october 2004csa4050: semantics iii1 csa4050: advanced topics in nlp semantics iii quantified...

October 2004 CSA4050: Semantics III 1

CSA4050:Advanced Topics in NLP

Semantics III

• Quantified Sentences


Outline

Language• Sentences• Determiners• Noun Phrases• Syntactic Structure

Logic• Generalised

Quantifiers• Higher order

functions• Translation into

Prolog

Syntax-Semantics Interface


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)))


Syntactic Shape vs.Semantic Shape

• John walks

semantics:walk(suzie).

• Every man talkssemantics:all(X, man(X) talk(X))

S

NP VPSuzie walks

S

NP VP

Det N talks Every man


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


Montague Solution

• Reorganising the semantic combination rules operating between VP and NP in rules such ass(S) --> np(NP), vp(VP).

• We will be considering [NP]([VP]) versus [VP]([NP]).

• NPs as higher order functions

• Analyse LF of quantified sentences


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


Examples

All lecturers are lazy

x lecturer(x) lazy(x)

• Restrictor = lecturers• Scope = lazy• Quantifier = All• Operator = implies• Bound Variable = x


Examples

There is a lazy lecturer

x lecturer(x) & lazy(x)

• Restrictor = lecturers• Scope = lazy• Quantifier = exist• Operator = and• Bound Variable = x


Anatomy of Quantified Sentences

Logic Q V R O S

x m(x) w(x) x m(x) w(x)

x d(x) & b(x) x d(x) & b(x)

x d(x) (h(x) & b(x))

x d(x) h(x) & b(x)


Generalized Quantifiers

• We adopt the following generalized quantifier representation for LF in which quantifier is a 3-place predicate:Q(<variable>,<restrictor>,<scope>)

• 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))


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))


Encoding in Prolog

• The VP remains as before, ieX^walks(X)

• The quantified NP every man will be of the formQ^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)}


DCG with QuantificationProgram 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].


Result?- s(X,[every,man,walks],[]).

X = all(_G397, man(_G397), _G405^walk(_G405))

all(x, man(x)=> y^walk(y))

• What is wrong with this?

• How can we fix it?


Result

?- s(X,[every,man,walks],[]).X = all(_G397, man(_G397), _G405^walk(_G405))

all(x, man(x)=> y^walk(y))• 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?

– We need to force the variables to be identical using reduce



% grammars(S) --> np(NP), vp(VP), {reduce(VP,NP,S)}vp(VP) --> v(V).

% lexiconv(X^walk(X)) --> [walks].np(Q^all(X,man(X) => P)) --> [every,man], {reduce(Q,X,P)}.


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


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 rulenp(NP) --> d(D), n(N).

• How does the semantics work?


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.


How NP is created

D

R^S^all(X,R,S)

every

N

Y^man(Y)

man

NP

S^all(X,man(X),S)


Fitting the Semantics Together

• Handle the quantified NPnp(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) }.



% grammars(S) --> np(NP), vp(VP), {reduce(NP,VP,S)}.np(NP) --> d(D), n(N), {reduce(D,N,NP) }.vp(VP) --> v(VP).

% lexiconv(X^walk(X)) --> [walks].n(X^man(X)) --> [man].d(RL^SL^all(X,R => S) --> [every], {reduce(RL,X,R), reduce(SL,X,S) }.


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)âll(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)âll(X, R, S), Z^man(Z), _L183)<: (9) reduce((X^man(X))^ (X^S)âll(X, man(X), S), X^man(X), (X^S)âll(X, man(X), S))<: (8) np((X^S)âll(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)âll(X, man(X), S), Y^walk(Y), _G510)<: (8) reduce((X^walk(X))âll(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], [])

october 2004csa4050: semantics iii1 csa4050: advanced topics in nlp semantics iii quantified...

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