acp advanced summer school on global constraints - samos, greece.1 global variables irit katriel...

ACP Advanced Summer School on Global Constraints - Samos, Greece.

1

Global Variables

Irit KatrielBRICS, University of Aarhus


2

Roles of Global Constraints Syntactic: Simplify models

nn

nnnn

n

n

xx

xxxx

xxxxxx

xxntAllDiffere

1

212

13121

1 vs.),...,(


3

Roles of Global Constraints Speed up solving (by propagation)

323121

321

321

vs.

),,(

}1,0{)()()(

xxxxxx

xxxntAllDiffere

xDxDxD


4

Global Variables

Assigned structured data, e.g., A set of scalar values A graph A function …

Same roles as global constraints Simplify models Improve propagation


5

Bin Packing

n items of different sizes: s1,…,sn

Bin size: b


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Bin Packing

Pack items in as few bins as possible


7

Bin Packing - Model 1

A variable for each item: x1,…,xn

At most n bins: D(x1) = {1,…,n}

Constraints:

Objective: ixi jj :min

ixjji

j

bs:


8

A variable for each item: x1,…,xn

At most n bins: D(x1) = {1,…,n}

Constraints:

Objective:


ixi jj :min

ixjji

j

bs:Complicated

constraintsSymmetry breaking?


9

A set for each bin: b1,…,bn

D(bi) = P({1,…,n})

Constraints:

Objective:


ibi :min

bs

xxbb

bb

ij bxji

nn

jiji

},,{ 11


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A set for each bin: b1,…,bn

D(bi) = P({1,…,n})

Constraints:

Objective:


ibi :min

bs

xxbb

bb

ij bxji

nn

jiji

},,{ 11

Simple constraints Symmetry:

lex order among sets.


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Shift Assignment (a generalization of GCC)

Shifts

Employees

[1,2] [1,1] [2,3] [0,1] [1,3] [1,2]

[1,1] [1,2] [0,1] [0,2] [1,3] [1,2] [1,1] [0,1]

Each employee is assigned a set of shifts!


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Domain Representation

Enumerate all subsets Pro: Accurate (any domain can be expressed) Con: Exponential blowup

D(X)= { {1,2,3}, {2,3}, {3} }


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Domain Representation

Solution: Approximate domains

D(X)= {5, 7, 9, 15, 23, 27, 39, 43}

Can be approximated by an interval of the integers

D(X)= [5, 43]


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Compact Set Representation Collection of domain variables

x1,…,xn, D(xi) = Universe AllDifferent(x1,…,xn) Break symmetry: x1< … <xn

If cardinality is unknown, add a dummy value (and replace AllDifferent by NValue)

Pro: Compact (in terms of space requirements) Con: Inaccurate, clumsy

Set variables are an abstraction of this representation


15

Set Bounds [Puget’92, Gervet’97]The domain is specified by two sets:

}|{],[)( UsLsULXD

L (lower bound)

U (upper bound)

Set inclusion lattice


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Compact Domain Representation D(X)= { {1,2,3}, {2,3}, {3} }

Becomes

D(X) = [ {3}, {1,2,3} ]


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Compact Domain Representation D(X)= { {1,2,3}, {2,3}, {3} }

Becomes

D(X) = [ {3}, {1,2,3} ]

Note: D(X) contains also {2,3} !


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Partial vs. Total Order

Note: L and U may not belong to the domain!

D(X)= { {1,2}, {1,3} }

Becomes

D(X) = [ {1}, {1,2,3} ]


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Filtering

Bound consistency: Narrow the domain as much as possible without losing any solutions

L (lower bound)

U (upper bound)


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Filtering

Bound consistency: Narrow the domain as much as possible without losing any solutions

L (lower bound)

U (upper bound)


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A Simple Example

Subset(S,X) (S is a subset of X) Where D(S)=[LS,US], D(X)=[LX,UX]

SXX

XSS

XS

LLL

UUU

UL

.3

.2

solution no is there\ If.1

:Algorithmy Consistenc Bound


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A Simple Example

Subset(S,X) (S is a subset of X) Where D(S)=[LS,US], D(X)=[LX,UX]

SXX

XSS

XS

LLL

UUU

UL

.3

.2

solution no is there\ If.1

:Algorithmy Consistenc Bound

Running time:O(|Us|+|Ux|)


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Lattice Domain Representation Pro

Compact (space and specification) Filtering time depends on size of representation and not

on cardinality of domain

Con Not all domains can be expressed accurately

Partial filtering (only bounds)


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Symmetric Cardinality Constraint [Kocjan, Kreuger – CPAIOR 2004]

Shifts

Employees

[1,2] [1,1] [2,3] [0,1]

[1,1] [1,2] [0,1] [0,2] [1,3]


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[1,2] [1,1] [2,3] [0,1]

[1,1]

[1,2] [0,1] [0,2][1,3]

[0,1]

s

t

[4,7]


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[1,2] [1,1] [2,3] [0,1]

[1,1]

[1,2] [0,1] [0,2][1,3]

[0,1]

s

t

[4,7]


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[0,1][0,1] [0,1]

[0,1] [0,1] [0,1][0,2]

[0,1]

s

t

[0,3]

Residual graph:


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[0,1][0,1] [0,1]

[0,1] [0,1] [0,1][0,2]

[0,1]

s

t

[0,3]

Strongly ConnectedComponents:


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[0,1][0,1] [0,1]

[0,1] [0,1] [0,1][0,2]

[0,1]

s

t

[0,3]

Non-flow edge betweenSCCs: Infeasible.


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[0,1][0,1] [0,1]

[0,1] [0,1] [0,1][0,2]

[0,1]

s

t

[0,3]

Flow edge betweenSCCs: Mandatory.


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Symetric Cardinality Constraint [Kocjan, Kreuger – CPAIOR 2004]Algorithm: Connect each employee with possible shifts

by edge of capacity [1,1] if the shift is mandatory by edge of capacity [0,1] otherwise


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Symetric Cardinality Constraint [Kocjan, Kreuger – CPAIOR 2004]Algorithm: Connect employee with all possible shifts

by edge of capacity [1,1] if the shift is mandatory by edge of capacity [0,1] otherwise

Find a flow, construct the residual graph Edges between SCCs are filtered

Non-flow edges are deleted Inconsistency if one of them was mandatory

Flow edges become mandatory If they were not already


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FixedCardinalityDisjoint(X1,…,Xn,C) Set Variables X1,…,Xn

A constant C

Semantics:

jiji

ii

XX

CX

[Sadler and Gevret, Techreport’04 / Bessiere et al. CP’04 / Implemented in ILOG solver]


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FixedCardinalityDisjoint(X1,…,Xn,C)

values

Set variables

[0,1] [0,1] [0,1] [0,1]

[C, C] [C,C] [C,C] [C,C] [C,C]

Special case of the symmetric cardinality constraint:


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FixedCardinalityDisjoint(X1,…,Xn,C)

values

Set variables

[0,1] [0,1] [0,1] [0,1]

[C,C] [C,C] [C,C] [C,C] [C,C]

Actually, this is an upside-down GCC:

values

Domain variables


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Multiset Variables [Walsh CP’03] Domain representations:

Domain Variables Same problems as with sets

Bounds: Generalization of Gervet/Puget’s set intervals D(X)=[L,U] where L and U are multisets

Occurrence representation D(X)=(a,{0,1}), (b,{3,4}), (c,{2,5})


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Multiset Variables [Walsh CP’03]Thm: Occurrence representation is more

expressive than bound representation

D(X) = { {a} , {a,a,a} }

Bounds: D(X)=[{a},{a,a,a}] Includes {a,a}

Occurrence: D(X)=[(a,{1,3})]


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Multiset Variables [Walsh CP’03]Thm: Occurrence representation is more

expressive than bound representation

This does not hold if occurrences are described as intervals!

(a,[1,3]) = [{a},{a,a,a}]


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Set vs. Multiset Variables

It’s all the same, but with repetitions. right?


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FixedCardinalityDisjoint(X1,…,Xn,C)It’s all the same, but with repetitions. right?

values

Multiset variables[ 0, occi ]

[C, C] [C,C] [C,C] [C,C] [C,C]


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values

Multiset variables

[ 0, occi ]

[C, C] [C,C] [C,C] [C,C] [C,C]

[ 0, max{occi} ]


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values

Multiset variables

[ 0, occi ]

[C, C] [C,C] [C,C] [C,C] [C,C]

[ 0, max{occi} ]

But how do we enforce disjointness?


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values

Multiset variables

{ 0, occi }

[C, C] [C,C] [C,C] [C,C] [C,C]

[ 0, max{occi} ]



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values

Multiset variables

{ 0, occi }

[C, C] [C,C] [C,C] [C,C] [C,C]

[ 0, max{occi} ]


Flow

becomes

NP-Hard


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Apparently not!

Thm [Bessiere et al. CP’04]: It is NP-hard to check feasibility for FCDisjoint on multiset variables.


46

Reduction from 3-SAT )()( 431321 xxxxxx

x11 x13 x23 x24 !x12 !x21

p1 p2 p3 p4 n1 n2 n3 n4

ValueSet variable


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x11 x13 x23 x24 !x12 !x21

ValueSet variable

Value for each non-negated occurrence

Value for each negated occurrenceVariable for each clause

Cardinality = 1

Clause variables ”take” one literal from each clause


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x11 x13 x23 x24 !x12 !x21

p1 p2 p3 p4 n1 n2 n3 n4

ValueSet variable

Variable for each non-negated occurrence

Cardinality = # occurrences

Variable for each negated occurrence

Cardinality = # occurrences

”True” and ”False” value for each variable

pi if xi is trueni if xi is false


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x11 x13 x23 x24 !x12 !x21

p1 p2 p3 p4n1 n2 n3 n4

ValueSet variable

Consistency variables

Cardinality = 1


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x11 x13 x23 x24 !x12 !x21

p1 p2 p3 p4n1 n2 n3 n4

ValueSet variable

X2=x4 =FALSE X1= X3 =TRUE

”0 times”


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x11 x13 x23 x24 !x12 !x21

p1 p2 p3 p4n1 n2 n3 n4

ValueSet variable

Multisets allow us to encode choices!


52

Graph Variables [Dooms et al. CP’05]Graph variable: Two set variables, V and E with VVE Conceptually:

L (lower bound)

U (upper bound)

graph inclusion lattice


53

Filtering With Graph Variables1.Useless

dashed nodes and edges are removed.

2. Mandatorydashed nodes and edges are made solid.


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Part 2

Different Domain Representations Length-Lex BDD

Exercises


56

Cardinality constraints

Often a solution is a collection of sets of fixed cardinality E.g., sports scheduling

D(X) = [{},{1,2,3,4}] and Cardinality(X) = 2 No filtering


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Lexicographic constraints

Symmetry breaking Enforce order within a set (mentioned) Enforce order between sets

Lexicographic constraints: X << Y: X is lexicographically smaller than Y.

D(X) = [{},{2,3,4,5}] and X >> {2,3,4} No filtering


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Length-Lex Domains [Gervet, Van Hentenryck AAAI’06] Total order of sets

first by length, then lexicographically Example: a, b, c, ab, ac, bc, abc

Domain is an interval of this total order [{b}, {a,c}] = {{b}, {c}, {a,b}, {a,c}}

Filtering: Bounds consistency Closer to the scalar domain case E.g., the bounds are consistent


59

Cardinality and Lex Constraints D(X) = [{},{1,2,3,4}] and Cardinality(X) = 2

Filter to D(X)=[{1,2},{3,4}]

D(X) = [{2},{1,2,3}], D(Y)=[{},{2,3}]and X << Y Filter to D(X)=[{2},{1,3}] and D(Y)=[{3},{2,3}]


60

Other Constraints

D(X) = [{1,3,6,7},{2,4,6,7}] Constraint: {3,4} is a subset of X

The new domain: First successor of {1,3,6,7} that contains {3,4} First predecessor of {2,4,6,7} that contains {3,4}


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Basic Successor Operation

U = {1,…,7} Succ({1,3,6,7}) = {1,4,5,6}

Location phase

Reconstruction phase

Linear time!


62

Complex Successor Operations Use variants of the location-reconstruction

algorithm Example:

U = {1,…,7} Find first successor of {1,3,6,7} that contains {3,4}

It’s {2,3,4,5}. Why?

For location phase: Subroutine LR(m,R,i,p)

I am in position i of m {R1…Rp-1 } are inside {m1 .. mi-1} can I reconstruct from i?


63

Location Phase

R is in m: nothing to do

No place left or too small:cannot reconstruct from iTry to see if you can reconstruct further

mi in Rp


64

Reconstruction Phase

start at i and need to cover {Rp,…}

Use elements from R when no more space

Use smallest possible elements


65

Length-Lex Domains

See other results in the AAAI’06 paper.

Brand-new idea, with many open problems In particular, global constraints/ filtering


66

Binary Decision Diagrams (BDDs) [Hawkins et al. - JAIR ’05] A completely different approach to set

domains BDD: A data structure that represents a

boolean function (set of sets) Accurate domain representation Exponential-size in the worst case Usually much smaller Used in other fields (e.g., formal verification)


67

Binary Decision Trees

x1

x2

x3 x3

x2

x3 x3

T T F T F F F T


68


x1

x2

x3 x3

x2

x3 x3

T T F T F F F T


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x1

x2

x3

x2

x3 x3T

F T F F F T


70


x1

x2

x3

x2

x3 x3T

F T F F F T


71


x1

x2

x3

x2

x3T

F T

F

F T


72


x1

x2

x3

x2

x3T

F T

F

F T


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Binary Decision Trees Diagrams

x1

x2

x3

x2

x3T

F T

F

F T


74


x1

x2

x3

x2

T

F T

F


75


x1

x2

x3

x2

FT


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Operations on BDDs

Given BDDs B1 and B2 representing functions f1 and f2, we can compute a BDD for the result of basic boolean operations in linear or quadratic time in the sizes of the BDDs.

Example: The BDD for the negation of is the same, with values flipped at the leaves.


77

Operations on BDDs: 21 ff

Case 1: if B1=T or B2=T then return T

Case 2: if B1=F then return B2

Case 3: if B2=F then return B1

Given BDDs B1 and B2, compute a BDD for B1 v B2


78

Operations on BDDs:

x1

21 ff

L1 R1

x2

L2 R2

Case 3: if x1=x2 then return x1

L1v L2 R1v R2

B1 B2

Recursive


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Operations on BDDs:

x1

21 ff

L1 R1

x2

L2 R2

Case 4: if x1<x2 then return x1

L1v B2 R1v B2

B1 B2

Recursive


80

Domains as Boolean Functions Let S be a subset of U={x1,…,xn}. B(S): a boolean formula describing S:

For a set of sets D,

iSi

iSi

xxAB)(

)()( SBDBDS


81

Constraints as Boolean Functions Sets v,w from universe {1,2,3} Boolean variables for v,w: <v1,v2,v3>, <w1,w2,w3>

The constraint is described by the boolean formula:

I.e., we can make a BDD for the constraint

wv

332211 wvwvwv


82

Filtering With BDDs

D(v) = {{1},{1,3},{2,3}}, D(w) = {{2},{1,2},{1,3}}

B(v) B(w)

v1

v2 v2

v3

FT

w1

w2 w2

w3

F

T

w3


83

Filtering With BDDs

BDD B(c) for the constraint wv

v1

F T

w1

v2

w2

v3

w3


84

Filtering With BDDs

Compute )()()( cBwBvB

v1

F T

w1

v2

w2

v3w3

w3


85

Filtering With BDDs

Project on {v1,v2,v3}

v1

F T

w1

v2

w2

v3w3

w3

v1

F T

v2

B’(v)


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Filtering With BDDs

Project on {w1,w2,w3}

v1

F T

w1

v2

w2

v3w3

w3F T

B’(w)w1

w2

w3w3


87

Bibliography Bessiere, Hnich, Hebrard, Walsh. Disjoint, Partition and

Itersection Constraints for Set and Multiset Variables, CP’04.

Dooms, Deville, Dupont. CP(Graph): Introducing a Graph Computation Domain in Constraint Programming, CP’05.

Gervet. Constraints over structured domains, Handbook of Constraint Programming, Elsevier’06.

Gervet. Interval Propagation to Reason about Sets: Definition and Implementation of a Practical Language, Constraints ’97.

Gervet, Van Hentenryck. Length-Lex Ordering for Set CSPs, AAAI’06.

Gervet, Van Hentenryck. Private Communication, ’06


88

Bibliography – cont.

Hawkins, Lagoon, Stuckley. Solving Set Constraint Satisfaction Problems using ROBDDs, JAIR’05.

Kocjan, Kreuger. Filtering Methods for Symmetric Cardinality Constraint, CPAIOR’04.

Puget. PECOS – a High Level Constraint Programming Language, Spicis’92.

Sadler, Gervet. Global Filtering for the Disjointness Constraint on Fixed Cardinality Sets. IC-PARC techreport, ’04.

Walsh. Consistency and Propagation with Multiset Constraints: A Formal Viewpoint, CP’03.

acp advanced summer school on global constraints - samos, greece.1 global variables irit katriel...

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