtracks for the IIS Chromatic number 9

70 70