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Published byDwayne Watkins Modified about 1 year ago

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Programming with SMs A new paradigm of problem representation with Logic Programming (Answer-Set Programming – ASP) –A problem is represented as (part of) a logic program (intentional database) –An instance of a problem is represented as a set of fact (extensional database) –Solution of the problems are the models of the complete program In Prolog –A problem is represented by a program –Instances are given as queries –Solutions are substitutions

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Finding subsets In Prolog subSet([],_). subSet([E|Ss],[_|S]) :- subSet([E|Ss],S). subSet([E|Ss],[E|S]) :- subSet(Ss,S). ?- subset(X,[1,2,3]). In ASP: –Program: in_sub(X) :- element(X), not out_sub(X). out_sub(X) :- element(X), not in_sub(X). –Facts: element(1). element(2). element(3). –Each stable model represents one subset. Which one do you find more declarative?

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Generation of Stable Models A pair of rules a :- not b b :- not a generates two stable models: one with a and another with b. Rules: a(X) :- elem(X), not b(X). b(X) :- elem(X), not a(X). with elem(X) having N solutions, generates 2 N stable models

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Small subsets From the previous program, eliminate stable models with more than one member –I.e. eliminate all stable models where in_sub(X), in_sub(Y), X ≠ Y Just add rule: foo :- element(X), in_sub(X), in_sub(Y), not eq(X,Y), not foo. %eq(X,X). Since there is no notion of query, it is very important to guarantee that it is possible to ground programs. –All variables appearing in a rule must appear in a predicate that defines the domains, and make it possible to ground it (in the case, the element(X) predicates.

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Restricting Stable Models A rule a :- cond, not a. eliminates all stable models where cond is true. In most ASP solvers, this is simply written as an integrity constraint :- cond. An ASP programs usually has: –A part defining the domain (and specific instance of the problem) –A part generating models –A part eliminating models

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N-Queens Place N queens in a NxN chess board so that none attacks no other. %Generating models hasQueen(X,Y) :- row(X), column(Y), not noQueen(Q,X,Y). noQueen(X,Y) :- row(X), column(Y), not hasQueen(Q,X,Y). %Eliminating models %No 2 queens in the same line or column or diagnonal :- row(X), column(Y), row(XX), hasQueen(X,Y), hasQueen(XX,Y), not eq(X,XX). :- row(X), column(Y), column(YY), hasQueen(X,Y), hasQueen(X,YY), not eq(Y,YY). :- row(X), column(Y), row(XX), column(YY), hasQueen(X,Y), hasQueen(XX,YY), not eq(abs(X-XX), abs(Y-YY)). %All rows must have at least one queen :- row(X), not hasQueen(X). hasQueen(X) :- row(X), column(Y), hasQueen(X,Y)

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The facts (in smodels) Define the domain of predicates and the specific program Possible to write in abbreviated form, and by resorting to constants const size=8. column(1..size). row(1..size). hide. show hasQueen(X,Y). Solutions by: > lparse –c size=4 | smodels 0

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N-Queens version 2 Generate less, such that no two queens appear in the same row or column. %Generating models hasQueen(X,Y) :- row(X), column(Y), not noQueen(Q,X,Y). noQueen(X,Y) :- row(X), column(Y), column(YY), not eq(Y,YY), hasQueen(X,YY). noQueen(X,Y) :- row(X), column(Y), rwo(XX), not eq(X,XX), hasQueen(XX,Y). This already guarantees that all rows have a queen. Elimination of models is only needed for diagonals: %Eliminating models :- row(X), column(Y), row(XX), column(YY), hasQueen(X,Y), hasQueen(XX,YY), not eq(abs(X-XX), abs(Y-YY)).

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Back to subsets in_sub(X) :- element(X), not out_sub(X). out_sub(X) :- element(X), not in_sub(X). Generate subsets with at most 2 :- element(X), element(Y), element(Z), not eq(X,Y), not eq(Y,Z), not eq(X,Z), in_sub(X), in_sub(Y), in_sub(Z). Generate subsets with at least 2 hasTwo :- element(X), element(Y), not eq(X,Y), in_sub(X), in_sub(Y). :- not hasTwo. It could be done for any maximum and minimum Smodels has simplified notation for that: 2 {in_sub(X): element(X) } 2.

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Simplified notation in Smodels Generate models with between N and M elements of P(X) that satisfy Q(X), given R. N {P(X):Q(X)} M :- R Example: %Exactly one hasQueen(X,Y) per model for each row(X) given column(Y) 1 {hasQueen(X,Y):row(X)} 1 :- column(Y). %Same for columns 1 {hasQueen(X,Y):column(Y)} 1 :- row(X). %Elimination in diagonal :- row(X), column(Y), row(XX), column(YY), hasQueen(X,Y), hasQueen(XX,YY), not eq(abs(X-XX), abs(Y-YY)).

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Graph colouring Problem: find all colourings of a map of countries using not more than 3 colours, such that neighbouring countries are not given the same colour. The predicate arc connects two countries. Use ASP rules to generate colourings, and integrity constraints to eliminate unwanted solutions

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Graph colouring arc(minnesota, wisconsin).arc(illinois, iowa). arc(illinois, michigan).arc(illinois, wisconsin). arc(illinois, indiana).arc(indiana, ohio). arc(michigan, indiana).arc(michigan, ohio). arc(michigan, wisconsin).arc(minnesota, iowa). arc(wisconsin, iowa).arc(minnesota, michigan). col(Country,Colour) ?? min wis ill iowind mic ohio

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Graph colouring %generate col(C,red) :- node(C), not col(C,blue), not col(C,green). col(C,blue) :- node(C), not col(C,red), not col(C,green). col(C,green) :- node(C), not col(C,blue), not col(C,red). %eliminate :- colour(C), con(C1,C2), col(C1,C), col(C2,C). %auxiliary con(X,Y) :- arc(X,Y). con(X,Y) :- arc(Y,X). node(N) :- con(N,C).

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min wis ill iowind mic ohio One colouring solution min wis ill iowind mic ohio Answer: 1 Stable Model: col(minnesota,blue) col(wisconsin,green) col(michigan,red) col(indiana,green) col(illinois,blue) col(iowa,red) col(ohio,blue)

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Hamiltonian paths Given a graph, find all Hamiltonian paths arc(a,b). arc(a,d). arc(b,a). arc(b,c). arc(d,b). arc(d,c). ab dc

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Hamiltonian paths % Subsets of arcs in_arc(X,Y) :- arc(X,Y), not out_arc(X,Y). out_arc(X,Y) :- arc(X,Y), not in_arc(X,Y). % Nodes node(N) :- arc(N,_). node(N) :- arc(_,N). % Notion of reachable reachable(X) :- initial(X). reachable(X) :- in_arc(Y,X), reachable(Y).

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Hamiltonian paths % initial is one (and only one) of the nodes initial(N) :- node(N), not non_initial(N). non_initial(N) :- node(N), not initial(N). :- initial(N1), initial(N2), not eq(N1,N2). % In Hamiltonian paths all nodes are reachable :- node(N), not reachable(N). % Paths must be connected subsets of arcs % I.e. an arc from X to Y can only belong to the path if X is reachable :- arc(X,Y), in_arc(X,Y), not reachable(X). % No node can be visited more than once :- node(X), node(Y), node(Z), in_arc(X,Y), in_arc(X,Z), not eq(Y,Z).

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Hamiltonian paths (solutions) ab dc {in_arc(a,d), in_arc(b,c), in_arc(d,b)} {in_arc(a,d), in_arc(b,a), in_arc(d,c)}

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