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Constraint Programming Overview based on Examples Marco Chiarandini - - PowerPoint PPT Presentation

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DM841 Discrete Optimization Part I Lecture 2 Constraint Programming Overview based on Examples Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark Outline 1. An Initial Example 2. Constraint

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DM841 Discrete Optimization Part I Lecture 2

Constraint Programming Overview based on Examples

Marco Chiarandini

Department of Mathematics & Computer Science University of Southern Denmark

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Outline

  • 1. An Initial Example
  • 2. Constraint
  • 3. Send More Money

Points to Remember Modeling in MILP

2

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Put a different number in each circle (1 to 8) such that adjacent circles cannot take consecutive numbers

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Constraint Programming An Introduction by example Patrick Prosser with the help of Toby Walsh, Chris Beck, Barbara Smith, Peter van Beek, Edward Tsang, ...

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A Puzzle

  • Place numbers 1 through 8 on nodes

– Each number appears exactly once

? ? ? ? ? ? ? ?

– No connected nodes have consecutive numbers

You have 8 minutes!

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Heuristic Search

Which nodes are hardest to number? ? ? ? ? ? ? ? ?

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Heuristic Search

? ? ? ? ? ? ? ?

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Heuristic Search

? ? ? ? ? ? ? ? Which are the least constraining values to use?

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Heuristic Search

? 1 ? ? 8 ? ? ? Values 1 and 8

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Heuristic Search

? 1 ? ? 8 ? ? ? Values 1 and 8

Symmetry means we don’t need to consider: 8 1

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Inference/propagation

? 1 ? ? 8 ? ? ? We can now eliminate many values for other nodes

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Inference/propagation

? 1 ? ? 8 ? ? ?

{1,2,3,4,5,6,7,8}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{2,3,4,5,6,7}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} By symmetry {3,4,5,6}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} {3,4,5,6} {1,2,3,4,5,6,7,8}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} {3,4,5,6} {2,3,4,5,6,7}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} {3,4,5,6} {3,4,5,6}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} By symmetry {3,4,5,6} {3,4,5,6} {3,4,5,6}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} {3,4,5,6,7} {3,4,5,6} {3,4,5,6} {3,4,5,6} {2,3,4,5,6}

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Inference/propagation

? 1 ? ? 8 ? ? ?

{3,4,5,6} {3,4,5,6,7} {3,4,5,6} {3,4,5,6} {3,4,5,6} {2,3,4,5,6} Value 2 and 7 are left in just one variable domain each

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Inference/propagation

? 1 ? ? 8 ? 2 7

{3,4,5,6} {3,4,5,6,7} {3,4,5,6} {3,4,5,6} {3,4,5,6} {2,3,4,5,6} And propagate …

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Inference/propagation

? 1 ? ? 8 ? 2 7

{3,4,5} {3,4,5,6,7} {3,4,5} {3,4,5,6} {3,4,5,6} {2,3,4,5,6} And propagate …

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Inference/propagation

? 1 ? ? 8 ? 2 7

{3,4,5} {3,4,5,6,7} {3,4,5} {4,5,6} {4,5,6} {2,3,4,5,6} And propagate …

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Inference/propagation

? 1 ? ? 8 ? 2 7

{3,4,5} {3,4,5} {4,5,6} {4,5,6} Guess a value, but be prepared to backtrack …

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Inference/propagation

3 1 ? ? 8 ? 2 7

{3,4,5} {3,4,5} {4,5,6} {4,5,6} Guess a value, but be prepared to backtrack …

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Inference/propagation

3 1 ? ? 8 ? 2 7

{3,4,5} {3,4,5} {4,5,6} {4,5,6} And propagate …

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SLIDE 28

Inference/propagation

3 1 ? ? 8 ? 2 7

{4,5} {5,6} {4,5,6} And propagate …

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Inference/propagation

3 1 ? ? 8 ? 2 7

{4,5} {5,6} {4,5,6} Guess another value …

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Inference/propagation

3 1 ? 5 8 ? 2 7

{4,5} {4,5,6} Guess another value …

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Inference/propagation

3 1 ? 5 8 ? 2 7

{4,5} {4,5,6} And propagate …

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Inference/propagation

3 1 ? 5 8 ? 2 7

{4} {4,6} And propagate …

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Inference/propagation

3 1 4 5 8 ? 2 7

{4} {4,6} One node has only a single value left …

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Inference/propagation

3 1 4 5 8 6 2 7

{6}

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Solution

3 1 4 5 8 6 2 7

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The Core of Constraint Computation

  • Modelling

– Deciding on variables/domains/constraints

  • Heuristic Search
  • Inference/Propagation
  • Symmetry
  • Backtracking
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Hardness

  • The puzzle is actually a hard problem

– NP-complete

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Constraint programming

  • Model problem by specifying constraints on

acceptable solutions

– define variables and domains – post constraints on these variables

  • Solve model

– choose algorithm

  • incremental assignment / backtracking search
  • complete assignments / stochastic search

– design heuristics

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Example CSP

  • Variable, vi for each node
  • Domain of {1, …, 8}
  • Constraints

– All values used allDifferent(v1 v2 v3 v4 v5 v6 v7 v8) – No consecutive numbers for adjoining nodes |v1 - v2 | > 1 |v1 - v3 | > 1 …

? ? ? ? ? ? ? ?

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Outline

  • 1. An Initial Example
  • 2. Constraint
  • 3. Send More Money

Points to Remember Modeling in MILP

3

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Constraint Programming - in a nutshell

◮ Declarative description of problems with

◮ Variables which range over (finite) sets of values ◮ Constraints over subsets of variables which restrict possible value

combinations

◮ A solution is a value assignment which satisfies all constraints

◮ Constraint propagation/reasoning

◮ Removing inconsistent values for variables ◮ Detect failure if constraint can not be satisfied ◮ Interaction of constraints via shared variables ◮ Incomplete

◮ Search

◮ User controlled assignment of values to variables ◮ Each step triggers constraint propagation

◮ Different domains require/allow different methods

4

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Constraint Programming

Constraint Programming: an alternative approach to imperative programming and object oriented programming.

◮ Variables each with a finite set of possible values (domain) ◮ Constraint on a sequence of variables: a relationship on their domains

Constraint Satisfaction Problem: finite set of constraints

5

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CP

Constraint Programming = model (representation) + propagation (reasoning, inference) + search (reasoning, inference)

6

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Basic Process

Problem Human Model Constraint Solver/Search Solution

Insight Centre for Data Analytics Slide 9 June 20th, 2016

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More Realistic

Problem Human Model Constraint Solver/Search Hangs Solution Wrong Solution

Insight Centre for Data Analytics Slide 10 June 20th, 2016

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Dual Role of Model

  • Allows Human to Express Problem
  • Close to Problem Domain
  • Constraints as Abstractions
  • Allows Solver to Execute
  • Variables as Communication Mechanism
  • Constraints as Algorithms

Insight Centre for Data Analytics Slide 11 June 20th, 2016

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Modelling Frameworks

  • MiniZinc (NICTA, Australia)
  • NumberJack (Insight, Ireland)
  • Essence (UK)
  • Allow use of multiple back-end solvers
  • Compile model into variants for each solver
  • A priori solver independent model(CP, MIP, SAT)

Insight Centre for Data Analytics Slide 12 June 20th, 2016

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Framework Process

Problem Human Model Compile/Reformulate CP MIP SAT Other Solution Solution Solution Solution

Insight Centre for Data Analytics Slide 13 June 20th, 2016

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Computational Models

Three main Computational Models to solve (combinatorial) constrained

  • ptimization problems:

◮ Mathematical Programming (LP, ILP, QP, SDP, ...) ◮ Constraint Programming (CSP as a model, SAT as a very special case) ◮ Local Search (... and Meta-heuristics) ◮ Others? Dynamic programming, dedicated algorithms, satisfiability

modulo theory, answer set programming, etc.

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Modeling

Modeling:

  • 1. identify:

◮ parameters ◮ variables ◮ domains ◮ constraints ◮ objective function

that formulate the problem

  • 2. express what in point 1) in a way that allows the solution by available

software

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Variables

In MILP: real and integer (mostly binary) variables In CP:

◮ finite domain integer (including Booleans), ◮ continuos with interval constraints ◮ structured domains: finite sets, multisets, graphs, ...

In LS: integer variables

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Constraint Programming vs MILP

◮ In MILP we formulate problems as a set of linear inequalities ◮ In CP we describe substructures (so-called global constraints) and

combine them with various combinators.

◮ Substructures capture building blocks often (but not always)

comptuationally tractable by special-purpose algorithms

◮ CP models can:

◮ be solved by the constraint engine ◮ be linearized and solved by their MIP solvers; ◮ be translated in CNF and solved by SAT solvers; ◮ be handled by local search

◮ In MILP the solver is often seen as a black-box

In CP and LS solvers leave the user the task of programming the search.

◮ CP = model + propagation + search

constraint propagation by domain filtering inference search = backtracking or branch and bound or local search

10

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Outline

  • 1. An Initial Example
  • 2. Constraint
  • 3. Send More Money

Points to Remember Modeling in MILP

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Aims

◮ Example of Finite Domain Constraint Problem ◮ Models and Programs ◮ Constraint Propagation and Search ◮ Some Basic Constraints:

linear arithmetic, alldifferent, disequality

◮ A Built-in search ◮ Visualizers for variables, constraints and search

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Problem: Send + More = Money

Send + More = Money You are asked to replace each letter by a different digit so that S E N D + M O R E = M O N E Y is correct. Because S and M are the leading digits, they cannot be equal to the 0 digit.

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Modelling

  • 1. Parameters
  • 2. Variables (ie, solution representation)
  • 3. Domains (ie, allowed values for the variables)
  • 4. Constraints

Later Objective Function

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Model

◮ Each character is a variable, which ranges over the values 0 to 9. ◮ An alldifferent constraint between all variables, which states that two

different variables must have different values. This is a very common constraint, which we will encounter in many other problems later on.

◮ Two disequality constraints (variable X must be different from value V )

stating that the variables at the beginning of a number can not take the value 0.

◮ An arithmetic equality constraint linking all variables with the proper

coefficients and stating that the equation must hold.

15

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Send More Money: CP model

SEND + MORE = MONEY

◮ Xi ∈ {0, . . . , 9} for all i ∈ I = {S, E, N, D, M, O, R, Y } ◮ Each letter takes a different digit 1 inequality constraint

alldifferent([X1, X2, . . . , X8]).

(it substitutes 28 inequality constraints: Xi = Xj, i, j ∈ I, i = j)

◮ XM = 0, XS = 0 ◮ Crypto constraint 1 equality constraint:

103X1 +102X2 +10X3 +X4 + 103X5 +102X6 +10X7 +X2 = 104X5 +103X6 +102X3 +10X2 +X8

16

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◮ This is one model, not the model of the problem ◮ Many possible alternatives ◮ Choice often depends on the constraint system available

Constraints available Reasoning attached to constraints

◮ Not always clear which is the best model

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Send More Money: CP model

Gecode-python

from gecode import * s = space() letters = s.intvars(8,0,9) S,E,N,D,M,O,R,Y = letters s.rel(M,IRT_NQ,0) s.rel(S,IRT_NQ,0) s.distinct(letters) C = [1000, 100, 10, 1, 1000, 100, 10, 1,

  • 10000, -1000, -100, -10, -1]

X = [S,E,N,D, M,O,R,E, M,O,N,E,Y] s.linear(C,X, IRT_EQ, 0) s.branch(letters, INT_VAR_SIZE_MIN, INT_VAL_MIN) for s2 in s.search(): print(s2.val(letters))

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Send More Money: CP model

MiniZinc

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Program Sendmory

:- module(sendmory). :- 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).

Insight Centre for Data Analytics Slide 21 June 20th, 2016

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Question

But how did the program come up with this solution?

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Constraint Setup

◮ Domain Definition ◮ Alldifferent Constraint ◮ Disequality Constraints ◮ Equality Constraint

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The following slides are taken from H. Simonis: H. Simonis’ demo, slides 33-134 and his tutorial at ACP2016.

22

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Domain Definition

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

Insight Centre for Data Analytics Slide 31 June 20th, 2016

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Domain Visualization

1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 32 June 20th, 2016

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Domain Visualization

Rows = Variables 1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 32 June 20th, 2016

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Domain Visualization

Columns = Values 1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 32 June 20th, 2016

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Domain Visualization

Cells= State 1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 32 June 20th, 2016

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Alldifferent Constraint

alldifferent(L),

  • Built-in of ic library
  • No initial propagation possible
  • Suspends, waits until variables are changed
  • When variable is fixed, remove value from domain of
  • ther variables
  • Forward checking

Insight Centre for Data Analytics Slide 33 June 20th, 2016

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Alldifferent Visualization

Uses the same representation as the domain visualizer 1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 34 June 20th, 2016

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Disequality Constraints

S #\= 0, M#\= 0, Remove value from domain S ∈ {1..9}, M ∈ {1..9} Constraints solved, can be removed

Insight Centre for Data Analytics Slide 35 June 20th, 2016

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Domains after Disequality

1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 36 June 20th, 2016

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Equality Constraint

  • Normalization of linear terms
  • Single occurence of variable
  • Positive coefficients
  • Propagation

Insight Centre for Data Analytics Slide 37 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D +1000*M+ 100*O+ 10*R+ E 10000*M+ 1000*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D +1000*M+ 100*O+ 10*R+ E 10000*M+ 1000*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D + 100*O+ 10*R+ E 9000*M+ 1000*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D + 100*O+ 10*R+ E 9000*M+ 1000*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D + 10*R+ E 9000*M+ 900*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ 10*N+ D + 10*R+ E 9000*M+ 900*O+ 100*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ D + 10*R+ E 9000*M+ 900*O+ 90*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

1000*S+ 100*E+ D + 10*R+ E 9000*M+ 900*O+ 90*N+ 10*E+ Y

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Normalization

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

Insight Centre for Data Analytics Slide 38 June 20th, 2016

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Simplified Equation

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

Insight Centre for Data Analytics Slide 39 June 20th, 2016

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Propagation

1000 ∗ S1..9 + 91 ∗ E0..9 + 10 ∗ R0..9 + D0..9 = 9000 ∗ M1..9 + 900 ∗ O0..9 + 90 ∗ N0..9 + Y 0..9

Insight Centre for Data Analytics Slide 40 June 20th, 2016

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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

Insight Centre for Data Analytics Slide 40 June 20th, 2016

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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

Insight Centre for Data Analytics Slide 40 June 20th, 2016

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SLIDE 89

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}

Insight Centre for Data Analytics Slide 40 June 20th, 2016

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SLIDE 90

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 Insight Centre for Data Analytics Slide 40 June 20th, 2016

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SLIDE 91

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 9000
  • Rest of lhs (left hand side) (91 ∗ E0..9 + 10 ∗ R0..9 + D0..9)

is atmost 918

  • S must be greater or equal to 9000−918

1000

= 8.082

  • otherwise lower bound of equation not reached by lhs
  • S is integer, therefore S ≥ ⌈ 9000−918

1000

⌉ = 9

  • S has upper bound of 9, so S = 9

Insight Centre for Data Analytics Slide 41 June 20th, 2016

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SLIDE 92

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 9918
  • Rest of rhs (right hand side) 900 ∗ O0..9 + 90 ∗ N0..9 + Y 0..9

is at least 0

  • M must be smaller or equal to 9918−0

9000

= 1.102

  • M must be integer, therefore M ≤ ⌊ 9918−0

9000 ⌋ = 1

  • M has lower bound of 1, so M = 1

Insight Centre for Data Analytics Slide 42 June 20th, 2016

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SLIDE 93

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 9918
  • Rest of rhs (right hand side) 9000 ∗ 1 + 90 ∗ N0..9 + Y 0..9 is

at least 9000

  • O must be smaller or equal to 9918−9000

900

= 1.02

  • O must be integer, therefore O ≤ ⌊ 9918−9000

900

⌋ = 1

  • O has lower bound of 0, so O ∈ {0..1}

Insight Centre for Data Analytics Slide 43 June 20th, 2016

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SLIDE 94

Propagation of equality: Result

1 2 3 4 5 6 7 8 9 S

  • ✹

E N D M ✹

  • O

✖ ✖ ✖ ✖ ✖ ✖ ✖ ✖ R Y

Insight Centre for Data Analytics Slide 44 June 20th, 2016

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SLIDE 95

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S

  • ✹

E N D M ✹

  • O

✖ ✖ ✖ ✖ ✖ ✖ ✖ ✖ R Y

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 96

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S ✹ E | N | D | M ✹ O R | Y |

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 97

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E | N | D | M ✹ O | R | Y |

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 98

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D M O ✹ R Y

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 99

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E | N | D | M O ✹ R | Y |

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 100

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D M O R Y O = 0, [E, R, D, N, Y] ∈ {2..8}

Insight Centre for Data Analytics Slide 45 June 20th, 2016

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SLIDE 101

Waking the equality constraint

  • Triggered by assignment of variables
  • or update of lower or upper bound

Insight Centre for Data Analytics Slide 46 June 20th, 2016

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SLIDE 102

Removal of constants

1000 ∗ 9 + 91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 = 9000 ∗ 1 + 900 ∗ 0 + 90 ∗ N2..8 + Y 2..8

Insight Centre for Data Analytics Slide 47 June 20th, 2016

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SLIDE 103

Removal of constants

1000 ∗ 9 + 91 ∗ E2..8 + 10 ∗ R2..8 + D2..8 = 9000 ∗ 1 + 900 ∗ 0 + 90 ∗ N2..8 + Y 2..8

Insight Centre for Data Analytics Slide 47 June 20th, 2016

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SLIDE 104

Removal of constants

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

Insight Centre for Data Analytics Slide 47 June 20th, 2016

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SLIDE 105

Propagation of equality (Iteration 1)

91 ∗ E2..8 + 10 ∗ R2..8 + D2..8

  • 204..816

= 90 ∗ N2..8 + Y 2..8

  • 182..728

Insight Centre for Data Analytics Slide 48 June 20th, 2016

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SLIDE 106

Propagation of equality (Iteration 1)

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

  • 204..728

Insight Centre for Data Analytics Slide 48 June 20th, 2016

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SLIDE 107

Propagation of equality (Iteration 1)

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

  • 204..728

N ≥ 3 = ⌈204 − 8 90 ⌉, E ≤ 7 = ⌊728 − 22 91 ⌋

Insight Centre for Data Analytics Slide 48 June 20th, 2016

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SLIDE 108

Propagation of equality (Iteration 2)

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

Insight Centre for Data Analytics Slide 49 June 20th, 2016

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SLIDE 109

Propagation of equality (Iteration 2)

91 ∗ E2..7 + 10 ∗ R2..8 + D2..8

  • 204..725

= 90 ∗ N3..8 + Y 2..8

  • 272..728

Insight Centre for Data Analytics Slide 49 June 20th, 2016

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SLIDE 110

Propagation of equality (Iteration 2)

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

  • 272..725

Insight Centre for Data Analytics Slide 49 June 20th, 2016

slide-111
SLIDE 111

Propagation of equality (Iteration 2)

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

  • 272..725

E ≥ 3 = ⌈272 − 88 91 ⌉

Insight Centre for Data Analytics Slide 49 June 20th, 2016

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SLIDE 112

Propagation of equality (Iteration 3)

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

Insight Centre for Data Analytics Slide 50 June 20th, 2016

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SLIDE 113

Propagation of equality (Iteration 3)

91 ∗ E3..7 + 10 ∗ R2..8 + D2..8

  • 295..725

= 90 ∗ N3..8 + Y 2..8

  • 272..728

Insight Centre for Data Analytics Slide 50 June 20th, 2016

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SLIDE 114

Propagation of equality (Iteration 3)

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

  • 295..725

Insight Centre for Data Analytics Slide 50 June 20th, 2016

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SLIDE 115

Propagation of equality (Iteration 3)

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

  • 295..725

N ≥ 4 = ⌈295 − 8 90 ⌉

Insight Centre for Data Analytics Slide 50 June 20th, 2016

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SLIDE 116

Propagation of equality (Iteration 4)

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

Insight Centre for Data Analytics Slide 51 June 20th, 2016

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SLIDE 117

Propagation of equality (Iteration 4)

91 ∗ E3..7 + 10 ∗ R2..8 + D2..8

  • 295..725

= 90 ∗ N4..8 + Y 2..8

  • 362..728

Insight Centre for Data Analytics Slide 51 June 20th, 2016

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SLIDE 118

Propagation of equality (Iteration 4)

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

  • 362..725

Insight Centre for Data Analytics Slide 51 June 20th, 2016

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SLIDE 119

Propagation of equality (Iteration 4)

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

  • 362..725

E ≥ 4 = ⌈362 − 88 91 ⌉

Insight Centre for Data Analytics Slide 51 June 20th, 2016

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SLIDE 120

Propagation of equality (Iteration 5)

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

Insight Centre for Data Analytics Slide 52 June 20th, 2016

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SLIDE 121

Propagation of equality (Iteration 5)

91 ∗ E4..7 + 10 ∗ R2..8 + D2..8

  • 386..725

= 90 ∗ N4..8 + Y 2..8

  • 362..728

Insight Centre for Data Analytics Slide 52 June 20th, 2016

slide-122
SLIDE 122

Propagation of equality (Iteration 5)

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

  • 386..725

Insight Centre for Data Analytics Slide 52 June 20th, 2016

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SLIDE 123

Propagation of equality (Iteration 5)

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

  • 386..725

N ≥ 5 = ⌈386 − 8 90 ⌉

Insight Centre for Data Analytics Slide 52 June 20th, 2016

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SLIDE 124

Propagation of equality (Iteration 6)

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

Insight Centre for Data Analytics Slide 53 June 20th, 2016

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SLIDE 125

Propagation of equality (Iteration 6)

91 ∗ E4..7 + 10 ∗ R2..8 + D2..8

  • 386..725

= 90 ∗ N5..8 + Y 2..8

  • 452..728

Insight Centre for Data Analytics Slide 53 June 20th, 2016

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SLIDE 126

Propagation of equality (Iteration 6)

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

  • 452..725

Insight Centre for Data Analytics Slide 53 June 20th, 2016

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SLIDE 127

Propagation of equality (Iteration 6)

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

  • 452..725

N ≥ 5 = ⌈452 − 8 90 ⌉, E ≥ 4 = ⌈452 − 88 91 ⌉ No further propagation at this point

Insight Centre for Data Analytics Slide 53 June 20th, 2016

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SLIDE 128

Domains after setup

1 2 3 4 5 6 7 8 9 S E N D M O R Y

Insight Centre for Data Analytics Slide 54 June 20th, 2016

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SLIDE 129

Outline

Problem Program Constraint Setup Search Step 1 Step 2 Further Steps Solution Points to Remember

Insight Centre for Data Analytics Slide 55 June 20th, 2016

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SLIDE 130

labeling built-in

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

  • Try variable is order given
  • Try values starting from smallest value in domain
  • When failing, backtrack to last open choice
  • Chronological Backtracking
  • Depth First search

Insight Centre for Data Analytics Slide 56 June 20th, 2016

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SLIDE 131

Search Tree Step 1

Variable S already fixed S E

9

Insight Centre for Data Analytics Slide 57 June 20th, 2016

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SLIDE 132

Step 2, Alternative E = 4

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

4 9

Insight Centre for Data Analytics Slide 58 June 20th, 2016

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SLIDE 133

Assignment E = 4

1 2 3 4 5 6 7 8 9 S E ✹

  • N

D M O R Y

Insight Centre for Data Analytics Slide 59 June 20th, 2016

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SLIDE 134

Propagation of E = 4, equality constraint

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

Insight Centre for Data Analytics Slide 60 June 20th, 2016

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SLIDE 135

Propagation of E = 4, equality constraint

91 ∗ 4 + 10 ∗ R2..8 + D2..8

  • 386..452

= 90 ∗ N5..8 + Y 2..8

  • 452..728

Insight Centre for Data Analytics Slide 60 June 20th, 2016

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SLIDE 136

Propagation of E = 4, equality constraint

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

  • 452

Insight Centre for Data Analytics Slide 60 June 20th, 2016

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SLIDE 137

Propagation of E = 4, equality constraint

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

  • 452

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

Insight Centre for Data Analytics Slide 60 June 20th, 2016

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SLIDE 138

Result of equality propagation

1 2 3 4 5 6 7 8 9 S E N ✹

  • D
  • ✹

M O R

  • ✹

Y ✹

  • Insight Centre for Data Analytics

Slide 61 June 20th, 2016

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SLIDE 139

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N ✹

  • D
  • ✹

M O R

  • ✹

Y ✹

  • Insight Centre for Data Analytics

Slide 62 June 20th, 2016

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SLIDE 140

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N ✹

  • D
  • ✹

M O R

  • ✹

Y ✹

  • Alldifferent fails!

Insight Centre for Data Analytics Slide 62 June 20th, 2016

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SLIDE 141

Step 2, Alternative E = 5

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

4

N

5 9

Insight Centre for Data Analytics Slide 63 June 20th, 2016

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SLIDE 142

Assignment E = 5

1 2 3 4 5 6 7 8 9 S E

  • ✹
  • N

D M O R Y

Insight Centre for Data Analytics Slide 64 June 20th, 2016

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SLIDE 143

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E

  • ✹
  • N

D M O R Y

Insight Centre for Data Analytics Slide 65 June 20th, 2016

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SLIDE 144

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E ✹ N | D | M O R | Y |

Insight Centre for Data Analytics Slide 65 June 20th, 2016

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SLIDE 145

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D M O R Y N = 5, N ≥ 6

Insight Centre for Data Analytics Slide 65 June 20th, 2016

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SLIDE 146

Propagation of equality

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

Insight Centre for Data Analytics Slide 66 June 20th, 2016

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SLIDE 147

Propagation of equality

91 ∗ 5 + 10 ∗ R2..8 + D2..8

  • 477..543

= 90 ∗ N6..8 + Y 2..8

  • 542..728

Insight Centre for Data Analytics Slide 66 June 20th, 2016

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SLIDE 148

Propagation of equality

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

  • 542..543

Insight Centre for Data Analytics Slide 66 June 20th, 2016

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SLIDE 149

Propagation of equality

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}

Insight Centre for Data Analytics Slide 66 June 20th, 2016

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SLIDE 150

Result of equality propagation

1 2 3 4 5 6 7 8 9 S E N ✹

  • D

✖ ✖ ✖ ✖ M O R

  • ✹

Y ✖ ✖ ✖ ✖

Insight Centre for Data Analytics Slide 67 June 20th, 2016

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SLIDE 151

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N ✹

  • D

✖ ✖ ✖ ✖ M O R

  • ✹

Y ✖ ✖ ✖ ✖

Insight Centre for Data Analytics Slide 68 June 20th, 2016

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SLIDE 152

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D | M O R ✹ Y

Insight Centre for Data Analytics Slide 68 June 20th, 2016

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SLIDE 153

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D ✹ M O R Y

Insight Centre for Data Analytics Slide 68 June 20th, 2016

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SLIDE 154

Propagation of alldifferent

1 2 3 4 5 6 7 8 9 S E N D M O R Y D = 7

Insight Centre for Data Analytics Slide 68 June 20th, 2016

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SLIDE 155

Propagation of equality

91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3

Insight Centre for Data Analytics Slide 69 June 20th, 2016

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SLIDE 156

Propagation of equality

91 ∗ 5 + 10 ∗ 8 + 7

  • 542

= 90 ∗ 6 + Y 2..3

  • 542..543

Insight Centre for Data Analytics Slide 69 June 20th, 2016

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SLIDE 157

Propagation of equality

91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3

  • 542

Insight Centre for Data Analytics Slide 69 June 20th, 2016

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SLIDE 158

Propagation of equality

91 ∗ 5 + 10 ∗ 8 + 7 = 90 ∗ 6 + Y 2..3

  • 542

Y = 2

Insight Centre for Data Analytics Slide 69 June 20th, 2016

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SLIDE 159

Last propagation step

1 2 3 4 5 6 7 8 9 S E N D M O R Y ✹

  • Insight Centre for Data Analytics

Slide 70 June 20th, 2016

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SLIDE 160

Further Steps: Nothing more to do

S E

4

N

5 9

Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 161

Further Steps: Nothing more to do

S E

4

N D

6 5 9

Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 162

Further Steps: Nothing more to do

S E

4

N D M

7 6 5 9

Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 163

Further Steps: Nothing more to do

S E

4

N D M O

1 7 6 5 9

Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 164

Further Steps: Nothing more to do

S E

4

N D M O R

1 7 6 5 9 Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 165

Further Steps: Nothing more to do

S E

4

N D M O R Y

8 1 7 6 5 9 Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 166

Further Steps: Nothing more to do

S E

4

N D M O R Y

2 8 1 7 6 5 9

Insight Centre for Data Analytics Slide 71 June 20th, 2016

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SLIDE 167

Complete Search Tree

S E

4

N D M O R Y

2 8 1 7 6 5 6 7 9

Insight Centre for Data Analytics Slide 72 June 20th, 2016

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SLIDE 168

Solution

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

Insight Centre for Data Analytics Slide 73 June 20th, 2016

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SLIDE 169

Outline

  • 1. An Initial Example
  • 2. Constraint
  • 3. Send More Money

Points to Remember Modeling in MILP

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SLIDE 170

Points to Remember

◮ Constraint models are expressed by:

variables + constraints + parameters

◮ Problems can have many different models, which can behave quite

  • differently. Choosing the best model is an art.

◮ Constraints can take many different forms. ◮ Propagation deals with the interaction of variables and constraints:

It removes some values that are inconsistent with a constraint from the domain of a variable.

◮ Constraints only communicate via shared variables.

24

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SLIDE 171

Points to Remember

◮ Propagation is data driven, and can be quite complex even for small

examples.

◮ Propagation usually is not sufficient, search may be required to find a

solution.

◮ The default search uses chronological depth-first backtracking,

systematically exploring the complete search space.

◮ The search choices and propagation are interleaved,

after every choice some more propagation may further reduce the problem.

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Applications

◮ Operation research (optimization problems) ◮ Graphical interactive systems (to express geometrical correctness) ◮ Molecular biology (DNA sequencing, 3D models of proteins) ◮ Finance ◮ Circuit verification ◮ Elaboration of natural languages (construction of efficient parsers) ◮ Scheduling of activities ◮ Configuration problem in form compilation ◮ Generation of coerent music programs [Anders and Miranda [2011]]. ◮ Data bases ◮ ... ◮ http://hsimonis.wordpress.com/

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SLIDE 173

Applications

Distribution of technology used at Google for optimization applications developed by the operations research team

[Slide presented by Laurent Perron on OR-Tools at CP2013]

27

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SLIDE 174

List of Contents

◮ Modeling with Finite Domain Integer Variables ◮ Introduction to Gecode ◮ Overview on global constraints ◮ Notions of local consistency ◮ Constraint propagation algorithms ◮ Filtering algorithms for global constraints ◮ Search ◮ Set variables ◮ Symmetries

28

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SLIDE 175

Outline

  • 1. An Initial Example
  • 2. Constraint
  • 3. Send More Money

Points to Remember Modeling in MILP

29

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SLIDE 176

Send More Money: ILP model 1

◮ xi ∈ {0, . . . , 9} for all i ∈ I = {S, E, N, D, M, O, R, Y } ◮ δij =

  • if xi < xj

1 if xj < xi

◮ Crypto constraint:

103x1 +102x2 +10x3 +x4 + 103x5 +102x6 +10x7 +x2 = 104x5 +103x6 +102x3 +10x2 +x8

◮ Each letter takes a different digit:

xi − xj − 10δij ≤ −1, for all i, j, i < j xj − xi + 10δij ≤ 9, for all i, j, i < j

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SLIDE 177

Send More Money: ILP model 2

◮ xi ∈ {0, . . . , 9} for all i ∈ I = {S, E, N, D, M, O, R, Y } ◮ yij ∈ {0, 1} for all i ∈ I, j ∈ J = {0, . . . , 9} ◮ Crypto constraint:

103x1 +102x2 +10x3 +x4 + 103x5 +102x6 +10x7 +x2 = 104x5 +103x6 +102x3 +10x2 +x8

◮ Each letter takes a different digit:

  • j∈J

yij = 1, ∀i ∈ I,

  • i∈I

yij ≤ 1, ∀j ∈ J, xi =

  • j∈J

jyij, ∀i ∈ I.

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SLIDE 178

Send More Money: ILP model

The quality of these formulations depends on both the tightness of the LP relaxations and the number of constraints and variables (compactness)

◮ Which of the two models is tighter?

project out all extra variables in the LP so that the polytope for LP is in the space of the x variables. By linear comb. of constraints:

Model 1 −1 ≤ xi − xj ≤ 10 − 1 Model 2

  • j∈J

xj ≥ |J| (|J| − 1) 2 , ∀J ⊂ I,

  • j∈J

xj ≤ |J| (2k − |J|) + 1 2 , ∀J ⊂ I.

◮ Can you find the convex hull of this problem?

Williams and Yan [2001] prove that model 2 is facet defining Suppose we want to maximize MONEY, how strong is the upper bound

  • btained with this formulation? How to obtain a stronger upper bound?

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SLIDE 179

Send More Money: ILP model (revisited)

◮ xi ∈ {0, . . . , 9} for all i ∈ I = {S, E, N, D, M, O, R, Y } ◮ Crypto constraint:

103x1 +102x2 +10x3 +x4 + 103x5 +102x6 +10x7 +x2 = 104x5 +103x6 +102x3 +10x2 +x8

◮ Each letter takes a different digit:

  • j∈J

xj ≥ |J| (|J| − 1) 2 , ∀J ⊂ I,

  • j∈J

xj ≤ |J| (2k − |J|) + 1 2 , ∀J ⊂ I. But exponentially many!

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SLIDE 180

Send More Money: CP model (revisited)

◮ Xi ∈ {0, . . . , 9} for all i ∈ I = {S, E, N, D, M, O, R, Y } ◮

103X1 +102X2 +10X3 +X4 + 103X5 +102X6 +10X7 +X2 = 104X5 +103X6 +102X3 +10X2 +X8

◮

alldifferent([X1, X2, . . . , X8]).

◮ Redundant constraints (5 equality constraints)

X4 + X2 = 10 r1 + X8, X3 + X7 + r1 = 10 r2 + X2, X2 + X6 + r2 = 10 r3 + X3, X1 + X5 + r3 = 10 r4 + X6, +r4 = X5. Can we do better? Can we propagate something?

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SLIDE 181

Send Most Money: CP model

Gecode-python

Optimization version: max

  • i∈I ′

CiXi, I ′ = {M, O, N, E, Y }

from gecode import * s = space() letters = s.intvars(8,0,9) S,E,N,D,M,O,T,Y = letters s.rel(M,IRT_NQ,0) s.rel(S,IRT_NQ,0) s.distinct(letters) C = [1000, 100, 10, 1, 1000, 100, 10, 1,

  • 10000, -1000, -100, -10, -1]

X = [S,E,N,D, M,O,S,T, M,O,N,E,Y] s.linear(C,X,IRT_EQ,0) money = s.intvar(0,99999) s.linear([10000,1000,100,10,1],[M,O,N,E,Y], IRT_EQ, money) s.maximize(money) s.branch(letters, INT_VAR_SIZE_MIN, INT_VAL_MIN) for s2 in s.search(): print(s2.val(money), s2.val(letters))

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Strengths

◮ CP is excellent to explore highly constrained combinatorial spaces quickly ◮ Math programming is particulary good at deriving lower bounds ◮ LS is particualry good at derving upper bounds

40

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Differences

◮ MILP models

◮ impose modelling rules: linear inequalities and objectives ◮ emphasis on tightness and compactness of LP, strength of bounds

(remove dominated constraints)

◮ CP models

◮ a large variety of algorithms communicating with each other: global

constraints

◮ more expressiveness ◮ emphasis on exploiting substructres, include redundant constraints 41

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SLIDE 184

Resume

◮ Constraint Satisfaction Problem ◮ Modelling in CP ◮ Examples, Send More Money, Sudoku

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SLIDE 185

References

Anders T. and Miranda E.R. (2011). Constraint programming systems for modeling music theories and composition. ACM Comput. Surv., 43(4), pp. 30:1–30:38. Hooker J.N. (2011). Hybrid modeling. In Hybrid Optimization, edited by P.M. Pardalos, P. van Hentenryck, and M. Milano, vol. 45 of Optimization and Its Applications, pp. 11–62. Springer New York. Smith B.M. (2006). Modelling. In Handbook of Constraint Programming, edited by

  • F. Rossi, P. van Beek, and T. Walsh, chap. 11, pp. 377–406. Elsevier.

Williams H. and Yan H. (2001). Representations of the all_different predicate of constraint satisfaction in integer programming. INFORMS Journal on Computing, 13(2), pp. 96–103.

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