The Thousand Faces of Constraint Propagation · Outline 1 Propagation and Constraint Programming 2 An Example of Propagator The Switch constraint 3 Complexity of Propagating vs. Complexity

The Thousand Faces of Constraint Propagation

Emmanuel Hebrard

Toulouse

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Outline

1 Propagation and Constraint Programming

2 An Example of PropagatorThe Switch constraint

3 Complexity of Propagating vs. Complexity of SolvingThe AtMostSeqCard constraint

4 NP-hard ConstraintsThe SoftAllEqual constraint

5 Assessing the TradeoffThe NValue constraint

6 Conclusion

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Constraint Programming

Solving hard combinatorial problems byDecomposition

I Good news: We can always decomposeF into known/easy/small subproblems

I Bad news: Solving each subproblem isnot enough

Constraint Progamming:I Constraint ⇔ “easy” SubproblemI Propagation to link the reasoning done

on each constraint

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on each constraint

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7

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3

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on each constraint

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Constraint Progamming:

I Constraint ⇔ “easy” SubproblemI Propagation to link the reasoning done

on each constraint

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15

12

7 7

11 10

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21

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7

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Constraint Progamming:I Constraint ⇔ “easy” Subproblem

I Propagation to link the reasoning doneon each constraint

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7 7

11 10

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7

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on each constraint

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7

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on each constraint

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Constraint Satisfaction Problem

Variables: x1, . . . , xnConstraints: C1, . . . ,Ck

I Relations = set of solutions

Domains: D1, . . . ,DnI Domain representation ' relaxation of the problem’s solutions

F D1 × D2 × . . .× Dn is a super set of the solutions

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Variables: x1, . . . , xn

Constraints: C1, . . . ,Ck




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Domain Representation (discrete domains)

4 Letters Palindrome City ProblemI Con. 1: It is a cityI Con. 2: It is a palindromeI Domain: any 2 letters (676 sol.)

L1 L2 L2 L1

A K K AA N N AI L L I

M A A MO T T O

{A,I,M,O} {A,K,L,N,T}

Minimal Relaxation (discrete domain)

A A A AA K K AA L L AA N N AA T T AI A A II K K II L L II N N II T T I

M A A MM K K MM L L MM N N MM T T MO A A OO K K OO L L OO N N OO T T O

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L1 L2 L2 L1


M A A MO T T O





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L1 L2 L2 L1


M A A MO T T O





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L1 L2 L2 L1


M A A MO T T O





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Domain Representation (bounds)


L1 L2 L2 L1


M A A MO T T O

[A,. . . ,O] [A,. . . ,T]

Minimal Relaxation (discrete bounds)

A A A AA B B AA C C AA D D AA E E AA F F AA G G AA H H AA I I AA J J AA K K AA L L AA M M AA N N AA O O AA P P AA Q Q AA R R AA S S A. . . .

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Constraint Propagation

Given a constraint C and a domain representation D

I Compute the intersection C ∩ DI Find the minimal relaxation D ′ of C ∩ D

F Find all solutions of C ∩ D (support)F Project on the domain representation (filtering)

Consistency: a domain is consistent iff it is a minimal relaxationI Discrete domain: Arc Consistency (ac)I Bounds: Bounds Consistency (bc)I Also:

F Multi-valued Decision Diagram MDD Consistency [Hooker, Hadzic, vanHoeve 2007]

F Set variables [Puget 1992, Gervet 1997], Length-Lex representation[Gervet and Van Hentenryck 2006]

F Graph variables, Function variables, ...

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Given a constraint C and a domain representation DI Compute the intersection C ∩ D

I Find the minimal relaxation D ′ of C ∩ DF Find all solutions of C ∩ D (support)F Project on the domain representation (filtering)





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Given a constraint C and a domain representation DI Compute the intersection C ∩ DI Find the minimal relaxation D ′ of C ∩ D






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Consistency: a domain is consistent iff it is a minimal relaxation

I Discrete domain: Arc Consistency (ac)I Bounds: Bounds Consistency (bc)I Also:




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Consistency: a domain is consistent iff it is a minimal relaxationI Discrete domain: Arc Consistency (ac)I Bounds: Bounds Consistency (bc)

I Also:F Multi-valued Decision Diagram MDD Consistency [Hooker, Hadzic, van

Hoeve 2007]F Set variables [Puget 1992, Gervet 1997], Length-Lex representation

[Gervet and Van Hentenryck 2006]F Graph variables, Function variables, ...

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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

∑6i=1 xi = 39

AllDifferent({x1, . . . , x6})

x1 : {

9

8 9}x2 : {1 2 6 7 8 9}x3 : {

9

8 9}x4 : {1 5 6 8 9}x5 : {1 2 6 7 8 9}x6 : {

9

4 5 8 9}

AllDifferent({x1, . . . , x6}) ∩ D888888888888888888888888888888888888999999999999999999999999999999999999

{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}

AllDifferent({x1, . . . , x6}) ∩ D8 8 9 9 {8, 9}6 7 6 7 {6, 7}9 9 8 8 {8, 9}5 5 5 5 {5}7 6 7 6 {6, 7}4 4 4 4 {4}

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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {1 2 6 7 8 9}x3 : {

9

8 9}x4 : {1 5 6 8 9}x5 : {1 2 6 7 8 9}x6 : {

9

4 5 8 9}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {1 2 6 7 8 9}x3 : {

9

8 9}x4 : {1 5 6 8 9}x5 : {1 2 6 7 8 9}x6 : {

9

4 5 8 9}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {1 2 6 7

8

8

9

}x3 : {

9

8 9}x4 : {1 5 6

8 9

}x5 : {1 2 6 7

8 9

}x6 : {

9

4 5

8 9

}

AllDifferent({x1, . . . , x6}) ∩ D888888888888888888888888888888888888999999999999999999999999999999999999 {8, 9}111111122222222222666666677777777777111111122222222222666666677777777777 {1, 2, 6, 7}999999999999999999999999999999999999888888888888888888888888888888888888 {8, 9}555666611115556666111155511115556666555666611115556666111155511115556666 {1, 5, 6}267227766771671177227712722661261122267227766771671177227712722661261122 {1, 2, 6, 7}444454545454444545454544445454444545444454545454444545454544445454444545 {4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {1 2 6 7

8

8

9

}x3 : {

9

8 9}x4 : {1 5 6

8 9

}x5 : {1 2 6 7

8 9

}x6 : {

9

4 5

8 9

}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {

1 2

6 7

8

8

9

}x3 : {

9

8 9}x4 : {

1

5 6

8 9

}x5 : {

1 2

6 7

8 9

}x6 : {

9

4 5

8 9

}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 {8, 9}6 6 6 6 6 6 7 7 7 7 6 6 6 6 7 {6, 7}8 8 8 9 9 9 8 8 8 9 8 8 8 9 8 {8, 9}5 6 6 5 5 6 5 5 6 5 5 5 6 5 5 {5, 6}7 6 7 6 7 6 6 7 6 6 6 7 6 6 6 {6, 7}5 5 4 5 4 4 5 4 4 4 5 4 4 4 4 {4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {

1 2

6 7

8

8

9

}x3 : {

9

8 9}x4 : {

1

5 6

8 9

}x5 : {

1 2

6 7

8 9

}x6 : {

9

4 5

8 9

}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Example: Kakuro

23 30 27 12 16

17

12

7 7

11 10

3916

17

35

21

6

7

16

24

8

5

3

39

∑6i=1 xi = 39


x1 : {

9

8 9}x2 : {

1 2

6 7

8

8

9

}x3 : {

9

8 9}x4 : {

1

5

6 8 9

}x5 : {

1 2

6 7

8 9

}x6 : {

9

4

5 8 9

}


{8, 9}

111111122222222222666666677777777777111111122222222222666666677777777777

{1, 2, 6, 7}

999999999999999999999999999999999999888888888888888888888888888888888888

{8, 9}

555666611115556666111155511115556666555666611115556666111155511115556666

{1, 5, 6}

267227766771671177227712722661261122267227766771671177227712722661261122

{1, 2, 6, 7}

444454545454444545454544445454444545444454545454444545454544445454444545

{4, 5}

∑6i=1 xi = 39 ∩ D

8 8 8 8 8 8 8 8 8 8 9 9 9 9 9

{8, 9}

6 6 6 6 6 6 7 7 7 7 6 6 6 6 7

{6, 7}

8 8 8 9 9 9 8 8 8 9 8 8 8 9 8

{8, 9}

5 6 6 5 5 6 5 5 6 5 5 5 6 5 5

{5, 6}

7 6 7 6 7 6 6 7 6 6 6 7 6 6 6

{6, 7}

5 5 4 5 4 4 5 4 4 4 5 4 4 4 4

{4, 5}


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Of course we do not want to enumerate solutions!

I Propagating is not necessarily harder than solvingF ac on AllDifferent in O(n1.5d), same as Matching

I In general:F Solving in O(K), then propagating in O(ndK)F Propagating in O(K), then solving in O(nK)

I In practice:F Same complexity class, but different algorithmsF Propagating is harder

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Of course we do not want to enumerate solutions!I Propagating is not necessarily harder than solving

F ac on AllDifferent in O(n1.5d), same as Matching



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An Example of Propagator: the Switch Constraint

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A set of n garments to embroidI Each garment require a subset of m available colors

A machine with k reels of thread

Replacing a reel costs time

What is the best sequence of garments?

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Embroidery Scheduling

A B C D E F G

Reel 1

Reel 2

Reel 3

1-st 2-nd 3-rd 4-th 5-th 6-th 7-th

A B C D E F G

switch switch switch switch

switch switch switch switch switch

AB CDEF G

switch switch

switch switch

switch

A,B,C,D,E,F,G: 9 switches

B,F,A,E,D,C,G: 5 switches

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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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A B C D E F G

Reel 1

Reel 2

Reel 3


A B C D E F G



AB CDEF G

switch switch

switch switch

switch



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The Switch Constraint

One Boolean variable per color j and position iI x ji = 1 iff the j-th color is on the buffer at time i

A variable M equal to the number of changes x ji < x ji+1

Not only useful for embroideryI Instruction scheduling (compilation)I Test sequencingI File transfer (observation satellites)

Switch turned up in 3/4 of the industrial projects I was involved in!

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Propagating the Switch Constraint

A B C D E F G

Reel 1

Reel 2

Reel 3


{B,F} {A} {B.F} {C ,D,E} {C ,D,E} {G} {C ,D,E}

?

?

?

? ?

? ?

?

?

How many switches at least? (compute a lower bound for M)

What are the possible colors (given the upper bound on M)

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A B C D E F G

Reel 1

Reel 2

Reel 3



?

?

?

? ?

? ?

?

?



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A B C D E F G

Reel 1

Reel 2

Reel 3



?

?

?

? ?

? ?

?

?



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A B C D E F G

Reel 1

Reel 2

Reel 3



?

?

?

? ?

? ?

?

?



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A B C D E F G

Reel 1

Reel 2

Reel 3



?

?

?

? ?

? ?

?

?



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S T


3capacity of the buffer

An edge for each color and each position

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S T


3

capacity of the buffer


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S T


3



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S T


3



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S T


3



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S

1-st 2-nd

c1

1

1

1

1

1

1

1

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S

1-st 2-nd

c1

1

1

1

1

1

1

1

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S

1-st 2-nd

c1

1

1

1

1

1

1

1

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S

1-st 2-nd

c1

1

1

1

1

1

1

1

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S T


3

c3 c5 c6

−11

OK OK

inc.

inc.

inc.

inc.

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S T


3

c3 c5 c6

−11

OK OK

inc.

inc.

inc.

inc.

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S T


3

c3 c5 c6

−11

OK OK

inc.

inc.

inc.

inc.

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S T


3

c3 c5

c6

−11

OK OK

inc.

inc.

inc.

inc.

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S T


3

c3 c5

c6

−11

OK OK

inc.

inc.

inc.

inc.

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S T


3

c3 c5

c6

−11

OK OK

inc.

inc.

inc.

inc.

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Propagation Algorithm (details at CP-AI-OR)

Find a maximum flow of minimum cost: O(nd)

Eliminate negative costs of the residual graph: O((nd)1.5)

Find all cycles of costs 0 and 1 (but not 2 or greater): O(n2d)

Total time complexity: O(n2d + nd1.5)

experiments

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experiments

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experiments

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experiments

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experiments

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Complexity of Propagating vs. Complexity of Solving

Propagating is often harderI E.g. Switch: solving in O(nd), propagating in O(n2d + nd1.5)

Sometimes, we can propagate “for free”, i.e., with the same timecomplexity as for solving

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Complexity of Propagating vs. Complexity of Solving

Propagating is often harderI E.g. Switch: solving in O(nd), propagating in O(n2d + nd1.5)

Sometimes, we can propagate “for free”, i.e., with the same timecomplexity as for solving

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Propagating for “Free”: the AtMostSeqCard Constraint

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The AtMostSeqCard Constraint

A sequence of variables [x1, . . . , xn]I xi = 1 iff the bolt is of the type that Charlot cares about

There is a given demand for bolts of this type

They should not occur alltogether (capacity constraints)I On each subsequence of size q, at most u are of this type

Example: u = 2, q = 4, d = 10, [x1, . . . , x22]

Definition: AtMostSeqCard(u, q, d , [x1, . . . , xn]) ⇐⇒

(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d)

∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Example: u = 2, q = 4, d = 10, [x1, . . . , x22]


(n∑

i=1

xi = d) ∧n−q∧i=0

(

q∑l=1

xi+l ≤ u)

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

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Solving the AtMostSeqCard Constraint

Finding a support (solving)

I We can be greedy and assign 1 whenever possible ⇒ leftmost

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Solving the AtMostSeqCard Constraint

Finding a support (solving)I We can be greedy and assign 1 whenever possible ⇒ leftmost

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Example: −→w = leftmost (u = 2, q = 4)

D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1

−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

Support: maximizing the cardinality while respecting capacitiesI Changing the value 1 to 0 has no impact on capacity constraints

I Greedily inserting 1 works

We can find a support in O(nq)

We can achieve ac in O(n2q)

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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ]

1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ]

1

0

0

1

1 0 0

0

1

0 1

0 0

1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ]

1

0

0

1

1 0 0

0

1

0 1 0 0 1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0

0

1

1 0 0

0

1

0 1 0 0 1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1

1 0 0

0

1

0 1 0 0 1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1

0 0

0

1

0 1 0 0 1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0

1

0 1 0 0 1

1 0 0 1 1 0 0

1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1





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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

Support: maximizing the cardinality while respecting capacitiesI Changing the value 1 to 0 has no impact on capacity constraintsI Greedily inserting 1 works



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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

Support: maximizing the cardinality while respecting capacitiesI Changing the value 1 to 0 has no impact on capacity constraintsI Greedily inserting 1 works



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Filtering the AtMostSeqCard Constraint

Can we propagate in the same time complexity?

I There is nothing to propagate unless |−→w | = dI Two runs of leftmost are enough

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Can we propagate in the same time complexity?I There is nothing to propagate unless |−→w | = d

I Two runs of leftmost are enough

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Can we propagate in the same time complexity?I There is nothing to propagate unless |−→w | = dI Two runs of leftmost are enough

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Example: −→w = leftmost (u = 2, q = 4, d = 10)

D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0

We can achieve ac in O(nq), that is, O(n2)

We can do better: leftmost can run in O(n) details

Total complexity of O(n), optimal!

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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1

←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 d

R[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 d

R[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 . 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 1 0 1 . . 1 . . . . . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 1 0 1 . . 1 . . . 1 . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 1 0 1 . . 1 . . . 1 . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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D(xi ) . 0 . 1 . . . 0 1 0 1 . . 1 . . . 1 . . . 1−→w [i ] 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 1 0 0 1←−w [i ] 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1

L[i ] 0 1 1 1 2 3 3 3 3 4 4 5 5 5 6 7 7 7 8 9 9 9 dR[i ] d d d 9 8 8 8 7 7 6 6 5 5 5 4 4 4 3 2 2 2 1 0




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NP-hard Constraints

If solving is NP-hard, then achieving ac is NP-hard

Should we decompose into simpler constraints?

27 / 54

NP-hard Constraints



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NP-hard Constraints: the SoftAllEqual Constraints

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The AllDifferent Constraint

Each variable should take a distinct value (Matching)I Propagation algorithm [Regin 1994, Costa 1994]

What about an “AllEqual” constraint?I Trivial to solve and propagateI However, the soft version is far from trivialI Try to make the variables as equal as possible

Two versionsI Maximise the number of equalitiesI Minimise the number of value skip

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What about an “AllEqual” constraint?

I Trivial to solve and propagateI However, the soft version is far from trivialI Try to make the variables as equal as possible


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What about an “AllEqual” constraint?I Trivial to solve and propagate

I However, the soft version is far from trivialI Try to make the variables as equal as possible


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What about an “AllEqual” constraint?I Trivial to solve and propagateI However, the soft version is far from trivial

I Try to make the variables as equal as possible


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The SoftAllEqual Constraint

Definition: SoftAllEqual(k , [x1, . . . , xn]) ⇐⇒

At least k pairs of equal variables in the set {x1, . . . , xn}

Meeting SchedulingI Each researcher states his/her availabilitiesI Each researcher goes to at most one meetingI Schedule meetings so that:

F The number of interactions is maximumF Interaction: a pair of researchers attend the same meeting

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Meeting Scheduling

I Each researcher states his/her availabilitiesI Each researcher goes to at most one meetingI Schedule meetings so that:


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Meeting SchedulingI Each researcher states his/her availabilities

I Each researcher goes to at most one meetingI Schedule meetings so that:


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Meeting SchedulingI Each researcher states his/her availabilitiesI Each researcher goes to at most one meeting

I Schedule meetings so that:F The number of interactions is maximumF Interaction: a pair of researchers attend the same meeting

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F The number of interactions is maximum

F Interaction: a pair of researchers attend the same meeting

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Alice

Bob

Charlie

Dave

Eve

Frank

Guy

Hugo

Ian

Joe

Tue.am

Tue.pm

Wed.am

Wed.pm

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

Alice

Charlie

Frank

Guy

Ian

Bob

Dave

Eve

Hugo

Joe

Alice Hugo

Bob

Charlie

DaveJoe

EveIan

Guy

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Alice

Bob

Charlie

Dave

Eve

Frank

Guy

Hugo

Ian

Joe

Tue.am

Tue.pm

Wed.am

Wed.pm

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

Alice

Charlie

Frank

Guy

Ian

Bob

Dave

Eve

Hugo

Joe

Alice Hugo

Bob

Charlie

DaveJoe

EveIan

Guy

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Alice

Bob

Charlie

Dave

Eve

Frank

Guy

Hugo

Ian

Joe

Tue.am

Tue.pm

Wed.am

Wed.pm

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

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X

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X

X

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X

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X

X

X

Alice

Charlie

Frank

Guy

Ian

Bob

Dave

Eve

Hugo

Joe

Alice Hugo

Bob

Charlie

DaveJoe

EveIan

Guy

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Finding an optimal solution is NP-hard (and so is Achieving ac)

Given an ordering (red, blue, green, ...), we can define a greedy algo:I Assign X to the first value whenever possible, then the second, etc.

If values are ordered by occurrences: 2-approximation

There exists such an ordering that gives an optimal solution

There is an algorithm exponential only in the number of values (V )

Exponential in something that is potentially smaller than the data ndI Fixed Parameter Tractable

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Achieving ac is NP-hard

What about bc?

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Achieving ac is NP-hard

What about bc?

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A bc Algorithm for SoftAllEqual

20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55

If we assign the value 55 first, we assign it to all variables

The problem is split into two smaller parts

I Polynomial size search tree: Dynamic Programming

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20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55




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20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55




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20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55


The problem is split into two smaller partsI Polynomial size search tree: Dynamic Programming

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Dynamic Programming

20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55

Ta,b = Maximum equalities on variables entirely contained in [a, b]

Given a value v ∈ [a, b], we must assign variables to v if possibleI Ta,b(v) = (Binomial coeff of) the number of variables hit by v +I table content on the splitted parts: Ta,v−1 + Tv+1,b

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Dynamic Programming

20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55


Given a value v ∈ [a, b], we must assign variables to v if possibleI Ta,b(v) = (Binomial coeff of) the number of variables hit by v +

I table content on the splitted parts: Ta,v−1 + Tv+1,b

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Dynamic Programming

20 60

60 90

0 40

50 80

10 40

0 90

20 60

50 80

0 90

55


Given a value v ∈ [a, b], we must assign variables to v if possibleI Ta,b(v) = (Binomial coeff of) the number of variables hit by v +I table content on the splitted parts: Ta,v−1 + Tv+1,b

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Dynamic Programming

Time complexity: O(nV 3)I Table of size V 2, for each cell, go through the V values and count

occurrences (O(n))

Since we consider interval domains, V can be very large!

I We can count occurrences of all values in O(n + V )I Only maximal cliques matterI We can reformulate in an equivalent problem with as many values as

maximal cliquesI Interval graph, we can find all (O(n)) cliques in O(n log n) time

Algorithm in O(n3) to find a solution and O(n4) for bc

36 / 54

Dynamic Programming


occurrences (O(n))

Since we consider interval domains, V can be very large!I We can count occurrences of all values in O(n + V )

I Only maximal cliques matterI We can reformulate in an equivalent problem with as many values as



36 / 54

Dynamic Programming


occurrences (O(n))

Since we consider interval domains, V can be very large!I We can count occurrences of all values in O(n + V )I Only maximal cliques matter

I We can reformulate in an equivalent problem with as many values asmaximal cliques

I Interval graph, we can find all (O(n)) cliques in O(n log n) time


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Dynamic Programming


occurrences (O(n))

Since we consider interval domains, V can be very large!I We can count occurrences of all values in O(n + V )I Only maximal cliques matterI We can reformulate in an equivalent problem with as many values as



36 / 54

Dynamic Programming


occurrences (O(n))

Since we consider interval domains, V can be very large!I We can count occurrences of all values in O(n + V )I Only maximal cliques matterI We can reformulate in an equivalent problem with as many values as



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Assessing the Tradeoff



There are alternativesI Achieve a weaker consistency (e.g., bc)I Approximate acI Use an exponential algorithm (good case: Fixed Parameter Tractable)

The best tradeoff might not be obviousI Theoretical comparisonI Empirical comparison

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There are alternatives

I Achieve a weaker consistency (e.g., bc)I Approximate acI Use an exponential algorithm (good case: Fixed Parameter Tractable)


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There are alternativesI Achieve a weaker consistency (e.g., bc)

I Approximate acI Use an exponential algorithm (good case: Fixed Parameter Tractable)


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There are alternativesI Achieve a weaker consistency (e.g., bc)I Approximate ac

I Use an exponential algorithm (good case: Fixed Parameter Tractable)


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Assessing the Tradeoff: The AtMostNValue Constraint

Definition: AtMostNValue([x1, . . . , xn],N) ⇐⇒

[x1, . . . , xn] assigned using at most N distinct values

NP-hard, equivalent to Minimum Hitting SetI Domains = collection of sets (D(xi ) = Si )I xi = an element in H ∩ SiI N = |H|

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NP-hard, equivalent to Minimum Hitting Set

I Domains = collection of sets (D(xi ) = Si )I xi = an element in H ∩ SiI N = |H|

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NP-hard, equivalent to Minimum Hitting Set

I Domains = collection of sets (D(xi ) = Si )I xi = an element in H ∩ SiI N = |H|

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NP-hard, equivalent to Minimum Hitting SetI Domains = collection of sets (D(xi ) = Si )

I xi = an element in H ∩ SiI N = |H|

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NP-hard, equivalent to Minimum Hitting SetI Domains = collection of sets (D(xi ) = Si )I xi = an element in H ∩ Si

I N = |H|

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NP-hard, equivalent to Minimum Hitting SetI Domains = collection of sets (D(xi ) = Si )I xi = an element in H ∩ SiI N = |H|

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Hitting Set

1

2

3

4

5

6

7

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

{2, 3}

{3, 4}

{1, 4, 5, 7}

{5, 6}

{6, 7}

{2, 3, 7}

3

5

6

39 / 54

Hitting Set

1

2

3

4

5

6

7

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

{2, 3}

{3, 4}

{1, 4, 5, 7}

{5, 6}

{6, 7}

{2, 3, 7}

3

5

6

39 / 54

Hitting Set

1

2

3

4

5

6

7

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

{2, 3}

{3, 4}

{1, 4, 5, 7}

{5, 6}

{6, 7}

{2, 3, 7}

3

5

6

39 / 54

Hitting Set

1

2

3

4

5

6

7

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

{2, 3}

{3, 4}

{1, 4, 5, 7}

{5, 6}

{6, 7}

{2, 3, 7}

3

5

6

39 / 54

Hitting Set

1

2

3

4

5

6

7

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

{2, 3}

{3, 4}

{1, 4, 5, 7}

{5, 6}

{6, 7}

{2, 3, 7}

3

5

6

39 / 54

Hitting Set, Clique Cover

Can we approximate Minimum Hitting Set to filterAtMostNValue?

I We need a lower bound; approximating MHS gives us an upper bound

I Another approach: Clique Cover of the Intersection Graph

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I We need a lower bound; approximating MHS gives us an upper bound

I Another approach: Clique Cover of the Intersection Graph

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I We need a lower bound; approximating MHS gives us an upper boundI Another approach: Clique Cover of the Intersection Graph

40 / 54

Intersection Graph: Discrete domains

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

77

6

7

2,3

3

A Minimum Clique Cover is a lower bound

However, finding a Minimum Clique Cover is NP-hard

41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3D(a) = {2, 3}D(b) = {3, 4}

4

5

77

6

7

2,3

3



41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

D(b) = {3, 4}D(c) = {1, 4, 5, 7}

5

77

6

7

2,3

3



41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

D(c) = {1, 4, 5, 7}D(d) = {5, 6}

77

6

7

2,3

3



41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

77

6

7

2,3

3



41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

77

6

7

2,3

3



41 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

77

6

7

2,3

3



41 / 54

Intersection Graph: Interval domains

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

4

5

77

6

7

2,3

3

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, . . . , 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, . . . , 7}

3

4

5, 6

6, 72, . . . , 7

6

6, 7

2, 3

3, 4

2, 3

5, 6

Finding a Minimum Clique Cover is polynomial (O(n log n))

A Minimum Clique Cover is an exact lower bound [Beldiceanu 2001]

42 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, . . . , 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, . . . , 7}

3

4

5, 6

6, 72, . . . , 7

6

6, 7

2, 3

3, 4

2, 3

5, 6



42 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, . . . , 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, . . . , 7}

3

4

5, 6

6, 72, . . . , 7

6

6, 7

2, 3

3, 4

2, 3

5, 6



42 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, . . . , 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, . . . , 7}

3

4

5, 6

6, 72, . . . , 7

6

6, 7

2, 3

3, 4

2, 3

5, 6



42 / 54


a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, . . . , 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, . . . , 7}

3

4

5, 6

6, 72, . . . , 7

6

6, 7

2, 3

3, 4

2, 3

5, 6



42 / 54

Independent Set

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

2,3

4

7

7

3

5

7

6

a

b

f

c

d

e

Find an Independent Set

Independance Number of G ≤ Intersection Number of θ(G )

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Independent Set

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

3

2,3

4

7

7

3

5

7

6

a

b

f

c

d

e



43 / 54

Independent Set

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

5

7

6

a

b

f

c

d

e



43 / 54

Independent Set

a

b c

d

ef

D(a) = {2, 3}D(b) = {3, 4}D(c) = {1, 4, 5, 7}D(d) = {5, 6}D(e) = {6, 7}D(f ) = {2, 3, 7}

a

b

f

c

d

e



43 / 54

Turan’s Lower Bound

There are even simpler lower boundsI [Turan 1941]

I [Favaron et al. 1993]

α(G ) ≤⌈

n2

2m + n

⌉

≤

2n − 2m

d 2mn e⌈

2mn

⌉+ 1

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Turan’s Lower Bound

There are even simpler lower boundsI [Turan 1941]I [Favaron et al. 1993]

α(G ) ≤⌈

n2

2m + n

⌉≤

2n − 2m

d 2mn e⌈

2mn

⌉+ 1

44 / 54

Linear Program

minimize∑v∈D

yv

subject to∑v∈xi

yv ≥ 1, xi ∈ X

yv ≥ 0 , v ∈ D

number of values

each variable takes a value

An integral solution is a Hitting Set

The linear relaxation gives a lower bound

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Linear Program

minimize∑v∈D

yv

subject to∑v∈xi

yv ≥ 1, xi ∈ X

yv ≥ 0 , v ∈ D

number of values




45 / 54

Linear Program

minimize∑v∈D

yv

subject to∑v∈xi

yv ≥ 1, xi ∈ X

yv ≥ 0 , v ∈ D

number of values




45 / 54

Tradeoff: Theory

Which solution is best?

I Better time complexity vs. better propagation

Comparing propagation φ against ψ, constraint C domain DI φ(C ,D) the domain representation / set of solutions after applying φI ψ(C ,D) the domain representation / set of solutions after applying ψ

φ is as strong as ψ is φ(C ,D) ⊆ ψ(C ,D)

Turan

MD

bc

LP ac6=

6=

Turan amortisedbc O(n log n)MD O(n2d)LP O(nd4)ac NP-hard

experiments

46 / 54

Tradeoff: Theory

Which solution is best?I Better time complexity vs. better propagation



Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Tradeoff: Theory


Comparing propagation φ against ψ, constraint C domain D

I φ(C ,D) the domain representation / set of solutions after applying φI ψ(C ,D) the domain representation / set of solutions after applying ψ


Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Tradeoff: Theory




Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Tradeoff: Theory




Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Tradeoff: Theory




Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Tradeoff: Theory




Turan

MD

bc

LP ac6=

6=


experiments

46 / 54

Conclusion

When solving a problem: find the algorithm with best complexity

I Here we also want to achieve the highest possible consistencyF ac? bc? Mdd-C? (SAC, MaxRPC, ...)F Even something undefined

When achieving a given consistency is NP-hardI Consider a lower consistencyI Consider approximationsI Consider parameterized complexity

Other related questions:I Is it decomposable?I What consistency/decomposition is the strongest?

Experimental tradeoff!

47 / 54

Conclusion

When solving a problem: find the algorithm with best complexityI Here we also want to achieve the highest possible consistency

F ac? bc? Mdd-C? (SAC, MaxRPC, ...)

F Even something undefined




47 / 54

Conclusion


F ac? bc? Mdd-C? (SAC, MaxRPC, ...)F Even something undefined




47 / 54

Conclusion


F ac? bc? Mdd-C? (SAC, MaxRPC, ...)F Even something undefined




47 / 54

The End

Questions?

48 / 54

Experimental Evaluation (120 garments, 3 machines, 7 colors)

120

140

160

180

200

220

0 10000 20000 30000 40000 50000 60000

objective (#switches)

cpu time (ms)

Decomposition (Impact)Decomposition (Wdeg)

Propagator (Impact)Propagator (Wdeg)

back

49 / 54

Experimental results

[label=expeamsc]

Table: Car-sequencing

ModelsG1 (70× 34× 5) G2 (4× 34× 5) G3 (5× 34× 5) G4 (7× 34× 5)

11900 680 850 1190#sol time #sol time #sol time #sol time

sum 8480 13.93 95 76.60 0 > 1200 64 43.81gsc 11218 3.60 325 110.99 31 276.06 140 56.61

amsc 10702 4.43 360 72.00 16 8.62 153 33.56amsc+gsc 11243 3.43 339 106.53 32 285.43 147 66.45

Sum: simple decomposition

GSC (Global Sequencing Constraint)I same as AtMostSeqCard for an option + demand for each type of

car requiring this optionI NP-hard, approximation

50 / 54


[label=expeamsc]


ModelsG1 (70× 34× 5) G2 (4× 34× 5) G3 (5× 34× 5) G4 (7× 34× 5)


sum 8480 13.93 95 76.60 0 > 1200 64 43.81gsc 11218 3.60 325 110.99 31 276.06 140 56.61

amsc 10702 4.43 360 72.00 16 8.62 153 33.56amsc+gsc 11243 3.43 339 106.53 32 285.43 147 66.45




50 / 54


[label=expeamsc]


ModelsG1 (70× 34× 5) G2 (4× 34× 5) G3 (5× 34× 5) G4 (7× 34× 5)


sum 8480 13.93 95 76.60 0 > 1200 64 43.81gsc 11218 3.60 325 110.99 31 276.06 140 56.61

amsc 10702 4.43 360 72.00 16 8.62 153 33.56amsc+gsc 11243 3.43 339 106.53 32 285.43 147 66.45


GSC (Global Sequencing Constraint)

I same as AtMostSeqCard for an option + demand for each type ofcar requiring this option

I NP-hard, approximation

50 / 54


[label=expeamsc]


ModelsG1 (70× 34× 5) G2 (4× 34× 5) G3 (5× 34× 5) G4 (7× 34× 5)


sum 8480 13.93 95 76.60 0 > 1200 64 43.81gsc 11218 3.60 325 110.99 31 276.06 140 56.61

amsc 10702 4.43 360 72.00 16 8.62 153 33.56amsc+gsc 11243 3.43 339 106.53 32 285.43 147 66.45




50 / 54


Table: Crew-Rostering

Heuristic Lexicographic

Modelsatifisable (1140) unsatisfiable (385)

#sol time avg bts #sol time avg btssum 0 - - 170 0.05 86gsc 25 308.93 74074 175 2.56 262

amsc 125 164.36 1828347 213 1.76 22621mamsc 534 87.29 685720 271 2.80 27150

MultiAtMostSeqCard: several capacity constraints together

back

51 / 54


Table: Crew-Rostering

Heuristic Lexicographic

Modelsatifisable (1140) unsatisfiable (385)

#sol time avg bts #sol time avg btssum 0 - - 170 0.05 86gsc 25 308.93 74074 175 2.56 262

amsc 125 164.36 1828347 213 1.76 22621mamsc 534 87.29 685720 271 2.80 27150

MultiAtMostSeqCard: several capacity constraints together

back

51 / 54

Linear time algorithm

We need to do each of the following in O(1):I access the cardinality of the jth subsequence containing xi : card(i , j)

F jump from xi to xi+1

F increment all subsequences containing xi

1 0 0 1 1 0 . 0 . 0 1 . . 1 . . . . . . . 1

card(8, 0) = 4−2 = 2

card(8, 1) = 4−3 = 1

card(8, 2) = 4−3 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 . 0 . 0 1 . . 1 . . . . . . . 1

card(8, 0) = 4−2 = 2

card(8, 1) = 4−3 = 1

card(8, 2) = 4−3 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(7, 0) = 2

card(8, 0) = 4−2 = 2

card(7, 1) = 1

card(8, 1) = 4−3 = 1

card(7, 2) = 0

card(8, 2) = 4−3 = 1

card(7, 3) = 0

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(7, 0) = 2

card(8, 0) = 4−2 = 2

card(7, 1) = 1

card(8, 1) = 4−3 = 1

card(7, 2) = 0

card(8, 2) = 4−3 = 1

card(7, 3) = 0

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(8, 0) = 1

card(8, 0) = 4−2 = 2

card(8, 1) = 0

card(8, 1) = 4−3 = 1

card(8, 2) = 0

card(8, 2) = 4−3 = 1

card(8, 3) = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(7, 0) = 2

card(8, 0) = 4−2 = 2

card(7, 1) = 1

card(8, 1) = 4−3 = 1

card(7, 2) = 0

card(8, 2) = 4−3 = 1

card(7, 3) = 0

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(7, 0) = 3−1 = 2

card(8, 0) = 4−2 = 2

card(7, 1) = 3−2 = 1

card(8, 1) = 4−3 = 1

card(7, 2) = 3−3 = 0

card(8, 2) = 4−3 = 1

card(7, 3) = 3−3 = 0

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(8, 0) = 3−2 = 1

card(8, 0) = 4−2 = 2

card(8, 1) = 3−3 = 0

card(8, 1) = 4−3 = 1

card(8, 2) = 3−3 = 0

card(8, 2) = 4−3 = 1

card(8, 3) = 3−2 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 . 0 1 . . 1 . . . . . . . 1

card(8, 0) = 3−2 = 1

card(8, 0) = 4−2 = 2

card(8, 1) = 3−3 = 0

card(8, 1) = 4−3 = 1

card(8, 2) = 3−3 = 0

card(8, 2) = 4−3 = 1

card(8, 3) = 3−2 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 1 0 1 . . 1 . . . . . . . 1

card(8, 0) = 3−2 = 1

card(8, 0) = 4−2 = 2

card(8, 1) = 3−3 = 0

card(8, 1) = 4−3 = 1

card(8, 2) = 3−3 = 0

card(8, 2) = 4−3 = 1

card(8, 3) = 3−2 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54





1 0 0 1 1 0 0 0 1 0 1 . . 1 . . . . . . . 1

card(8, 0) = 4−2 = 2

card(8, 1) = 4−3 = 1

card(8, 2) = 4−3 = 1

card(8, 3) = 4−2 = 2

card(i , j)=∑i

k=1 +c[i + j mod q]

back

52 / 54


We need to do each of the following in O(1):I check whether there exists a subsequence of cardinality u: #sub(i , k)


→ OK using card(i , 0) and card(i + 1, 4)


→ decrement pointer

0 0 0 0 0 0 3 1 2 0

back

53 / 54




→ OK using card(i , 0) and card(i + 1, 4)



0 0 0 0 0 0 3 1 2 0

0 1 2 3

back

53 / 54



F jump from xi to xi+1 → OK using card(i , 0) and card(i + 1, 4)F increment all subsequences containing xi


0 0 0 0 0 0 3 1 2 0

0 1 2 3

back

53 / 54



F jump from xi to xi+1 → OK using card(i , 0) and card(i + 1, 4)F increment all subsequences containing xi → decrement pointer

0 0 0 0 0 0 3 1 2 0

0 1 2 3

back

53 / 54



F jump from xi to xi+1 → OK using card(i , 0) and card(i + 1, 4)F increment all subsequences containing xi → decrement pointer

0 0 0 0 0 0 3 1 2 0

0 1 2 3

back

53 / 54

Tradeoff: In Practice

Random instances

Dominating set of the Queens Graph of order nI One vertex per square of the chessboard (n2)I An edge iff if the two squares are attacked (same row, column or

diagonal)

minimum number of Queens to attack the whole chessboard

problemTuran bc MD LP

time bkt time bkt time bkt time bkt

Random

7.2 90786 6.9 72773 2.5 18885 15.8 17866

n = 50, d = 15

Queens

- - - - 203 880669 16 2243

(9× 9)

54 / 54


Random instances


diagonal)




Random

7.2 90786 6.9 72773 2.5 18885 15.8 17866

n = 50, d = 15

Queens

- - - - 203 880669 16 2243

(9× 9)

54 / 54


Random instances


diagonal)




Random7.2 90786 6.9 72773 2.5 18885 15.8 17866

n = 50, d = 15

Queens

- - - - 203 880669 16 2243

(9× 9)

54 / 54


Random instances


diagonal)




Random7.2 90786 6.9 72773 2.5 18885 15.8 17866

n = 50, d = 15

Queens- - - - 203 880669 16 2243

(9× 9)

54 / 54

Documents

The Thousand Faces of Constraint Propagation · Outline 1 Propagation and Constraint Programming 2 An Example of Propagator The Switch constraint 3 Complexity of Propagating vs. Complexity