October 2004CSA4050: Semantics III1 CSA4050: Advanced Topics in NLP Semantics III Quantified Sentences

October 2004CSA4050: Semantics III2 Outline Language Sentences Determiners Noun Phrases Syntactic Structure Logic Generalised Quantifiers Higher order functions Translation into Prolog Syntax-Semantics Interface

October 2004CSA4050: Semantics III3 Determiners and Quantifiers in Language and Logic A dog barked  x dog(x) & bark(x) Every dog barked  x dog(x)  bark(x) Fido chased a cat  x cat(x) & chase(fido,x) Every dog chased a cat  x dog(x)  (  y cat(x) & chase(x,y)))

October 2004CSA4050: Semantics III4 Syntactic Shape vs. Semantic Shape John walks semantics: walk(suzie). Every man talks semantics: all(X, man(X)  talk(X)) S NP VP Suzie walks S NP VP Det N talks Every man

October 2004CSA4050: Semantics III5 Problem Similar syntactic shape Dissimilar semantic shape How is this possible if the syntax drives the combination of semantic fragments as per rule-to-rule hypothesis? Answer: be creative about logical forms and semantic combination rules

October 2004CSA4050: Semantics III6 Montague Solution Reorganising the semantic combination rules operating between VP and NP in rules such as s(S) --> np(NP), vp(VP). We will be considering [NP]([VP]) versus [VP]([NP]). NPs as higher order functions Analyse LF of quantified sentences

October 2004CSA4050: Semantics III7 LF of Quantified Sentences LF of quantified sentences has a general shape involving –a restrictor predicate R –a scope predicate S R restricts the set of things we are talking about S says something further about set element(s) –a logical quantifier Q –a bound variable V –a logical operator O connecting R and S

October 2004CSA4050: Semantics III8 Anatomy of Quantified Sentences LogicQVROS  x m(x)  w(x)  xm(x)  w(x)  x d(x) & b(x)  xd(x)&b(x)  x d(x)  (h(x) & b(x))  xd(x)  h(x) & b(x)

October 2004CSA4050: Semantics III9 Generalized Quantifiers We adopt the following generalized quantifier representation for LF in which quantifier is a 3- place predicate: Q(,, ) Operator is omitted. Examples all(X,man(X),walk(X)) exist(X,man(X),walk(X)) the(X,man(X),climbed(X,everest)) most(X,lecturer(X),poor(X))

October 2004CSA4050: Semantics III10 NP as higher order function NP Q^all(X,man(X),Q) every man VP Y^walk(Y) walks S all(X,man(X),walk(X))

October 2004CSA4050: Semantics III11 Encoding in Prolog The VP remains as before, ie X^walks(X) The quantified NP every man will be of the form Q^all(X,man(X) => Q) The semantic rule for S now ensures that the NP function is applied to the VP function. s(S)--> np(NP),vp(VP), {reduce(NP,VP,S)}

October 2004CSA4050: Semantics III12 DCG with Quantification Program 1 % grammar s(S) --> np(NP), vp(VP), {reduce(NP,VP,S)} vp(VP) --> v(V). % lexicon v(X^walk(X)) --> [walks]. np(Q^all(X,man(X),Q)) --> [every,man].

October 2004CSA4050: Semantics III13 Result ?- s(X,[every,man,walks],[]). X = all(_G397, man(_G397)=>_G405^walk(_G405)) What is wrong with this? How can we fix it?

October 2004CSA4050: Semantics III14 Result ?- s(X,[every,man,walks],[]). X = all(_G397, man(_G397), _G405^walk(_G405)) What is wrong with this? –The variables _G397 and _G405 are distinct. They should be identical. –The consequent of the implication is a λ expression How can we fix it?

October 2004CSA4050: Semantics III15 Result ?- s(X,[every,man,walks],[]). X = all(_G397, man(_G397), G405^walk(_G405)) What is wrong with this? –The variables _G397 and _G405 are distinct. They should be identical. –The consequent of the implication is a λ expression How can we fix these problems? –We need to force the variables to be same using reduce.

October 2004CSA4050: Semantics III16 DCG with Quantification Program 2 % grammar s(S) --> np(NP), vp(VP), {reduce(VP,NP,S)} vp(VP) --> v(V). % lexicon v(X^walk(X)) --> [walks]. np(Q^all(X,man(X) => P)) --> [every,man], {reduce(Q,X,P)}.

October 2004CSA4050: Semantics III17 Result ?- s(X,[every,man,walks],[]). X = all(_G397, man(_G397),walk(_G397)) The effect of the reduce clause is –to identify the appropriate variables –to remove the λ variable

October 2004CSA4050: Semantics III18 Handling Quantified NPs Before we cheated by having every man as a lexical item. np(Q^all(X,man(X) => P)) --> [every,man], { reduce(Q,X,P)}. Now we see what is involved in analysing the NP from its parts. Step 1 is to write a new syntactic rule np(NP) --> d(D), n(N). How does the semantics work?

October 2004CSA4050: Semantics III19 LF of determiners Key idea is determiner has LF of a 2 argument function corresponding to R and S which become bound during processing. λR.λS.Q(V,R,S) where Q is associated with the particular determiner When we apply this function to the adjacent noun, we obtain the LF of the NP.

October 2004CSA4050: Semantics III20 How NP is created D R^S^all(X,R,S) every N Y^man(Y) man NP S^all(X,man(X),S)

October 2004CSA4050: Semantics III21 Fitting the Semantics Together Handle the quantified NP np(NP) --> d(D), n(N), {reduce(D,N,NP)}. Add lexical entry for every d(RL^SL^all(X,R => S)) -->[every], {reduce(RL,X,R), reduce(SL,X,S) }.

October 2004CSA4050: Semantics III22 DCG with Quantification Program 3 % grammar s(S) --> np(NP), vp(VP), {reduce(NP,VP,S)}. np(NP) --> d(D), n(N), {reduce(D,N,NP) }. vp(VP) --> v(VP). % lexicon v(X^walk(X)) --> [walks]. n(X^man(X)) --> [man]. det(RL^SL^all(X,R => S) --> [every], {reduce(RL,X,R), reduce(SL,X,S) }.

October 2004CSA4050: Semantics III23 Trace >: (7) s(_G510, [every, man, walks], []) >: (8) np(_L183, [every, man, walks], _L184) >: (9) d(_L205, [every, man, walks], _L206) <: (9) d((X^R)^ (X^S)^all(X, R, S), [every, man, walks], [man, walks]) >: (9) n(_L207, [man, walks], _L208) <: (9) n(Z^man(Z), [man, walks], [walks]) >: (9) reduce((X^R)^ (X^S)^all(X, R, S), Z^man(Z), _L183) <: (9) reduce((X^man(X))^ (X^S)^all(X, man(X), S), X^man(X), (X^S)^all(X, man(X), S)) <: (8) np((X^S)^all(X, man(X), S), [every, man, walks], [walks]) >: (8) vp(_L185, [walks], _L186) >: (9) v(_L185, [walks], _L186) <: (9) v(Y^walk(Y), [walks], []) <: (8) vp(Y^walk(Y), [walks], []) >: (8) reduce((X^S)^all(X, man(X), S), Y^walk(Y), _G510) <: (8) reduce((X^walk(X))^all(X, man(X), walk(X)), X^walk(X), all(X, man(X), walk(X))) <: (7) s(all(X, man(X), walk(X)), [every, man, walks], [])