Solutions to the subject: Formal languages theory Date: 26 August 2014

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Cursus: M1, computer-science Code UE: JEIN8602

Solutions to the subject: Formal languages theory Date: 26 August 2014

Duration: 3H

Documents: authorized

Lectures by: Mr G´eraud S´enizergues

Exercice 1 _[/4]

e := ((abb) ^∗ a(cb) ^∗ ) ∪ (ab) ^∗ (ba) ^∗ Let us apply Glushkov’s method.

The locally testable language associated to e is:

L ^′ := ((a 1 b 1 b 2 ) ^∗ a 2 (c 1 b 3 ) ^∗ ) ∪ (a 3 b 4 ) ^∗ (b 5 a 4 ) ^∗ We compute

Ini(L ^′ ) = {a 1 , a 2 , a 3 , b 5 }, F in(L ^′ ) = {a 2 , b 3 , b 4 , a 4 },

Dig(L ^′ ) = {a 1 b 1 , b 1 b 2 , b 2 a 1 , b 2 a 2 , a 2 c 1 , c 1 b 3 , b 3 c 1 , a 3 b 4 , b 4 a 3 , b 4 b 5 , b 5 a 4 , a 4 b 5 } This gives the finite automaton represented on figure 1, where the set of states is

{0, a 1 , a 2 , a 3 , a 4 , b 1 , b 2 , b 3 , b 4 , b 5 , c 1 } 0 is the unique initial state and a 2 , b 3 , b 4 , a 4 are the final states.

a

1

b

1

b

0 b

2

a

2

c

1

b

3

a

3

b

4

b

5

a

4

b a

a

a a

b b

c b

c

a

b b

Figure 1: finite automaton for L _e

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Exercice 2 _[/4] We transform A into a normalized extended f.a. A 1 where i (resp. t)

0

1 2 a

b

a

c

b b

a c

b ∪ cb

^∗

a b ∪ ab

^∗

c

c ∪ ab

^∗

a

a ∪ cb

^∗

c i

i ε

ε t

0 1 t

t

i 0

ε

b ∪ cb

^∗

a ∪ (( a ∪ cb

^∗

c )( b ∪ ab

^∗

c )

^∗

( c ∪ ab

^∗

a )) ( a ∪ cb

^∗

c )( b ∪ ab

^∗

c )

^∗

[ b ∪ cb

^∗

a ∪ (( a ∪ cb

^∗

c )( b ∪ ab

^∗

c )

^∗

( c ∪ ab

^∗

a ))]

^∗

[( a ∪ cb

^∗

c )( b ∪ ab

^∗

c )

^∗

]

i t

ε ε

Figure 2: sequence of extended automata, ex. 2

is the initial (resp. terminal) state. We then eliminate successively states in the ordering:

2, 1, 0. We obtain the sequence of extended automata shown on figure 2. It follows that:

[b ∪ cb ^∗ a ∪ ((a ∪ cb ^∗ c)(b ∪ ab ^∗ c) ^∗ (c ∪ ab ^∗ a))] ^∗ [(a ∪ cb ^∗ c)(b ∪ ab ^∗ c) ^∗ ]

is a regular expression for L A .

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Exercice 3 _[/5] The language L _e is recognized by the following finite automaton A:

Making A complete and taking its complement, we obtain a finite automaton B recognizing

0 a 1 b 2

c

Figure 3: a non-det automaton for L _e

{a, b, c} ^∗ − L _e . We transform B into a normalized extended f.a. B 1 where i (resp. t) is the

0 a 1 b 2

c

a ∪ c a ∪ b b ∪ c

Figure 4: a det automaton for {a, b, c} ^∗ − L _e

initial (resp. terminal) state. We then eliminate successively states in the ordering: 1, 2, 3, 0.

We obtain the sequence of extended automata shown on figure 5. It follows that:

(abc) ^∗ (b ∪ c ∪ aa ∪ ac ∪ aba ∪ abb)(a ∪ b ∪ c) ^∗ is a regular expression for {a, b, c} ^∗ − L _e .

Exercice 4 [/4] We consider the context-free grammar G := (A, N, R) where A = {a, b}, N = {S, T 1 , T 2 } and R consists of the following rules:

r1: S → T 1

r2: S → T 2

r3: T 1 → aT 1

r4: T 1 → aT 1 b r5: T 1 → a r6: T 2 → T 2 b r7: T 2 → aT 2 b r8: T 2 → b

The start symbol of G is S.

1- L(G, T ) = {a | ≥ 0}

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3- L(G, S) = {a ^p b ^q | p ≥ 0, q ≥ 0, p 6= q}

4- The word aaab has two different leftmost derivations:

S → T 1 → aT 1 b → aaT 1 b → aaab S → T 1 → aT 1 → aaT 1 b → aaab This shows that G is ambiguous.

5- The ambiguity is due to the fact that rules r3, r4 are commuting and, as well r6, r7 are commuting. Hence a word w is generated by G with exactly one derivation-tree iff w is generated by a derivation tree T such that:

- either T uses r4 but not r3 - or T uses r7 but not r6 - or T uses r3 but not r4 - or T uses r6 but not r7

The two first types of derivation-trees give rise to the set of words:

{a ^p b ^q | p ≥ 1, q ≥ 1, |p − q| = 1}

The third type of derivation-trees gives rise to the set of words: {a ^p | p ≥ 2}

The fourth type of derivation-trees gives rise to the set of words: {b ^q | q ≥ 2}.

Consequently:

L ^′ = {a ^p b ^q | p ≥ 1, q ≥ 1, |p − q| = 1} ∪ {a ^p | p ≥ 2} ∪ {b ^q | q ≥ 2}

6- A non-ambigyous c.f. grammar could be, for example: G ^′ := (A, N ^′ , R ^′ ) where A = {a, b}, N ^′ = {S, T 1 , T 2 , E} and R ^′ consists of the following rules:

r’1: S → T 1

r’2: S → T 2

r’3: T 1 → aT 1

r’4: T 1 → aE r’5: T 1 → a r’6: T 2 → T 2 b r’7: T 2 → Eb r’8: T 2 → b r’9: E → aEb r’10: E → ab

Exercice 5 [/4] We consider the context-free grammar G := (A, N, R) where A = {a, b, c}, N = {S 1 , S 2 , S 3 , S 4 , S 5 } and R consists of the following 12 rules:

S 1 → aS 1 S 1 S 1 → bS 3 S 1 S 1 → S 2 c S 2 → S 2 S 1 S 2 → aS 3 S 2 → S 1 S 2 S 1

S 3 → a S 3 → S 1 S 3 S 4 → cS 4

S 4 → aS 4 S 5 S 5 → aS 5 S 5 → aS 4 a The start symbol of G is S 1 .

1- We compute the subset of productive non-terminals of G by the fixpoint technique explained

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in the lecture:

V 1 = {S 3 }, V 2 = {S 3 , S 2 }.V 3 = {S 3 , S 2 , S 1 }, V 4 = V 3 . Hence the set of productive non-terminals is {S 1 , S 2 , S 3 }.

2- We compute the subset of useful non-terminals of G by the fixpoint technique explained in the lecture:

N 1 = {S 1 }, N 2 = {S 1 , S 2 , S 3 }, N 3 = N 2

Hence the set of useful non-terminals is {S 1 , S 2 , S 3 }.

3- We can thus transform the grammar G into an equivalent grammar G ^′ where every non- terminal is productive and useful, just by restricting both the non-terminal alphabet and the rules to the subset N 2 :

G ^′ := (A, N ^′ , R ^′ ) where A = {a, b, c}, N ^′ = {S 1 , S 2 , S 3 } and R ^′ consists of the following 8 rules:

S 1 → aS 1 S 1 S 1 → bS 3 S 1 S 1 → S 2 c S 2 → S 2 S 1 S 2 → aS 3 S 2 → S 1 S 2 S 1

S 3 → a S 3 → S 1 S 3

4- The language L(G, S 1 ) is not empty, since the non-terminal S 1 is productive.

5- We observe that:

S 1 → aS 1 S 1 → ^∗ aS 1 aac and S 1 → ^∗ aac

hence {a ⁿ (aac) ⁿ ⁺¹ } ⊆ L(G, S 1 ), which shows that the language L(G, S 1 ) is infinite.

Exercice 6 [/4] We consider the two following context-free grammars G 1 := (A, N 1 , R 1 ), G 2 :=

(A, N 2 , R 2 ) where A = {a, b, c}, N 1 = {S, T }, N 2 = {U } ,R 1 consists of the rules:

r1: S → aSbT r2: S → cT r3: T → aT T b r4: T → c

and R 2 consists of the rules:

r4: U → U U b r5: U → a

1- The following context-free grammar generates the language L(G 1 , S) · L(G 2 , U ):

G · := (A, N, R · ) where A = {a, b, c}, N := N 1 ∪ N 2 ∪ {σ} ,R · := R 1 ∪ R 2 ∪ {σ → ST }, 2- The following context-free grammar generates the language L(G 1 , S) ∪ L(G 2 , U ):

G ∪ := (A, N, R ∪ ) where A = {a, b, c}, N := N 1 ∪ N 2 ∪ {σ} ,R ∪ := R 1 ∪ R 2 ∪ {σ → S, σ → T}, 3- The following context-free grammar generates the language L(G 1 , S) ^∗ :

G ∗ := (A, N ∗ , R ∗ ) where A = {a, b, c}, N ∗ := N 1 ∪ {σ} ,R ∗ := R 1 ∪ {σ → σS, σ → ε},

4- The following context-free grammar H 1 generates ϕ(L(G 1 , S)): H 1 := (B, N 1 , R 1,ϕ ) where B = {x, y}, R 1 ,ϕ consists of the rules:

S → xySyxT, S → yT, T → xyT T yx, T → y

The following context-free grammar H 2 generates ϕ(L(G 2 , U )): H 2 := (B, N 2 , R 2 ,ϕ ) where

Solutions to the subject: Formal languages theory Date: 26 August 2014

Cursus: M1, computer-science Code UE: JEIN8602

Solutions to the subject: Formal languages theory Date: 26 August 2014

Duration: 3H

Documents: authorized

Lectures by: Mr G´eraud S´enizergues

Exercice 1 [/4]

e := ((abb) ∗ a(cb) ∗ ) ∪ (ab) ∗ (ba) ∗ Let us apply Glushkov’s method.

The locally testable language associated to e is:

L ′ := ((a 1 b 1 b 2 ) ∗ a 2 (c 1 b 3 ) ∗ ) ∪ (a 3 b 4 ) ∗ (b 5 a 4 ) ∗ We compute

Ini(L ′ ) = {a 1 , a 2 , a 3 , b 5 }, F in(L ′ ) = {a 2 , b 3 , b 4 , a 4 },

Dig(L ′ ) = {a 1 b 1 , b 1 b 2 , b 2 a 1 , b 2 a 2 , a 2 c 1 , c 1 b 3 , b 3 c 1 , a 3 b 4 , b 4 a 3 , b 4 b 5 , b 5 a 4 , a 4 b 5 } This gives the finite automaton represented on figure 1, where the set of states is

{0, a 1 , a 2 , a 3 , a 4 , b 1 , b 2 , b 3 , b 4 , b 5 , c 1 } 0 is the unique initial state and a 2 , b 3 , b 4 , a 4 are the final states.

a

b

b

0

b

a

c

b

a

b

b

a

b a

a

a

a

a a

b b

c b

c

a

b b

Figure 1: finite automaton for L e

Exercice 2 [/4] We transform A into a normalized extended f.a. A 1 where i (resp. t)

0

1

2 a

b

a

c

c

b b

a c

b ∪ cb

a b ∪ ab

c

c ∪ ab

a

a ∪ cb

c i

i ε

ε t

0 1 t

t

i 0

ε

b ∪ cb

a ∪ (( a ∪ cb

c )( b ∪ ab

c )

( c ∪ ab

a )) ( a ∪ cb

c )( b ∪ ab

c )

[ b ∪ cb

a ∪ (( a ∪ cb

c )( b ∪ ab

c )

( c ∪ ab

a ))]

[( a ∪ cb

c )( b ∪ ab

c )

]

i t

ε ε

Figure 2: sequence of extended automata, ex. 2

Exercice 1 _[/4]

e := ((abb) ^∗ a(cb) ^∗ ) ∪ (ab) ^∗ (ba) ^∗ Let us apply Glushkov’s method.

L ^′ := ((a 1 b 1 b 2 ) ^∗ a 2 (c 1 b 3 ) ^∗ ) ∪ (a 3 b 4 ) ^∗ (b 5 a 4 ) ^∗ We compute

Ini(L ^′ ) = {a 1 , a 2 , a 3 , b 5 }, F in(L ^′ ) = {a 2 , b 3 , b 4 , a 4 },

Dig(L ^′ ) = {a 1 b 1 , b 1 b 2 , b 2 a 1 , b 2 a 2 , a 2 c 1 , c 1 b 3 , b 3 c 1 , a 3 b 4 , b 4 a 3 , b 4 b 5 , b 5 a 4 , a 4 b 5 } This gives the finite automaton represented on figure 1, where the set of states is

Figure 1: finite automaton for L _e

Exercice 2 _[/4] We transform A into a normalized extended f.a. A 1 where i (resp. t)

[b ∪ cb ^∗ a ∪ ((a ∪ cb ^∗ c)(b ∪ ab ^∗ c) ^∗ (c ∪ ab ^∗ a))] ^∗ [(a ∪ cb ^∗ c)(b ∪ ab ^∗ c) ^∗ ]

Exercice 3 _[/5] The language L _e is recognized by the following finite automaton A:

Figure 3: a non-det automaton for L _e

{a, b, c} ^∗ − L _e . We transform B into a normalized extended f.a. B 1 where i (resp. t) is the

Figure 4: a det automaton for {a, b, c} ^∗ − L _e

(abc) ^∗ (b ∪ c ∪ aa ∪ ac ∪ aba ∪ abb)(a ∪ b ∪ c) ^∗ is a regular expression for {a, b, c} ^∗ − L _e .

3- L(G, S) = {a ^p b ^q | p ≥ 0, q ≥ 0, p 6= q}

{a ^p b ^q | p ≥ 1, q ≥ 1, |p − q| = 1}

The third type of derivation-trees gives rise to the set of words: {a ^p | p ≥ 2}

The fourth type of derivation-trees gives rise to the set of words: {b ^q | q ≥ 2}.

L ^′ = {a ^p b ^q | p ≥ 1, q ≥ 1, |p − q| = 1} ∪ {a ^p | p ≥ 2} ∪ {b ^q | q ≥ 2}

6- A non-ambigyous c.f. grammar could be, for example: G ^′ := (A, N ^′ , R ^′ ) where A = {a, b}, N ^′ = {S, T 1 , T 2 , E} and R ^′ consists of the following rules:

3- We can thus transform the grammar G into an equivalent grammar G ^′ where every non- terminal is productive and useful, just by restricting both the non-terminal alphabet and the rules to the subset N 2 :

G ^′ := (A, N ^′ , R ^′ ) where A = {a, b, c}, N ^′ = {S 1 , S 2 , S 3 } and R ^′ consists of the following 8 rules:

S 1 → aS 1 S 1 → ^∗ aS 1 aac and S 1 → ^∗ aac

hence {a ⁿ (aac) ⁿ ⁺¹ } ⊆ L(G, S 1 ), which shows that the language L(G, S 1 ) is infinite.

G ∪ := (A, N, R ∪ ) where A = {a, b, c}, N := N 1 ∪ N 2 ∪ {σ} ,R ∪ := R 1 ∪ R 2 ∪ {σ → S, σ → T}, 3- The following context-free grammar generates the language L(G 1 , S) ^∗ :