Introduction to Constraint ProgrammingConstraint Programming - in a nutshell Declarative description...

Introduction to Constraint Programming

Helmut Simonis

Cork Constraint Computation CentreComputer Science Department

University College CorkIreland

CP Meets CAV, Turunc, June 25th, 2012

Helmut Simonis Introduction to CP 1

Licence

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Acknowledgments

The author is partially supported by Science Foundation Ireland(Grant Number 05/IN/I886). This material was developed aspart of the ECLiPSe ELearning course:http://4c.ucc.ie/~hsimonis/ELearning/index.htmSupport from Cisco Systems and the Silicon Valley CommunityFoundation is gratefully acknowledged.


http://4c.ucc.ie/~hsimonis/ELearning/index.htm

Objectives

Overview of Core Constraint ProgrammingThree Main Concepts

Constraint PropagationGlobal ConstraintsCustomizing Search

Based on Examples, not Formal Description


Outline

Why Constraint Programming?Constraint PropagationGlobal ConstraintsCustomizing SearchWhat is missing?


Examples in ECLiPSe

Open sourced constraint programming languageDevelopment goes back to 1985ECRC, ICL, IC-Parc, PTL, Ciscohttp://www.eclipse-clp.org/

SpecialitiesDevelop new solvers for specific domainsIntegration with MIP


http://www.eclipse-clp.org/

ECLiPSe ELearning Course

Self-study course in constraint programmingSupported by Cisco Systems and Silicon ValleyCommunity FoundationMulti-media format, video lectures, slides, handout etchttp://4c.ucc.ie/~hsimonis/ELearning/index.htm


http://4c.ucc.ie/~hsimonis/ELearning/index.htm

Constraint Programming - in a nutshell

Declarative description of problems withVariables which range over (finite) sets of valuesConstraints over subsets of variables which restrict possiblevalue combinationsA solution is a value assignment which satisfies allconstraints

Constraint propagation/reasoningRemoving inconsistent values for variablesDetect failure if constraint can not be satisfiedInteraction of constraints via shared variablesIncomplete

SearchUser controlled assignment of values to variablesEach step triggers constraint propagation

Different domains require/allow different methods


Basic Process

Problem

Human

Model

Constraint Solver/Search

Solution


More Realistic

Problem

Human

Model


Hangs Solution Wrong Solution


Dual Role of Model

Allows Human to Express ProblemClose to Problem DomainConstraints as Abstractions

Allows Solver to ExecuteVariables as Communication MechanismConstraints as Algorithms


ProblemProgram

Constraint SetupSearch

Points to Remember

Part I

Basic Constraint Propagation


ProblemProgram


Points to Remember

Example 1: SEND+MORE=MONEY

Example of Finite Domain Constraint ProblemModels and ProgramsConstraint Propagation and SearchSome Basic Constraints: linear arithmetic, alldifferent,disequalityA Built-in searchVisualizers for variables, constraints and search


ProblemProgram


Points to Remember

Outline

1 Problem

2 Program

3 Constraint Setup

4 Search

5 Points to Remember


ProblemProgram


Points to Remember

Problem Definition

A Crypt-Arithmetic PuzzleWe begin with the definition of the SEND+MORE=MONEYpuzzle. It is often shown in the form of a hand-written addition:

S E N D+ M O R EM O N E Y


ProblemProgram


Points to Remember

Rules


Each character stands for a digit from 0 to 9.Numbers are built from digits in the usual, positionalnotation.Repeated occurrence of the same character denote thesame digit.Different characters denote different digits.Numbers do not start with a zero.The equation must hold.


ProblemProgram


Points to Remember

Outline

1 Problem

2 Program

3 Constraint Setup

4 Search



ProblemProgram


Points to Remember

Model

Each character is a variable, which ranges over the values0 to 9.An alldifferent constraint between all variables, whichstates that two different variables must have differentvalues. This is a very common constraint, which we willencounter in many other problems later on.Two disequality constraints (variable X must be differentfrom value V ) stating that the variables at the beginning ofa number can not take the value 0.An arithmetic equality constraint linking all variables withthe proper coefficients and stating that the equation musthold.


ProblemProgram


Points to Remember

Program Sendmory

:- module(sendmory).é Define Module:- export(sendmory/1).:- lib(ic).sendmory(L):-

L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember

Program Sendmory

:- module(sendmory).:- export(sendmory/1).é Make predicate visible:- lib(ic).sendmory(L):-



ProblemProgram


Points to Remember

Program Sendmory

:- module(sendmory).:- export(sendmory/1).:- lib(ic).é Use ic librarysendmory(L):-



ProblemProgram


Points to Remember

Program Sendmory

:- module(sendmory).:- export(sendmory/1).:- lib(ic).sendmory(L):-é Predicate definition



ProblemProgram


Points to Remember

Program Sendmory

:- module(sendmory).:- export(sendmory/1).:- lib(ic).sendmory(L):-

L = [S,E,N,D,M,O,R,Y], é Define listL :: 0..9,alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember

Program Sendmory


L = [S,E,N,D,M,O,R,Y],L :: 0..9,é Define integer domain 0..9alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember

Program Sendmory


L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L), é Digits must be differentS #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember

Program Sendmory



L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L),S #\= 0, M #\= 0,é Numbers don’t start with 01000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember

Program Sendmory





ProblemProgram


Points to Remember

Program Sendmory


L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).é built-in search routine


ProblemProgram


Points to Remember

Program Sendmory




ProblemProgram


Points to Remember

General Program Structure


L = [S,E,N,D,M,O,R,Y],é VariablesL :: 0..9,alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember



L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L),é ConstraintsS #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).


ProblemProgram


Points to Remember



L = [S,E,N,D,M,O,R,Y],L :: 0..9,alldifferent(L),S #\= 0, M #\= 0,1000*S + 100*E + 10*N + D +1000*M + 100*O + 10*R + E #=10000*M + 1000*O + 100*N + 10*E + Y,labeling(L).é Search


ProblemProgram


Points to Remember

Choice of Model

This is one model, not the model of the problemMany possible alternativesChoice often depends on your constraint system

Constraints availableReasoning attached to constraints

Not always clear which is the best modelOften: Not clear what is the problem


ProblemProgram


Points to Remember

Running the program

To run the program, we have to enter the querysendmory:sendmory(L).

ResultL = [9, 5, 6, 7, 1, 0, 8, 2]yes (0.00s cpu, solution 1, maybe more)


ProblemProgram


Points to Remember

Question

But how did the program come up with this solution?


ProblemProgram


Points to Remember

Domain DefinitionAlldifferent ConstraintDisequality ConstraintsEquality Constraint

Outline

1 Problem

2 Program

3 Constraint SetupDomain DefinitionAlldifferent ConstraintDisequality ConstraintsEquality Constraint

4 Search

5 Points to RememberHelmut Simonis Introduction to CP 36

ProblemProgram


Points to Remember


Domain Definition

L = [S,E,N,D,M,O,R,Y],L :: 0..9,

[S,E ,N,D,M,O,R,Y ] ∈ {0..9}


ProblemProgram


Points to Remember


Domain Visualization

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember



Rows =Variables

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember



Columns = Values0 1 2 3 4 5 6 7 8 9

SENDMORY


ProblemProgram


Points to Remember



Cells= State

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember


Alldifferent Constraint

alldifferent(L),

Built-in of ic libraryNo initial propagation possibleSuspends, waits until variables are changedWhen variable is fixed, remove value from domain of othervariablesForward checking


ProblemProgram


Points to Remember


Alldifferent Visualization

Uses the same representation as the domain visualizer

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember


Disequality Constraints

S #\= 0, M#\= 0,

Remove value from domain

S ∈ {1..9},M ∈ {1..9}

Constraints solved, can be removed


ProblemProgram


Points to Remember


Domains after Disequality

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember


Equality Constraint

Normalization of linear termsSingle occurence of variablePositive coefficients

Propagation


ProblemProgram


Points to Remember


Normalization

1000*S+ 100*E+ 10*N+ D+1000*M+ 100*O+ 10*R+ E

10000*M+ 1000*O+ 100*N+ 10*E+ Y


ProblemProgram


Points to Remember


Normalization

1000*S+ 100*E+ 10*N+ D+ 100*O+ 10*R+ E

9000*M+ 1000*O+ 100*N+ 10*E+ Y


ProblemProgram


Points to Remember


Normalization

1000*S+ 100*E+ 10*N+ D+ 10*R+ E

9000*M+ 900*O+ 100*N+ 10*E+ Y


ProblemProgram


Points to Remember


Normalization

1000*S+ 100*E+ D+ 10*R+ E

9000*M+ 900*O+ 90*N+ 10*E+ Y


ProblemProgram


Points to Remember


Normalization

1000*S+ 91*E+ D+ 10*R

9000*M+ 900*O+ 90*N+ Y


ProblemProgram


Points to Remember


Simplified Equation

1000∗S+91∗E +10∗R+D = 9000∗M +900∗O+90∗N +Y


ProblemProgram


Points to Remember


Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9 =

9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9


ProblemProgram


Points to Remember


Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸1000..9918

=

9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..89919


ProblemProgram


Points to Remember


Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

=

9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918


ProblemProgram


Points to Remember


Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

=

9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918

Deduction:M = 1,S = 9,O ∈ {0..1}


ProblemProgram


Points to Remember


Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

=

9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918

Deduction:M = 1,S = 9,O ∈ {0..1}

Why? Skip


ProblemProgram


Points to Remember


Consider lower bound for S

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

= 9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918

Lower bound of equation is 9000Rest of lhs (left hand side) (91 ∗E0..9 + 10 ∗R0..9 +D0..9) isatmost 918S must be greater or equal to 9000−918

1000 = 8.082otherwise lower bound of equation not reached by lhs

S is integer, therefore S ≥ d9000−9181000 e = 9

S has upper bound of 9, so S = 9


ProblemProgram


Points to Remember


Consider upper bound of M

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

= 9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918

Upper bound of equation is 9918Rest of rhs (right hand side) 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9

is at least 0M must be smaller or equal to 9918−0

9000 = 1.102

M must be integer, therefore M ≤ b9918−09000 c = 1

M has lower bound of 1, so M = 1


ProblemProgram


Points to Remember


Consider upper bound of O

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9︸︷︷︸9000..9918

= 9000 ∗M1..9 + 900 ∗O0..9 + 90 ∗ N0..9 + Y 0..9︸︷︷︸9000..9918

Upper bound of equation is 9918Rest of rhs (right hand side) 9000 ∗ 1 + 90 ∗ N0..9 + Y 0..9 isat least 9000O must be smaller or equal to 9918−9000

900 = 1.02

O must be integer, therefore O ≤ b9918−9000900 c = 1

O has lower bound of 0, so O ∈ {0..1}


ProblemProgram


Points to Remember


Propagation of equality: Result

0 1 2 3 4 5 6 7 8 9S - - - - - - - - Y

ENDM Y - - - - - - - -O 6 6 6 6 6 6 6 6

RY


ProblemProgram


Points to Remember


Propagation of alldifferent

0 1 2 3 4 5 6 7 8 9S - - - - - - - - Y

ENDM Y - - - - - - - -O 6 6 6 6 6 6 6 6

RY


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9S Y

E |N |D |M Y

OR |Y |


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SE |N |D |M Y

O |R |Y |


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SENDMO Y

RY


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SE |N |D |MO Y

R |Y |


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SENDMORY

O = 0, [E ,R,D,N,Y ] ∈ {2..8}


ProblemProgram


Points to Remember


Waking the equality constraint

Triggered by assignment of variablesor update of lower or upper bound


ProblemProgram


Points to Remember


Removal of constants

1000 ∗ 9 + 91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 =

9000 ∗ 1 + 900 ∗ 0 + 90 ∗ N2..8 + Y 2..8


ProblemProgram


Points to Remember



1000 ∗ 9 + 91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 =

9000 ∗ 1 + 900 ∗ 0 + 90 ∗ N2..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 = 90 ∗ N2..8 + Y 2..8


ProblemProgram


Points to Remember


Propagation of equality (Iteration 1)

91 ∗ E2..8 + 10 ∗ R2..8 + D2..8︸︷︷︸204..816

= 90 ∗ N2..8 + Y 2..8︸︷︷︸182..728


ProblemProgram


Points to Remember



91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 = 90 ∗ N2..8 + Y 2..8︸︷︷︸204..728


ProblemProgram


Points to Remember



91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 = 90 ∗ N2..8 + Y 2..8︸︷︷︸204..728

N ≥ 3 = d204− 890

e,E ≤ 7 = b728− 2291

c


ProblemProgram


Points to Remember



91 ∗ E2..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E2..7 + 10 ∗ R2..8 + D2..8︸︷︷︸204..725

= 90 ∗ N3..8 + Y 2..8︸︷︷︸272..728


ProblemProgram


Points to Remember



91 ∗ E2..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8︸︷︷︸272..725


ProblemProgram


Points to Remember



91 ∗ E2..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8︸︷︷︸272..725

E ≥ 3 = d272− 8891

e


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8︸︷︷︸295..725

= 90 ∗ N3..8 + Y 2..8︸︷︷︸272..728


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8︸︷︷︸295..725


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N3..8 + Y 2..8︸︷︷︸295..725

N ≥ 4 = d295− 890

e


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8︸︷︷︸295..725

= 90 ∗ N4..8 + Y 2..8︸︷︷︸362..728


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8︸︷︷︸362..725


ProblemProgram


Points to Remember



91 ∗ E3..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8︸︷︷︸362..725

E ≥ 4 = d362− 8891

e


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8︸︷︷︸386..725

= 90 ∗ N4..8 + Y 2..8︸︷︷︸362..728


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8︸︷︷︸386..725


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N4..8 + Y 2..8︸︷︷︸386..725

N ≥ 5 = d386− 890

e


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8︸︷︷︸386..725

= 90 ∗ N5..8 + Y 2..8︸︷︷︸452..728


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8︸︷︷︸452..725


ProblemProgram


Points to Remember



91 ∗ E4..7 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8︸︷︷︸452..725

N ≥ 5 = d452− 890

e,E ≥ 4 = d452− 8891

e

No further propagation at this point


ProblemProgram


Points to Remember


Domains after setup

0 1 2 3 4 5 6 7 8 9SENDMORY


ProblemProgram


Points to Remember

Step 1Step 2Further StepsSolution

Outline

1 Problem

2 Program

3 Constraint Setup

4 SearchStep 1Step 2Further StepsSolution

5 Points to RememberHelmut Simonis Introduction to CP 100

ProblemProgram


Points to Remember


labeling built-in

labeling([S,E,N,D,M,O,R,Y])

Try variable is order givenTry values starting from smallest value in domainWhen failing, backtrack to last open choiceChronological BacktrackingDepth First search


ProblemProgram


Points to Remember


Search Tree Step 1

Variable S already fixed

S

E9


ProblemProgram


Points to Remember


Step 2, Alternative E = 4

Variable E ∈ {4..7}, first value tested is 4

S

E4

9


ProblemProgram


Points to Remember


Assignment E = 4

0 1 2 3 4 5 6 7 8 9SE Y - - -NDMORY


ProblemProgram


Points to Remember


Propagation of E = 4, equality constraint

91 ∗ 4 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ 4 + 10 ∗ R2..8 + D2..8︸︷︷︸386..452

= 90 ∗ N5..8 + Y 2..8︸︷︷︸452..728


ProblemProgram


Points to Remember



91 ∗ 4 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8︸︷︷︸452


ProblemProgram


Points to Remember



91 ∗ 4 + 10 ∗ R2..8 + D2..8 = 90 ∗ N5..8 + Y 2..8︸︷︷︸452

N = 5,Y = 2,R = 8,D = 8


ProblemProgram


Points to Remember


Result of equality propagation

0 1 2 3 4 5 6 7 8 9SEN Y - - -D - - - - - - Y

MOR - - - - - - Y

Y Y - - - - - -


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SEN Y - - -D - - - - - - Y

MOR - - - - - - Y

Y Y - - - - - -


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9S |E |N Y - - |D - - - - - - Y

M |O |R - - - - - - Y

Y Y - - - - - |

Alldifferent fails!


ProblemProgram


Points to Remember


Step 2, Alternative E = 5

Return to last open choice, E , and test next value

S

E4

N5

9


ProblemProgram


Points to Remember


Assignment E = 5

0 1 2 3 4 5 6 7 8 9SE - Y - -NDMORY


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SE - Y - -NDMORY


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SE Y

N |D |MOR |Y |


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SENDMORY

N 6= 5,N ≥ 6


ProblemProgram


Points to Remember


Propagation of equality

91 ∗ 5 + 10 ∗ R2..8 + D2..8 = 90 ∗ N6..8 + Y 2..8


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ R2..8 + D2..8︸︷︷︸477..543

= 90 ∗ N6..8 + Y 2..8︸︷︷︸542..728


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ R2..8 + D2..8 = 90 ∗ N6..8 + Y 2..8︸︷︷︸542..543


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ R2..8 + D2..8 = 90 ∗ N6..8 + Y 2..8︸︷︷︸542..543

N = 6,Y ∈ {2,3},R = 8,D ∈ {7..8}


ProblemProgram


Points to Remember


Result of equality propagation

0 1 2 3 4 5 6 7 8 9SEN Y - -D 6 6 6 6

MOR - - - - - Y

Y 6 6 6 6


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SEN Y - -D 6 6 6 6

MOR - - - - - Y

Y 6 6 6 6


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SEND |MOR Y

Y


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SEND Y

MORY


ProblemProgram


Points to Remember



0 1 2 3 4 5 6 7 8 9SENDMORY

D = 7


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ 8 + 7︸︷︷︸542

= 90 ∗ 6 + Y 2..3︸︷︷︸542..543


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3︸︷︷︸542


ProblemProgram


Points to Remember



91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3︸︷︷︸542

Y = 2


ProblemProgram


Points to Remember


Last propagation step

0 1 2 3 4 5 6 7 8 9SENDMORY Y -


ProblemProgram


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Complete Search Tree

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ProblemProgram


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Solution

9 5 6 7+ 1 0 8 51 0 6 5 2


ProblemProgram


Points to Remember

Outline

1 Problem

2 Program

3 Constraint Setup

4 Search



ProblemProgram


Points to Remember

Points to Remember

Constraint models are expressed by variables andconstraints.Problems can have many different models, which canbehave quite differently. Choosing the best model is an art.Constraints can take many different forms.Propagation deals with the interaction of variables andconstraints.It removes some values that are inconsistent with aconstraint from the domain of a variable.Constraints only communicate via shared variables.


ProblemProgram


Points to Remember

Points to Remember

Propagation usually is not sufficient, search may berequired to find a solution.Propagation is data driven, and can be quite complex evenfor small examples.The default search uses chronological depth-firstbacktracking, systematically exploring the complete searchspace.The search choices and propagation are interleaved, afterevery choice some more propagation may further reducethe problem.


ProblemInitial Propagation (Forward Checking)

Improved ReasoningSearch

Part II

Global Constraints




Example 2: Sudoku

Global ConstraintsPowerful modelling abstractionsNon-trivial propagationDifferent consistency levels

Example: Sudoku puzzle




Outline

6 Problem

7 Initial Propagation (Forward Checking)

8 Improved Reasoning

9 Search




Problem Definition

SudokuFill in numbers from 1 to 9 so that each row, column and blockcontain each number exactly once

41 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 1 7

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9 8 2 5

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9 3 7 6

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 6

1 2 34 5 67 8 9 4




Problem Definition

SudokuFill in numbers from 1 to 9 so that each row, column and blockcontain each number exactly once

41 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 1 7

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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4 2 8 5 6 3 1 7 93 5 9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6 3 9 8 2 5 75 9 2 7 4 1 3 8 68 3 7 6 2 5 9 4 12 7 4 9 5 6 8 1 36 8 3 2 1 4 7 9 59 1 5 8 3 7 6 2 4




Model

A variable for each cell, ranging from 1 to 9A 9x9 matrix of variables describing the problemPreassigned integers for the given hintsalldifferent constraints for each row, column and 3x3block




Reminder: alldifferent

Argument: list of variablesMeaning: variables are pairwise differentReasoning: Forward Checking (FC)

When variable is assigned to value, remove the value fromall other variablesIf a variable has only one possible value, then it is assignedIf a variable has no possible values, then the constraint failsConstraint is checked whenever one of its variables isassignedEquivalent to decomposition into binary disequalityconstraints




Main Program

model(Matrix):-Matrix[1..9,1..9] :: 1..9,(for(I,1,9),param(Matrix) do

alldifferent(Matrix[I,1..9]),alldifferent(Matrix[1..9,I])

),(multifor([I,J],[1,1],[7,7],[3,3]),param(Matrix) do

alldifferent(flatten(Matrix[I..I+2,J..J+2]))),flatten_array(Matrix,List),labeling(List).




Domain Visualizer

41 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 1 7

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9 3 2

1 2 34 5 67 8 9

1 2 34 5 67 8 9 6

1 2 34 5 67 8 9

1 2 34 5 67 8 9 8 2 5

1 2 34 5 67 8 9

1 2 34 5 67 8 9 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9 3 7 6

1 2 34 5 67 8 9

1 2 34 5 67 8 9 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

2 71 2 34 5 67 8 9

1 2 34 5 67 8 9 5

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 1 4

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 6

1 2 34 5 67 8 9 4

Problem shown as matrixEach cell corresponds to a variableInstantiated: Shows integer value (large)Uninstantiated: Shows values in domain




Outline

6 Problem



9 Search




Initial State (Forward Checking)

41 2 34 5 67 8 9 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9 1 4

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9 4




Propagation Steps (Forward Checking)

41 2 34 5 67 8 9 8

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1 2 34 5 67 8 9

1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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1 2 34 5 67 8 9

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Skip Animation





41 2 34 5 67 8 9 8

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Back to Start Skip Animation





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7 9 4Back to Start Skip Animation





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After Setup (Forward Checking)

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Domain ConsistencyComparison

Outline

6 Problem


8 Improved ReasoningDomain ConsistencyComparison

9 Search





Can we do better?

The alldifferent constraint is missing propagationHow can we do more propagation?Do we know when we derive all possible information fromthe constraint?

Constraints only interact by changing domains of variables





A Simpler Example

:-lib(ic).

top:-X :: 1..2,Y :: 1..2,Z :: 1..3,alldifferent([X,Y,Z]),writeln([X,Y,Z]).





Using Forward Checking

No variable is assignedNo reduction of domainsBut, values 1 and 2 can be removed from ZThis means that Z is assigned to 3





Visualization of alldifferent as Graph

X

Y

Z

1

2

3Show problem as graph with two types of nodes

Variables on the leftValues on the right

If value is in domain of variable, show link between themThis is called a bipartite graph





A Simpler Example

Value Graph for

X :: 1..2,

Y :: 1..2,

Z :: 1..3

X

Y

Z

1

2

3





A Simpler Example

Check interval [1,2]X

Y

Z

1

2

3





A Simpler Example

Find variables completelycontained in intervalThere are two: X and YThis uses up the capacity of theinterval

X

Y

Z

1

2

3





A Simpler Example

No other variable can use thatinterval

X

Y

Z

1

2

3





A Simpler Example

Only one value left in domain of Z,this can be assigned

X

Y

Z

1

2

3





Idea (Hall Intervals)

Take each interval of possible values, say size NFind all K variables whose domain is completely containedin intervalIf K > N then the constraint is infeasibleIf K = N then no other variable can use that intervalRemove values from such variables if their bounds changeIf K < N do nothingRe-check whenever domain bounds change





Implementation

Problem: Too many intervals (O(n2)) to considerSolution:

Check only those intervals which update boundsEnumerate intervals incrementallyStarting from lowest(highest) valueUsing sorted list of variables

Complexity: O(n log(n)) in standard implementationsImportant: Only looks at min/max bounds of variables





Bounds Consistency

DefinitionA constraint achieves bounds consistency, if for the lower andupper bound of every variable, it is possible to find values for allother variables between their lower and upper bounds whichsatisfy the constraint.





Can we do better?

Bounds consistency only considers min/max boundsIgnores “holes” in domainSometimes we can improve propagation looking at thoseholes





Another Simple Example

:-lib(ic).

top:-X :: [1,3],Y :: [1,3],Z :: 1..3,alldifferent([X,Y,Z]),writeln([X,Y,Z]).






Value Graph for

X :: [1,3],

Y :: [1,3],

Z :: 1..3

X

Y

Z

1

2

3






Check interval [1,2]No domain of a variablecompletely contained in intervalNo propagation

X

Y

Z

1

2

3






Check interval [2,3]No domain of a variablecompletely contained in intervalNo propagation

X

Y

Z

1

2

3






But, more propagation is possible,there are only two solutions

X

Y

Z

1

2

3






Solution 1: assignment in blueX

Y

Z

1

2

3






Solution 2: assignment in greenX

Y

Z

1

2

3






Solution 1 Solution 2 CombinedX

Y

Z

1

2

3

X

Y

Z

1

2

3

X

Y

Z

1

2

3

Combining solutions shows that Z=1 andZ=3 are not possible.






Can we deduce this without enumeratingsolutions?





Solutions and maximal matchings

A Matching is subset of edges which do not coincide in anynodeNo matching can have more edges than number ofvariablesEvery solution corresponds to a maximal matching andvice versaIf a link does not belong to some maximal matching, then itcan be removed





Implementation

Possible to compute all links which belong to somematching

Without enumerating all of them!

Enough to compute one maximal matchingRequires algorithm for strongly connected componentsExtra work required if more values than variablesAll links (values in domains) which are not supported canbe removedComplexity: O(n1.5d)





Domain Consistency

DefinitionA constraint achieves domain consistency, if for every variableand for every value in its domain, it is possible to find values inthe domains of all other variables which satisfy the constraint.

Also called generalized arc consistency (GAC)or hyper arc consistency





Can we still do better?

NO! This extracts all information from this one constraintWe could perhaps improve speed, but not propagationBut possible to use different modelOr model interaction of multiple constraints





Should all constraints achieve domain consistency?

Domain consistency is usually more expensive thanbounds consistency

Overkill for simple problemsNice to have choices

For some constraints achieving domain consistency isNP-hard

We have to live with more restricted propagation





Initial State (Domain Consistency)

41 2 34 5 67 8 9 8

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Propagation Steps (Domain Consistency)

41 2 34 5 67 8 9 8

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Skip Animation






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1 35 6

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41 25 6 8

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7 8 9

1 24 5 68

1 2 34 5

9

2 34 57 8 9 1 4

1 34 57 8

1 2 34 5 67 8 9

1 2 34 5 67 8 9

1 35 6

7 8 9

1 24 5 68

1 2 34 5

9

2 34 57 8 9

2 34 6

9

1 2 35 6

7 9 61 2 34 5 67 8 9 4







41 25 6 8

2 35

7 9

2 369

1 2 35 6

7 9

1 35

7

1 2 35 6

7 9

1 2 35 6

7 93

5 68 9

24 5 68

2 34 5

9 1 72 35 69

34 58

2 34 5 68 9

2 34 5 68 9

15 6

7 9

14 5 6

14 5

94 57 9 8

15 6

7 9

14 57 3 2

1 3

7 9

14 6

347 9

34

9 8 2 51 347 9

1 35 6

7 91 2 34 5

2 34 57

2 34 6

1 2 35 6

7

1 34 57 8

1 2 34 5 67

158 3 7 6

24

1 25 9

1 24 58

1 24 58

2 71 34

9

348 9 5

1 369

1 348

1 34 68 9

1 34 68 9

35 6

7 8 9

25 68

2 359

2 35

7 8 9 1 43

57 8

2 35 6

7 8 9

2 35 6

7 8 91 35 6

7 8 9

1 24 5 68

1 2 34 5

9

2 34 57 8 9

2 34 6

9

1 2 35 6

7 9 61 2 34 5 67 8 9 4







41 25 6 8

2 35

7 9

2 369

1 2 35 6

7 9

1 35

7

1 26

7 9

1 2 35 6

7 93

5 68 9

24 5 68

2 34 5

9 1 72 35 69

34 58

24 6

9

2 34 5 68 9

15 6

7 9

14 5 6

14 5

94 57 9 8

15 6

7 9

14 57 3 2

1 3

7 9

14 6

347 9

34

9 8 2 51 347 9

1 35 6

7 91 2 34 5

2 34 57

2 34 6

1 2 35 6

7

1 34 57 8

1 2 34 5 67

158 3 7 6

24

1 25 9

1 24

1 24 58

2 71 34

9

348 9 5

1 369

1 348

14 6

9

1 34 68 9

35 6

7 8 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

2 35 6

7 8 91 35 6

7 8 9

1 24 5 68

1 2 34 5

9

2 34 57 8 9

2 34 6

9

1 2 35 6

7 9 61 24 67 9 4







41 25 6 8

2 35

7 9

2 369

1 2 35 6

7 9

1 35

7

1 26

7 9

1 2 35 6

7 93

5 68 9

24 5 68

2 34 5

9 1 72 35 69

34 58

24 6

9

2 34 5 68 9

15 6

7 9

14 5 6

14 5

94 57 9 8

15 6

7 9

14 57 3 2

1 3

7 9

14 6

347 9

34

9 8 2 51 347 9

1 35 6

7 91 2 34 5

2 34 57

2 34 6

1 2 35 6

7

1 34 57 8

1 2 34 5 67

158 3 7 6

24

1 25 9

1 24

1 24 58

2 71 34

9

348 9 5

1 369

1 348

14 6

9

1 34 68 9

35 6

7 8 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

2 35 6

7 8 91 35

7 8 9

1 258

1 2 359

2 35

7 8 9

2 3

9

1 2 35

7 9 61 2







41 25 6 8

2 35

7 9

2 369

1 2 35 6

7 9

1 35

7

1 26

7 9

1 35 6

7 93

5 68 9

24 5 68

2 34 5

9 1 72 35 69

34 58

24 6

9

35 68 9

714 5 6

14 5

94 57 9 8

15 6

7 9

14 57 3 2

1 3

7 9 4 63

47 9

34

9 8 2 51 3

7 91 35 6

7 91 2 34 5

2 34 57

2 34 6

1 2 35 6

7

1 34 57 8

1 35 6

7158 3 7 6

24

1 25 9

1 24

158

2 71 34

9

348 9 5

1 369

1 348

14 6

9

1 36

8 93

5 67 8 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

35 6

7 8 91 35

7 8 9

1 258

1 2 359

2 35

7 8 9

2 3

9

1 2 35

7 9 61 2







41 25 6 8

2 35

7 9

2 369

1 2 35 6

7 9

1 35

7

1 26

7 9

1 35 6

7 93

5 69

25 6

2 359 1 7

2 35 69

34 58

24 6

9

35 68 9

715 6

159 4 8

15 69

14 5 3 2

1 3

9 4 63

7 9

3

9 8 2 51 3

7 91 35 6 9

1 2 34 5

2 34 57

2 34 6

1 2 35 6

7

1 34 57 8

1 35 6

7158 3 7 6

24

1 25 9

1 24

158

2 71 34

9

348 9 5

1 369

1 348

14 6

9

1 36

8 93

5 68 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

35 6

7 8 91 358 9

1 258

1 2 359

2 35

7 8 9

2 3

9

1 2 35

7 9 61 2







41 25 6 8

2 359

2 369

2 35 69

1 35

7

1 26

7 9

1 35 6

7 93

5 69

25 6

2 359 1 7

2 35 69

34 58

24 6

9

35 68 9

715 6

159 4 8 5 6

9

15 3 2

1 3

9 4 63

7 9

3

9 8 2 51 3

7 91 35 6 9

1 2 34 5

2 35

7

2 34 6

1 2 35 6

7

1 34 57 8

1 35 6

7158 3 7 6

24

1 25 9

1 24

158

2 71 34

9

3

8 9 51 3

69

1 348

14 6

9

1 36

8 93

5 68 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

35 6

7 8 91 358 9

1 258

1 2 359

2 35

7 8 9

2 3

9

1 2 35

7 9 61 2







41 25 6 8

2 359

2 369

2 35 69

15

7

16

7 9

15 6

7 93

5 69

25 6

2 359 1 7

2 35 694 58

4 69

5 68 9

715 6

159 4 8 5 6

9

15 3 2

1 4 63

7 9

3

9 8 2 51 3

7 9

5 9 22 35

7

2 34 6

1 2 35 6

7

1 34 57 8

1 35 6

7

8 3 7 62

41 25 9

1 24

158

2 7 43

8 9 51 3

69

1 348

14 6

9

1 36

8 93

5 68 9

25 68

2 359

2 35

7 8 9 1 43

57 8

26

7 9

35 6

7 8 91 358 9

1 258

1 2 359

2 35

7 8 9

2 3

9

1 2 35

7 9 61 2







4 2 8 5 6 3 11

67 9

15 6

7 9369 5

359 1 7 2 4 6 8

7 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7 9

5 9 23

7 4 11 347 8 6

8 3 7 6 2 5 9 4 12 7 4

3

8 9 5 6 8 11 3

68 9

6 83

59 2 1 4

35

7 8

26

7 9 53

9 1 5 82 3

9 7 6 2 4Back to Start Skip Animation






4 2 8 5 6 3 1 7 9 7 93

9 53

9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7 9

5 9 23

7 4 13

7 8 68 3 7 6 2 5 9 4 12 7 4

3

9 5 6 8 13

9

6 83

9 2 1 43

7 7 9 53

9 1 5 83







4 2 8 5 6 3 1 7 9 7 93

9 53

9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7

5 9 23

7 4 13

7 8 68 3 7 6 2 5 9 4 12 7 4

3

9 5 6 8 13

9

6 83

9 2 1 43

7 7 9 53

9 1 5 83







4 2 8 5 6 3 1 7 9 7 93

9 53

9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7

5 9 23

7 4 13

7 8 68 3 7 6 2 5 9 4 12 7 4

3

9 5 6 8 13

9

6 83

9 2 1 43

7 7 9 53

9 1 5 83

9 7 6 2 4Back to Start





After Setup (Domain Consistency)

4 2 8 5 6 3 1 7 9 7 93

9 53

9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7

5 9 23

7 4 13

7 8 68 3 7 6 2 5 9 4 12 7 4

3

9 5 6 8 13

9

6 83

9 2 1 43

7 7 9 53

9 1 5 83

9 7 6 2 4





Comparison

Forward Checking Bounds Consistency Domain Consistency4

1 25 6 8

2 359

369

2 369

15

7

16

7 95 6

7 9369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

41 2

8 5 62 3 1

7

1

7 9 7 93

9

25

3

9 1 72 3

4 58

69

58 9

7 615 4 8 9

15 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

7

34 1

3

7 8 68 3 7 6 2 5 9 4 12 7 4

3

8 9 5 61 3

8

1

9

3

8 9

6 58

3

9

2 3

8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

15

2

8

3

9 7 61 2

4

4 2 8 5 6 3 1 7 9 7 93

9 53

9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6

3

7 9

3

9 8 2 53

7

5 9 23

7 4 13

7 8 68 3 7 6 2 5 9 4 12 7 4

3

9 5 6 8 13

9

6 83

9 2 1 43

7 7 9 53

9 1 5 83

9 7 6 2 4





Typical?

This does not always happenSometimes, two methods produce same amount ofpropagationPossible to predict in certain special casesIn general, tradeoff between speed and propagationNot always fastest to remove inconsistent values earlyBut often required to find a solution at all




Outline

6 Problem



9 Search




Simple search routine

Enumerate variables in given orderTry values starting from smallest one in domainComplete, chronological backtracking




Search Tree (Forward Checking)4

1 25 6 8

2 359

369

2 369

15

7

16

7 95 6

7 9369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

Skip Animation




Search Tree (Forward Checking)4 1 8

2 359

369

2 369

15

7

16

7 95 6

7 9369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

41





Search Tree (Forward Checking)4 1 8 2

369

2 369

57

67 9

5 67 9

369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

4

52

1






369

369

57

67 9

5 67 9

369

25 6

359 1 7

3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

3

2

4

53

2

1






369

369

57

67 9

5 67 9

369

25 6

359 1 7

3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

6

2

4

53 6

2

1






369

369

57

67 9

5 67 9

369

25 6

359 1 7

3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

9

2

4

53 6 9

2

1






369

2 369

57

67 9

5 67 9

369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

4

53 6 9

2

53

1





Search Tree (Forward Checking)4 1 8 3 6

9

269

57

67 9

5 67 9

369

25 6

359 1 7

2694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 6 7 9

3

9 8 2 53

7

5 9 2 47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9 8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2

7 8 9

3

9

2 3

7 9 61 2

7 9 4

6

2

4

53 6 9

2

56

3

1






9

269

57

67 9

5 67 9

369

25 6

359 1 7

2694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 6 7 9

3

9 8 2 53

7

5 9 2 47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9 8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2

7 8 9

3

9

2 3

7 9 61 2

7 9 4

9

2

4

53 6 9

2

56 9

3

1






2 359

369

2 369

57

67 9

5 67 9

369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

5

2

4

53 6 9

2

56 9

3 5

1






2 359

369

2 369

57

67 9

5 67 9

369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

5 6 594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

958

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

9

2

4

53 6 9

2

56 9

3 5 9

1






2 359

369

2 369

15

7

16

7 95 6

7 9369

25 6

359 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

4

53 6 9

2

56 9

3 5 9

1

42






369

369

15

7

16

7 95 6

7 9369

5 63

59 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

4

53 6 9

2

56 9

3 5 9

1

4

53

2






969

15

7

16

7 95 6

7 9369

5 63

59 1 7

2694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 6 7 9

3

9 8 2 53

7

5 9 2 47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9 8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2

7 8 9

3

9

2 3

7 9 61 2

7 9 4

6

2

4

53 6 9

2

56 9

3 5 9

1

4

56

3

2






969

15

7

16

7 95 6

7 9369

5 63

59 1 7

2694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 6 7 9

3

9 8 2 53

7

5 9 2 47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9 8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2

7 8 9

3

9

2 3

7 9 61 2

7 9 4

9

2

4

53 6 9

2

56 9

3 5 9

1

4

56 9

3

2






369

369

15

7

16

7 95 6

7 9369

5 63

59 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594 5

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

8 3 7 6 2 5 9 4 12 7

1 34

9

3

8 9 5369

1 3

8

1

9

3

8 9369

5 68

359

2 3

7 8 9 1 43

57 8

2

7 9

35

7 8 93

9

158

1 359

2 3

7 8 9

3

9

2 3

7 9 61 2

7 9 4

2

4

53 6 9

2

56 9

3 5 9

1

4

56 9

3

55

2






369

369

1

7

16

7 96

7 9369

5 63

59 1 7

2 3694 58

69

5 68 9

67 9

15 6

1594

9 8 69

14 57 3 2

1 4 63

7 9

3

9 8 2 53

7

5 9 23

47

34

1 3

7

3

7 836

7

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Search Tree (Forward Checking)4 2 8 5 6 3 1 7 93 5 9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6 3 9 8 2 5 75 9 2 7 4 1 3 8 68 3 7 6 2 5 9 4 12 7 4 9 5 6 8 1 36 8 3 2 1 4 7 9 59 1 5 8 3 7 6 2 4

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Search Tree (Bounds Consistency)4

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Search Tree (Bounds Consistency)4 2 8 5 6 3 1 7 93 5 9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6 3 9 8 2 5 75 9 2 7 4 1 3 8 68 3 7 6 2 5 9 4 12 7 4 9 5 6 8 1 36 8 3 2 1 4 7 9 59 1 5 8 3 7 6 2 4

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Search Tree (Domain Consistency)4 2 8 5 6 3 1 7 9 7 9

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Search Tree (Domain Consistency)4 2 8 5 6 3 1 7 93 5 9 1 7 2 4 6 87 6 1 4 8 9 5 3 21 4 6 3 9 8 2 5 75 9 2 7 4 1 3 8 68 3 7 6 2 5 9 4 12 7 4 9 5 6 8 1 36 8 3 2 1 4 7 9 59 1 5 8 3 7 6 2 4

87

Back to Start




Tradeoff: nodes and effort

How many nodes do we need to explore?How much effort do we spend in each node?Extreme 1: SAT, do very little reasoning in each node, butdo many nodes very rapidlyExtreme 2: MIP, do a lot of reasoning in root node, and ineach node, reduce number of nodes to exploreConstraint Programming: Choice of balance tuneable forproblem




Global Constraint Catalog

http://www.emn.fr/z-info/sdemasse/gccat/index.html

Description of 399 global constraints, 3250 pagesNot all of them are widely usedDetailed, meta-data description of constraints in Prolog






Key Global Constraints

alldifferent

cumulative

cycle

diffn

element

global_cardinality

minimum_weight_alldifferent

nvalue

sort




alldifferent(L)

A collection of variables L are pairwise differentAlgorithm: Flow, refinementsUse: EverywhereSimilar: permutation, alldifferent_except_0,lex_alldifferent




cumulative(Tasks, Limit)

A set of tasks with start times si , durations di and resourcerequirements ri do not exceed resource limit Limit at anytimeAlgorithm: compulsory parts, energy, edge-finding,not-first/not-lastUse: Scheduling, PlacementSimilar: disjunctive




cycle(N, Succ)

A graph given by the successors si of nodes i contains NcyclesAlgorithm: alldifferent, strongly connectedcomponentsUse: Transportation, SchedulingSimilar: circuit, tree




diffn(Obj)

n-dimensional objects given by orign < xi1, ..., xin > andsize < di1, ...,din > do not overlapAlgorithm: sweep, compulsory partsUse: placementSimilar: geost




element(X, L, C)

C is the xth element of LAlgorithm: basicUse: functional dependencies, costSimilar: table




global_cardinality(L, Values)

Count how often certain values occur in the collection ofvariables LAlgorithm: Flow, refinementsUse: TImetablingSimilar: generalizes alldifferent, among_seq




minimum_weight_alldifferent(L, Matrix,Cost)

Cost is the cost of the assignment of the variables in L, thecost of each entry is given by the Matrix of cost values.The entries in L are pairwise different.Algorithm: Hungarian Method, Flow, SimplexUse: Resource allocationSimilar: global_cardinality_with_costs




nvalue(N, L)

Count the number N of distinct values in a collection ofvariables LAlgorithm: specific, bounds-consistency onlyUse: Assignment problemsSimilar: same




sort(L,K)

K is the sorted collection of the variables in LAlgorithm: specificUse: Building block for reformulation of other constraints




Why are there so many global constraints?

Algorithmic aspectMore specific restrictions allow more refined algorithms

Modelling aspectCapture exactly the properties of the problem we are after

Families of constraints, restrictions and generalizations




Constraint Programming to the rescue!

Constraint Seeker tool (Beldiceanu, Simonis 2011)http://seeker.mines-nantes.fr/

Given positive and negative examples, produce ranked listof possible matching global constraintsItself a collection of constraint programs


http://seeker.mines-nantes.fr/



Basis of modelling tool

Model Seeker (Beldiceanu, Simonis 2012)From positive sample solutions find potential models forproblemExpressed as conjunction of global constraintsFor highly structured problemsDo you have sample solutions?


ProblemProgram

Naive SearchImprovements

Part III

Customizing Search


ProblemProgram


What we want to introduce

Importance of search strategy, constraints alone are notenoughDynamic variable ordering exploits information frompropagationVariable and value choiceHard to find strategy which works all the timesearch builtin, flexible search abstractionDifferent way of improving stability of search routine


ProblemProgram


Example Problem

N-Queens puzzleRather weak constraint propagationMany solutions, limited number of symmetriesEasy to scale problem size


ProblemProgram


Outline

10 Problem

11 Program

12 Naive Search

13 Improvements


ProblemProgram


Problem Definition

8-QueensPlace 8 queens on an 8× 8 chessboard so that no queenattacks another. A queen attacks all cells in horizontal, verticaland diagonal direction. Generalizes to boards of size N × N.


ProblemProgram


Problem Definition

8-QueensPlace 8 queens on an 8× 8 chessboard so that no queenattacks another. A queen attacks all cells in horizontal, verticaland diagonal direction. Generalizes to boards of size N × N.

Solution for board size 8× 8


ProblemProgram


Model

Outline

10 Problem

11 ProgramModel

12 Naive Search

13 Improvements


ProblemProgram


Model

Basic Model

Cell based ModelA 0/1 variable for each cell to say if it is occupied or notConstraints on rows, columns and diagonals to enforceno-attackN2 variables, 6N − 2 constraints

Column (Row) based ModelA 1..N variable for each column, stating position of queen inthe columnBased on observation that each column must containexactly one queenN variables, N2/2 binary constraints


ProblemProgram


Model

Model

assign [X1,X2, ...XN ]

s.t.

∀1 ≤ i ≤ N : Xi ∈ 1..N∀1 ≤ i < j ≤ N : Xi 6= Xj

∀1 ≤ i < j ≤ N : Xi 6= Xj + i − j∀1 ≤ i < j ≤ N : Xi 6= Xj + j − i


ProblemProgram


Outline

10 Problem

11 Program

12 Naive Search

13 Improvements


ProblemProgram


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ProblemProgram


First Solution

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ProblemProgram


Observations

Even for small problem size, tree can become largeNot interested in all detailsIgnore all automatically fixed variablesFor more compact representation abstract failed sub-trees


ProblemProgram


Compact Representation

Number inside triangle: Number of choicesNumber under triangle: Number of failures1

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ProblemProgram


Exploring other board sizes

How stable is the model?Try all sizes from 4 to 100Timeout of 100 seconds


ProblemProgram


Naive Stategy, Problem Sizes 4-100

0

5

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0 5 10 15 20 25 30

Tim

e[s]

Problem Size

"naive/all.txt"


ProblemProgram


Observations

Time very reasonable up to size 20Sizes 20-30 times very variableNot just linked to problem sizeNo size greater than 30 solved within timeout


ProblemProgram


Dynamic Variable ChoiceImproved HeuristicsMaking Search More Stable

Outline

10 Problem

11 Program

12 Naive Search

13 ImprovementsDynamic Variable ChoiceImproved HeuristicsMaking Search More Stable


ProblemProgram



Possible Improvements

Better constraint reasoningRemodelling problem with 3 alldifferent constraintsGlobal reasoning as described beforeNot explored here

Better control of searchStatic vs. dynamic variable orderingBetter value choiceNot using complete depth-first chronological backtracking


ProblemProgram



Static vs. Dynamic Variable Ordering

Heuristic Static OrderingSort variables before search based on heuristicMost important decisionsSmallest initial domain

Dynamic variable orderingUse information from constraint propagationDifferent orders in different parts of search treeUse all information available


ProblemProgram



First Fail strategy

Dynamic variable orderingAt each step, select variable with smallest domainIdea: If there is a solution, better chance of finding itIdea: If there is no solution, smaller number of alternativesNeeds tie-breaking method


ProblemProgram



Modification of Program

:-module(nqueen).:-export(top/0).:-lib(ic).

top:-nqueen(8,L), writeln(L).

nqueen(N,L):-length(L,N),L :: 1..N,alldifferent(L),noattack(L),labeling(L).é replace with


ProblemProgram



Modification of Program

:-module(nqueen).:-export(top/0).:-lib(ic).

top:-nqueen(8,L), writeln(L).

nqueen(N,L):-length(L,N),L :: 1..N,alldifferent(L),noattack(L),search(L,0,first_fail,indomain,complete,[]).


ProblemProgram



Variable Choice

Determines the order in which variables are assignedinput_order assign variables in static order givenfirst_fail select variable with smallest domain firstmost_constrained like first_fail, tie break based onnumber of constraints in which variable occursOthers, including programmed selection


ProblemProgram



Value Choice

Determines the order in which values are tested forselected variablesindomain Start with smallest value, on backtracking trynext larger valueindomain_max Start with largest valueindomain_middle Start with value closest to middle ofdomainindomain_random Choose values in random order


ProblemProgram



Comparison

Board size 16x16Naive (Input Order) StrategyFirst Fail variable selection


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FirstFail Strategy (Size 16)

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Back to Start


ProblemProgram



Comparing Solutions

Naive First Fail


ProblemProgram



Comparing Solutions

Naive First Fail

Solutions are different!


ProblemProgram



FirstFail, Problem Sizes 4-100

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"first_fail/all.txt"


ProblemProgram



Observations

This is much betterBut some sizes are much harderTimeout for sizes 88, 91, 93, 97, 98, 99


ProblemProgram



Can we do better?

Improved initial orderingQueens on edges of board are easier to assignDo hard assignment first, keep simple choices for laterBegin assignment in middle of board

Matching value choiceValues in the middle of board have higher impactAssign these early at top of search treeUse indomain_middle for this


ProblemProgram



Start from Middle (Size 16)

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ProblemProgram



Comparing Solutions

Naive First Fail Middle


ProblemProgram



Comparing Solutions

Naive First Fail Middle

Again, solutions are different!


ProblemProgram



Middle, Problem Sizes 4-100

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ProblemProgram



Observations

Not always better than first failFor size 16, trees are similar sizeTimeout only for size 94But still, one strategy does not work for all problem sizesThere are ways to resolve this!


ProblemProgram



Approach 1: Heuristic Portfolios

Try multiple strategies for the same problemWith multi-core CPUs, run them in parallelOnly one needs to be successful for each problem


ProblemProgram



Approach 2: Restart with Randomization

Only spend limited number of backtracks for a searchattemptWhen this limit is exceeded, restart at beginningRequires randomization to explore new search branchRandomize variable choice by random tie breakRandomize value choice by shuffling valuesNeeds strategy when to restart


ProblemProgram



Approach 3: Partial Search

Abandon depth-first, chronological backtrackingDon’t get locked into a failed sub-treeA wrong decision at a level is not detected, and we have toexplore the complete subtree below to undo that wrongchoiceExplore more of the search treeSpend time in promising parts of tree


ProblemProgram



Example: Credit Search

Explore top of tree completely, based on creditStart with fixed amount of creditEach node consumes one credit unitSplit remaining credit amongst childrenWhen credit runs out, start bounded backtrack searchEach branch can use only K backtracksIf this limit is exceeded, jump to unexplored top of tree


ProblemProgram



Credit, Search Tree Problem Size 9448

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ProblemProgram



Credit, Problem Sizes 4-200

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ProblemProgram



Points to Remember

Choice of search can have huge impact on performanceDynamic variable selection can lead to large reduction ofsearch spacePackaged search can do a lot, but programming searchadds even moreDepth-first chronologicial backtracking not always bestchoiceHow to control this explosion of search alternatives?


Part IV

What is missing?


Many Specialized Topics

How to design efficient core engineHybrids with LP/MIP toolsHybrids with SATSymmetry breakingUse of MDD/BDD to encode sets of solutionsHigh level modelling toolsDebugging/visualization


Reformulation

Just because the user has modelled it this way, it doesn’tmean we have to solve it that way

Replace some constraint(s) by other, equivalent constraintsBecause we don’t have that constraint in our systemFor performance


Learning

While solving the problem we can learn how to strengthenthe model/search

Understand which constraints/method contribute topropagation and change scheduleLearn no-good constraints by explaining failureAdapt search strategy based on search experience


Refined Process

Problem

Human

User Model

Reformulation

Implementation Model


SolutionHelmut Simonis Introduction to CP 432

What is CP actually used for?

http://hsimonis.wordpress.com


http://hsimonis.wordpress.com

Introduction to Constraint ProgrammingConstraint Programming - in a nutshell Declarative description...

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