CONSTRAINT LOGIC PROGRAMMING DECLARATIVE MODELING … · • Integrates various Logic Programming extensions • It contains: • several constraint solver libraries • a high-level

1

Costin Bădică

University of Craiova

Department of Computers and Information Technology

CONSTRAINT LOGIC PROGRAMMING:

FROM DECLARATIVE MODELING TO

PROBLEM SOLVING

01.08.2018 Problem Solving in Real World Systems and Games. Frankfurt am Main, 2018

2

UNIVERSITY OF CRAIOVA


3

IDS RESEARCH GROUP

• Intelligent Distributed Systems Research

Group

• http://ids.software.ucv.ro/

• Synergies between:

– Intelligent computing

– Distributed computing


http://ids.software.ucv.ro/


4

OVERVIEW

I. Basics of CSP, LP & CLP

CSP

Prolog

ECLiPSe-CLP

II. Examples

Basic Combinatorics

Logic Puzzles

Logistics Scheduling


5

EXAMPLE: MAP COLORING

• Variables: color of each region; 5 variables

• Domains: set of colors (green, red, yellow, blue)

• Constraints: any two neighbor regions have a different color;

e.g. color of region 3 ≠ color of region 2



6

CONSTRAINT SATISFACTION PROBLEM – CSP

01.08.2018

• CSP: triple 𝒫 = ⟨𝑋, 𝐷, 𝐶⟩ such that:

– 𝑋 = (𝑋1, 𝑋2, … , 𝑋𝑛) an 𝑛-tuple of (decision) variables

– 𝐷 = (𝐷1, 𝐷2, … , 𝐷𝑛) an 𝑛-tuple of domains such that

domain of 𝑋𝑖 is 𝐷𝑖

– 𝐶 = (𝐶1, 𝐶2, … , 𝐶𝑡) an 𝑡-tuple of constraints (relations)

such that:

𝐶𝑗 = (𝑅𝑆𝑗 , 𝑆𝑗)

where:

𝑆𝑗 = 𝑠𝑐𝑜𝑝𝑒 𝐶𝑗

are the variables occurring in 𝐶𝑗.

Problem Solving in Real World Systems and Games. Frankfurt am Main, 2018

7

CSP FEASIBLE SOLUTION

01.08.2018

• Feasible solution of a CSP 𝒫 = ⟨𝑋, 𝐷, 𝐶⟩ is an 𝑛-

tuple 𝐴 = (𝑋1 = 𝑎1, 𝑋2 = 𝑎2, … , 𝑋𝑛 = 𝑎𝑛) such that:

– 𝑎𝑖 ∈ 𝐷𝑖

– For each constraint 𝐶𝑗 the projection of tuple 𝐴 on

𝑠𝑐𝑜𝑝𝑒(𝐶𝑗) satisfies 𝑅𝑆𝑗.

• Note that a feasible solution:

𝐴 = 𝑋1 = 𝑎1, 𝑋2 = 𝑎2, … , 𝑋𝑛 = 𝑎𝑛

is often written as:

𝐴 = (𝑎1, 𝑎2, … , 𝑎𝑛)


8

CSP OPTIMAL SOLUTION

01.08.2018

• Let 𝒫 = ⟨𝑋, 𝐷, 𝐶⟩ be a CSP. We define a cost

function:

𝑓: 𝐷𝑖 → ℝ𝑛𝑖<1

• Optimal solution of a CSP 𝒫 = ⟨𝑋, 𝐷, 𝐶⟩ according

to cost function 𝑓 is an 𝑛-tuple 𝐴 = (𝑋1 = 𝑎1, 𝑋2 =𝑎2, … , 𝑋𝑛 = 𝑎𝑛) such that:

– 𝐴 is a feasible solution of 𝒫

– 𝑓(𝐴) is minimum (optimal)


9

TRIANGLE AREAS

01.08.2018

𝑌

𝑋 𝑂

𝑀

𝑁

𝐴

2

• There are given 3 integers: 𝑁,𝑀, 𝐴 ∈ ℕ.

• Find a triangle 𝑇 with vertices 𝑋𝑖 , 𝑌𝑖 , 𝑖 = 1,2,3 such that:

– Vertices satisfy 0 ≤ 𝑋𝑖 ≤ 𝑁 and 0 ≤ 𝑌𝑖 ≤ M

– Area of 𝑇 is 𝐴

2.


10

SIMPLIFY PROBLEM

We can assume

that one vertex is

0,0 !

01.08.2018

𝑌

𝑋 𝑂

𝑀

𝑁

𝐴

2

𝑌

𝑋 𝑂

𝑀

𝑁

𝐴

2

𝑌

𝑋 𝑂

𝑀

𝑁

𝐴

2

𝑌

𝑋 𝑂

𝑀

𝑁

𝐴

2

Flip horizontal

Translate

Flip vertical


11

TRIANGLE AREAS AS CSP • Problem variables are coordinates of triangle vertices:

𝑋1, 𝑌1 = 0,0 , 𝑋2, 𝑌2 , (𝑋3, 𝑌3)

• Domains:

𝑋2, 𝑋3 ∈ ,0,𝑁- 𝑌2, 𝑌3 ∈ ,0,𝑀-

• Constraints:

Triangle area =𝐴

2 i.e.

1

2det

0 0 1𝑋2 𝑌2 1𝑋3 𝑌3 1

=𝐴

2 i.e.

𝑋2𝑌3 − 𝑋3𝑌2 = 𝐴


12

CRYPT-ARITHMETIC PUZZLES

• Variables: digits 𝑇,𝑊,𝑂, 𝐹, 𝑈, 𝑅 ∈ *0,1,2,… , 9+.

• Constraints:

𝑎𝑙𝑙𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡( 𝑇,𝑊,𝑂, 𝐹, 𝑈, 𝑅 )

𝑇 ≠ 0

𝐹 ≠ 0

200𝑇 + 20𝑊 + 2𝑂 = 1000𝐹 + 100𝑂 + 10𝑈 + 𝑅

• Additional variables: carries 𝐶1, 𝐶2, 𝐶3 ∈ *0,1+ • Replace last constraint with 4 constraints:

2𝑂 = 𝑅 + 10𝐶1

2𝑊 + 𝐶1 = 𝑈 + 10𝐶2

2𝑇 + 𝐶2 = 𝑂 + 10𝐶3

𝐹 = 𝐶3

01.08.2018

T W O +

T W O

F O U R Global constraint


13

4-QUEENS PROBLEM

• Variables: 𝑄1, 𝑄2, 𝑄3, 𝑄4 such that 𝑄𝑖 is the row of the queen

placed on column 𝑖.

• Domains: *1,2,3,4+

• Constraints: no two distinct queens can attack on rows, on

main diagonals, on secondary diagonals.

– On rows: 𝑄𝑖 ≠ 𝑄𝑗 for all 𝑖 ≠ 𝑗

– On main diagonals: 𝑄𝑖 − 𝑄𝑗 ≠ 𝑖 − 𝑗

– On secondary diagonals: 𝑄𝑖 − 𝑄𝑗 ≠ 𝑗 − 𝑖

– On whatever diagonals: 𝑄𝑖 − 𝑄𝑗 ≠ |𝑖 − 𝑗|

01.08.2018

4 Q

3 Q

2 Q

1 Q

1 2 3 4 𝑄1 = 3


14

LOGIC PROGRAMMING

01.08.2018

Logic Program =

Facts + Rules +

Queries

Knowledge

Base

Logic Data =

Constants + Variables +

Compound Terms Terms

123 𝑝𝑟𝑜𝑙𝑜𝑔 𝑎 𝑁𝑢𝑚𝑏𝑒𝑟 𝐿𝑎𝑛𝑔𝑢𝑎𝑔𝑒 𝑋 _𝑥

𝑙𝑎𝑛𝑔𝑢𝑎𝑔𝑒 𝑝𝑟𝑜𝑙𝑜𝑔 1,2,3 𝑓(𝑎, 𝑔(𝑋))


15

PROLOG PROGRAMMING should_i_learn(Language) :-

curious_about_programming_languages,

already_own_book_covering(Language).

curious_about_programming_languages :-

reading(seven_languages_book).

already_own_book_covering(Language) :-

covers(Book, Language),

own(Book).

reading(seven_languages_book).

own(seven_languages_book).

covers(seven_languages_book, clojure).

covers(seven_languages_book, haskell).

covers(seven_languages_book, io).

covers(seven_languages_book, prolog).

covers(seven_languages_book, scala).

covers(seven_languages_book, erlang).

covers(seven_languages_book, ruby).

01.08.2018

“If you're curious about

programming languages and

you already own a book that

covers whatever language

you're asking about then yes,

you should learn it“

(Jeff Dallien, Quora, 2011)

| ?- should_i_learn(prolog).

yes

Rules

Facts Query


16

CRYPT-ARITHMETIC IN PROLOG sol([T,W,O,F,U,R]) :-

digit(T),

T \= 0,

digit(W),

W \= T,

digit(O),

O \= T, O \= W,

TWO is 100*T+10*W+O,

digit(F),

F \= 0, F \= T, F \= W, F \= O,

digit(U),

U \= T, U \= W, U \= O, U \= F,

digit(R),

R \= T, R \= W, R \= O, R \= F, R \= U,

FOUR is 1000*F+100*O+10*U+R,

TWO+TWO =:= FOUR.

01.08.2018

digit(0).

digit(1).

digit(2).

digit(3).

digit(4).

digit(5).

digit(6).

digit(7).

digit(8).

digit(9).

Can be replaced with: \+ U = F

OR not U = F


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LISTS • A list is a compound term representing sequence of

elements (terms) 𝑡1, 𝑡2, … , 𝑡𝑛:

,𝑡1, 𝑡2, … , 𝑡𝑛-

• A list is either a compound term written ,𝐻|𝑇- composed of:

– Head = first element 𝐻

– Tail = list of rest of elements after removing the head 𝑇

OR an empty list denoted by constant , -.

• For example the list:

,𝑎, 𝑏, 2-

represents terms:

,𝑎| 𝑏 2 - and ,𝑎| 𝑏, 2 - and ,𝑎|,𝑏| 2 -- and ,𝑎, 𝑏|,2--

01.08.2018

All 5 versions are equivalent !


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RECURSION

• I know you if

We met

OR

I met someone that

knows you

01.08.2018

met(alice,bob).

met(bob,carol).

met(alice,john).

knows_by_meeting(A,B) :-

met(A,B);met(B,A).

knows(A,B) :-

knows_by_meeting(A,B).

knows(A,B) :-

knows_by_meeting(A,C),

knows(C,B).

top :-

knows(A,B),

printf("%q knows %q\n",[A,B]),

fail.

top.

Failure driven loop

Infinite loop !

WHY?


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SOLUTION – LIMIT SEARCH ...

knows(A,B,N) :-

N>=0,

knows_by_meeting(A,B).

knows(A,B,N) :-

N>0,

M is N-1,

knows_by_meeting(A,C),

knows(C,B,M).

knows(A,B) :-

knows(A,B,3).

...

01.08.2018

• I know you if

We met

OR

I know you through at

most 𝑁 intermediaries


20

CONSTRAINT LOGIC PROGRAMMING

• CLP program = set of facts and rules built with:

i. Normal Prolog predicates handled by Prolog's search

engine.

ii. Constraints handled by specialized constraint solvers.

• CLP program structure:

i. Definition of variables and domains

ii. Definition of constraints

iii. Definition of cost variable

iv. Search for optimal solution


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ECLIPSE-CLP

• Open-source software system with support from

CISCO

• Integrates various Logic Programming extensions

• It contains:

• several constraint solver libraries

• a high-level modeling and control language – extension

of Prolog

• interfaces to third-party solvers

• API for embedding into host environments

• Tk/Tcl-based IDE


22

CARTESIAN PRODUCTS

• Let 𝐴1, 𝐴2, … , 𝐴𝑚 be 𝑚 finite sets:

𝐴𝑖 = 1, 2, … , 𝑛𝑖 , 1 ≤ 𝑖 ≤ 𝑚

• Generate all the Π𝑖<1𝑚 𝑛𝑖 elements of the Cartesian

product 𝑃 = 𝐴1 × 𝐴2 ×⋯× 𝐴𝑚.

• Decision variables: a vector 𝐸 of 𝑚 elements.

• Domains: 𝐸𝑖 ∈ 𝐴𝑖

• Constraints: ∅


23

CARTESIAN PRODUCTS IN CLP :-lib(ic).

no_of_sets(3).

top :-

Max = [](2,3,2),

no_of_sets(NSets),

dim(E,[NSets]),

(for(I,1,NSets), param(E,Max) do

arg(I,E,EI),

EI :: 1..Max[I]

),

labeling(E),

printf(output,"Solution: %q \n",[E]),

fail.

top.

01.08.2018

There are 3 sets

No. of. elems. of each set

Declarative loop to define

domains of each decision

variable

Default search

Define decision variables

E and Max are global

to the loop


24

ORDER OF SEARCH

• Variables are instantiated in the order of appearance in the collection:

𝐸1, 𝐸2, 𝐸3

• Values are considered according to their domain ordering: 1,2,3

01.08.2018

1 2

1 3 2

1 2 1 2 1 2

1 3 2

1 2 1 2 1 2


25

LIMITED DISCREPANCY SEARCH

• Replace labeling(E) with search(E,0,input_order,indomain,lds(2),[]).

• lds(Disc) = Disc “discrepancies” from the heuristic / first value chosen.

• Incomplete search !

01.08.2018

2

1 2

1 3 2

1 2 1 2 1

1

1 2 1

1 2 3 4 5 6 7 8

Order of generating

solutions


26

SUBSETS • Let 𝐴 be a set with 𝑛 elements:

𝐴 = *1,2,… , 𝑛+

• Subsets 𝑆 ⊆ 𝐴 can be represented using a vector 𝑋 =(𝑥1, 𝑥2, … , 𝑥𝑛) with elements in *0,1+ such that:

𝑥𝑖 = 1 𝑖𝑓 𝑖 ∈ 𝑆0 𝑖𝑓 𝑖 ∉ 𝑆

• For example the subsets of *1,2,3+ are 0,0,0 , (0,0,1), (0,1,0), 0,1,1 , (1,0,0), (1,0,1), (1,1,0), 1,1,1 .

• Subsets with this representation can be generated by

adapting the method for Cartesian products. HOW?


27

SUBSETS-2 • Let 𝐴 be a set with 𝑛 elements:

𝐴 = *1,2,… , 𝑛+

• Subsets 𝑆 ⊆ 𝐴 can be represented using a vector 𝑋 =(𝑥1, 𝑥2, … , 𝑥𝑛) with elements in *0,1,… , 𝑛+ such that:

𝑆 = 𝑥𝑖 𝑥𝑖 > 0, 1 ≤ 𝑖 ≤ 𝑛+

• To ensure uniqueness of representation of each set:

– Vector 𝑋 is increasingly sorted: 𝑥1 ≤ 𝑥2 ≤ ⋯ ≤ 𝑥𝑛

– Each element of 𝑆 occurrs uniquely in 𝑋

• For example the subsets of *1,2,3+ are 0,0,0 , (0,0,1), (0,0,2), 0,0,3 , (0,1,2), (0,1,3), (0,2,3), 1,2,3 .


28

SUBSETS-2 IN CLP :-lib(ic).

no_of_elems(4).

top :-

no_of_elems(NElems),

dim(S,[NElems]),

S :: 0..NElems,

(for(I,2,NElems), param(S) do

S[I-1] #=< S[I],

(S[I] #> 0) => (S[I-1] #< S[I])

),

labeling(S),

printf(output,"Solution: %q \n",[S]),

fail.

top.

01.08.2018

Ensures that:

𝑆 1 ≤ 𝑆 2 ≤ ⋯ ≤ 𝑆,𝑁-

Reified constraint.

Ensures that any two

non-zero elements are

distinct.


29

COMBINATIONS

• Let 𝐴 be a set with 𝑛 elements:

𝐴 = *1,2, … , 𝑛+

• Subsets 𝑆 ⊆ 𝐴 with 𝑘 elements can be represented

using a vector 𝑋 = (𝑥1, 𝑥2, … , 𝑥𝑘) with elements in

*1, … , 𝑘+ such that:

𝑥𝑖;1 < 𝑥𝑖 for all 2 ≤ 𝑖 ≤ 𝑛

• Think about an alternative representation using the

characteristic vector ! HOW?


30

COMBINATIONS IN CLP :-lib(ic).

no_of_elems(4). % N

no_of_selected_elems(2). % K

top :-

no_of_elems(N),

no_of_selected_elems(K),

dim(S,[K]),

S :: 1..N,

(for(I,2,K), param(S) do

S[I-1] #< S[I]

),

labeling(S),

printf(output,"Subset: %q \n",[S]),

fail.

top.

01.08.2018

Ensures that:

𝑆 1 < 𝑆 2 < ⋯ < 𝑆,𝐾-


31

PERMUTATIONS

• Let 𝐴 be a set with 𝑛 elements:

𝐴 = *1,2, … , 𝑛+

• A permutation is a one-to-one mapping:

𝜎: 𝐴 → 𝐴

• A permutation can be represented using a vector

𝑋 = (𝑥1, 𝑥2, … , 𝑥𝑛) with elements in *1, … , 𝑛+ such

that:

𝑎𝑙𝑙𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡(,𝑥1, 𝑥2, … , 𝑥𝑛-)


32

PERMUTATIONS IN CLP :-lib(ic).

no_of_elems(4).

top :-

no_of_elems(N),

dim(P,[N]),

P :: 1..N,

alldifferent(P),

labeling(S),

printf(output,"Solution: %q \n",[P]),

fail.

top.


33

TRIANGLE AREAS IN CLP :- lib(ic).

model(N, M, A, [X2, Y2, X3, Y3]) :-

[X2, X3] :: 0..N,

[Y2, Y3] :: 0..M,

A #= abs(X2 * Y3 - X3 * Y2).

do_case(Case_num, N, M, A) :-

printf("Case #%w: ", [Case_num]),

( model(N, M, A, Points), labeling(Points) ->

printf("0 0 %w %w %w %w", Points)

;

write("IMPOSSIBLE")

),

nl.

top :-

read_token(C,integer),

( for(Case_num, 1, C) do

read_token(N,integer), read_token(M,integer),

read_token(A,integer),

do_case(Case_num, N, M, A) ).

01.08.2018

Input:

1000

1 1 1

1 2 64

10 10 1

1 1 1

...

No.of.cases (C)

Case 1 (N,M,A)


34

CRYPT-ARITHMETIC IN CLP :- lib(ic). % library for arithmetic constraints

twotwofour(Digits) :-

Digits = [T,W,O,F,U,R], % Variables ... digits

Digits :: [0..9], % Their domains {0,1,...,9}

Carries = [C1,C2,C3], % Other variables ... carries

Carries :: [0..1], % Their domains {0,1}

alldifferent(Digits), % Global constraint

T #\= 0, % Constraint “not equal”

F #\= 0,

C3 #= F, % Constraint “equal”

C2 + 2*T #= O + 10*C3,

C1 + 2*W #= U + 10*C2,

2*O #= R + 10*C1,

labeling(Carries), % Search for assignments

labeling(Digits).


35

LOGIC PUZZLES • On a street there are five houses of different colors, occupied by five

individuals of different nationalities, who own five different pets, drink five

different beverages, and smoke five different brands of cigarettes. Also: 1. The occupant of the red house is English.

2. The Spaniard owns a dog.

3. The coffee drinker lives in a green house.

4. The Ukrainian drinks tea.

5. The ivory house is to the left of the green one.

6. The snails owner smokes Winston’s.

7. The person in the yellow house smokes Kool’s.

8. The occupant in the middle house drinks milk.

9. The Norwegian lives in the leftmost house.

10. The Chesterfield smoker lives next to the fox owner.

11. The Kool smoker lives next to the horse owner.

12. The orange juice drinker smokes Lucky Strike’s.

13. The Parliament smoker is Japanese.

14. The Norwegian lives next to a blue house.

• Who owns the zebra? Who drinks water?


36

LOGIC PUZZLES IN CLP :- lib(ic).

top :-

% Define variables and their domains

...

% Define constraints

...

% Search for solutions

...

% Decode and print out solutions

...


37

DEFINE VARIABLES AND DOMAINS

% Variables: we use 5 arrays of 5 variables each

Nat = [](English, Spaniard, Ukrainian, Norwegian,

Japanese),

Color = [](Red, Green, Ivory, Yellow, Blue),

Cigarettes = [](Winston, Kool, Chesterfield, Lucky,

Parliament),

Pet = [](Dog, Snails, Fox, Horse, Zebra),

Drink = [](Coffee, Tea, Milk, Juice, Water),

% Domains: all variables range over house numbers 1 to 5

Nat :: 1..5,

Color :: 1..5,

Cigarettes :: 1..5,

Pet :: 1..5,

Drink :: 1..5,


38

DEFINE CONSTRAINTS % Constraints: the values in each list are exclusive

alldifferent(Nat), alldifferent(Color), alldifferent(Cigarettes),

alldifferent(Pet),alldifferent(Drink),

% The actual 14 constraints

English = Red, % 1

Spaniard = Dog, % 2

Coffee = Green, % 3

Ukrainian = Tea, % 4

Green #= Ivory + 1, % 5

Snails = Winston, % 6

Yellow = Kool, % 7

Milk = 3, % 8

Norwegian = 1, % 9

Dist1 #= Chesterfield - Fox, Dist1 :: [-1, 1], % 10

Dist2 #= Kool - Horse, Dist2 :: [-1, 1], % 11

Juice = Lucky, % 12

Parliament = Japanese, % 13

Dist3 #= Norwegian - Blue, Dist3 :: [-1, 1], % 14


39

SEARCH AND PRINT OUT SOLUTIONS

% Search: label all variables with values

labeling([](Nat, Color, Cigarettes, Pet, Drink)),

% Print out solutions: we need to do some decoding

NatNames = [English-english, Spaniard-spaniard, Japanese-japanese,

Ukrainian-ukrainian, Norwegian-norwegian],

member(Zebra-ZebraNat, NatNames),

member(Water-WaterNat, NatNames),

printf("The %w owns the zebra%n", [ZebraNat]),

printf("The %w drinks water%n", [WaterNat]).


40

FREIGHT TRANSPORTATION EXCHANGES

• Opportunity:

– need of transporting goods

– availability of free vehicles

• Goal:

– capturing transportation

opportunities

– matchmaking of owners of

goods with freight

transportation providers

01.08.2018

• New business model: virtual logistics platforms


41

MAS FOR FREIGHT BROKERING


42

AGENT INTERACTION


43

FREIGHT BROKER AGENT

• Declarative optimization model of the

FBAgent:

– Defined as type of Vehicle Routing with

Pickup and Delivery Problem – VR-PDP

– Computing optimal schedules using

Constraint Logic Programming – CLP.


44

VR-PDP

• Well-known problem in operation research:

Given:

– A set of customers

– A vehicle pool to service deliveries

Determine:

– A minimum cost set of vehicle routes that services

all customers

• Service:

– Pickup = freight loading point

– Delivery = freight unload point


45

DECLARATIVE MODELING

• AIM: The definition of a declarative optimization

agent – OAgent.

– Declarative constraint model as a special type of

vehicle routing with pickup and delivery problem.

– Augmenting this model with a declarative

representation of optimal transport schedules.

– Computing optimal schedules using Constraint Logic

Programming – ECLiPSe-CLP.


46

PROBLEM Tuple ℒ, 𝒪, 𝒯, Δ s.t.:

1. ℒ = locations of interest = 𝒫 ∪ℋ, 𝒫 = *1,2,… , 𝑘+ = pickup

and delivery points, 𝑘 > 0. ℋ = *𝑘 + 1,… , 𝑘 + 𝑕+ = truck

home locations, 𝑕 ≥ 0

2. 𝒪 = customer orders, |𝒪| = 𝑛. 𝑂𝑖 = (𝑂𝑆𝑖 , 𝑂𝐷𝑖 , 𝐶𝑖) s.t.

𝑂𝑆𝑖 , 𝑂𝐷𝑖 ∈ 𝒫, 𝑂𝑆𝑖 ≠ 𝑂𝐷𝑖 are pickup and delivery points,

𝐶𝑖 > 0 = requested capacity of order 𝑖. Obs: 2 ≤ 𝑘 ≤ 2𝑛.

3. 𝒯 = set of trucks, |𝒯| = 𝑡. 𝑇𝑖 = (𝐻𝑖 , Γ𝑖) s.t. 𝐻𝑖 ∈ ℋ and Γ𝑖 > 0

are home location and provided capacity of truck 𝑖.

4. Δ = 𝑘 + 𝑕 × (𝑘 + 𝑕) real positive matrix s.t. Δ𝑖𝑗 > 0 are

distances between locations 1 ≤ 𝑖 ≠ 𝑗 ≤ 𝑘 + 𝑕.


47

EXAMPLE PROBLEM • 𝑘 = 3 pickup and delivery points

• 𝑛 = 3 orders: 𝒪 = * 1 → 2,5 , 1 → 3,2 , 2 → 3,6 +

• 𝑡 = 2 trucks: 𝒯 = *(4,7), (4,5)+. Trucks home location is

4, so there are 𝑕 = 1 home locations

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4

1

2

3

2

5

6


48

SCHEDULE

Tuple ⟨𝑋,𝑀, 𝑆, 𝐷⟩ s.t.:

1. 𝑋 ∈ 1,2,… , 𝑘 𝑚 = hops sequence of truck routes. Each

location is visited, each order needs two hops, 𝑘 ≤ 𝑚 ≤ 2𝑛.

2. 𝑀 ∈ *0,1,… ,𝑚+𝑡 s.t. 𝑀𝑙 is the number of hops of each truck

𝑙. Total number of hops: 𝑚 = ∑𝑙<1𝑡 𝑀𝑙.

– 𝑀𝑙 ≥ 0 allows solutions of “at most” 𝑡 trucks.

– 𝑀𝑙 ≥ 1 constraints solutions to use “exactly” 𝑡 trucks.

3. 𝑆, 𝐷 ∈ *0,1+𝑚×𝑛 are two Boolean matrices s.t. 𝑆𝑖𝑗 = 1 iff 𝑋𝑖 is

the pickup point of order 𝑗, otherwise 𝑆𝑖𝑗 = 0. 𝐷𝑖𝑗 = 1 iff 𝑋𝑖 is

the delivery point of order 𝑗, otherwise 𝐷𝑖𝑗 = 0.


49

EXAMPLE SCHEDULE • 𝑘 = 3 pickup and delivery points

• 𝑛 = 3 orders: 𝒪 = * 1 → 2,5 , 1 → 3,2 , 2 → 3,6 +

• 𝑡 = 2 trucks: 𝒯 = *(4,7), (4,5)+. Trucks home location is

4, so there are 𝑕 = 1 home locations


50

STATE SPACE SIZE

• 𝑋 = 𝑚, 𝑋𝑖 ∈ 1,2, … , 𝑘 , 𝑘 ≤ 𝑚 ≤ 2𝑛

• |𝑆| = 2𝑚𝑛, 𝐷 = 2𝑚𝑛

• 𝑆𝑖𝑧𝑒 = ∑ 2𝑚𝑛2𝑚𝑛𝑚𝑘2𝑛𝑚<𝑘

• 𝑛 = 5, 𝑘 = 6

• 𝑆𝑖𝑧𝑒 = 26066 + 27076 + 28086 + 29096 + 2100106 >𝟏𝟎𝟑𝟔 !!


51

PICKUP & DELIVERY DEFINITION

• Constraints (1), (2). For each hop 𝑖 and order 𝑗 the

pickup point of order 𝑗 is 𝑂𝑆𝑗 and the delivery point

of order 𝑗 is 𝑂𝐷𝑗 so:

(∀𝑖, 𝑗)((𝑆𝑖𝑗 = 1) ⇒ (𝑋𝑖 = 𝑂𝑆𝑗))

(∀𝑖, 𝑗)((𝐷𝑖𝑗 = 1) ⇒ (𝑋𝑖 = 𝑂𝐷𝑗))


52

NON-REDUNDANT HOPS

• Constraint (3). There exists at least one load or

unload in each hop 𝑖 so:

∀𝑖 ∑𝑖<1𝑛 𝑆𝑖𝑗 + 𝐷𝑖𝑗 ≥ 1


53

PICKUP PRECEDES DELIVERY

• Constraint (4). For all hops 𝑖, 𝑘 if there exists an

order 𝑗 such that 𝑖 is the pickup point of 𝑗 and 𝑘 is

the delivery point of 𝑗 then 𝑖 precedes 𝑘, so:

(∀𝑖, 𝑘) ( ∃𝑗 ((𝑆𝑖𝑗 = 1) ∧ (𝐷𝑖𝑗 = 1)) ⇒ 𝑖 < 𝑘 )


54

SERVICE COMPLETENESS

• Constraint (5). For each order 𝑗 there is a unique

load point and a unique unload point, so:

(∀𝑗)((∑𝑖<1𝑚 𝑆𝑖𝑗 = 1) ∧ (∑𝑖<1

𝑚 𝐷𝑖𝑗 = 1))


55

TRUCK ASSIGNMENTS • Constraints (6). Each order is completely served by a

unique truck 𝑙, so order with load point assigned to 𝑙 has the

unload point assigned to 𝑙 :

∀𝑙, 𝑗 ∑𝑖<𝐴𝑙

𝐵𝑙 𝑆𝑖𝑗 = ∑𝑖<𝐴𝑙

𝐵𝑙 𝐷𝑖𝑗

• where

𝐴𝑙 = 1 + ∑𝛼< 1𝑙;1 𝑀𝛼 for 1 ≤ 𝑙 ≤ 𝑡

𝐵𝑙 = ∑𝛼<1𝑙 𝑀𝛼 for 1 ≤ 𝑙 ≤ 𝑡

• Constraints (7). Non-redundancy constraint:

∀𝑖 ∈ 𝐴𝑙 , 𝐵𝑙 − 1 𝑋𝑖≠ 𝑋𝑖:1


56

CAPACITY CONSTRAINTS

• Constraints (8). For each truck hop the

transportation capacity does not overflow, so:

𝑇0 = 0

𝑇𝑖 = 𝑇𝑖;1 + ∑𝑗 <1𝑛 𝑆𝑖𝑗 − 𝐷𝑖𝑗 𝐶𝑗 for 1 ≤ 𝑙 ≤ 𝑡 and

𝐴𝑙 ≤ 𝑖 ≤ 𝐵𝑙

𝑇𝑖 ≤ Γ𝑖 for 1 ≤ 𝑙 ≤ 𝑡


57

OPTIMIZATION COST • Cost. The total distance 𝐷𝑙 travelled by truck 𝑙 = 1,2, … , 𝑡, as well as the

total distance 𝐷𝐼𝑆𝑇 travelled by all trucks, is:

𝐷𝑙 =

0 if 𝐼𝑙 = 𝐴𝑙 , 𝐵𝑙 = ∅

Δ𝐻𝑙𝑋𝐴𝑙+ Δ𝑋𝐵𝑙𝐻𝑙

+ ∑ Δ𝑋𝑙𝑋𝑙+1𝐵𝑙;1𝑖<𝐴𝑙

otherwise

𝐷𝐼𝑆𝑇 = ∑𝑙<1𝑡 𝐷𝑙

• For CLP mapping, variables Δ𝐻𝑙𝑋𝐴𝑙, Δ𝑋𝐵𝑙𝐻𝑙

, and Δ𝑋𝑖𝑋𝑖+1 are rewritten:

Δ𝐻𝑙𝑋𝐴𝑙= ∑𝑥<1

𝑘 Δ𝐻𝑙𝑥(𝑥 = 𝑋𝐴𝑙)

Δ𝑋𝐵𝑙𝐻𝑙= ∑ Δ𝐻𝑙𝑥(𝑥 = 𝑋𝐵𝑙)

𝑘𝑥<1

Δ𝑋𝑖𝑋𝑖+1 = ∑ Δ𝑥𝑦(𝑥 = 𝑋𝑖)(𝑦 = 𝑋𝑖:1)𝑘𝑥,𝑦<1


58

OPTIMIZATION AGENT – OAGENT


59

SCHEDULING PROBLEM AS PROLOG FACTS number_of_orders(3).

number_of_cities(3).

number_of_trucks(3).

number_of_homes(1).

% order(OrderIdx, LoadIdx, UnloadIdx, Capacity)

order(1,1,2,5).

order(2,1,3,2).

order(3,2,3,6).

% truck(TruckIdx, HomeIdx, MaxCapacity)

truck(1,4,7).

truck(2,4,4).

truck(3,4,5).

% distance(LocationI, LocationJ, DistanceIJ)

distance(1,2,10).

distance(2,1,10).

% ...

distance(4,3,10).


60

SOLUTION PREDICATE

solution(_m,_n,_t,M,S,D,X,Ds) :-

domains_and_variables(_m,_n,_t,M,S,D,X,Delta),

constraints(_m,_n,_t,M,S,D,X),

search_query(X,S,D),

compute_distances(_m,_t,M,Delta,X,Ds).


61

DECLARATIVE LOOPS

( 𝐼𝑡𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑆𝑝𝑒𝑐𝑠 do 𝐺𝑜𝑎𝑙 )

• Iteration specifiers:

– for(𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 , 𝑀𝑖𝑛𝐸𝑥𝑝𝑟 , 𝑀𝑎𝑥𝐸𝑥𝑝𝑟)

– param(𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒1 , 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒2 , …)

– fromto(𝐹𝑖𝑟𝑠𝑡 , 𝐼𝑛 , 𝑂𝑢𝑡 , 𝐿𝑎𝑠𝑡)


62

IMPLEMENTATION OF CONSTRAINTS

• Using 𝑝 ⇒ 𝑞 ≡ ≦𝑞 ⇒ ≦𝑝, constraint:

(∀𝑖, 𝑘) ( ∃𝑗 ((𝑆𝑖𝑗 = 1) ∧ (𝐷𝑖𝑗 = 1)) ⇒ 𝑖 < 𝑘 )

is transformed to equivalent form:

(∀𝑖, 𝑗, 𝑘)( 𝑖 ≥ 𝑘 ⇒ ((𝑆𝑖𝑗 = 0) ∨ (𝐷𝑖𝑗 = 0))

• In ECLiPSe-CLP:

constraint_2(M,N,S,D) :-

( for(J,1,N), param(M,S,D) do

( for(I,1,M), param(J,S,D) do

( for(K,1,I), param(I,J,S,D) do

S[I,J] #= 0 or D[K,J] #= 0

)

)

).


63

SEARCH PROCESS

01.08.2018

• Top level search

– Instantiates number of hops 𝑚 and vector 𝑀 of

route lengths of each truck such that:

∑ 𝑀𝑙 = 𝑚𝑡𝑙<1

• Main search

– Searches for problem solutions 𝑋, 𝑆, 𝐷 for fixed

values of 𝑚 and 𝑀.


64

SEARCH QUERY

search_query(X,S,D) :-

search([](X,S,D),0,

most_constrained,

indomain_random,

lds(2),

[nodes(28000)]

).

01.08.2018

Limited

discrepancy

search

Stochastic

search


65

DATA SET & SEARCH CONFIGURATION


66

SOLUTIONS OF 10-4 PROBLEM


67

REFERENCES • Costin Bǎdicǎ, Florin Leon, Amelia Bǎdicǎ. Freight transportation broker

agent based on constraint logic programming. Evolving Systems.

Springer (2018). doi: 10.1007/s12530-018-9230-3

• Sergii Dymchenko, Mariia Mykhailova. Declaratively solving tricky

google code jam problems with prolog-based ECLiPSe CLP system.

Proc.30th Annual ACM Symposium on Applied Computing (SAC '15).

2122-2124, ACM (2015). doi: 10.1145/2695664.2696032. Extended

version: https://arxiv.org/abs/1412.2304v2

• Hector J. Levesque. Thinking as Computation. A First Course. MIT

Press (2012)

• Antoni Niederliński. A gentle guide to constraint logic programming via

ECLiPSe, 3rd ed. Jacek Skalmierski Computer Studio, Gliwice (2014)

• ECLiPSe Web page: http://eclipseclp.org. Follow the ‘Examples’ link.


68 01.08.2018 Problem Solving in Real World Systems and Games. Frankfurt am Main, 2018

Documents

CONSTRAINT LOGIC PROGRAMMING DECLARATIVE MODELING … · • Integrates various Logic Programming extensions • It contains: • several constraint solver libraries • a high-level