English grammar/Determiners

Articles ("a", "the"), demonstrative determiners ("this", "that", "same" etc.), interrogative determiners ("which", "what") and quantifiers ("all", "any" etc.) are represented in UNL as attributes: "the book" > book.@def, "this book" > book.@proximal, "all books" > book.@all, "which book" > book.@wh. Numbers are represented by the relation "qua": "two books" > qua(book,2), "second book" > qua(book,2.@ordinal). Possessive determiners ("my", "your" etc) and possessive pronouns ("mine", "yours" are represented by the relation "pos" and the personal pro-forms of UNL (00.@1, 00.@2 and 00.@3): "my book" > pos(book,00.@1), "book of mine" > pos(book,00.@1).

UNLization

There are basically two types of UNLization rules dealing with determiners: attribute rules and relation rules. Attribute rules normally make use of the attribute "att" and its corresponding features assigned in the dictionary.

the book

INPUT (ENG): the book OUTPUT (UNL): book.@def DICTIONARY:

• [the]{}"" (LEX=D,POS=ART,att=@def)<eng,255,0>;
• [book]{}"book" (LEX=N,POS=NOU,NUM=SNG)<eng,0,0>;
• [ ]{}""(BLK)<eng,0,0>;

T-GRAMMAR:

1. (D,att,%x)(N,%y):=(%y,+att=%x);
2. (BLK):=;

D-GRAMMAR: not necessary
TRACE:
INPUT: the book

• STATE#1: [the][ ][book] (tokenization)
• STATE#2: [the][book] (blank space is deleted)
• STATE#3: [book.@def] ("the" is deleted and its attribute is copied to "book"]
• OUTPUT: book.@def

Note that Rule#1 will delete the determiner because its node (%x) is not reproduced in the right side of the rule. In list rules, according to the UNL Grammar Specs, nodes that do not appear in the right side are deleted. The feature "+att=%x" creates an attribute "att" and assigns the value of the same attribute in %x in to it. Note that [book] has no "att" in the dictionary, but [the] does. The feature "+att=%x" creates an attribute "att" for [book] and assigns to it the value of the same attribute in [the]. This kind of rule will only work if %x has "att"; if not, the attribute "att" will be created, but without any value.
In the English grammar, the rule is actually:

1. (D,att,%x)(NB,%y):=(%y,+att=%x);

Note that %y has the feature NB instead of N. The reason for this is that, in English, there can be other words between the determiner and the noun: "a beautiful book", for instance. And it's important to assign the value of the determiner to the head of the noun phrase (which is "book", and not "beautiful"). In that case, we first process the noun phrase, generating all its intermediate projections (NB's). Only then we resolve the determiners.
The single rule (D,att,%x)(NB,%y):=(%y,+att=%x); will apply for all the determiners having the feature "att", i.e., [a], [an], [this], [that], [same], [other] etc.

two books

Numbers are not represented as attributes, but as relations. In this case, we cannot avoid some syntactic processing. We present below the trace and the details. INPUT: two books DICTIONARY:

The rule for dealing with cardinals is the following:

(DIGIT,^ORD,%y)(NB,%x):=(XB(%x;%y,+spec,+qua),+LEX=N,+XB=NB,%z); two books > XB(books,two)

This rules states that, if there is a digit, which is not ordinal, before an intermediate projection of the noun phrase (NB), then we should create another intermediate projection XB between the NB and the digit, i.e.: two books will be analyzed as:

    XB
   /  \
  NB   two
  |
books

Note that we are not only converting a list: [two][books] into a relation: XB(books;two). We are actually creating a hyper-node which contains this relation. Compare:

• (%x)(%y):=XB(%y;%x); transforms a list into a relation
• (%x)(%y):=(XB(%y;%x),%z); replaces two nodes (%x and %y) by a third node (%z) and creates an XB relation between %x and %y inside this node.

Why is this necessary? There are at least two reasons for operating this way:

1. If we simply replace the nodes %x and %y by a relation, we will loose the relations that they could have with other nodes in the sentence. For instance: in case of [two][books][about][Paris], if we replace [two][books] by XB(books;two) we will loose the relation between [books] and [Paris], because nodes inserted in graphs are removed from the list structure, i.e., we will have two isolated data structures: the graph, which contains only XB(books;two), and the list with [about][Paris]. In order to preserve this relation, we have to replace [two][books] by a new node [XB(books;two)] so that we could have [XB(books;two)][about][Paris]. We would loose the relation between [book] and [about], but there will be still a linear relation between the XB(books;two) and [about].
2. It is important to deal with general categories, such as XB (any intermediate projection), instead of specific categories, such as NB (intermediate projection of a noun phrase). In the dearborization phase of the grammar, we will transform this tree-like structures into head-driven structures. If we have a general relation such as XB, the number of rules is considerably smaller:

• XB(XB(%x;%y);%z):=XB(%x;%y)XB(%x;%z);
• XP(XB(%x;%y);%z):=XB(%x;%y)XS(%x;%z);
• XP(%x;%y):=XS(%x;%y);

Note that these rules will transform hyper-relations into simple relations. If we use specific categories (such as NB,VB,PB etc) instead of the general category XB, we will have to repeat this for all different types of relations. However, we do need to preserve the information that this intermediate projection (XB) is a projection of a noun. This is done by assigning the features +LEX=N,+XB=NB to the hyper-node so that, later on, we will be able to convert XB into NB.

d

Note that this is done inside a node

Ordinals (ORD)

(DIGIT,ORD,%y)(NB,%x):=(XB(%x;%y,+spec,+mod),+LEX=N,+XB=NB,%z); first book > XB(book,first)

Quantifiers (DIGIT)

Posessive determiners (POD)

(POD,%x)(NB,%y):=(XP(%y;%x,+spec,+pos),+LEX=N,+XP=NP,%z); my book > XP(book,I)

Possessive pronouns (SPR)

(N,%x)([of])(SPR,%y):=(XB(%x;%y,+spec,+pos),+LEX=N,+XB=NB,%z); book of mine > XP(book,I)

Genitive (GNT)

(N,%x)(PTC,GNT)(N,%y):=(XB(%y;%x,+spec,+pos),+LEX=N,+XB=NB,%z); John's book > XB(book,John)

Indefinite pronouns (NPR)

(NPR,%x)([of])(NP,%y):=(XB(%y;%x,+spec,+qua),+LEX=N,+XB=NB,%z); none of them > XB(them, none)

Interrogative determiners

([whose],%x)(N,%y):=(XB(%y;%x,+spec,+pos),+LEX=N,+XB=NB,%z);