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Constraint Satisfaction Problems and Consistency Restoration Florent Launay Undergraduate student at FIT

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Summary Focused interest in time point-sequencing problem. Focused interest in time point-sequencing problem. What are CSP’s ? What are CSP’s ? Crossword Problem. Crossword Problem. Dealing with consistency. Dealing with consistency. 1. Finding and storing consistent relations 1. Finding and storing consistent relations 2. Finding a “minimal set to remove inconsistency” 2. Finding a “minimal set to remove inconsistency” Linking the algorithms Linking the algorithms Questions, comments. Questions, comments. The Program The Program

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Focus interest Relation from {, =, <>, =, } P1P2P3 Here, we can see that some constraints between the set of points {P1, P2, P3} exist : P1 < P2 P2 < P3

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What are CSP’s ? Constraints Constraints X >= 5X <> 9X < 12

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What are CSP’s ? Example : Example : Common names from English vocabulary beginning with a ‘c’, ending with a ‘t’, but that doesn’t contain any ‘e’… Valid set : {‘count’, ‘cart’, ‘chat’, ‘court’, …}

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What are CSP’s ? Valid regions and ‘Box’ Valid regions and ‘Box’ 5912 X >= 5X <> 9X < 12 Valid regions = {[5,5], [5,9], [9,12]} X < 15X > 3 153

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What are CSP’s ? Inconsistencies Inconsistencies 5912 X >= 5X <> 9X < 12X > 13 13 No valid region => The system is inconsistent

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The crossword problem Constraints Valid Set Inconsistency Definition, Size of the world, across or down, eventual existing letters. Definition 13 : Part of the human body, beginning with an ‘H’ (ambiguous) E A D ?A N D ?A I RH O W Definition 14 : Negation {[HEAD], [HAND], [HAIR]} NOTNOT A N D

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Dealing with consistency Within a set of relations, some relation can conflict with other relations in a system, making the whole system inconsistent. Within a set of relations, some relation can conflict with other relations in a system, making the whole system inconsistent. Let’s have a look to the following example.

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A simple example Let s consider a point P1 on a time line basis :

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A simple example Valid region P2 P2>P1 P3 P2 P1 No valid region can be found for P3 !!!

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Dealing with consistency With that kind of inconsistency, two interesting questions arise: With that kind of inconsistency, two interesting questions arise: Which are the relations that make the system inconsistent ? Which are the relations that make the system inconsistent ? What would be a minimal set to remove to make the system consistent ? What would be a minimal set to remove to make the system consistent ?

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1. Finding and storing inconsistent relations System of relations Consistent YES NO Finish Store inconsistent relations Finish (Call the 2nd algorithm)

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Inconsistent relations FIND MULTIPLE EQUALITY FIND MULTIPLE EQUALITY =>Store inconsistent relations VERIFY CONSISTENCY IN CASE OF UNIQUE EQUALITY VERIFY CONSISTENCY IN CASE OF UNIQUE EQUALITY =>Store inconsistent relations FIND THE BOX: PRE-PROCESSING FIND THE BOX: PRE-PROCESSING =>Store inconsistent relations FIND VALID REGIONS FIND VALID REGIONS =>If no inconsistency has been found

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Inconsistent relations FIND MULTIPLE EQUALITY FIND MULTIPLE EQUALITY P2P1P3P4P5P6 If Pnew = P3 Pnew And Pnew = P6 The system is inconsistent since P3<>P6

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Inconsistent relations VERIFY CONSISTENCY IN CASE OF UNIQUE EQUALITY VERIFY CONSISTENCY IN CASE OF UNIQUE EQUALITY P2P1P3P4P5P6 Pnew If Pnew = P3 And Pnew < P2 The system is inconsistent

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Inconsistent relations FIND THE BOX: PRE-PROCESSING FIND THE BOX: PRE-PROCESSING P2P1P3P4P5P6 If If we found that a valid region for Pnew was {[P3,P4], [P4,P5]} Then, no other relation should forbid that : ex Pnew > P6, Or the system becomes inconsistent

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Inconsistent relations FIND VALID REGIONS FIND VALID REGIONS =>If no inconsistency has been found P2P1P3P4P5P6 A valid region region could be : {[P2,P3], [P3,P3], [P3,P4], [P4,P5], [P5,P5], [P5,P6]} {[P2,P3], [P3,P3], [P3,P4], [P4,P5], [P5,P5], [P5,P6]}

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Inconsistent relations End of the first algorithm…

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2. Finding a minimal set to remove to make the system consistent SET OF RELATIONS MAKING THE SYSTEM INCONSISTENT Algorithm MINIMUM SET OF RELATIONS MAKING THE SYSTEM INCONSISTENT MAXIXMUM SET OF CONSISTENT RELATIONS

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2. Finding a minimal set to remove to make the system consistent This short algorithm finds a minimal set. We will illustrate this with an other example. But first, we have to introduce the notion of “degree of inconsistency”.

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Degree of inconsistency Here, if we want to add a new point Pnew such that: Pnew P5 Pnew P6 Pnew < P3, Pnew < P4, P2P1P3P4P5P6

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Degree of inconsistency From here, we can see that : Pnew P5 Pnew P5 Pnew P6Pnew P6 Pnew P5Pnew P5 Pnew P6Pnew P6 P2P1P3P4P5P6

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Questions before we move to the Algorithm ? Questions before we move to the Algorithm ?

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Algorithm Var: Var: S = set of all relations making the system inconsistent n S = number of relations in S R = Temporary relation with highest degree of inconsistency in S m R = degree of inconsistency of R I = a set of minimal relation to remove in order to make the system consistent

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Algorithm Initialize I to be empty; Initialize I to be empty; While (n S != 0) { * Find the relation R with highest inconsistent degree m; * n S = n S – m R ; * Remove 1 from every inconsistent degree from every relation conflicting with R; * Record R in I; * Remove every inconsistent relation conflicting with R from S; * Remove R from S; }

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Example Let us consider the last example. Here are the variables of the algorithm. Let us consider the last example. Here are the variables of the algorithm. S= {P1,P2, P3, P4, P5, P6} 2 Pinc P5 4 2 Pinc P5 4 2Pinc P6 4 2 Pinc P6 4 2Pinc P5 4 2 Pinc P5 4 2Pinc P6 4 2 Pinc P6 4 2 Pinc P5 4 2 Pinc P5 4 2 Pinc P6 4 2 Pinc P6 4 n S = 8 R = {} m R = 0 I = {} S S n S relations Degree of inconsistency

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Example One relation with highest inconsistency degree is : One relation with highest inconsistency degree is : Pinc > P5. S= {P1,P2, P3, P4, P5, P6} 2 Pinc P5 4 2 Pinc P5 4 2Pinc P6 4 2 Pinc P6 4 2Pinc P5 4 2 Pinc P5 4 2Pinc P6 4 2 Pinc P6 4 2 Pinc P5 4 2 Pinc P5 4 2 Pinc P6 4 2 Pinc P6 4 n S = 8 R = {} m R = 0 I = {} S 8 – 4 = 4 {Pinc>P5} 4 {Pinc>P5} 1 1 1 1 Find the relation R with highest inconsistency degreen S = n S – m R Remove 1 from every inconsistent degree from every relation conflicting with R Record R in IRemove every inconsistent relation conflicting with R from SRemove R from S 1 1 1 1

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Example The variables are now : The variables are now : S= {P1,P2, P3, P4, P6} 1Pinc P6 4 1 Pinc P6 4 n S = 4 R = {Pinc>P5} m R = 4 I = {Pinc>P5} S The relation with highest inconsistency degree is now Pinc>P6. 4 {Pinc>P6} 4 – 4 = 0 {Pinc>P5, Pinc>P6} n = 0, the system is now consistent, and a minimal set to remove is found : I = {Pinc>P5, Pinc>P6} is found : I = {Pinc>P5, Pinc>P6} Find the relation R with highest inconsistency degree n S = n S – m R Remove 1 from every inconsistent degree from every relation conflicting with R Record R in IRemove every inconsistent relation conflicting with R from SRemove R from S 0 0 0 0

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Linking the algorithms

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Consistent ? Store Input Yes No Find Set of inconsistent relations Find minimal set to remove to make the system consistent Set constraints Choose input S I Perform algorithm

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”Constraint programming represents one of the closest approaches computer science has yet made to the Holy Grail of programming: the user states the problem, the computer solves it.” ”Constraint programming represents one of the closest approaches computer science has yet made to the Holy Grail of programming: the user states the problem, the computer solves it.” Eugene C. Freuder, CONSTRAINTS, April 1997 CONSTRAINTS

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Questions, comments ? Questions, comments ?

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The Program… The Program…

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