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UW-Madison CS 731 - Inductive Logic Programming - The Problem Specification

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Inductive Logic Programming: The Problem SpecificationILP Specification (Continued)A Common ApproachA DifficultySubsumption for LiteralsSubsumption for ClausesPowerPoint PresentationLeast Generalization of TermsLeast Generalization of Terms (Continued)Least Generalization of LiteralsLattice of LiteralsLattice of Literals (Continued)Slide 13Slide 14Slide 15Slide 16Least Generalization of ClausesExampleLattice of ClausesLattice of Clauses for the Given Hypothesis LanguageIncorporating Background Knowledge: SaturationSaturation (Continued)Slide 23Alternative Derivation of SaturationSlide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Overview of Some ILP AlgorithmsAlgorithms (Continued)Inductive Logic Programming: The Problem Specification•Given:–Examples: first-order atoms or definite clauses, each labeled positive or negative.–Background knowledge: in the form of a definite clause theory.–Language bias: constraints on the form of interesting new clauses.ILP Specification (Continued)•Find:–A hypothesis h that meets the language constraints and that, when conjoined with B, entails (implies) all of the positive examples but none of the negative examples.•To handle real-world issues such as noise, we often relax the requirements, so that h need only entail significantly more positive examples than negative examples.A Common Approach•Use a greedy covering algorithm.–Repeat while some positive examples remain uncovered (not entailed):•Find a good clause (one that covers as many positive examples as possible but no/few negatives).•Add that clause to the current theory, and remove the positive examples that it covers.•ILP algorithms use this approach but vary in their method for finding a good clause.A Difficulty•Problem: It is undecidable in general whether one definite clause implies another, or whether a definite clause together with a logical theory implies a ground atom.•Approach: Use subsumption rather than implication.Subsumption for Literalsb).p(f(a),notbut a)p(f(a), subsumes X)p(f(X), :Example.L Lsuch that on substituti aexists thereifonly and if L subsumes L Literal2121Subsumption for Clausesc}.{Xon substituti theusingb)p(Y, c)p(a, subsumes X)p(a, W}.ZW,Y W,{X on substituti theusingW),p(W, subsumes Z)p(Y, Y)p(X, :Examplesliterals). its ofset theas viewedis clause a(where C Csuch that on substituti a exists thereifonly and if C clause subsumes C Clause2121Least Generalization of Terms).(return ,)).lgg(),...,(lgg(Return terms.are and and 0 is f ofarity thewhere),( and )( then f, symbolfunctionprimary same thefrombuilt are tand tor in appear not do that variablesto termsof pairs (ordered) frombijection a be φLet ).,lgg(tion generalizaleast :. and termsTwo :211111121121212121,ttφ,su,suf,...,ss,...,uun,...,ssft,...,uuft.ttttttnnnnnnOtherwiseIfOutputInputLeast Generalization of Terms (Continued)•Examples:–lgg(a,a) = a–lgg(X,a) = Y–lgg(f(a,b),g(a)) = Z–lgg(f(a,g(a)),f(b,g(b))) = f(X,g(X))•lgg(t1,t2,t3) = lgg(t1,lgg(t2,t3)) = lgg(lgg(t1,t2),t3): justifies finding the lgg of a set of terms using the pairwise algorithm.Least Generalization of Literals.)lgg(),...,lgg(Return . form thehas and form thehas Otherwise, TOP.return then unnegated)other theand negated is (one signsdifferentor predicatesdifferent have and If).,lgg(tion generalizaLeast :. and Literals :111211212121),su,sup(),...,sp(sL),...,up(uLLLLLLLnnnnOutputInputLattice of Literals•Consider the following partially ordered set.•Each member of the set is an equivalence class of literals, equivalent under variance.•One member of the set is greater than another if and only if one member of the first set subsumes one member of the second (can be shown equivalent to saying: if and only if every member of the first set subsumes every member of the second).Lattice of Literals (Continued)•For simplicity, we now will identify each equivalence class with one (arbitrary) representative literal.•Add elements TOP and BOTTOM to this set, where TOP is greater than every literal, and every literal is greater than BOTTOM.•Every pair of literals has a least upper bound, which is their lgg.Lattice of Literals (Continued)•Every pair of literals has a greatest lower bound, which is their greatest common instance (the result of applying their most general unifier to either literal, or BOTTOM if no most general unifier exists.)•Therefore, this partially ordered set satisfies the definition of a lattice.Least Generalization of Clausesclause. resulting theReturn ).C,lgg(C of literal a as lgg thisadd thenTOPnot is lgg their if ,C from one and Cfrom one literals, ofpair every For set.empty the to)C,lgg(Cin literals ofset theInitialize).C,lgg(Ction generalizaLeast l...lCand l...lC clauses, Two :21212121m.2,2,12n1,1,11 :OutputInputExampleT h e l g g o f t h e f o l l o w i n g t w o c l a u s e s a n di s :p a f a p b b p b f bp f a f a p f a b p a f ap X f a p X Y p Z Z p Z b p U f U( , ( ) ) ( , ) ~ ( , ( ) )( ( ) , ( ) ) ( ( ) , ) ~ ( , ( ) )( , ( ) ) ( , ) ( , ) ( , ) ~ ( , ( ) )     Lattice of Clauses•We can construct a lattice of clauses in a manner analogous to our construction of literals.•Again, the ordering is subsumption; again we group clauses into variants; and again we add TOP and BOTTOM elements.•Again the least upper bound is the lgg, but the greatest lower bound is just the union (clause containing all literals from each).Lattice of Clauses for the Given Hypothesis Language active(X) active(X) :- active(X) :- active(X) :- has-hydrophobic(X,A) has-donor(X,A) has-acceptor(X,A) active(X) :- active(X) :- active(X) :- has-hydrophobic(X,A), has-donor(X,A), has-acceptor(X,A), has-donor(X,B), has-donor(X,B), has-donor(X,B), distance(X,A,B,5.0) distance(X,A,B,4.0) distance(X,A,B,6.0) . . .Incorporating Background Knowledge: Saturation•Recall that we wish to find a hypothesis clause h that together with the background knowledge B will entail the positive examples but not the negative examples.•Consider an arbitrary positive example e. Our hypothesis h together with B should entail e: Bh ⊨ e. We can also write this as h ⊨ B  e.Saturation (Continued)•If e is an atom (atomic formula), and we


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