Fully Lexicalized Pregroup Grammars

Transcription

1 Fully Lexicalized Pregroup Grammars Annie Foret joint work with Denis Béchet and LINA Nantes University, FRANCE IRISA University Rennes1, FRANCE Wollic 2007 p.1

2 PLAN Introduction Transformation : hopes Background : PG First facts A Preliminary Fact Trials in the restricted case Encodings as morphisms Construction what is the full construction : illustration how does it work? key properties (soundness is easy, completeness is not) Conclusion and some related questions Wollic 2007 p.2

3 Introduction The categories are types of a free pregroup generated by a set of primitive types (P r) with a partial order on P r. This partial order is not a lexicalized information, not independent of the language that corresponds to a PG. q s for a yes-or-no question Question : Is it possible to find a PG that is equivalent to a given PG but where the partial order on primitive types is universal? Such that : (hopes) the computed PG is not too big polynomial transformation it works in a similar way as the source PG a homomorphism,... the same number of types to a word k-valued. Wollic 2007 p.3

4 Transformation Wollic 2007 p.4 language equivalence = lexicon not too big k-valued + length similar parse

5 Pregroup : definitions A pregroup is a structure (P,,, l, r, 1) s. t. (P,,, 1) is a partially ordered monoid a in which l, r are unary operations on P that satisfy: (P RE) a l.a 1 a.a l and a.a r 1 a r.a or equivalently: a.b c a c.b l b a r.c Some equations follow from the def. a rl = 1 = a lr we also have: (a.b) r = b r.a r, (a.b) l = b l.a l, 1 r = 1 = 1 l but not, in general: a rr a a ll iterated adjoints:... a ( 2) =a ll, a ( 1) =a l, a (0) =a, a (1) =a r, a (2) =a rr... a A monoid is a structure < M,, 1 >, such that is associative and has a neutral element 1 A partially ordered monoid is a monoid (M,, 1) with a partial order that satisfies a, b, c: a b c a c b and a c b c. Wollic 2007 p.5

6 Background Pregroup def [Free pregroup] Let (P, ) be an ordered set of primitive types, T (P, ) = {p (i 1) 1 p (i n) n 0 k n, p k P and i k } is the set of types. the empty sequence is denoted by 1. For X and Y T (P, ), X Y iff this relation is deductible in the following system where p, q P, n, k and X, Y, Z T (P, ) : X X (Id) X Y Y Z (Cut) X Z XY Z Xp (n) p (n+1) Y Z (A L ) X Y Z X Y p (n+1) p (n) Z (A R ) Xp (k) Y Z (INDL ) Xq (k) Y Z X Y q (k) Z (INDR ) X Y p (k) Z q p if k is even, and p q if k is odd Wollic 2007 p.6

7 Background Let (P, ) be a finite partially ordered set. Pregroup grammar def A pregroup grammar based on (P, ) is a lexicalized a grammar G = (Σ, I, s) such that s T (P, ) ; G assigns a type X to a string v 1,..., v n of Σ iff for 1 i n, X i I(v i ) such that X 1 X n X in the free pregroup T (P, ). The language L(G) is the set of strings in Σ that are assigned s by G. a a lexicalized grammar is a triple (Σ, I, s): Σ is a finite alphabet, I assigns a finite set of categories (or types) to each c Σ, s is a category (or type) associated to correct sentences. Wollic 2007 p.7

8 Background Pregroup example Our example is taken from Lambek, with the primitive types: π 2 = second person, p 2 = past participle, o = object, q = yes-or-no question, q = question q q This sentence gets type q (q s): whom have you seen q o ll q l qp l 2 πl 2 π 2 p 2 o l Wollic 2007 p.8

9 Background Pregroup example Our example is taken from Lambek, with the primitive types: π 2 = second person, p 2 = past participle, o = object, q = yes-or-no question, q = question q q This sentence gets type q (q s): whom have you seen q o ll q l qp l 2 πl 2 π 2 p 2 o l Wollic 2007 p.8

10 Background Pregroup example Our example is taken from Lambek, with the primitive types: π 2 = second person, p 2 = past participle, o = object, q = yes-or-no question, q = question q q This sentence gets type q (q s): whom have you seen q o ll q l qp l 2 πl 2 π 2 p 2 o l Wollic 2007 p.8

11 Background Pregroup example Our example is taken from Lambek, with the primitive types: π 2 = second person, p 2 = past participle, o = object, q = yes-or-no question, q = question q q This sentence gets type q (q s): whom have you seen q o ll q l qp l 2 πl 2 π 2 p 2 o l Wollic 2007 p.8

12 PLAN Introduction Transformation : hopes Background : PG First facts A Preliminary Fact Trials in the restricted case Encodings as morphisms Construction what is the full construction : illustration how does it work? key properties (soundness is easy, completeness is not) Conclusion and some related questions Wollic 2007 p.9

13 Proposition. A Preliminary Fact from Buszkowski 01 [PG equivalence and generative capacity] For any PG G on (P, ), we can construct a PG G based on (P, =), s.t. G and G have the same language. Wollic 2007 p.10

14 Proposition. A Preliminary Fact from Buszkowski 01 [PG equivalence and generative capacity] For any PG G on (P, ), we can construct a PG G based on (P, =), s.t. G and G have the same language. Obvious : Pregroups Lang. = Context-free Lang. = order 1 AB-Categorial Or duplicate the lexical types for each occurrence involved in some an ordering p i p j However this does not preserve the size of the lexicon in general order(a/b)=order(b\a)=max(order(a, 1+order(B))), 0 on P r if X = X p 2n 0 Y {X p 2n j Y p j p 0 } ; if X = X p0 2n+1 Y {X pj 2n+1 Y p j p 0 } Wollic 2007 p.10

15 Trials in a restricted case Let (P, ) be reduced to p 0 p 1, with as =, (and q = p 1 ). Consider the homomorphism from (P, ) to (P, ) h γ : h γ (X (n) ) = h γ (X) (n) h γ (p 0 ) = q γ γ (1) h γ (X.Y ) = h γ (X).h γ (Y ) h γ (p 1 ) = q h γ (1) = 1 h γ (p i ) = p i if i {0, 1} h(p 0 ) h(p 1 ), since γ γ (1) 1 h(p 1 ) h(p 0 ), since γ γ (1) 1 but... it is not "order-reflecting" : X = p (2n 2) 0 p (2n 1) 0 p (2n) 1 p (2n 1) 1 such that X 1, whereas h γ (X) 1 Wollic 2007 p.11

16 Trials in a restricted case Let (P, ) be reduced to p 0 p 1, with as =, (and q = p 1 ). Consider the homomorphism from (P, ) to (P, ) h γ : h γ (X (n) ) = h γ (X) (n) h γ (p 0 ) = q γ γ (1) h γ (X.Y ) = h γ (X).h γ (Y ) h γ (p 1 ) = q h γ (1) = 1 h γ (p i ) = p i if i {0, 1} but... it is not "order-reflecting" : X = p (2n 2) 0 p (2n 1) 0 p (2n) 1 p (2n 1) 1 such that X 1, whereas h γ (X) 1 : h γ (X) = p 1 γγ (1) γ (2) γ (1) p (1) 1 p(2) 1 p(1) 1 Wollic 2007 p.11

17 Towards the general case various partial order schemas Difficulties p i1 p i2 p i3 p i4... p j1 p j2 p j3 p j4... p k4 p k5 p k1 pk2 pk6... p k3 p k7 from group images [h(x)] = [h(y )] implies [X] = [Y ] thus ensure different equivalence classes (q Pr ) for different connex components [X] = [Y ] implies [h(x)] = [h(y )] thus ensure same equivalence class (q) inside a connex component involving s, upper and/or lower Wollic 2007 p.12

18 Morphism-based encodings definitions A mapping h from the free pregroup on (P, ) to the free pregroup on (P, ), is a pregroup homomorphism iff 1. X T (P, ) : h(x (n) ) = h(x) (n) 2. X, Y T (P, ) : h(x.y ) = h(x).h(y ) 3. h(1) = 1 4. X, Y T (P, ) : if X Y then h(x) h(y ) [Monotonicity] A mapping from a poset (P, ) to a poset (P, ) is said partial-order-preserving iff 4b. p i, p j P : if p i p j then h(p i ) h(p j ). to ensure, for (4) Wollic 2007 p.13

19 A similar parse Morphism-based encodings An example revisited π 2 = second person, p 2 = past participle, o = object, q = yes-or-no question, q q q = question This sentence gets type q (q s): whom have you seen q o ll q l qp l 2 πl 2 π 2 p 2 o l h(q )h(o) ll h(q) l h(q)h(p 2 ) l h(π 2 ) l h(π 2 ) h(p 2 )h(o) l Wollic 2007 p.14

20 Simulation :a converse of monotonicity Def-Prop. [Order-reflecting homomorphism] Every homomorphism h from the free pregroup on (P, ) to the free pregroup on (P, ) is such that (1) and (2) are equivalent and define order-reflecting homomorphisms: (1). X, Y T (P, ) if h(x) h(y ) then X Y (2). X T (P, ) if h(x) 1 then X 1 to ensure (2), rather than (1) Wollic 2007 p.15

21 Simulation :a converse of monotonicity Def-Prop. [Order-reflecting homomorphism] Every homomorphism h from the free pregroup on (P, ) to the free pregroup on (P, ) is such that (1) and (2) are equivalent and define order-reflecting homomorphisms: (1). X, Y T (P, ) if h(x) h(y ) then X Y (2). X T (P, ) if h(x) 1 then X 1 Proof. (1) is a corollary of (2) as follows: suppose (2) holds for a pregroup-morphism h with (3) h(x) h(y ), we get (4) h(x Y r ) = h(x) h(y ) r h(y ) h(y ) r 1 property (2) then gives (5) X Y r 1, hence X Y (by adding Y on the right of (5) ). to ensure (2), rather than (1) Wollic 2007 p.15

22 PLAN Introduction Transformation : hopes Background : PG First facts A Preliminary Fact Trials in the restricted case Encodings as morphisms Construction what is the full construction : illustration how does it work? key properties (soundness is easy, completeness is not) Conclusion and some related questions Wollic 2007 p.16

23 Construction on one component. Let P = P r P r, P r a connex component q and β k, γ k, are new letters (no postulate) for each p k of P r a We take as poset P = P r {q} {β k, γ k p k P r}, is the restriction of on P r ( is identity if P r is empty). Definition. [Simulation-morphism h for P r] h(x (n) ) = h(x) (n) h(p i ) = λ (0) i q (0) δ (0) i }{{} h(1) = 1 h(x.y ) = h(x).h(y ) for p i P r h(p i ) = p i if p i P r a the symbol q can also be written q P r if necessary w.r.t. P r Wollic 2007 p.17

24 Construction on one component. Let P = P r P r, P r a connex component q and β k, γ k, are new letters (no postulate) for each p k of P r Definition. [Simulation-morphism h for P r] h(x (n) ) = h(x) (n) h(p i ) = λ (0) i q (0) δ (0) i }{{} h(1) = 1 h(x.y ) = h(x).h(y ) for p i P r h(p i ) = p i if p i P r where λ i = α ( (γ (1) for k =n..1 if p k p i k γ k )) and δ i = ( (β k β (1) k for k=1..n, if p k p i )) α the symbol q can also be written q P r if necessary w.r.t. P r Wollic 2007 p.17

25 Restricted case : p 0 p 1... h(p i ) = λ (0) i q (0) δ (0) i }{{} δ (1) j q (1) λ (1) j }{{} = h(p j) (1) (note the inversion) for p 0 p 1 where (new symbols): λ 0 = α (γ (1) 0 γ 0 ) δ 0 = (β 0 β (1) 0 ) (β 1 β (1) 1 )α λ 0 λ 1 λ 1 = α (γ (1) 1 γ 1 ) (γ (1) 0 γ 0 ) δ 1 = (β 0 β (1) 0 ) α δ 0 δ 1 we have : λ 0 λ 1, because (β 1 β (1) 1 ) 1 therefore h(p 0 ).h(p 1 ) (1) 1 with h(p 1 )h(p 0 ) (1) 1 δ 0 δ 1, because 1 (γ (1) 1 γ 1 ) a subterm per lower atom ( 1) in λ, per upper ( 1) atom in δ Wollic 2007 p.18

26 Restricted case : p 0 p 1... h(p i ) = λ (0) i q (0) δ (0) i }{{} δ (1) j q (1) λ (1) j }{{} = h(p j) (1) (note the inversion) for p 0 p 1 where (new symbols): λ 0 = α (γ (1) 0 γ 0 ) δ 0 = (β 0 β (1) 0 ) (β 1 β (1) 1 )α λ 0 λ 1 λ 1 = α (γ (1) 1 γ 1 ) (γ (1) 0 γ 0 ) δ 1 = (β 0 β (1) 0 ) α δ 0 δ 1 we have : λ 0 λ 1, because (β 1 β (1) 1 ) 1 therefore h(p 0 ).h(p 1 ) (1) 1 = δ 0 δ 1, because 1 (γ (1) 1 γ 1 ) α (γ (1) 1 γ 1 )q (0) (β 1 β (1) 1 ) (β 0 β (1) 0 )α.α (1) (β (1+1) 0 β (1) 0 )q(1) (γ (1) 1 γ (1+1) 1 ) (γ (1) 0 γ (1+1) 0 ) α (1) 1 with h(p 1 )h(p 0 ) (1) 1 a subterm per lower atom ( 1) in λ, per upper ( 1) atom in δ Wollic 2007 p.18

27 Restricted case : p 0 p 1... h(p i ) = λ (0) i q (0) δ (0) i }{{} for p 0 p 1 p 2 p 1 where (new symbols): λ 0 = α (γ (1) 2 γ 2 ) δ 0 = (β 0 β (1) 0 ) (β 1 β (1) 1 )α λ 2 = α (γ (1) 2 γ 2 ) δ 2 = (β 1 β (1) 1 ) (β 2 β (1) 2 )α λ 1 = α (γ (1) 2 γ 2 ) (γ (1) 1 γ 1 ) (γ (1) 0 γ 0 ) δ 1 = (β 1 β (1) 1 ) α a subterm per lower atom ( 1) in λ, per upper ( 1) atom in δ if "reflexive subterm" were dropped : h(p 0 ) = h(p 2 )! Wollic 2007 p.18

28 The Unrestricted Case on P = P r }{{} P r. Proposition. [h is order-preserving] (easy by construction) if p i p j then h(p i ) h(p j ). It is then a pregroup homomorphism. Proposition. [Monotonicity of h] (corollary) X, Y T (P, ) : if X Y then h(x) h(y ) Proposition. [h is order-reflecting] (hard part) X, Y T (P, ) : if h(x) h(y ) then X Y Proposition. [Incrementality] The construction holds for a connex component P r or several connex components, possibly whole P Wollic 2007 p.19

29 How h(x) 1 X 1 Reasoning with Left Derivations h(x) =... q u 1... q u 2... q u 3... links...? Derivation D from 1 1 ending in h(x) 1 Let k denote the step of the leftmost introduction of q, as q (n) q (n+1) : 0 = 1. k 1 = Γ 0 Γ 0 with k 1 1 k = Γ 0 q (n) q (n+1) Γ 0 with k 1. }{{} m = Γ m k q (n) Γm k q (n+1) Γ m k with h(x) = m 1 }{{}}{{}}{{} =h(x) Γ 0 1 Γ 0 by derivation structure Wollic 2007 p.20

30 How h(x) 1 X 1 Technical Lemmas On a kind of parenthesizing in the encodings Lemma. let X = X 1 α (2u1) Y 1 α (2u 1+1) }{{}... X n α (2un) Y n α (2u n+1) }{{} X n+1 where all X k, Y k have no α, and α is not related by ; (i) if X 1 then : (1) k : Y k 1 (2) X 1 X 2... X k... X n X n+1 1 X 1 α (2u 1) Y 1 α (2u 1+1)... X n α (2u n) Y n α (2u n+1) X n+1 variant Wollic 2007 p.21

31 How h(x) 1 X 1 Technical Lemmas On a kind of parenthesizing in the encodings Lemma. let X = X 1 α (2u1) Y 1 α (2u 1+1) }{{}... X n α (2un) Y n α (2u n+1) }{{} X n+1 where all X k, Y k have no α, and α is not related by ; (i) if X 1 then : (1) k : Y k 1 (2) X 1 X 2... X k... X n X n+1 1 (ii) if Y 0 α (2v+1) X α (2v+2) 1 where Y 0 has no α, then : (1) k : Y k 1 (2) X 1 X 2... X k... X n X n+1 1 variant Wollic 2007 p.21

32 Central Lemma Lemma. Let I P r denote the set of indices of elements in P r. Let Γ = h(x 1 ) Y 1 }{{}... h(x k) Y k }{{}... h(x m) Y m }{{} h(x m+1) (m 0) where all Y k have the form: λ 2u i λ 2u+1 j or δ 2u 1 i δj 2u with i, j I P r where some X k may be empty (then considered as 1, with h(1) = 1) (1) If Γ 1 then k 1 m : ( k k 1 : h(x k ) has no q) ( k k 1 : Y k 1) (2) If Γ 1 then X 1 X 2... X m X m+1 1 (3) If δ (2k) i Γδ (2k+1) j 1, or λ (2k+1) i Γλ (2k+2) j 1 then k 1 m : ( k k 1 : h(x k ) has no q) ( k k 1 : Y k 1) (4) If δ (2k) i Γδ (2k+1) j 1, or λ (2k+1) i Γλ (2k+2) j 1 then X 1 X 2... X m X m+1 1 The proof is technical : a key-point is that some derivations are impossible ( leftmost introductions + parenthesizing lemma). Some complications in the formulation (1) and (3) simplify the proof discussion we get later p i p j in the first form, p j p i in the second form Wollic 2007 p.22

33 Main results Proposition. [Equivalence property] X, Y T (P, ) : h(x) h(y ) iff X Y Proposition. [Pregroup Grammar Simulation] Given a pregroup grammar G = (Σ, I, s) on (P, ) we define h from a PG on (P, ) to a PG on (P, ) ; we construct a grammar G = (Σ, h(i), h(s)), on (P, ) where h(i) is the assignment of h(x i ) to a i for X i I(a i ), as a result we have : L(G) = L(G ) This proposition applies the transformation to the whole set of primitive types, thus providing a fully lexicalized grammar G (no order postulate). A similar result holds to a fragment P r of P = P r P r Wollic 2007 p.23

34 Conclusion Pregroups were introduced as a simplification of Lambek calculus. The order on primitive types has been introduced in PG to simplify the calculus for simple types. The consequence is that PG is not fully lexicalized. We have proven that this restriction is not so important because a PG using an order on primitive types can be transformed into a PG based on a simple free pregroup using a pregroup morphism, s.t. : its size is bound by the size of the initial PG times the number of primitive types (times a constant which is approximatively 4), the size of the resulting PG is often much less that this bound; moreover, this transformation does not change the number of types that are assigned to a word (a k-valued PG is transformed into a k-valued PG). Wollic 2007 p.24

35 Transformation language equivalence = similar parse lexicon not too big k-valued + length Related questions... add a preorder to other calculus, similar property?... hierarchies with/without order postulates?... learning with/without order postulates? Wollic 2007 p.25

36 Some other slides for some details and links. Wollic 2007 p.26

37 A schema L AB (G) L NL (G) L L (G) L NL (G) L L (G) L P G,= (G) L P G, (G 0 ) Wollic 2007 p.27

38 Detailed writing and properties of the auxiliary types The Central Lemma uses the following facts on auxiliary types : if p i p j then λ 2u i λ 2u+1 j 1 if p i p j then λ 2u i λ 2u+1 j 1 if p j p i then δ 2u 1 i δj 2u 1 if p j p i then δ 2u 1 i δj 2u 1 This can be summarized as : the only Y k, s.t. Y k 1 of central Lemma are λ 2u i λ 2u+1 j for p i p j and δ 2u 1 i δj 2u for p j p i Wollic 2007 p.28

39 Detailed writing and properties of the auxiliary types The only Y k 1 of Central Lemma are λ 2u i λ 2u+1 j for p i p j and δ 2u 1 i for p j p i. This can be checked using the following detailed writing λ 2u i λ 2u+1 j (i = j): α (2u) ( for k =n..1 if p k p i (γ (2u+1) k γ (2u) k )) ( for k=1..n if p k p i (γ (2u+1) k γ (2u+1+1) k )) α (2u+1) δ 2u j δ 2u 1 i δj 2u (i = j): α (2u 1) ( for k =n..1 if p k p i (β (2u 1+1) k β (2u 1) k )) ( (β (2u) k for k=1..n, if p k p i β (2u+1) k )) α (2u) λ 2u i λ 2u+1 j (i j): α (2u) ( for k =n..1 if p k p i (γ (2u+1) k γ (2u) k )) ( for k=1..n if p k p j (γ (2u+1) k γ (2u+1+1) k )) α (2u+1) δ 2u 1 i δj 2u (i j): α (2u 1) ( for k =n..1 if p k p i (β (2u 1+1) k β (2u 1) k )) ( (β (2u) k for k=1..n, if p k p j β (2u+1) k )) α (2u) Wollic 2007 p.29

40 Morphism-based encodings Proposition. Each order-preserving mapping from (P, ) to (P, ) can be uniquely extended to a unique pregroup homomorphism on T (P, ) to T (P, ). i.e. from monotonicity on P r : p i p j implies h(p i ) h(p j ) to monotonicity on T (P, ) : if X Y then h(x) h(y ) to ensure : p i p j implies h(p i ) h(p j ) Wollic 2007 p.30

41 Morphism-based encodings Proposition. Each order-preserving mapping from (P, ) to (P, ) can be uniquely extended to a unique pregroup homomorphism on T (P, ) to T (P, ). i.e. from monotonicity on P r : p i p j implies h(p i ) h(p j ) to monotonicity on T (P, ) : if X Y then h(x) h(y ) Proof. The unicity comes from the three first points of the definition of pregroup homomorphism. The last point is a consequence of order-preservation which is easy by induction on a derivation D for X Y to ensure : p i p j implies h(p i ) h(p j ) Wollic 2007 p.30

42 Lemma variant Variants. A similar result holds in the odd case (called Bis version ) for X = X 1 α (2u1 1) Y 1 α (2u 1) }{{}... X k α (2uk 1) Y k α (2u k) }{{}... X n α (2un 1) Y n α (2u n) }{{} X n+1 Wollic 2007 p.31

43 Order 1 Lemma (Buszkowski LACL 01) If p is atomic and o(a i ) 1 for 1 i n : A 1,..., A n p is derivable in AB iff it is derivable in L1 iff T (A 1 )... T (A n ) p in free pregroups where T (A/B) = T (A).T (B) l T (B\A) = T (B) r.t (A) the analogue of : is A/B, B A T (A/B).T (B) = T (A).T (B) l.t (B) T (A) and the analogue of : B, B\A A is T (B).T (B\A) = T (B).T (B) r T (A) T (A) Wollic 2007 p.32

44 On Types for Correct Sentences. Usually, type s associated to correct sentences must be a primitive type (s P ). here s can be any type in T (P, ). not a significant modification of PG because X 1 X n X is equivalent to X 1 X n X r 1. if Y X then Y X r XX r 1 ; if Y X r 1 then Y Y X r X X Wollic 2007 p.33

45 On Types for Correct Sentences. Usually, type s associated to correct sentences must be a primitive type (s P ). here s can be any type in T (P, ). not a significant modification of PG because X 1 X n X is equivalent to X 1 X n X r 1. if Y X then Y X r XX r 1 ; if Y X r 1 then Y Y X r X X PG G using a composed type S into a PG G using s, by a right wall S r s (the type can be seen as the type of the final point of a sentence) because X S XS r s s. Wollic 2007 p.33

46 Main Results Proposition. [Order-reflecting property] The simulation-morphism h from the free pregroup on (P, ) to the free pregroup on (P, ) satisfies (1) and (2): (1). X, Y T (P, ) if h(x) h(y ) then X Y (2). X, Y T (P, ) if h(x) 1 then X 1 by (P, ). In fact, (1) can be shown from the central lemma (2) (case m = 0) and (2) above is equivalent to (1) as explained before. As a corollary of monotonicity and previous proposition, we get: Proposition. [Pregroup Order Simulation] The simulation-morphism h from the free pregroup on (P, ) to the free pregroup on (P, ) enjoys the following property: X, Y T (P, ) h(x) h(y ) iff X Y Wollic 2007 p.34