Non-projective axioms for pregroup grammars as cut eliminations Denis B´ echet Denis.Bechet@univ-nantes.fr LINA, University of Nantes Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 1
Non-projective axioms with pregroup Pregroup analysis of “quand il l’avait ramené...” (when he took her home...): l’ is assigned π r 3 so ll s l π 3 rather than o ll A better analysis : ⇒ We need non-projective axioms = Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 2
Non-projective axioms with pregroup How can we introduce “non projective axioms” ? 1. Using products of free pregroups (Kobele 2005) ⇒ NP complete (proof here) = 2. Using an external product as with Categorial Dependency Grammars (Dikovsky 2004) (demo here) = ⇒ Not a pregroup 3. Using “cut elimination”, non-projective axioms are created from projective axioms (demo here) = ⇒ Complex annotation system and less powerful Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 3
Free pregroup X ≤ Y Y ≤ Z ( CUT ) ( Id ) p ( n ) ≤ p ( n ) X ≤ Z XY ≤ Z X ≤ Y Z ( A G ) ( A D ) Xp ( n ) p ( n +1) Y ≤ Z X ≤ Y p ( n +1) p ( n ) Z Xp ( k ) Y ≤ Z ( IND G ) X ≤ Y p ( k ) Z ( IND D ) Xq ( k ) Y ≤ Z X ≤ Y q ( k ) Z q ≤ Pr p if k is even or p ≤ Pr q if k is odd Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 4
Pregroup grammars and languages A grammar G = (Σ , ( Pr, ≤ Pr ) , I, s ) : Σ a finite alphabet ( Pr, ≤ ) a finite partially ordered set (primitive types) that defines free pregroup ( Tp, ≤ Tp ) I ⊂ Σ × Tp , a lexicon, assigns a finite set of types to each c ∈ Σ s ∈ Pr is a primitive type for correct sentences The language L ( G ) ∈ Σ + : v 1 · · · v n ∈ L ( G ) iff for 1 ≤ i ≤ n , ∃ X i ∈ I ( v i ) such that X 1 · · · X n ≤ Tp s Can be adapted to any pregroup (not only free pregroup) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 5
(not free) Pregroup grammars A grammar G = (Σ , P, I, s ) : Σ a finite alphabet P = ( Tp, • , 1 , ≤ Tp , l , r ) a (not free) pregroup I ⊂ Σ × Tp , a lexicon, assigns a finite set of types to each c ∈ Σ s ∈ Tp is a type for correct sentences The language L ( G ) ∈ Σ + : v 1 · · · v n ∈ L ( G ) iff for 1 ≤ i ≤ n , ∃ X i ∈ I ( v i ) such that X 1 • · · · • X n ≤ Tp s Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 6
Non-projective axioms with pregroup How can we introduce “non projective axioms” ? 1. Using products of free pregroups (Kobele 2005) ⇒ NP complete (proof here) = 2. Using an external product as with Categorial Dependency Grammars (Dikovsky 2004) (demo here) = ⇒ Not a pregroup 3. Using “cut elimination”, non-projective axioms are created from projective axioms (demo here) = ⇒ Complex annotation system and less powerful Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 7
Product of pregoups (Kobele 2005) For 2 pregroups P 1 = ( Tp 1 , • 1 , 1 1 , ≤ Tp 1 , l 1 , r 1 ) and P 2 = ( Tp 2 , • 2 , 1 2 , ≤ Tp 2 , l 2 , r 2 ) : P 1 × P 2 = ( Tp 1 × Tp 2 , ◦ , (1 1 , 1 2 ) , ≤ , L , R ) 1. ( x, y ) ≤ ( x ′ , y ′ ) iff x ≤ Tp 1 x ′ and y ≤ Tp 2 y ′ 2. ( x, y ) ◦ ( x ′ , y ′ ) = ( x • 1 x ′ , y • 2 y ′ ) 3. ( x, y ) L = ( x l 1 , y l 2 ) and ( x, y ) R = ( x r 1 , y r 2 ) Kobele 2005 : “Cross product” over lexicalized grammars L ( G 1 × G 2 ) = L ( G 2 ) ∩ L ( G 1 ) But, grammars using products of free pregroups are NP complete (proof here) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 8
Product of pregroup : NP hard Proof : we encode any SAT problem The proof uses the product of three free pregroups on { t, f } : ( P, ∆ 1 , ∆ 2 ) (in fact P = ∆ 1 = ∆ 2 ) P : for formula inferences ∆ 1 and ∆ 2 for the propagation of the boolean values of variables A formula is transformed into a string. The formula can be satisfied iff the string is in the language generated by a fixed grammar G SAT based on ( P, Delta 1 , Delta 2 ) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 9
NP hard (formula transformation) A formula F that contains (at most) n variables v 1 , . . . , v n is transformed into a string T n ( F ) : T n ( F ) = a · · · a [ F ] e · · · e � �� � � �� � n n [ v i ] n = b · · · b c b · · · b d · · · d � �� � � �� � � �� � n i − 1 n − i [ F 1 ∨ F 2 ] n = ∨ [ F 1 ] n [ F 2 ] n [ F 1 ∧ F 2 ] n = ∧ [ F 1 ] n [ F 2 ] n [ ¬ F ] n = ¬ [ F ] n For instance : T 2 ( v 1 ∨ ( v 1 ∧ v 2 )) = aa ∨ cbdd ∧ cbdd bcdd ee ���� ���� ���� for v 1 for v 1 for v 2 Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 10
NP hard (grammar) G SAT is defined by the following lexicon : a �→ (1 , t l , 1) or (1 , f l , 1) b �→ (1 , t, t l ) or (1 , f, f l ) c �→ ( t, t, t l ) or ( f, f, f l ) d �→ (1 , t l , t ) or (1 , f l , f ) e �→ (1 , t, 1) or (1 , f, 1) ∧ �→ ( tt l t l , 1 , 1) or ( ff l t l , 1 , 1) or ( ft l f l , 1 , 1) or ( ff l f l , 1 , 1) ∨ �→ ( tt l t l , 1 , 1) or ( tf l t l , 1 , 1) or ( tt l f l , 1 , 1) or ( ff l f l , 1 , 1) ¬ �→ ( tf l , 1 , 1) or ( ft l , 1 , 1) The types corresponding to correct strings are ≤ ( t, 1 , 1) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 11
NP hard (example 1) T 1 ( v 1 ∧ v 1 ) = a ∧ cd cd e For v 1 = t , the formula is true. There exists an assignement of the symbols of T 1 ( v 1 ∧ v 1 ) through G SAT whose product is ≤ ( t, 1 , 1) : (1 , t l , 1) ( tt l t l , 1 , 1) ( t, t, t l ) (1 , t l , t ) ( t, t, t l ) (1 , t l , t ) (1 , t, 1) � �� � � �� � � �� � � �� � � �� � � �� � � �� � for a for ∧ for c for d for c for d for e ≤ ( t, 1 , 1) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 12
NP hard (example 2) T 2 ( v 1 ∧ ¬ v 2 ) = aa ∧ cbdd ¬ bcdd ee For v 1 = t and v 2 = f , the formula is true. There exists an assignement of the symbols of T 2 ( v 1 ∧ ¬ v 2 ) through G SAT whose product is ≤ ( t, 1 , 1) : (1 , f l , 1) (1 , t l , 1) ( tt l t l , 1 , 1) ( t, t, t l ) (1 , f, f l ) (1 , f l , f ) (1 , t l , t ) � �� � � �� � � �� � � �� � � �� � � �� � � �� � for a for a for ∧ for c for b for d for d ( tf l , 1 , 1) (1 , t, t l ) ( f, f, f l ) (1 , f l , f ) (1 , t l , t ) (1 , t, 1) (1 , f, 1) � �� � � �� � � �� � � �� � � �� � � �� � � �� � for ¬ for b for c for d for d for e for e ≤ ( t, 1 , 1) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 13
Product of pregroup : NP complete A grammar using the product of N free pregroups is in NP because, to test if a string is in the language, we just have to know the right assignment through the grammar and test that the N pregroup type components are less (or equal) than the corresponding pregroup type components associated to correct sentences. ⇒ Product of N free pregroups for N ≥ 3 is NP complete Remark: This is also true for N = 2 (the proof is similar but the construction is more complex) Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 14
Non-projective axioms with pregroup How can we introduce “non projective axioms” ? 1. Using products of free pregroups (Kobele 2005) ⇒ NP complete (proof here) = 2. Using an external product as with Categorial Dependency Grammars (Dikovsky 2004) (demo here) = ⇒ Not a pregroup 3. Using “cut elimination”, non-projective axioms are created from projective axioms (demo here) = ⇒ Complex annotation system and less powerful Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 15
External product with a free pregroup Is it possible to have a product of a free pregroup with a simpler structure (i.e. with a polynomial complexity) ? ⇒ Pregroup with potential Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 16
Free pregroup with potential Structure like Categorial Dependency Grammar (Dikovsky 2004) but with a free pregroup rather than a set of flat categorial types : P × ∆ 1 × · · · × ∆ n P : a (free) pregroup used by the grammar as a pregroup grammar ∆ i : strings of parentheses used by the grammar as a Dyck language (with only one couple of parentheses) Proprety : - Parsing is polynomial - Some languages are context-sensitive languages Problems : - Strings of parentheses do not form a pregroup - We need to introduce “anchors” Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 17
Free pregroup with potential Strings of parentheses do not form a pregroup because : We need at the same time a left adjoint and a right adjoint for “ տ o ” If ( տ o ) l = ( տ o ) r = ւ o then the structure is not a Dyck language If ( տ o ) l � = ( տ o ) r then the structure has at least three generators = ⇒ A free pregroup with potential is not a pregroup Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 18
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