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CSA4050: Advanced Topics in NLP Semantics IV Partial Execution Proper Noun Adjective

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Partial Execution Partial execution involves the replacing of certain runtime computations with changes to the source of the Prolog program itself. For example, we can replace the rule: s(S) --> np(NP), vp(VP) { reduce(NP,VP,S) }. with s(S) --> np(VP^S), vp(VP). This is because the computation of reduce involves only the mutual binding of several variables.

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How it works s(S) --> np(NP), vp(VP) {reduce(NP,VP,S)}. 1. match reduce(A^F,A,F) 2. rewrite s(F) --> np(A^F), vp(A). 3. rename s(S) --> np(VP^S), vp(VP).

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Exercise 1 What is the result of eliminating the reduce clause by partial execution in the following rule? np(NP) --> d(D), n(N), {reduce(D,N,NP)}.

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Answer np(NP) --> d(D), n(N) { reduce(D,N,NP)}. 1. match reduce(A^F,A,F) 2. rewrite np(F) --> d(A^F), n(A). 3. rename np(NP) --> d(N^NP), n(N).

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DCG with Quantification Program 4 % grammar s(S) --> np(VP^S), vp(VP). np(NP) --> d(N^NP), n(N). vp(VP) --> v(VP). % lexicon v(X^walk(X)) --> [walks]. n(X^man(X)) --> [man]. n(suzie)--> [‘Suzie’]. det(RL^SL^all(X,R,S) --> [every], {reduce(RL,X,R), reduce(SL,X,S) }.

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Handling Proper Nouns The grammar handles every man walks X = all(_G, man(_G), walk(_G)) Will this grammar parse Suzie walks? Let’s try it! ?- s(X,['Suzie',walks],[ ]).

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October 20048 ?- s(X,['Suzie',walks],[ ]). Call: (8) s(_G492, ['Suzie', walks], []) ? Call: (9) np(_L183, ['Suzie', walks], _L184) ? Call: (10) pn(_L183, ['Suzie', walks], _L184) ? Exit: (10) pn(suzie, ['Suzie', walks], [walks]) ? Exit: (9) np(suzie, ['Suzie', walks], [walks]) ? Call: (9) vp(_L185, [walks], _L186) ? Call: (10) iv(_L185, [walks], _L186) ? Exit: (10) iv(_G556^walk(_G556), [walks], []) ? Exit: (9) vp(_G556^walk(_G556), [walks], []) ? Call: (9) reduce(suzie, _G556^walk(_G556), _G492) ? Fail: (9) reduce(suzie, _G556^walk(_G556), _G492)?........ suzie is not a function

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Handling Proper Nouns Problem is with the “type” of LF of Suzie. We require that LF of Suzie has the same type as any other NP - such as every man, i.e. –As a lambda expression it would be λp.p(suzie). –In Prolog we can regard this as a function from [VP] to [S] such that reduce(VP,suzie,S) holds.

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DCG with Quantification Program 4 % grammar s(S) --> np(VP^S), vp(VP). np(NP) --> n(NP). np(NP) --> d(N^NP), n(N). vp(VP) --> v(VP). % lexicon v(X^walk(X)) --> [walks]. n(X^man(X)) --> [man]. n(VP^S) --> [suzie],{reduce(VP,suzie,S)}. d(RL^SL^all(X,R => S)) --> [every], {reduce(RL,X,R), reduce(SL,X,S) }.

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Exercise 2 Using partial execution, eliminate the reduce clause in pn(VP^S) --> [‘Suzie’],{reduce(VP,suzie,S)}.

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s(X, ['Suzie', walks], [ ]) ? Call: (7) s(_G292, ['Suzie', walks], []) ? ↓ Exit: (9) pn((suzie^_G357)^_G357, ['Suzie', walks], [walks]) ? ↓ Exit: (9) iv(_G362^walk(_G362), [walks], []) ? Exit: (8) vp(_G362^walk(_G362), [walks], []) ? : Call: (8) reduce((suzie^_G357)^_G357, _G362^walk(_G362), _G292) ? : Exit: (8) reduce((suzie^walk(suzie))^walk(suzie), suzie^walk(suzie), walk(suzie)) ? Exit: (7) s(walk(suzie), ['Suzie', walks], []) ? creep X = walk(suzie)

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Adjectives In english, adjectives combine with nouns, e.g. big dog. We will call the an adjective/noun combination (without the article) N1. Here is a syntactic rule for a simplified version of N1: N1 → A N N1 NA big dog

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Semantics of N1 Adjectives and common nouns translate to one-place predicates. big = x.big(x) dog = x.dog(x) To get the semantics of the N1 we need to join the predicates together with & (and), i.e. big dog = x.big(x) & dog(x)

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N1 in Prolog %grammar n1(X^(P&Q)) --> a(X^P), n(X^Q). %lexicon n(X^dog(X)) --> [dog]. a(X^big(X)) --> [big]. Behaviour is as follows: ?- n1(X, [big,dog],[]). X = _G343^ (big(_G343)&dog(_G343))

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