Non-projective axioms for pregroup grammars as cut eliminations - - PowerPoint PPT Presentation

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 πr

3sollslπ3 rather than oll

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

(Id) p(n) ≤ p(n) X ≤ Y Y ≤ Z (CUT) X ≤ Z XY ≤ Z (AG) Xp(n)p(n+1)Y ≤ Z X ≤ Y Z (AD) X ≤ Y p(n+1)p(n)Z Xp(k)Y ≤ Z (INDG) Xq(k)Y ≤ Z X ≤ Y p(k)Z (INDD) 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+:

v1 · · · vn ∈ L(G)

iff for 1 ≤ i ≤ n, ∃Xi ∈ I(vi) such that X1 · · · Xn ≤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+:

v1 · · · vn ∈ L(G)

iff for 1 ≤ i ≤ n, ∃Xi ∈ I(vi) such that X1 • · · · • Xn ≤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 P1 = (Tp1, •1, 11, ≤Tp1,l1 ,r1 ) and

P2 = (Tp2, •2, 12, ≤Tp2,l2 ,r2 ) : P1 × P2 = (Tp1 × Tp2, ◦, (11, 12), ≤,L ,R )

• 1. (x, y) ≤ (x′, y′) iff x ≤Tp1 x′ and y ≤Tp2 y′
• 2. (x, y) ◦ (x′, y′) = (x •1 x′, y •2 y′)
• 3. (x, y)L = (xl1, yl2) and (x, y)R = (xr1, yr2)

Kobele 2005 : “Cross product” over lexicalized grammars

L(G1 × G2) = L(G2) ∩ L(G1)

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 GSAT based on

(P, Delta1, Delta2)

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 v1, . . . , vn is transformed into a string Tn(F) :

Tn(F) = a · · · a

n

[F] e · · · e

n

[vi]n = b · · · b

i−1

c b · · · b

n−i

d · · · d

n

[F1 ∨ F2]n = ∨[F1]n[F2]n [F1 ∧ F2]n = ∧[F1]n[F2]n [¬F]n = ¬[F]n

For instance : T2(v1 ∨ (v1 ∧ v2)) = aa ∨ cbdd

• for v1

∧ cbdd

• for v1

bcdd

• for v2

ee

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 10

NP hard (grammar)

GSAT is defined by the following lexicon : a → (1, tl, 1)or(1, fl, 1) b → (1, t, tl)or(1, f, fl) c → (t, t, tl)or(f, f, fl) d → (1, tl, t)or(1, fl, f) e → (1, t, 1)or(1, f, 1) ∧ → (ttltl, 1, 1)or(ffltl, 1, 1)or(ftlfl, 1, 1)or(fflfl, 1, 1) ∨ → (ttltl, 1, 1)or(tfltl, 1, 1)or(ttlfl, 1, 1)or(fflfl, 1, 1) ¬ → (tfl, 1, 1)or(ftl, 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)

T1(v1 ∧ v1) = a ∧ cd cd e

For v1 = t, the formula is true. There exists an assignement of the symbols of T1(v1 ∧ v1) through GSAT whose product is ≤ (t, 1, 1) :

(1, tl, 1)

for a

(ttltl, 1, 1)

• for ∧

(t, t, tl)

for c

(1, tl, t)

for d

(t, t, tl)

for c

(1, tl, t)

for d

(1, t, 1)

for e

≤ (t, 1, 1)

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 12

NP hard (example 2)

T2(v1 ∧ ¬v2) = aa ∧ cbdd ¬ bcdd ee

For v1 = t and v2 = f, the formula is true. There exists an assignement of the symbols of T2(v1 ∧ ¬v2) through GSAT whose product is ≤ (t, 1, 1) :

(1, fl, 1)

• for a

(1, tl, 1)

for a

(ttltl, 1, 1)

• for ∧

(t, t, tl)

for c

(1, f, fl)

• for b

(1, fl, f)

• for d

(1, tl, t)

for d

(tfl, 1, 1)

• for ¬

(1, t, tl)

for b

(f, f, fl)

• for c

(1, fl, f)

• for d

(1, tl, t)

for d

(1, t, 1)

for e

(1, f, 1)

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 “տo” If (տo)l = (տo)r = ւo then the structure is not a Dyck language If (տo)l = (տo)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

Free pregroup with potential

We need to introduce “anchors” to control the end position

• f a “non projective axiom”

Accept also :

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 19

Free pregroup with potential

Without “anchors” : With “anchors” :

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 20

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. 21

Free pregroup with cut annotations

We introduce cut annotations [· · · ] on types

l′ → [πr

3[sollsl]π3]

After “annotated cut elimination” of [πr

3[sollsl]π3] :

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 22

Free pregroup with cut annotations

Non-projective axioms introduced at a final step The complexity is polynomial Does not extend the class of languages Needs (not natural) annotations on types

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 23

Conclusion

Non-projective axioms in pregroup are not easy to introduce Here, 3 propositions (2 are implemented) that are not completely satisfying : Must be polynomial Uses a (not free) pregroup Extends the class of languages beyong the class of context free languages Better solution ?

Joachim Lambek – Mathematics, Logic and Language – Chieti, 8–9 July 2011 – p. 24