Formal Languages-Course 5.

(1)

Formal Languages-Course 5.

Géraud Sénizergues

Bordeaux university

18/05/2020

Master computer-science MINF19, IEI, 2019/20

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1 Context-free grammars : ambiguity

2 Context-free languages :closure properties

3 Context-free grammars :transformations Cleaning context-free grammars A simple transformation

Quadratic normal form

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(3)

Context-free grammars : ambiguity

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(4)

context-free gramars : ambiguity

Definition

A context-free grammar G = h A, N, R, σi is called ambiguous iff there exist derivations trees T

₁

, T

₂

, with root σ, such that T

₁

6= T

₂

and fr (T

1

) = fr (T

2

) ∈ A

^∗

.

1- The grammar G is called non-ambiguous iff it is not ambiguous.

2- In words : G is ambiguous iff there exist two different derivation-trees T

1

, T

2

with the same frontier w ∈ A

^∗

.

3- Equivalently, G is ambiguous iff one of the following is true : - there are two different leftmost-derivations D

₁

, D

₂

that end in the same terminal word w

- there are two different rightmost-derivations D

₁

, D

₂

that end in the same terminal word w

- there are two different construction-trees that have the same yield w .

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context-free gramars : ambiguity

Example

Let G = hA, N, R, S i with

A = {a, b}, N = {S },

R : S −→ aS , S −→ aSb, S −→ ab.

L = L (G , S ) = {a

ⁿ

b

^m

| n ≥ m ≥ 1}.

D

₁

: S

_ℓ

−→

R

aS

_ℓ

−→

R

aaSb

_ℓ

−→

R

aaabb D

₂

: S

_ℓ

−→

R

aSb

_ℓ

−→

R

aaSb

_ℓ

−→

R

aaabb

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context-free gramars : ambiguity

Example

Let G = hA, N, R, S i with A = {a, b}, N = {S }, R : S −→ aS , S −→ aSb, S −→ ab.

L = L (G , S ) = {a

ⁿ

b

^m

| n ≥ m ≥ 1}.

T2

a a a b b

S

a a a a b b

S

T1

Figure – Two derivation-trees

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(7)

context-free gramars : ambiguity

Example

Let H = hA, N

^′

, R

^′

, Si with

A = {a, b}, N

^′

= {S , T }, R

^′

: S −→ aS , S −→ T .

T −→ aTb, T −→ ab.

L = L (H, S ) = {a

ⁿ

b

^m

| n ≥ m ≥ 1}.

This new grammar for the same language L is non-ambiguous.

Let w = a

ⁿ

b

^m

with n ≥ m ≥ 1.

The only leftmost derivation for a

^m+r

b

^m

is D(r, m) : S −→

^rR^′

a

^r

S −→

¹R^′

a

^r

T −→

^m−1R^′

a

^r

a

^m−1

Tb

^m−1

−→

¹R^′

a

^r

a

^m−1

abb

^m−1

= a

^m+r

b

^m

.

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(8)

context-free gramars : ambiguity

T S

S T

a^r a^m−¹ ab b^m−¹

Figure – The derivation-tree

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(9)

context-free gramars : ambiguity

Example

Let H = hA, N

^′

, R

^′

, Si be the previous grammar. Let us name the rules :

r1 : S −→ aS, r2 : S −→ T . r3 : T −→ aTb, r4 : T −→ ab.

The only construction-tree for a

^m+r

b

^m

is : r

₁^r

· r

₂

· r

₃^m−1

· r

₄

.

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context-free gramars : ambiguity

yd(T) =a^ra^mb^m r1

r1

.. .

r1

r2

r3

.. .

r3

r4

rnodes

m−1nodes

T

Figure – The construction-tree

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context-free gramars : ambiguity

Example

H = hA, N, P i whith set of rules :

C −→ CsC | bCe | wC | i

We already considered :

D

1

: C

_ℓ

−→ wC

_ℓ

−→ wbCe

_ℓ

−→ wbCsC e

_ℓ

−→ wbwCsCe

ℓ

−→ wbwisCe

_ℓ

−→ wbwisi e But there is also :

D

₂

: C

_ℓ

−→ wC

_ℓ

−→ wbCe

_ℓ

−→ wbwCe

_ℓ

−→ wbwCsCe

_ℓ

−→

wbwisCe

_ℓ

−→ wbwisi e

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context-free gramars : ambiguity

The two corresponding derivation-trees are :

C C

C

b C e b C e

w w

i i

T1 T2

w C s C w C s C

i i

C

Figure – The two derivation-trees

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context-free gramars : ambiguity

wbwisie can be seen, according to T

₁

as : while cond

––begin ––while cond ––––i

––;

––i

––end or, according to T

₂

as : while cond

––begin ––while cond ––––i

––––;

––––i

––end

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context-free gramars : ambiguity

Example

H = hA, {C }, P, C i whith set of rules : C −→ CsC |bCe |wC |i

Another grammar (for the same language) could be : K = hA, {C , D}, R, C i whith set of rules :

C −→ DsC|D D −→ wD|bCe|i

Intuition : D generates the “ prime” instructions (those which cannot be factored as usv with u, v instructions)

C generates all instructions : c = d

₀

s d

₁

· · · s d

_n−1

where each d

_i

is a prime instruction.

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context-free gramars : ambiguity

K = hA, {C , D}, R, C i whith set of rules :

C −→ DsC | D D −→ wD | bCe | i

The leftmost derivation of wbwisie in this grammar is : C −→

³

wbC e , wbC e −→

²

wbDsDe −→

³

wbwisi e, producing the structure :

while cond ––begin ––while cond ––––i

––;

––i ––end

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Context-free languages :closure properties

Context-free languages : closure properties

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Closure properties

Theorem

Let A be a finite alphabet.

1- The set of context-free languages over A

^∗

is closed under the operations : union, product, star, cross.

2- If Card (A) ≥ 2, then the set of context-free languages over A

^∗

is not closed under complement ; it is also not closed under

intersection.

We give a constructive proof of point 1. Hence, from c.f. grammars generating languages L

₁

, L

₂

, one can construct c.f. grammars for the languages L

₁

∪ L

₂

, L

₁

· L

₂

, L

^∗₁

.

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Closure properties

sketch of proof of point 1

Let G

₁

= hA, N

₁

, R

₁

, S

₁

i, G

₂

= hA, N

₁

, R

₁

, S

₁

i. We assume that N

₁

∩ N

₂

= ∅.

union :

H = hA, N

₁

∪ N

₂

∪ {σ}, R, σi

where R = R

₁

∪ R

₂

∪ {σ→S

1

, σ→S

2

}

L (H, σ) = L (G

₁

, S

₁

) ∪ L (G

₂

, S

₂

).

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Closure properties

Let G

₁

= hA, N

₁

, R

₁

, S

₁

i, G

₂

= hA, N

₁

, R

₁

, S

₁

i.

product :

K = hA, N

₁

∪ N

₂

∪ {σ}, R, σi where R = R

₁

∪ R

₂

∪ {σ→S

1

S

₂

}

L (H, σ) = L (G

1

, S

₁

) · L (G

2

, S

₂

).

star :

K

_∗

= hA, N

₁

∪ {σ}, R, σi where R = R

₁

∪ {σ→σS

1

, σ→ε}

L (K

∗

, σ) = L (G

1

, S

₁

)

^∗

.

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Closure properties

point 2 :Let

L = {a

ⁿ

b

ⁿ

c

ⁿ

| n ≥ 1}.

L is not context-free.

Intuitive idea : if it was, there would be a repetition on some branch of a derivation-tree for w = a

^N

b

^N

c

^N

, leading, by pumping, to a larger derivation-tree with frontier w

^′

= a

^N+k

b

^N+ℓ

c

^N

∈ / L or w

^′′

= a

^N

b

^N+k

c

^N+ℓ

∈ / L.

Let L

₁

= {a

^p

b

^q

c

^r

| p, q , r ∈ N \ {0}, p = q}, L

₂

= {a

^p

b

^q

c

^r

| p , q, r ∈ N \ {0}, q = r}.

L

1

, L

2

are context-free (exercise).

L = L

₁

∩ L

₂

is not context-free.

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Closure properties

point 2 :Let

L = {a

ⁿ

b

ⁿ

c

ⁿ

| n ≥ 1}.

L is not context-free.

L

₃

= {a

^p

b

^q

c

^r

| p, q, r ∈ N \ {0}, p = q},

L

₄

= {a

^p

b

^q

c

^r

| p, q, r ∈ N \ {0}, q = r}.

L

₃

, L

₄

are context-free (exercise).

L = L

₃

∩ L

₄

.

This shows that CF ({a, b, c }

^∗

) is not closed under intersection.

L = C(CL

3

∪ CL

4

).

This shows that CF ({a, b, c }

^∗

) is not closed under complement.

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Closure properties

A corollary of the above theorem, point 1,is Proposition

Let A be a finite alphabet. Then every regular subset of A

^∗

is also context-free.

One can also give a direct construction of a c.f. grammar for every finite automaton : let A = hQ, A, δ, q

₀

, F i be some nfa .

Let G = hA, N, R, S

_q₀

i with

N = {S

q

| q ∈ Q},

R : S

_p

−→ xS

_q

, for every (p , x, q) ∈ δ S

_q

−→ ε for every p ∈ F .

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(23)

Closure properties

One can prove, by induction on |u | that : S

p

−→

∗R

u · S

q

⇔ p −→

^uA

q, showing that

L (G , S

_q₀

) = L (A).

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Closure properties

Theorem

Let A be a finite alphabet.

For every languages L ∈ CF (A

^∗

) and R ∈ REG (A

^∗

), L ∩ R is context-free.

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(25)

Closure properties

proof of the theorem :

Let G = hA, N, P , Si be a c.f. grammar and A = hQ, A, δ, q

₀

, F i be some nfa .

We build the c.f. grammar : H = hA, N

^′

, P

^′

, σi with N

^′

= {(p, v, q) | p , q ∈ Q, v ∈ N}.

For every rule U −→ w

₁

U

₁

· w

₂

U

₂

· · · w

_k

U

_k

w

_k+1

in P we insert in P

^′

the set of all rules

(p, U , q) −→ w

₁

(p

1

, U

₁

, q

₁

)w

2

(p

2

, U

₂

, q

₂

) · · · w

_k

(p

k

, U

_k

, q

_k

)w

_k+1

where

p −→

^w¹A

p

₁

, q

₁

−→

^w²A

p

₂

. . . q

_k−1

−→

^w^kA

p

_k

, q

_k

−→

^w^k+1A

q.

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Closure properties

We add in P

^′

all the rules

σ −→ (q

0

, S , q)

for q ∈ F . One can prove that, for every v ∈ N, p, q ∈ Q L (H, (p, v, q )) = L (G , v) ∩ {u ∈ A

^∗

| p −→

^uA

q}.

It follows that :

L (H, σ) = L (G , S ) ∩ L (A).

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(27)

Closure properties

Exercice

Is CF (A

^∗

) closed under reversal ?

Is CF (A

^∗

) closed under regular substitution ?

Is CF (A

^∗

) closed under context-free substitution ? (i.e.

substitutions σ such that, for every x ∈ A, σ({x}) ∈ CF (A

^∗

))

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Context-free grammars :transformations

Context-free grammars : transformations

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Context-free grammars :transformations Cleaning context-free grammars

clean grammars

Let G = hA, N, R, σi be a context-free grammar. A non-terminal S ∈ N is said productive iff

∃u ∈ A

^∗

, S −→

^∗ R

u.

A non-terminal S is said useful iff

∃α, β, u ∈ A

^∗

, σ −→

^∗ R

αS β and S −→

^∗R

u.

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clean grammars

Proposition

Let G = h A, N, R, σi be a context-free grammar. The set N

^′

of productive non-terminals of G can be computed.

1- If σ / ∈ N

^′

, then L (G , σ) = ∅.

2- If σ ∈ N

^′

, then the grammar G

^′

= h A, N

^′

, R

^′

, σi with

R

^′

= R ∩ N

^′

× (A ∪ N

^′

)

^∗

has productive non-terminals only and generates the same language as G .

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clean grammars

Let G = hA, N, R, σi.

Point 1 :Let

N

₁

= {S ∈ N | ∃w ∈ A

^∗

, S −→

¹ R

w }.

N

_i+1

= {S ∈ N | ∃w ∈ (A ∪ N

i

)

^∗

, S −→

¹R

w } ∪ N

i

. One can show that

N

^′

= [

i≥1

N

_i

.

The sequence N

_i

is increasing and included in the finite set N, hence there exists k ≤ Card (N) such that

N

₁

⊂ N

₂

⊂ . . . ⊂ N

_k

= N

_k+1

. It follows that

N

^′

= [

i≥1

N

_i

= N

_k

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(32)

clean grammars

Point 2 : If σ −→

^∗R^′

u, since R

^′

⊆ R, also σ −→

^∗ R

u.

If σ −→

^∗R

u, every non-terminal used in this derivation is productive, hence σ −→

^∗R^′

u.

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clean grammars :example

Consider the context-free grammar G := hA, N, R, S

₁

i where A = {a, b, c}, N = {S

1

, S

₂

, S

₃

, S

₄

, S

₅

, S

₆

} and R consists of the following 15 rules :

S

1

→ aS

3

S

5

S

1

→ bS

2

S

2

S

1

→ aS

4

S

2

→ aS

3

S

3

S

2

→ bS

6

c

S

₃

→ bS

₂

S

₃

→ aS

₃

S

₆

S

₃

→ cS

₆

S

₄

→ bS

₃

S

₄

S

₄

→ aS

₄

S

₄

S

₄

→ b S

₅

→ aS

₄

S

₅

S

₅

→ bS

₁

S

₅

→ aS

₅

.

S

6

→ aS

3

.

N

₁

= {S

4

}, N

₂

= {S

1

, S

₄

}, N

₃

= {S

1

, S

₄

, S

₅

}, N

₄

= N

₃

. Hence the set of productive non-terminals is

N

^′

= N

3

= {S

1

, S

4

, S

5

}.

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clean grammars :example

N

^′

= {S

1

, S

4

, S

5

}. In grammar G the productions using N \ N

^′

must be removed :

S

1

→ aS

3

S

5

S

1

→ bS

2

S

2

S

1

→ aS

4

S

2

→ aS

3

S

3

S

2

→ bS

6

c

S

3

→ bS

2

S

3

→ aS

3

S

6

S

3

→ cS

6

S

⁴

→ bS

³

S

⁴

S

⁴

→ aS

⁴

S

⁴

S

⁴

→ b S

5

→ aS

4

S

5

S

5

→ bS

1

S

5

→ aS

5

.

S

6

→ aS

3

.

The new grammar G

^′

has the set of productions R

^′

: S

₁

→ aS

₄

S

₄

→ aS

₄

S

₄

S

₄

→ b S

5

→ aS

4

S

5

S

5

→ bS

1

S

5

→ aS

5

.

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clean grammars

Proposition

Let G

^′

= h A, N

^′

, R

^′

, σi be a context-free grammar with only productive non-terminals.

1- The set N

^′′

of useful non-terminals can be computed.

2- The grammar G

^′′

= h A, N

^′′

, R

^′′

, σi with

R

^′′

= R

^′

∩ (N

^′′

× (A ∪ N

^′′

)

^∗

) has useful non-terminals only and generates the same language as G

^′

.

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(36)

clean grammars

Let G

^′

= hA, N

^′

, R

^′

, σi.

Point 1 :Let M

₁

= {σ}.

M

_i+1

= {S ∈ N

^′

| ∃α, β ∈ (A ∪ N

^′

)

^∗

, ∃T ∈ M

i

, T −→

¹R^′

αSβ} ∪ M

i

. One can show that

N

^′′

= [

i≥1

M

_i

.

The sequence M

_i

is increasing and included in the finite set N, hence there exists k ≤ Card (N) such that

M

₁

⊂ M

₂

⊂ . . . ⊂ M

_k

= M

_k+1

. It follows that

N

^′′

= [

i≥1

M

_i

= M

_k

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(37)

clean grammars

Point 2 : If σ −→

^∗R^′′

u , since R

^′′

⊆ R

^′

, also σ −→

^∗R^′

u.

If σ −→

^∗R^′

u, every non-terminal used in this derivation is useful, hence σ −→

^∗R^′′

u.

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(38)

clean grammars :example

G

^′

= hA, N

^′

, R

^′

, S

₁

i where R

^′

is the set of rules : S

₁

→ aS

₄

S

₄

→ aS

₄

S

₄

S

₄

→ b S

₅

→ aS

₄

S

₅

S

₅

→ bS

₁

S

₅

→ aS

₅

.

M

1

= {S

1

}, M

2

= {S

1

, S

4

}, M

3

= M

2

N

^′′

= {S

1

, S

4

}

G

^′′

= hA, N

^′′

, R

^′′

, S

₁

i where R

^′′

is the set of rules : S

1

→ aS

4

S

4

→ aS

4

S

4

S

4

→ b

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Context-free grammars :transformations A simple transformation

A simple transformation

Proposition

Let G = hA, N, Ri be a context-free grammar and U ∈ N such that U does not occur in the right-hand side of the rules with left-hand side U.

One can transform G into a grammar G

^′

= hA, N \ {U }, R

^′

i such that, for every S ∈ N \ {U }, L (G

^′

, S ) = L (G, S).

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(40)

A simple transformation

Notation : given languages L

1

, L

2

we note L

1

+ L

2

the union L

₁

∪ L

₂

. A set of grammar rules with same left-hand side :

S −→ w

1

, S −→ w

2

, . . . , S −→ w

n

is denoted by :

S −→ w

1

+ w

2

+ . . . + w

n

Given a c.f. grammar G = hA, N, R, σi, where

N = {S

1

, . . . , S

_i

, . . . , S

_n

}, its set of rules R can always be written under the form :

S

₁

−→

ℓ1

X

j=1

w

_1,j

, S

_i

−→

ℓi

X

j=1

w

_i,j

, S

_n

−→

ℓn

X

j=1

w

_n,j

.

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(41)

A simple transformation

Let G = hA, N, Ri, be a grammar of the previous form and U = S

_n

fulfills the assumption that :

∀j ∈ [1, ℓ

_n

], w

_n,j

∈ (A ∪ {S

1

, · · · , S

_n−1

})

^∗

. We define the substitution σ : P (A ∪ N)

^∗

→P(A ∪ N)

^∗

by :

∀x ∈ A, σ(x) = x, ∀S ∈ N \ {U }, σ(S) = S, σ(U) =

ℓn

X

j=1

w

_n,j

.

Let G

^′

= hA, N \ {U}, R

^′

, σi be defined by : for every 1 ≤ i ≤ n − 1 :

S

i

−→

ℓi

X

j=1

σ(w

i,j

)

(i.e. R

^′

consists of all the rules above with lhs and rhs in N

^′

= {S

1

, . . . , S

_n−1

}).

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(42)

A simple transformation

This grammar does not use U = S

_n

anymore. Let us show that, for every i ∈ [1, n − 1] :

L (G

^′

, S

_i

) = L (G , S

_i

).

1-

∀w ∈ σ(U ), U −→

¹R

w Hence,

∀v ∈ (A ∪ N)

^∗

, ∀w ∈ σ(v), v −→

^∗R

w this implies that

∀i ∈ [1, n − 1], ∀w ∈

ℓi

X

j=1

σ(w

i,j

), S

i

−→

∗ R

w . (1)

By (1) R

^′

⊆ −→

^∗R

, implying that, for every i ∈ [1, n − 1]

L (G

^′

, S

i

) ⊆ L (G , S

i

).

_{42 / 51}

(43)

A simple transformation

2- Let us prove by induction on d that :for every i ∈ [1, n − 1], w ∈ A

^∗

:

S

i d

−→

R

w ⇒ S

i ∗

−→

R^′

w . Let us suppose that :

S

_i

−→

¹R

w

_i,j

= α

1

S

_n

α

2

S

_n

· · · α

k

S

_n

α

_k+1

−→

^d−1R

w , (2) where α

_i

∈ (A ∪ {S

1

, . . . , S

_n−1

})

^∗

.

Base : d = 1 i.e. w

_i,j

= w .

Then w

_i,j

= w hence w ∈ σ(w

i,j

), hence S

_i

−→

¹R^′

w .

43 / 51

(44)

A simple transformation

Induction :d ≥ 2. Derivation (2) shows that

w = ˆ α

1

w

1

α ˆ

2

w

2

· · · α ˆ

_k

w

_k

α ˆ

_k+1

for some words α ˆ

i

, w

i

∈ A

^∗

such that α

_i

−→

^aⁱR

α ˆ

_i

, S

_n

−→

^dⁱR

w

_i

and ( P

k

i=1

a

_i

+ d

_i

) + a

_k+1

= d − 1.Hence S

n

−→

1R

w

n,ji

di−1

−→

R

w

i

44 / 51

(45)

A simple transformation

By induction hypothesis, since w

_n,j_k

∈ (A ∪ {S

1

, . . . , S

_n−1

})

^∗

and a

i

< d , d

i

< d

α

i

−→

∗R^′

α ˆ

i

, w

_n,j_i

−→

^∗R^′

w

_i

(3) Note that

w

^′

= α

₁

w

_n,j₁

α

₂

w

_n,j₂

· · · α

_k

w

_n,j_k

α

_k+1

∈ σ(w

_i_,j

) Combining the rule S

_i

−→

¹R^′

w

^′

with derivations (3) give a derivation

S

_i

−→

¹R^′

w

^′

= α

1

w

_n,j₁

α

2

w

_n,j₂

· · · α

k

w

_n,j_k

α

_k+1

−→

^∗R^′

w . Hence

L (G , S

i

) ⊆ L (G

^′

, S

i

).

45 / 51

(46)

A simple transformation : example

Example

Let G = hA, N, R, S

₁

i with

A = {a, b, c, d }, N = {S

1

, S

₂

, S

₃

, S

₄

}, and R consists of the rules :

S

₁

→ aS

₂

bS

₂

+ ε S

₂

→ cS

₃

+ S

₃

d S

₃

→ cS

₃

+ ε S

4

→ S

4

d + ε

Since S

₂

does not appear in its right-hand side, we can remove S

₂

and obtain : G = hA, N

^′

, R

^′

, S

1

i with N

^′

= {S

1

, S

3

, S

4

},

46 / 51

(47)

A simple transformation : example

Example (continued)

R

^′

consists of the rules :

S

1

→ a(cS

3

+ S

3

d )b(cS

3

+ S

3

d ) + ε S

3

→ cS

3

+ ε

S

₄

→ S

₄

d + ε

which can be written as :

S

1

→ acS

3

bcS

3

+ acS

3

bS

3

d + aS

3

dbcS

3

+ aS

3

dbS

3

d + ε S

3

→ cS

3

+ ε

S

₄

→ S

₄

d + ε

_{47 / 51}

(48)

Context-free grammars :transformations Quadratic normal form

w-Quadratic normal form

Proposition

Every c.f. grammar G = hA, N, R, σi can be transformed into a c.f.

grammar G

^′

= hA, N

^′

, R

^′

, σi such that 1- ∀(S , m) ∈ R

^′

, |m| ≤ 2

2- L (G , σ) = L (G

^′

, σ

^′

).

Some remarks :

such a grammar is called in weak quadradic normal form (quadratic would mean that : ∀(S, m) ∈ R

^′

, |m| = 2 or (S = σ and m = ε)).

48 / 51

(49)

w-Quadratic normal form

step 1 :Let G = hA, N, R, σi such that one rule S → w is such that |w | ≥ 3. We can decompose

w = w

1

· w

2

with |w

1

| < |w |, |w

2

| < |w |.

Let G

^′

= hA, N

^′

, R

^′

, σi be defined by : N

^′

= N ∪ {W

1

, W

2

}, R

^′

= R \ { S → w

₁

· w

₂

}

∪ {S → W

₁

· W

₂

, W

₁

→ w

₁

, W

₂

→ w

₂

}.

Since W

1

(resp. W

2

) does not appear in w

1

(resp. w

2

), applying twice the above transformation, G

^′

is mapped into G . Hence

L (G , σ) = L (G

^′

, σ).

49 / 51

(50)

w-Quadratic normal form

step 2 : As long as there remains some rhs with length ≥ 3, we apply step 1 :

G = G

₀

G

₁

G

₂

. . . G

n

until we obtain a c.f. grammar G

n

where all rhs have length ≤ 2.

Since for every i ∈ [0, n − 1], the grammars G

_i

, G

_i₊₁

generate the same language, we have obtained a new grammar G

_n

in weak quadradic normal form for the language L (G , σ).

50 / 51

(51)

w-Quadratic normal form : example

Example

Let G = hA, N, R, S i where A = {a, b, c , d }, N = {S } and R consists of the following rules :

S → aSbSc + d We obtain successively :

N = {S, W

1

, W

2

}, S → W

1

W

2

+ d , W

1

→ aSb, W

2

→ Sc , N = {S, W

₁

, W

₂

, W

₃

, W

₄

}, S → W

₁

W

₂

+ d , W

₁

→ W

₃

W

₄

,

W

₂

→ Sc, W

₃

→ aS, W

₄

→ b.

51 / 51

Formal Languages-Course 5.