Institute for Software Technology · 2019. 7. 25. · Alexander Felfernig & Gerald Steinbauer...

Alexander Felfernig & Gerald Steinbauer

Institute for Software Technology

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Alexander Felfernig und Gerald SteinbauerInstitut für Softwaretechnologie

Inffeldgasse 16b/2A-8010 Graz

Austria

Wissensverarbeitung- Constraint Satisfaction -



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References• Skriptum (TU Wien, Institut für Informationssysteme,

Thomas Eiter et al.) ÖH-Copyshop, Studienzentrum

• Stuart Russell und Peter Norvig. Artificial Intelligence - A Modern Approach. Prentice Hall. 2003.

• Knowledge-based Configuration - From Research to Business Cases: Alexander Felfernig, Lothar Hotz, Claire Bagley, and Juha Tiihonen (ab Mai 2014).

• Vorlesungsfolien TU Graz (teilweise basierend auf den Folien der TUWien)



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References1. S. Brailsford, C. Potts, and B. Smith. Constraint Satisfaction Problems: Algorithms

and Applications, Europ. Jrnl. of Operations Research, 119:557-581, 1999. [OR1999]

2. A. Felfernig, G. Friedrich, K. Isak, K. Shchekotykhin, E. Teppan, and D. Jannach. Automated Debugging of Recommender User Interface Descriptions, Journal ofApplied Intelligence, Springer Verlag, 31(1):1-14, 2009. [AI2009]

3. A. Felfernig, G. Friedrich, D. Jannach, and M. Zanker, An Environment for the Development of Knowledge-based Recommender Applications, International Journal of Electronic Commerce (IJEC), 11(2), pp. 11-34, 2006. [IJEC2006]

4. A. Felfernig, Reducing Development and Maintenance Efforts for Web-based Recommender Applications, International Jrnl. of Web Engineering and Technology, 3(3):329-351, 2007. [WebEng2007]

5. E. Teppan, A. Felfernig: Minimization of Product Utility Estimation Errors in Recommender Result Set Evaluations. Web Intelligence 2009: 20-27, 2009. [WebIntelligence2009]

6. A. Felfernig, G. Friedrich, D. Jannach, and M. Stumptner, Consistency-basedDiagnosis of configuration knowledge bases, AI Journal, 152(2):213–234, 2004. [AIJ2004]



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References6. A. Felfernig, G. Friedrich, M. Schubert, M. Mandl, M. Mairitsch, and E. Teppan.

Plausible Repairs for Inconsistent Requirements, 21st International Joint Conference on Artificial Intelligence (IJCAI'09), Pasadena, California, USA, pp. 791-796, 2009. [IJCAI2009]

7. A. Felfernig and R. Burke. Constraint-based Recommender Systems: Technologies and Research Issues, ACM International Conference on Electronic Commerce (ICEC'08), Innsbruck, Austria, Aug. 19-22, pp. 17-26, 2008. [ICEC2008]

8. E.C. Freuder. In Pursuit of the Holy Grail. Constraints, 2(1): 57-61, 1997. [Constraints1997]

9. U. Junker. QUICKXPLAIN: Preferred Explanations and Relaxations for Over-Constrained Problems. 19th National Conference on AI (AAAI04), pages 167–172, 2004. [AAAI2004]

10. R. Reiter. A theory of diagnosis from first principles, AI Journal, 23(1):57–95, 1987. [AIJ1987]

11. S. Russel and P. Norvig. Artificial Intelligence - A Modern Approach, Prentice Hall, 2nd edition, 2002, see also: http://aima.cs.berkeley.edu [ArtInt2002]

12. A. Felfernig, M. Schubert, and C. Zehentner. An Efficient Diagnosis Algorithm for Inconsistent Constraint Sets, AI for Engineering Design, Analysis, and Manufacturing (AIEDAM), Cambridge University Press, 25(2):175—184, 2011. [AIEDAM2011]



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Goals

Modeling Constraint Satisfaction Problems (CSPs).Applications of constraint technologies.Solving CSPs.Debugging CSPs.



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Constraint Technologies

Constraint technologies are one of the closest approaches computer science has yet made to the Holy Grail of programming: a user states the problem, the computer solves it[Constraints1997].Example application areas:

- Configuration (e.g., cars, telecommunication switches)

- Scheduling (e.g., job shop scheduling, course scheduling)

- Recommendation (e.g., computers, financial services)

- Test case generation (e.g., in model-based testing)



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Constraint Satisfaction Problem (CSP)

Specific type of decision problem where different alternatives constrain each other

Alternatives are typically represented in the form of finite domain variables

Alternatives should be selected in a way that none of the constraints is violated

Users could have preferences on the inclusion of certain alternatives, e.g., prefer(Rome, Paris)

Optimization requirements could exist, e.g., minimizecosts in a building design, maximize number of used slots in a module



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A Constraint Satisfaction Problem (or CSP) is defined by a set of variables V = {v1,v2, …, vm}, and a set of constraints C = {c1, c2, …, cn}. Each variable vi has a nonempty domain dvi of possible values. Each constraint cj involves some subset of the variables and specifies the allowable combinations of values for that subset.

A state of the problem is defined by an assignment of values to some or all of the variables.

An assignment that does not violate any constraints is called a consistent or legal assignment.

A complete assignment is one in which every variable is mentioned, and a solution to a CSP is a complete assignment that satisfies all the constraints.

Let d be the domain size → O(d m)

Constraint Satisfaction Problem (CSP)



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V={v1, v2, v3}dv1=[1,2]dv2=[1,2,3]dv3=[1,2]

C={c1,c2}c1:v1>v2c2:v2=v3

Solutions …

CSPs: A Simple Example



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Types of ConstraintsUnary Constraints

– restrict the values of a single variable– e.g., v1 <> 5

Binary Constraints– binary constraints relate two variables– e.g., v1 > v2

Higher order constraints – involve three or more variables– e.g., (v1 = 5 or v2 = 3) and v4 > v5

Extensional vs. intensional– dom(vi)={1,2}– intensional: v1 = v2– extensional: {(1,1), (2,2)}



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Boolean CSPsVariable domains with exactly two valuesAny CSP is convertible into a Boolean CSPSAT encoding

– a boolean variable indicates whether a certain value of a variable is set

Conventional representation– V={va, vb}; dva={1,2}; dvb={1,2}– C={c1: va>vb}

Corresponding SAT encoding– V={va1, va2, vb1, vb2}; dom(vai)=dom(vbi)={true, false}– C={c1: va2=true vb1=true va1=false vb2=false}



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Conjunctive Queries

SELECT v1.a, v2.a, v3.aFROM v1,v2,v3WHERE v1.a > v2.a and v2.a = v3.a;

variables and

domains

constraints

Table v1 (a)12

Table v2 (a)123

Table v3 (a)12

solution



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PROLOG

– v1(1).– v1(2).– v2(1).– v2(2).– v2(3).– v3(1).– v3(2).

– c1(V1,V2):-V1>V2.– c2(V2,V3):-V2=V3.

– q(V1,V2,V3):-v1(V1),v2(V2),v3(V3),c1(V1,V2),c2(V2,V3).

variables and

domains

constraints

all constraints must be satisfied…

solutions …?- q(X,Y,Z).

X = 2Y = 1Z = 1 ;



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Map Coloring

– Map Coloring Problem: assign colors to each region s.t. no neighboring regions have the same color.

– Representation as CSP:• V = {WA, NT, SA, Q, NSW, V, T}; dvi = {red, green, blue};

• C = {WA≠NT, WA≠SA, NT≠SA, NT≠Q, SA≠Q,

SA≠NSW, SA≠V, Q≠NSW, NSW≠V}



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– Example solution:• WA = red, NT = green, SA = blue, Q = red, NSW = green, V = red,

T = red ;

Map Coloring



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CHOCO

CHOCO: JAVA library for CSP solving.The central element of a CHOCO program is the Problem

object: Problem pb = new Problem();

Definition of variables: IntDomainVar WA = pb.makeEnumIntVar("WA", 0, 2);

Definition of constraints: pb.post(pb.neq(Map[0], Map[1]));

Operators, e.g.:neq(IntExp v1, IntExp v2) : v1 != v2. eq(IntExp v1, IntExp v2) : v1 = v2. leq(IntExp v1, IntExp v2) : v1 <= v2. lt(IntExp v1, IntExp v2) : v1 < v2. minus(IntExp exp1, IntExp exp2) : exp1 - exp2. plus(IntExp exp1, IntExp exp2) : exp1 + exp2.

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package chocosamples;import choco.Problem;import choco.integer.IntDomainVar;import java.util.logging.Logger;

public class map {…

public static void main(String[] args) {System.out.println(demo());

}

public static String demo() {…logger.fine("CHOCO Sample: Map Coloring");Problem pb = new Problem();

IntDomainVar WA = pb.makeEnumIntVar("WA", 0, 2);IntDomainVar NT = pb.makeEnumIntVar("NT", 0, 2);IntDomainVar SA = pb.makeEnumIntVar("SA", 0, 2);IntDomainVar Q = pb.makeEnumIntVar("Q", 0, 2);IntDomainVar NSW = pb.makeEnumIntVar("NSW", 0, 2);IntDomainVar V = pb.makeEnumIntVar("V", 0, 2);IntDomainVar T = pb.makeEnumIntVar("T", 0, 2);

IntDomainVar[] Map = {WA, NT, SA, Q, NSW, V, T};

pb.post(pb.neq(Map[0], Map[1]));pb.post(pb.neq(Map[0], Map[2]));pb.post(pb.neq(Map[1], Map[2]));pb.post(pb.neq(Map[1], Map[3]));pb.post(pb.neq(Map[2], Map[3]));pb.post(pb.neq(Map[2], Map[4]));pb.post(pb.neq(Map[2], Map[5]));pb.post(pb.neq(Map[3], Map[4]));pb.post(pb.neq(Map[4], Map[5]));

if (pb.solve() == Boolean.TRUE){

do {System.out.println("WA:" + WA.getVal());System.out.println("NT:" + NT.getVal());System.out.println("SA:" + SA.getVal());

System.out.println("Q:" + Q.getVal());System.out.println("NSW:" + NSW.getVal());

System.out.println("V:" + V.getVal());System.out.println("T:" + T.getVal());

System.out.println("\n");}

while(pb.nextSolution() == Boolean.TRUE);}

…}}

problemobject

variables

array ofvariables

constraints

output results

CHOCO



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a b c d e f g h8 87 76 65 54 43 32 21 1

a b c d e f g h

8 queens: positioning 8 queens on a 8x8 board (b): none of the queens

endangers any of the other queens.

Constraints (example):b11 = true

b1j = false (j=2..8)b11 = true bk1 = false (k=2..8)

b11 = true blm = false (8 ≥ l,m ≥ 2 l=m)

8 queens



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Magic Square

Magic Square: n2 numbers are arranged in a way s.t. the values in rows, columns, and diagonals result in the same sum.

Constraints (example):alldifferent(v11, v12, v13,

v21, v22, v23, v31, v32, v33)

sum(v11..v13) = …sum(v31..v33) = …sum(v11..v31) = …

2 7 6 159 5 1 154 3 8 1515 15 15 15



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Resource Allocation

Task:The following items should be distributed between three different containers:

1 barrel of fuel1 paper roll1 box of fireworks1 palette with 6 personal computers1 palette with computer games1 barell of old oil1 palette of roof tiles1 palette of aluminium rain pipes

Constraints:The following constraints have to be taken into account: c1: fuel and oil must not be combined

with fireworks in one container c2: personal computers and computer

games must be stored in one container c3: aluminium rain pipes must be stored

in container #3 c4: each container is allowed to have at

maximum 3 items c5: container #1 is not allowed to carry

fuel c6: if fuel is in container #3 then

container #2 must not store roof tiles



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Resource Allocation

V={fuel, paper, fireworks, computer, games, oil, rooftiles, rainpipes, c1fuel, c2fuel, c3fuel, c1paper, …, c3rainpipes}dfuel={1..3}dpaper={1..3}dfireworks={1..3}dcomputer={1..3}dgames={1..3}doil={1..3}drooftiles={1..3}drainpipes={1..3}dc1fuel={0,1}, …, dc3rainpipes={0,1}

C={c1, c2, c3, c4, c5, c6}c1: fuel <> fireworks and oil <>

fireworksc2: computer = games

c3: rainpipes = 3c4a: c1fuel=1 fuel=1 and

c2fuel=1 fuel=2 andc3fuel=1 fuel=3 … ?

c4b: c1fuel + c1paper + c1fireworks + c1computer + c1games + c1oil + c1rooftiles + c1rainpipes <=3 …

c5: fuel <> 1c6: not(fuel = 3 ) or not(rooftiles=2)



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Configuration Knowledge Base (CKB)Constraint Satisfaction Problem (CSP)– Variables V: {v1, v2, …, vm}

– Domains D: {dv1, dv2, …, dvm}

– Constraints C: {c1, c2, …, cn}

CSP Solution– assignment (S) of variables

in V consistent with the constraints ci C

Example: Configuration Problem V: Customer and product

properties VC={vc1, vc2, …, vck} VPROD={vp1, vp2, …, vpl}

D: Domains of VC and VPROD D={dvc1, …, dvck, dvp1, …, dvpl}

C: Compatibility (CC), filter (CF), product constraints (CPROD), and customer requirements (CR) CC={cc1,…, ccp} CF={cf1,…, cfq} CPROD={cp1,…, cpr} CR={cr1,…, crs}

[ICEC2008]



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Configuration: Simple Example– rr:exp-return-rate– wr:willingness-to-take-risks– ip:investment-period

– id:product-identifier– pr:return-rate– prl:risk-level– pip:investment-period

– dom(rr)={‘1-3%’,’>3-6%’,’>6-9%’,’>9%’}

– dom(wr)={low, medium, high}– dom(ip)={short-t, medium-t,

long-t}

– dom(id)=integer– dom(pr)=dom(rr)– dom(prl)=dom(wr)– dom(pip)=dom(ip)

cc1: wr=medium ip<>short-t cc2: wr=high ip=long-t cc3: ip=long-t rr=>3-6%

rr=>6-9% cc4: rr=>9% wr=high cc5: rr=>6-9% wr<>low

cf1: rr=pr cf2: wr=prl cf3: ip=pip

cr1: rr=1-3% cr2: wr=low cr3: ip=short-t

cp1: id=1 pr=‘1-3%’ prl=low pip=short-t … … cpr:id=r pr=‘>9%’ prl=high pip=long-t

VC

VPROD

D

CC

CF

CR

CPROD

Solution S: {rr=1-3%, wr=low, ip=short-t, id=1, pr=1-3%, prl=low, pip=short-t}



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CSP Solving

Goal: find an assignment for the variables which is consistent and complete

Basic approach:– Select variable and assign a corresponding value– Test consistency between instantiated variables and

constraints– Backtrack in the case of an inconsistency

Backtracking is a complete search algorithm – every existing solution is found



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Backtracking

• „Uninformed“ algorithm which is very inefficient for large problems• Improvements: e.g., forward checking, local consistency



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Backtracking

Variable and value orderings



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Forward Checking

Whenever a variable vi is assigned, forward checking looks at each unassigned variable vj connected to vivia a constraint and deletes from vjs domain any value that is inconsistent with the value chosen from vi.



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Local Consistency

Node consistency: Node consistency requires that every unary constraint on a variable is satisfied by all values in the domain of the variable.

Arc consistency: A variable of a constraint satisfaction problem is arc-consistent with another one if each of its admissible values is consistent with some admissible value of the second variable. A CSP is arc consistent if every variable is arc consistent with any other one.



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Local Consistency



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Local Consistency



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Performance



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Exercisesa. Machen Sie das folgende CSP arc-consistent: V={v1, v2, v3}dv1=[1,2]dv2=[1,2,3]dv3=[1,2]

C={c1,c2}c1:v1>v2c2:v2=v3

b. Geben Sie Beispiele für Constraints zur Unterstützung der persönlichen LV Planung



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Configuration Knowledge Base (CKB)Constraint Satisfaction Problem (CSP)– Variables V: {v1, v2, …, vm}

– Domains D: {dv1, dv2, …, dvm}

– Constraints C: {c1, c2, …, cn}

CSP Solution– assignment (S) of variables

in V consistent with the constraints ci C

Example: Configuration Problem V: Customer and product

properties VC={vc1, vc2, …, vck} VPROD={vp1, vp2, …, vpl}

D: Domains of VC and VPROD D={dvc1, …, dvck, dvp1, …, dvpl}

C: Compatibility (CC), filter (CF), product constraints (CPROD), and customer requirements (CR) CC={cc1,…, ccp} CF={cf1,…, cfq} CPROD={cp1,…, cpr} CR={cr1,…, crs}



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Configuration: Simple Example– rr:exp-return-rate– wr:willingness-to-take-risks– ip:investment-period

– id:product-identifier– pr:return-rate– prl:risk-level– pip:investment-period

– dom(rr)={‘1-3%’,’>3-6%’,’>6-9%’,’>9%’}

– dom(wr)={low, medium, high}– dom(ip)={short-t, medium-t,

long-t}

– dom(id)=integer– dom(pr)=dom(rr)– dom(prl)=dom(wr)– dom(pip)=dom(ip)

cc1: wr=medium ip<>short-t cc2: wr=high ip=long-t cc3: ip=long-t rr=>3-6%

rr=>6-9% cc4: rr=>9% wr=high cc5: rr=>6-9% wr<>low

cf1: rr=pr cf2: wr=prl cf3: ip=pip

cr1: rr=1-3% cr2: wr=low cr3: ip=short-t

cp1: id=1 pr=‘1-3%’ prl=low pip=short-t … … cpr:id=r pr=‘>9%’ prl=high pip=long-t

VC

VPROD

D

CC

CF

CR

CPROD

Solution S: {rr=1-3%, wr=low, ip=short-t, id=1, pr=1-3%, prl=low, pip=short-t}



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Engineering of CKBs: Scenario

Knowledge Engineers are responsible for constraint acquisition & maintenance.

Domain experts articulate domain knowledge (often in terms of examples).

Constraints are changed very frequently.

The correctness of the knowledge base is crucial: faulty definitions would cause follow-up costs.

New constraints have to be integrated and tested as soon as possible.



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Engineering of CKBs

Goal: Effective testing & debugging of constraint setsApproach:

– induce conflicts in the constraint set– resolve the conflict sets using model-based diagnosis

Result: minimal diagnoses for constraint sets.

?

?

?



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Diagnosing Constraint Sets: Approach

c1

c2

c3

c4

cn

Constraint Set (C)

t1

t2

tp

Test Cases (T)+

should be:consistentbut are:

inconsistent

Diagnoses (i)

Conflict Set CS = {c1, c2, …, cq} C s.t. tiT: CS {ti} inconsistent. Minimal (CS): CS’ with CS’ CS.

…

…

c1

c2

c3

Conflict Sets (CSi)

c4

c2

c1

CS1

CSr

CS2 …

…

c1

c4

Diagnosis C: C - {ti} consistent ti T.

1

…

c2

c3

k

…

…



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Diagnosing Constraint Sets: Example

c1

c2

c3

c4

c5

Constraint Set (C)

t1

t2

Test Cases (T)+

should be:consistent

Diagnoses (i)

t1: wr=high rr=‘>9%’

t2: rr =>6-9% wr=medium

c1: wr=medium ip<>short-tc2: wr=high ip=long-tc3: ip=long-t rr=>3-6% rr=>6-9%c4: rr=>9% wr=highc5: rr=>6-9% wr<>low wr<>medium

c3

c3(t1 ok)

2={c5, c3}

but are: inconsistent

CS1c5

c5 (t2 ok)

c2

CS2

1={c5, c2}

c2 (t1 ok)

Constraints:

Test cases:

c2

c3

CS2

Conflict Sets (CSi)

c5

CS1



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Diagnosing Constraint Sets: Example

V={v1,v2,v3,v4}dv1=[1..4], dv2=[1..4], dv3=[1..3], dv4=[1..2]c1: v1 > v2

c2: v3 > v1

c3: v3 = v1

c4: v4 < v3

c5: v3 > v2

c6: v4 < v1

c7: v4 <> v2

Conflict Set CS = {c1, c2, …, cq} C s.t. CS inconsistent.

Minimal (CS): CS’ with CS’ CS.

Diagnosis C: C - consistent.

Solution?



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Determining Conflict Sets: Simple Algorithm (Sketch)V={v1,v2,v3,v4}; dv1=[1..4], dv2=[1..4], dv3=[1..3], dv4=[1..2]c1: v1 > v2, c2: v3 > v1, c3: v3 = v1, c4: v4 < v3, c5: v3 > v2, c6: v4 < v1, c7: v4 <> v2

v1 v2 v3 v4

1 1 1 1

2 2 2 2

3 3 3

4 4

v1 v2 v3 v4

1 1 1 1

2 2 2 2

3 3 3

4 4

v1 v2 v3 v4

1 1 1 1

2 2 2 2

3 3 3

4 4

v1 v2 v3 v4

1 1 1 1

2 2 2 2

3 3 3

4 4

c1 c2 c3 c3 c1 c2c3 c2 conflict set CS1: {c2, c3}

c1 c2 c4 c5 c6 c7c1 c2 c4 c5 c6c7 c6 c1 c2c7

c6 c2 c1c7 conflict set CS2: {c1,c2,c6,c7}



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Determining Diagnoses:Simple Algorithm (Sketch)

CS1:{c2, c3}

diagnosis 1 {c2}

c2 c3

CS2:{c1, c2, c6, c7}c1

c2 c6c7diagnosis 2 {c1,c3}

¬diagnosis {c2,c3} (not minimal)

diagnosis 3 {c3,c6} diagnosis 4 {c3,c7}



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Example Knowledge Representation

SELECT *FROM v1, v2, v3, v4WHERE v1.v1>v2.v2 And /*c1*/v3.v3>v1.v1 And /*c2*/v3.v3=v1.v1 And /*c3*/v4.v4<v3.v3 And /*c4*/v3.v3>v2.v2 And /*c5*/v4.v4<v1.v1 And /*c6*/v4.v4<>v2.v2; /*c7*//* original query …

no solution*/

variables and

domains

Table v1 (v1)

1234

Table v2 (v2)

1234

Table v3 (v3)

123

Table v4 (v4)

12

SELECT *FROM v1, v2, v3, v4WHERE v1.v1>v2.v2 And /*c1*/v3.v3>v1.v1 And /*c2*/v3.v3=v1.v1 And /*c3*/v4.v4<v3.v3 And /*c4*/v3.v3>v2.v2 And /*c5*/v4.v4<v1.v1 And /*c6*/v4.v4<>v2.v2; /*c7*//* query taking into

account 1*/

SELECT *FROM v1, v2, v3, v4WHERE v1.v1>v2.v2 And /*c1*/v3.v3>v1.v1 And /*c2*/v3.v3=v1.v1 And /*c3*/v4.v4<v3.v3 And /*c4*/v3.v3>v2.v2 And /*c5*/v4.v4<v1.v1 And /*c6*/v4.v4<>v2.v2; /*c7*//* query taking into

account 4*/

…



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Conflict Sets and Diagnoses: Add. Examples

V={v1,v2,v3}dv1=dv2=dv3=[1..3]c1:v1>v2

c2:v2>v3

c3:v3>v1

Conflict sets:– CS1={c1,c2,c3}

Diagnoses:– 1={c1}– 2={c2}– 3={c3}

V={v1,v2,v3} dv1=dv2=dv3=[1..3] c1:v1>v2 c2:v1<v2 c3:v2=v1 Conflict sets:

- CS1={c1,c2}- CS2={c1,c3}- CS3={c2,c3}

Diagnoses:- 1={c1,c2}- 2={c1,c3}- 3={c2,c3}



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Simple Conflict Detection

C {c1, c2, …, cn}; /* C: all constraints*/B ; /* B: constraints part of the minimal conflict */IF inconsistent(C)

REPEAT B; /* : currently investigated constraint set */REPEAT

c element(C - ) /* c: one constraint */ {c};

UNTIL inconsistent()B B {c};

UNTIL inconsistent(B)RETURN (B);

#consistency checks: worst case: O(n2/2+n)

consistency checks best case: 3



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Example

V={v1,v2,v3}; dv1=dv2=dv3=[1..3]; C={c1,c2,c3}c1:v1>v2; c2:v2>v3; c3:v3>v1B: ; :

– c:c1; : {c1}– c:c2; : {c1, c2}– c:c3; : {c1, c2, c3}

B:{c3}; : {c3}– c:c1; : {c3,c1}– c:c2; : {c3,c1,c2}

B:{c3,c2}; : {c3,c2}– c:c1; : {c3,c2,c1}

B:{c3,c2,c1} /* minimal conflict set */



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Example

V={v1,v2,v3}; dv1=dv2=dv3=[1..3]; C={c1,c2,c3}c1:v1>v2; c2:v2=v1; c3:v3<>v1

B: ; : – c:c1; : {c1}– c:c2; : {c1, c2}

B:{c2}; : {c2}– c:c1; : {c2,c1}

B:{c2,c1} /* minimal conflict set */



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Example

V={v1,v2,v3}; dv1=dv2=dv3=[1..3]; C={c1,c2,c3}c1:v1>v2; c2:v2>v2; c3:v3<>v1

B: ; : – c:c1; : {c1}– c:c2; : {c1, c2}

B:{c2} /* minimal conflict set */



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QuickXPlain

QuickXPlain(C={c1,c2, …, cm})IF (is empty(C) or consistent(C)) return /* no inconsistency */ELSE return QX(, , C); /* C inconsistent */

QX(B, , C={c1,c2, …, cn})IF ( <> AND inconsistent(B)) return(); /* pruning of C */IF (C = {c}) return({c}); /* conflict element c detected */k=n/2;C1 {c1, …, ck}; C2 {ck+1, …, cn}; /* B still consistent */2 QX(B C1, C1, C2);1 QX(B 2, 2, C1);return (2 1)

#consistency checks: worst case: O(2k*log(m/k)+2k) best case: O(log(m/k)+2k



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Example

V={v1,v2,v3}; dv1=dv2=dv3=[1..3]; C={c1,c2,c3}c1:v1>v2; c2:v2>v3; c3:v3>v1

QuickXPlain(C={c1,c2,c3})

B=, C={c1,c2,c3}C1={c1} C2={c2,c3}

2={c3, c2}

{c3, c2,c1}

B={c1}, C={c2,c3}C1={c2} C2={c3}

B={c1,c2}, C={c3}2={c3}

B={c1,c3}, C={c2}

1={c2}

B={c3,c2}, C={c1}

1={c1}



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ExampleV={v1,v2,v3}; dv1=dv2=dv3=[1..3]; C={c1,c2,c3}

c1:v1>v2; c2:v2=v2; c3:v2>v1;c4:3=3; c5:7=7;c6:v3=v3;c7:2=2;c8:3=3,c9:1=1,c10:2=2

QuickXPlain(C={c1..c10})

B=, C={c1..c10}C1={c1..c5} C2={c6..c10}

B={c1..c5}, C={c6..c10}C1={c6..c7} C2={c8..c10}

2= B=, C={c1..c5}

C1={c1..c2} C2={c3..c5}

B= {c1..c2}, C={c3..c5}C1={c3} C2={c4..c5}

B= {c1..c3},C={c4..c5}

2= B={c1..c2}, C={c3}

1= {c3}

2= {c3}B= {c3}, C={c1..c2}

C1={c1} C2={c2}

1= {c1}

B= {c3,c1}, C={c2}2=

B= {c3}, C={c1}

1={c1}

1= {c3,c1}

{c3,c1}



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Exercisesa. Berechnen Sie alle minimalen Konflikte im folgenden

CSP mit Hilfe von „Simple Conflict Detection“: V={v1, v2, v3}dv1=[1,2]; dv2=[1,2]; dv3=[1,2]

C={c1,c2,c3}c1:v1>v2c2:v2>v3

c3:v3>v1

b. Repräsentieren Sie das o.a. CSP als conjunctive query.

Institute for Software Technology · 2019. 7. 25. · Alexander Felfernig & Gerald Steinbauer...

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Transcript of Institute for Software Technology · 2019. 7. 25. · Alexander Felfernig & Gerald Steinbauer...