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

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

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.

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).

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)}.

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).

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) }.

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],[ ]).

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

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.

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) }.

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

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)

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

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)

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))