event scheduling using constraint programming november 22, 2002 martin henz, school of computing,...

Event Scheduling Using Constraint Programming

November 22, 2002

Martin Henz, School of Computing, NUSwww.comp.nus.edu.sg/~henz/sma2002

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Overview

Constraint programming in a nutshell Elements of constraint programming Sport tournament scheduling Production scheduling Assessment

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Overview


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Constraint Programming in a Nutshell

SEND MORE MONEY

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Constraint Programming in a Nutshell

SEND + MORE = MONEY

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SEND + MORE = MONEY

Assign distinct digits to the letters

S, E, N, D, M, O, R, Y

such that

S E N D

+ M O R E

= M O N E Y holds.

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SEND + MORE = MONEY

Assign distinct digits to the letters

S, E, N, D, M, O, R, Y

such that S E N D

+ M O R E

= M O N E Y

holds.

Solution

9 5 6 7

+ 1 0 8 5

= 1 0 6 5 2

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Modeling

Formalize the problem as a constraint problem:

n variables m constraints: c1,…,cm

n

problem: Find a = (v1,…,vn) n such

that a ci , for all 1 i m

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A Model for MONEY

8 variables: {S,E,N,D,M,O,R,Y} 5 constraints: c1 = {(S,E,N,D,M,O,R,Y) 8 | 0 S,…,Y 9 }

c2 = {(S,E,N,D,M,O,R,Y) 8 |

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

+ 1000*M + 100*O + 10*R + E

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

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A Model for MONEY (continued) more constraints

c3 = {(S,E,N,D,M,O,R,Y) 8 | S 0 }

c4 = {(S,E,N,D,M,O,R,Y) 8 | M 0 }

c5 = {(S,E,N,D,M,O,R,Y) 8 | S…Y all different}

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Solution for MONEY c1 = {(S,E,N,D,M,O,R,Y) 8 | 0S,…,Y9 }

c2 = {(S,E,N,D,M,O,R,Y) 8 |

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

+ 1000*M + 100*O + 10*R + E

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

c3 = {(S,E,N,D,M,O,R,Y) 8 | S 0 }

c4 = {(S,E,N,D,M,O,R,Y) 8 | M 0 }

c5 = {(S,E,N,D,M,O,R,Y) 8 | S…Y all different}

Solution: (9,5,6,7,1,0,8,2) 8

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Constraint Programming

Choose propagation algorithms all different: “singleton wakeup” sum: “interval consistency”

Choose branching algorithms “first-fail”

Choose exploration algorithms “depth-first search”

Exploiting constraints during tree search

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S E N D+ M O R E

= M O N E Y

0S,…,Y9 S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S E N D M O R Y

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S E N D+ M O R E

= M O N E Y

0S,…,Y9

S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S {0..9}E {0..9}N {0..9}D {0..9}M {0..9}O {0..9}R {0..9}Y {0..9}

Propagate

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S E N D+ M O R E

= M O N E Y

0S,…,Y9 S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S {1..9}E {0..9}N {0..9}D {0..9}M {1..9}O {0..9}R {0..9}Y {0..9}

Propagate

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S E N D+ M O R E

= M O N E Y

0S,…,Y9 S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

Propagate

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S E N D+ M O R E

= M O N E Y

0S,…,Y9 S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 4 E 4

Branching

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S E N D+ M O R E

= M O N E Y

0S,…,Y9 S 0M 0

S…Y all different

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

+ 1000*M + 100*O + 10*R + E

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

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 4 E 4

Propagate

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S E N D+ M O R E

= M O N E Y

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 4 E 4

Branching

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 5 E 5

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

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S E N D+ M O R E

= M O N E Y

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 4 E 4

Propagate

S {9}E {6..7}N {7..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 5 E 5S {9}E {5}N {6}D {7}M {1}O {0}R {8}Y {2}

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S E N D+ M O R E

= M O N E Y

CompleteSearchTree

S {9}E {4..7}N {5..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 4 E 4

S {9}E {6..7}N {7..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}

E = 5 E 5S {9}E {5}N {6}D {7}M {1}O {0}R {8}Y {2}

E = 6 E 6

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The Art of Constraint Programming

Choose model Choose propagation algorithms Choose branching algorithms Choose exploration algorithms

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Demo: SEND + MORE = MONEY

Click here for MONEY

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Constraint Programming Systems

• Constraint programming languages• CONSTRAINTS (Steele, Sussman 1980)• CHIP (Dincbas, Hentenryck, Simonis, Aggoun 1988)• CLP(R) (Jaffar, Maher, Stuckey, Yap 1992)• Oz (Smolka and others 1995)• OPL (Hentenryck 1998)

• Constraint programming library• ILOG Solver (Puget 1993)

support constraint programming withhigh-level constructs

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Using OPL Syntax

enum Letter {S,E,N,D,M,O,R,Y};var int l[Letter] in 0..9;solve { alldifferent(l); l[S] <> 0; l[M] <> 0; l[S]*1000 + l[E]*100 + l[N]*10 + l[D] + l[M]*1000 + l[O]*100 + l[R]*10 + l[E] = l[M]*10000 + l[O]*1000 + l[N]*100 + l[E]*10 + l[Y] };search { forall(i in Letter ordered by increasing dsize(l[i])) tryall(v in 0..9) l[i] = v; };

All Solutions; Execution Run

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Overview


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Elements of Constraint Programming

The Approach Propagation Branching Exploration

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Constraint Solving

Given: a satisfiable constraint C and a new constraint C’.

Constraint solving is deciding whether C C’ is satisfiable.

Example: C: n > 2C’: an + bn = cn

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Completeness

General arithmetic constraints are undecidable (Hilbert’s Tenth Problem).

Constraint solving is not always possible. Example:

c1: n > 2

c2: an + bn = cn

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The Constraint Programming Approach

Constraint programming separates constraints into

Basic constraints: constraint solving (complete) Non-basic constraints: propagation

(incomplete)

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Basic Constraints vs. Propagators

Basic constraints are conjunctions of constraints of the form x S, where S is a finite set of integers

enjoy complete constraint solving Propagators

can be arbitrarily expressive (arithmetic, symbolic) implementation typically fast but incomplete

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Basic Constraints in Constraint Programming

Basic constraints are conjunctions of constraints of the form X S, where S is a finite set of integers.

Constraint solving is done by intersecting domains.

Example:

C = X{1..10}, Y{9..20}, C’ = X{9..15}, Y{14..30}. In practice, we keep a solved form, storing the

current domain of every variable.

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Basic Constraints and Propagators

S {1..9}E {0..9}N {0..9}D {0..9}M {1..9}O {0..9}R {0..9}Y {0..9}

all different(S,E,N,D, M,O,R,Y)

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

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S {1..9}E {0..9}N {0..9}D {0..9}M {1}O {0..9}R {0..9}Y {0..9}


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

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S {2..9}E {0,2..9}N {0,2..9}D {0,2..9}M {1}O {0,2..9}R {0,2..9}Y {0,2..9}


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

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S {2..9}E {0,2..9}N {0,2..9}D {0,2..9}M {1}O {0}R {0,2..9}Y {0,2..9}


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

and so on and so on

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S {9}E {5..7}N {6..8}D {2..8}M {1}O {0}R {2..8}Y {2..8}


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

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Issues in Propagation

Expressivity: What kind of information can be expressed as propagators?

Completeness: What behavior can be expected from propagation?

Efficiency: How much computational resources does propagation consume?

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Domain-Consistent Propagation

Given: Basic constraint C and propagator P. Propagation for P is domain-consistent, if

for every variable x and every value v in the domain of x, there is an assignment, in which x=v, that satisfies C and P.

Domain-consistent propagation is also called arc-consistent propagation.

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Current DomainLet c be basic constraint. For a given variable x,

the maximal set S such that c x S is

consistent is called the current domain of x in c,

denoted domc(x).

Example:

C: x{1..10}, y{9..20}, x{8..12}

domc(y) = {9..20}, domc(x)={8..10}

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Domain Consistency

Assume a constraint d over variables x and y.

A basic constraint c is domain consistent in x

with respect to d, if for every i domc(x)

there is a j domc(y) such that (i,j) d.

Example:

d: x + y = 10

c: x {7,9}, y {1..20}

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Domain Consistency


A basic constraint c is domain consistent in x

with respect to d, if for every i domc(x)

there is a j domc(y) such that (i,j) d.

Example:

d: x + y = 10

c: x {7,9}, y {1,3}

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Domain vs. Interval Consistency

Domain consistency: Check all elements of the domains of all variables known to the propagator

Interval consistency: Check only the boundaries of the domains of the variables

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Interval Consistency


A basic constraint c is interval consistent in x with

respect to d, if for i=min(domc(x)) and for

i=max(domc(x)) there is a j domc(y)

such that (i,j) d.

Example:

d: x + y = 10

C: x {7,9}, y {1..20}

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Interval Consistency


A basic constraint c is interval consistent in x with

respect to d, if for i=min(domc(x)) and for

i=max(domc(x)) there is a j domc(y)

such that (i,j) d.

Example:

d: x + y = 10

C: x{7,9}, y{1..3}

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Domain and Interval Consistency

OPL syntax for all-different

interval consistency: alldifferent(l);

domain consistency: alldifferent(l) domain;

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Propagation vs Branching

trade-off:

complex propagation algorithms incur

fewer, but more expensive nodes in tree

Example: MONEY with

alldiff and sum: only test fixed

assignment

alldiff: wait for fixed variables sum: interval cons.

alldiff and sum: domain

consistency

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Branching Algorithms

Constraint programming systems come with libraries of predefined branching algorithms programming support for user-defined

branching algorithms

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Branching for MONEY enum Letter {S,E,N,D,M,O,R,Y};var int l[Letter] in 0..9;solve { alldifferent(l); l[S] <> 0; l[M] <> 0; l[S]*1000 + l[E]*100 + l[N]*10 + l[D] + l[M]*1000 + l[O]*100 + l[R]*10 + l[E] = l[M]*10000 + l[O]*1000 + l[N]*100 + l[E]*10 + l[Y] };search { forall(i in Letter ordered by increasing dsize(l[i])) tryall(v in 0..9) l[i] = v; };All Solutions; Execution Run

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Basic Choice Points

try x < y | x >= y endtry;

x < y x >= y

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Choice Point Sequences

try x < y | x >= y endtry;

try z = 1 | z = 2 endtry;

x < y x >= y

z = 1 z = 2 z = 1 z = 2

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Abbreviation: tryall

tryall(i in 1..5) x = i;

stands for

try x=1|x=2|x=3|x=4|x=5 endtry;

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Examples of Branching Algorithms Enumeration: Choose variable, choose value

naive enumeration: choose variables and values in a fixed sequence

first-fail enumeration: choose a variable with minimal domain size

Domain-splitting:try x < mid | x >= mid endtry;

Task sequencing for scheduling

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Exploration for MONEY enum Letter {S,E,N,D,M,O,R,Y};var int l[Letter] in 0..9;solve { alldifferent(l); l[S] <> 0; l[M] <> 0; l[S]*1000 + l[E]*100 + l[N]*10 + l[D] + l[M]*1000 + l[O]*100 + l[R]*10 + l[E] = l[M]*10000 + l[O]*1000 + l[N]*100 + l[E]*10 + l[Y] };search { forall(i in Letter ordered by increasing dsize(l[i])) tryall(v in 0..9) l[i] = v; };


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Exploration

Depth-first search (default in OPL) Best-first search Limited discrepancy search

[Harvey/Ginsberg 95] user-defined explorations

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Specifying Exploration in OPL

enum Letter {S,E,N,D,M,O,T,Y};var int l[Letter] in 0..9;solve { alldifferent(l); l[S] <> 0; l[M] <> 0; l[S]*1000 + l[E]*100 + l[N]*10 + l[D] + l[M]*1000 + l[O]*100 + l[S]*10 + l[T] = l[M]*10000 + l[O]*1000 + l[N]*100 + l[E]*10 + l[Y] };search { LDSearch(4) forall(i in Letter ordered by increasing dsize(l[i])) tryall(v in 0..9) l[i] = v; };

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Other Search Components

Optimization Interaction Visualization

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Optimization

SEND + MOST = MONEY

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Optimization in OPL

enum Letter {S,E,N,D,M,O,T,Y};var int l[Letter] in 0..9;maximize money subject to { money = l[M]*10000+l[O]*1000+l[N]*100+l[E]*10+l[Y] alldifferent(l); l[S] <> 0; l[M] <> 0; l[S]*1000 + l[E]*100 + l[N]*10 + l[D] + l[M]*1000 + l[O]*100 + l[S]*10 + l[T] = l[M]*10000 + l[O]*1000 + l[N]*100 + l[E]*10 + l[Y] };search { forall(i in Letter ordered by increasing dsize(l[i])) tryall(v in 0..9) l[i] = v; };


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Interaction

First-solution search All solution search Last solution search Search with user interaction

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Visualization

Example: Oz Explorer [Schulte 1997].

Oz Explorer combines visualization first/all solution / user interaction branch-and-bound optimization depth-first search

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Overview


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Sport Tournament Scheduling

CP Approach Propagators Example

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Round Robin Tournament Planning Problems

n teams, each playing a fixed number of times r against every other team

r = 1: single, r = 2: double round robin. Each match is home match for one and

away match for the other Dense round robin:

At each date, each team plays at most once. The number of dates is minimal.

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CP Approach to Round Robins

1 2 3 4 5

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

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

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

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

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

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

Rounds

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

{1, 6} {1,3,4,5,6} {1,3,4,5,6} {1,3,4,5,6} {1,3,4,5,6}

{1,4,5} {1,2,4,5,6} {1,2,4,5,6} {1,2,4,5,6} {1,2,4,5,6}

{3,5} {1,2,3,5,6} {1,2,3,5,6} {1,2,3,5,6} {1,2,3, 5,6}

{3,4} {1,2,3,4,6} {1,2,3,4,6} {1,2,3,4,6} {1,2,3,4,6}

{1,2} {1,2,3,4,5} {1,2,3,4,5} {1,2,3,4,5} {1,2,3,4,5}

add given

constraints

Rounds

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

{6} {1} {3,4,5} {3,4,5} {3,4,5}

{1} {4,5} {2,4,5,6} {2,4,5,6} {2,4,5,6}

{5} {3, 6} {1,2,3, 6} {1,2,3, 6} {1,2,3, 6}

{4} {3, 6} {1,2,3, 6} {1,2,3, 6} {1,2,3,6}

{2} {4,5} {1, 3,4,5} {1, 3,4,5} {1, 3,4,5}

run filtering

algorithms

until fixpoint

is reached

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CP Approach to Round Robins{3} {2} {4,5,6} {4,5,6} {4,5,6}

{6} {1} {3,4,5} {3,4,5} {3,4,5}

{1} {4,5} {2,4,5,6} {2,4,5,6} {2,4,5,6}

{5} {3, 6} {1,2,3, 6} {1,2,3, 6} {1,2,3, 6}

{4} {3, 6} {1,2,3, 6} {1,2,3, 6} {1,2,3,6}

{2} {4,5} {1, 3,4,5} {1, 3,4,5} {1, 3,4,5}

perform

branching

step

team 3 plays against

team 4 in round 2

team 3 plays against

team 5 in round 2

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

{6} {1} {3,4,5} {3,4,5} {3,4,5}

{1} {5} {2,4, 6} {2,4, 6} {2,4, 6}

{5} {6} {1,2,3} {1,2,3} {1,2,3}

{4} {3} {1,2, 6} {1,2, 6} {1,2, 6}

{2} {4} {1, 3, 5} {1, 3, 5} {1, 3,5}

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

{6} {1} {3,4,5} {3,4,5} {3,4,5}

{1} {4} {2,5,6} {2,5,6} {2,5,6}

{5} {3} {1,2,6} {1,2,6} {1,2,6}

{4} {6} {1,2,3} {1,2,3} {1,2,3}

{2} {5} {1, 3,4} {1, 3,4} {1, 3,4}

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The All-Different Constraint

Team 1 2 3 4 5

France England,USA,Italy,Brasil,Germany

England,USA,Italy,Brasil,Germany

England,USA,Italy

England,USA,Italy

England,USA,Italy

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1

5

4

2

3

England

Germany

Brasil

USA

Italy

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1

5

4

2

3

England

Germany

Brasil

USA

Italy

Remove all edgesthat do not belongto a maximum matching in bipartite graph[Regin 94]

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1

5

4

2

3

England

Germany

Brasil

USA

Italy

...by applyingmaximum matching algorithm in bipartite graphs[Hopcroft, Karp 1973] O(|X|2 dmax

2)

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All-Different in OPL

weak but fast O(|X|): alldifferent(E,U,I,B,G) onValue domain-consistent O(|X|2 dmax

2) : alldifferent(E,U,I,B,G) onDomain

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The One-Factor Constraint

Team Possible Opponents

France Italy, USA, Germany

Italy France, Germany

USA France, England, Brasil

England USA, Brasil

Brasil USA, England

Germany France, Italy

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France

Italy USA

England

Germany

Brasil

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France

Italy USA

England

Germany

Brasil

Remove all edgesthat do not belongto a maximum matching[Regin 99]

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France

Italy USA

England

Germany

Brasil

... by modifyingEdmond’s maximum matching algorithm[Edmonds 65]

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ACC 1997/98: A Success Story of Constraint Programming

Integer programming + enumeration, 24 hoursNemhauser, Trick: Scheduling a Major College

Basketball Conference, Operations Research, 1998, 46(1)

Constraint programming, less than 1 minute.Henz: Scheduling a Major College Basketball

Conference - Revisited, Operations Research, 2001,

49(1)

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The ACC 1997/98 Problem

9 teams participate in tournament Dense double round robin:

there are 2 * 9 dates at each date, each team plays either home, away

or has a “bye” Alternating weekday and weekend matches

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The ACC 1997/98 Problem (cont’d) No team can play away on both last dates. No team may have more than two away matches

in a row. No team may have more than two home matches

in a row. No team may have more than three away

matches or byes in a row. No team may have more than four home matches

or byes in a row.

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The ACC 1997/98 Problem (cont’d) Of the weekends, each team plays four at home,

four away, and one bye. Each team must have home matches or byes at

least on two of the first five weekends. Every team except FSU has a traditional rival.

The rival pairs are Clem-GT, Duke-UNC, UMD-UVA and NCSt-Wake. In the last date, every team except FSU plays against its rival, unless it plays against FSU or has a bye.

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The ACC 1997/98 Problem (cont’d) The following pairings must occur at least once

in dates 11 to 18: Duke-GT, Duke-Wake, GT-UNC, UNC-Wake.

No team plays in two consecutive dates away against Duke and UNC. No team plays in three consecutive dates against Duke UNC and Wake.

UNC plays Duke in last date and date 11. UNC plays Clem in the second date. Duke has bye in the first date 16.

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The ACC 1997/98 Problem (cont’d) Wake does not play home in date 17. Wake has a bye in the first date. Clem, Duke, UMD and Wake do not play away

in the last date. Clem, FSU, GT and Wake do not play away in

the fist date. Neither FSU nor NCSt have a bye in the last

date. UNC does not have a bye in the first date.

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Some Literature Global Constraints for Round Robin Tournament

Scheduling. Martin Henz, Tobias Müller, Sven Thiel. European Journal of Operational Research (EJORS), 2002 (to appear)

Scheduling a Major College Basketball Conference-Revisited. Martin Henz. Operations Research, 49(1), Jan/Feb 2001.

Constraint-based Round Robin Tournament Planning. Martin Henz, ICLP 1999.

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Sports Scheduling in Practice Non-round-robin formats Optimization criteria / soft constraints Commercial aspects:

niche market research-intensive flexible business models needed

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Production Scheduling Problems

Scheduling Problems Propagation Algorithms for Resource

Constraints Branching Algorithms Exploration Algorithms Literature

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Production Scheduling Problems


Constraints Branching Algorithms Exploration Algorithms Literature

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Scheduling Problems

Assign starting times (and sometimes durations) to tasks, subject to

resource constraints, precedence constraints, and various other constraints, and

an optimization function, typically to minimize overall schedule duration.

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Example: Building a Bridge

Show Problem

Resource constraintsExample: a1 and a2 use excavator, cannot overlap in

time Precedence constraints

Example: p1 requires a3, a3 must end before p1 starts

Animate Solution Gantt Chart

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Modeling Indices: Task = {begin,a1,a2,a3,a4,a5,a6,p1,...,end}

Constants: duration of tasks duration = #[begin:0, a1:4, a2:2,...]#

Variables: represent each task with a finite domain variable representing its starting time.

Activity a[t in Task](duration[t])

a[t].start is finite domain variable representing the starting time of a[t]

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Precedence ConstraintsFor each two tasks t1, t2, where t1 must precede t2, introduce a constraint

a[t1].start + a[t1].duration a[t2].start

In OPL, just write a[t1] precedes a[t2]

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Resource ConstraintsFor each two tasks t1, t2 that require a unary resource r, we

have the constraint a[t1].start + a[t1].duration a[t2].start \/ a[t2].start + a[t2].duration a[t1].start

But: many constraints and weak propagation.Thus, introduce global resource constraints.

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Global Resource Constraints in OPL Declare resource

UnaryResource excavator; Require resource

a[a1] requires excavator;

a[a2] requires excavator;... All “requires” constraints on a resource

together form a global resource constraint.

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Propagation: Disjunctive Resource

For all tasks t1, t2 using the same resource:

a[t1].start + a[t1].duration

a[t2].start

\/ a[t2].start + a[t2].duration a[t1].start

Weakest but fastest form of propagation for unary resources.

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Propagation: Edge finding

For a given task t and set of tasks S, all sharing the same unary resource , find out whether t can occur

before all tasks in S, after all tasks in S, between two tasks in S,

and infer corresponding basic constraints.

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Which Edge Finder?As usual, trade-off between run-time of propagation

algorithm and strength of propagation. Edge finding can be expensive depending on what tasks are considered.

Recent edge finders have complexity n2 for one propagation step, where n is the number of tasks using the resource.

Edge finding based on task intervals can be stronger than these, but typically have complexity n3 .

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Specifying Propagation Algorithms

OPL fixes propagation for unary resources to an (undisclosed) edge-finding algorithm.

For discrete (non-unary) resources, the user can choose between default, disjunctive and edgeFinder when declaring a resource. Example:

DiscreteResource crane(3) using edgeFinder;

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Branching Algorithms: Serializers Simple enumeration techniques such as first-fail are

hopeless for scheduling. Use unary resources to guide branching. For two tasks t1, t2 sharing the same resource, use the

constraints

t1.start + t1.duration t2.start and t2.start + t2.duration t1.start for branching.

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Which Tasks To Serialize? Resource-oriented serialization: Serialize all tasks of

one resource completely, before tasks of another resource are serialized. Most-used-resource serialization Global slack Local slack

Task-oriented serialization: Choose two suitable tasks at a time, regardless of resources. Slack-based task serialization (see later)

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Most-used-resource Serialization“Most-used-resource” serialization: serialize first the

resource that is used the most.

Let T be a set of tasks running on resource r.

demand(T) = tT t.durationLet S be the set of all tasks using resource r.demand(r) := demand(S),

Serialize the resource r with maximal demand(r) first.

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Slack-based Resource SerializationLet T be a set of tasks running on resource r.supply(T) = lct(T) - est(T)

demand(T) = tT t.durationslack(T) = supply(T) - demand(T)

Let S be the set of all tasks using resource r.slack(r) := slack(S), S is set of all task running

on r.

Global slack serialization: Serialize resource with smallest slack first

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Local-Slack Resource Serialization

Let Ir be all task intervals on r. The local slack is defined as min {slack(I) | I Ir}

Local slack serialization: Serialize resource with smallest local slack first. Use global slack for tie-breaking.

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Which Tasks Are Serialized?

Ideas: Use edge finding to look for possible first tasks (and last

tasks) Choose tasks according to their est, lst (lct, ect).

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Task-oriented SerializationAmong all tasks using all resources, select a pair of

tasks according to local/global slack criteria and other considerations, regardless what resources are serialized already.

Provides more fine-grained control at the expense of runtime for finding candidate task pairs among all tasks.

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Programming Branching Algorithms in OPL

Scheduling-specific try constructs: tryRankFirst(u,a): activity a comes first tryRankLast(u,a): activity a comes last rank(u): serialize all tasks using u

Reflective functions for unary resources: isRanked(Unary): 1 if the resource is ranked nbPossibleFirst(Unary):

number of activities that can come first globalSlack(Unary): global slack of u localSlack(Unary): local slack of u

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Example: Task-oriented Serialization in OPL

while not isRanked(tool) do

select(r in Resources: not isRanked(tool[r]))

select(t in tasks[r] :

not isRanked(tool[r],a[t])

tryRankFirst(tool[r],a[t]);

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Pointers to Literature

Van Hentenryck: The OPL optimization programming language, 1999

Constraint Programming Tutorial of Mozart, 2000 (www.mozart-oz.org)

Various papers by Laburthe, Caseau, Baptiste, Le Pape, Nuijten,

see “Overview of Finite Domain Constraint Programming” in www.comp.nus.edu.sg/~henz/publications

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Assessment: Don’t Use It!Don’t use constraint programming for: Problems for which there are known efficient algorithms

or heuristics. Example: Traveling salesman. Problems for which integer programming works well.

Example: Many discrete assignment problems. Problems with weak constraints and a complex

optimization function. Example: many timetabling problems.

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Assessment: Do Use It!Use constraint programming for: Problems for which integer programming does not work

(linear models too large). Problems for which there are no efficient solutions

available. Problems with tight constraints, where propagation can be

employed. Example: ACC 97/98. Problems for which strong branching algorithms exist.

Example: Scheduling with unary resources.

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Myths Debunked

A fad! Constraint programming has been used successfully in a number of application areas, most spectacularly in scheduling

Universal! More failure stories than success stories. All new! Many ideas come from AI search and Operations

Research. In particular, the important ideas for scheduling come from OR.

Artificial Intelligence! All quite earthly algorithms. Constraint programming systems provide framework for different kinds of algorithms to interact.

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The Future

In OR, constraint programming is being added as a standard technique for solving combinatorial problems, along with local search (“heuristic approach”).

Constraint programming techniques will be tightly integrated with integer programming and local search.

Look at the Workshop Series on Integration of AI and OR Techniques for Combinatorial Optimization Problems (CP-AI-OR)

event scheduling using constraint programming november 22, 2002 martin henz, school of computing,...

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