Grammars Automata

8/20/2019 Grammars Automata

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Introduction to Formal Languages,Automata and Computability

K. Krithivasan and R. Rama

Introduction to Formal Languages, Automata and Computability – p.1/74


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Language

Strings are defined over an alphabet which is finite.Alphabet may vary depending upon the application.

Elements of an alphabet are called symbols. Usuallywe denote the basic alphabet set either as Σ or T . Forexample, the following are a few examples of analphabet set.

Σ1 = {a, b}

Σ2 = {0, 1, 2}

Σ3 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.



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contd.

A string or word is a finite sequence of symbols from

the alphabet, usually written as concatenated symbols

and not separated by gaps or commas. For example if

Σ = {a, b}, a string abbab is a string or word over Σ.

If w is a string over an alphabet Σ, then the length of w written as len(w) or |w| is the number of symbols it

contains. If |w| = 0

, then w

is called as empty string

denoted either as λ or .



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contd.

For any word w, w = w = w. For any string

w = a1 . . . an of length n, the reverse of w is writ-

ten as wR which is the string anan−1 . . . a1, where each

symbol ai belongs to the basic alphabet Σ. A string z

that is appearing consecutively within another string wis called a substring or subword of w. For example aab

is a substring of baabb

.



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contd.

The set of all strings over an alphabet Σ is denoted byΣ∗ which includes the empty string . For example for

Σ = {0, 1}, Σ∗

= {, 0, 1, 00, 01, 10, 11, . . . }. Notethat Σ∗ is a countably infinite set. Also Σn denotes theset of all strings over Σ whose length is n. Hence

Σ∗

= Σ0

∪ Σ1

∪ Σ2

∪ Σ3

. . . andΣ+ = Σ1 ∪ Σ2 ∪ Σ3 . . . . Subsets of Σ∗ are calledlanguages. For example if Σ = {a, b}

L1 = {,a,b}

L2 = {ab, aabb, aaabbb, . . . }

L3 = {w ∈ Σ∗/|w|a = |w|b}

In the above example, L1 is finite, L2 and L3 areinfinite languages. φ denotes an empty language.



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contd.

DefinitionA set is an unordered collection of objects.



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contd.

DefinitionA set is an unordered collection of objects.Example Let W denote the set of well formed

parentheses. It can be defined inductively as follows:Basis clause : [ ] ∈ W Inductive clause : if x, y ∈ W , xy ∈ W and [x] ∈ W

Extremal clause : No object is a member of W unlessits being so follows from a finite number of applica-

tions of the basis and the inductive clauses.



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Language

Definition Let Σ be any alphabet set. Σ+ is a set of nonempty

strings over Σ defined as follows:

1. Basis : If a ∈ Σ, then a ∈ Σ+.

2. Induction : If α ∈ Σ+ and a ∈ Σ, αa, aα are in Σ+.

3. No other element belong to Σ

+

.Clearly the set Σ+ contains all strings of length n, n ≥ 1.



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Language

Definition Let Σ be any alphabet set. Σ+ is a set of nonempty

strings over Σ defined as follows:

1. Basis : If a ∈ Σ, then a ∈ Σ+.

2. Induction : If α ∈ Σ+ and a ∈ Σ, αa, aα are in Σ+.

3. No other element belong to Σ

+

.Clearly the set Σ+ contains all strings of length n, n ≥ 1.

Definition Let Σ be any alphabet set. Σ∗ is defined as follows:

1. Basis : ∈ Σ∗.

2. Induction : If α ∈ Σ∗, a ∈ Σ, then aα, αa ∈ Σ∗.

3. No other element is in Σ∗.



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contd.

Since languages are sets, one can define the set-

theoretic operations of union, intersection, difference,

complement in the usual fashion. The following oper-

ations are also defined for languages. If x = a1 . . . an,

y = b1 . . . bm, the concatenation of x and y is definedas xy = a1 . . . anb1 . . . bm. The catenation (or con-

catenation) of two languages L1

and L2

is defined by,

L1L2 = {w1w2/w1 ∈ L1 and w2 ∈ L2}. Note that

concatenation of languages is associative because con-

catenation of strings is associative. Also L0 = {} and

L = L = , L = L = L. Introduction to Formal Languages, Automata and Computability – p.8/74


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contd.

The concatenation closure (Kleene closure) of alanguage L, in symbols L∗ is defined to be the union

of all powers of L:

L∗ =∞

i=0

Li

Also L

+

=

∞

i=1 L

i

.



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contd.

The right quotient and right derivative are thefollowing sets respectively.

L1\L2 = {y|yz ∈ L1 for some z ∈ L2}

∂ rzL = L/{z } = {y/yz ∈ L}

Similarly left quotient of a language L1 by a languageL2 is defined by

L2/L1 = {z |yz ∈ L1 for some y ∈ L2}.

The left derivative of a language L with respect to a

word y is denoted as ∂ yL which is equal to {z |yz ∈ L}.Introduction to Formal Languages, Automata and Computability – p.10/74


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contd.

The mirror image (or reversal) of a language is thecollection of the mirror images of its words and

mir(L) = {mir(w)/w ∈ L} or LR

= {wR

/w ∈ L}.The operations substitution and homomorphism aredefined as follows.

For each symbol a of an alphabet Σ, let σ(a) be alanguage over Σa. Also σ() = , σ(αβ ) = σ(α).σ(β )

for α, β ∈ Σ+. σ is a mapping from Σ∗ to 2V ∗

where

V is the union of the alphabets Σa, is called asubstitution. For a language L over Σ, we define

σ(L) = {α/α ∈ σ(β ) for some β ∈ L}.



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contd.

A substitution is -free if and only if none of thelanguage σ(a) contains . A family of languages is

closed under substitution if and only if whenever L isin the family and σ is a substitution such that σ(a) isin the family, then σ(L) is also in the family.

A substitution σ such that σ(a) consists of a singleword wa is called a homomorphism. It is called -freehomomorphism if none of σ(a) is .

Algebraically, one can see that Σ∗ is a free semigroupwith as its identity.



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contd.

The homomorphism which is defined above agreeswith the customary definition of homomorphism of

one semigroup into another.Inverse homomorphism can be defined as follows:

h−1(w) = {x|h(x) = w}

h−1(L) = {x|h(x) is in L}.

It should be noted that h(h−1

(L)) need not be equal toL. Generally h(h−1(L)) ⊆ L and h−1(h(L)) ⊇ L.



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Grammar

Definition A phrase-structure grammar or a type 0

grammar is a 4-tuple G = (N , T , P , S ), where N is a

finite set of nonterminal symbols called thenonterminal alphabet, T is a finite set of terminalsymbols called the terminal alphabet, S ∈ N is thestart symbol and P is a set of productions (also calledproduction rules or simply rules) of the form u → v,where u ∈ (N ∪ T )∗N (N ∪ T )∗ and v ∈ (N ∪ T )∗.Derivations are defined as follows:

If αuβ is a string in (N ∪ T )∗ and u → v is a rule in

P , from αuβ we get αvβ by replacing u by v. This

is denoted as αuβ ⇒ αvβ . ⇒ is read as ‘directly

derives’. Introduction to Formal Languages, Automata and Computability – p.14/74


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contd.

If α1 ⇒ α2, α2 ⇒ α3, . . . , αn−1 ⇒ αn, the derivation

is denoted as α1 ⇒ α2 ⇒ · · · ⇒ αn or α1∗

⇒ αn. ∗⇒ is

the reflexive, transitive closure of ⇒.Definition The language generated by a grammarG = (N , T , P , S ) is the set of terminal strings

derivable in the grammar from the start symbol.

L(G) = {w/w ∈ T ∗, S ∗⇒ w}



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Example

Consider the grammar G = (N , T , P , S ) whereN = {S, A}, T = {a,b,c}, production rules in P are

S → aSc, S → aAc, A → bA typical derivation in the grammar is

S ⇒ aSc

⇒ aaScc

⇒ aaaAccc

⇒ aaabccc

The language generated is L(G) = {anbcn/n ≥ 1}.



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Lengths

DefinitionIf the rules are of the form αAβ → αγβ ,α, β ∈ (N ∪ T )∗, A ∈ N , γ ∈ (N ∪ T )+, the

grammar is called context-sensitive grammar.



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Lengths


grammar is called context-sensitive grammar.DefinitionIf in the rule u → v, |u| |v|, the grammaris called length increasing grammar.



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Lengths


grammar is called context-sensitive grammar.DefinitionIf in the rule u → v, |u| |v|, the grammaris called length increasing grammar.

Example Let G = (N , T , P , S ) where N = {S, B},T = {a,b,c},P has the following rules:

1. S → aSBc2. S → abc

3. cB → Bc

4. bB → bbIntroduction to Formal Languages, Automata and Computability – p.17/74


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contd..

Let us consider the language generated. The numberappearing above ⇒ denotes the rule being used.

S 2⇒ abc; here abc ∈ L(G)

S 1⇒ aSBc2⇒ aabcBc3

⇒ aabBcc4

⇒ aabbcc, a2b2c2 ∈ L(G)



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contd.

Similarly

S

1

⇒ aSBc1⇒ aaSBcBc2

⇒ aaabcBcBc3⇒ aaabBccBc3

⇒ aaabBcBcc

3⇒ aaabBBccc4

⇒ aaabbBccc4

⇒ aaabbbccc, a3

b3

c3

∈ L(G)Introduction to Formal Languages, Automata and Computability – p.19/74


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contd.

In general any string of the form anbncn will begenerated.

S ∗⇒ an−1S (Bc)n−1 (by applying rule 1 (n-1) times)

⇒ anbc(Bc)n−1 (rule 2 once)

∗⇒ anbBn−1cn (by applying rule 3 n(n − 1)2

times)

∗⇒ anbncn (by applying rule 4,(n − 1) times)

Hence L(G) = {anbncn/n ≥ 1}. This is a type 1

language.Introduction to Formal Languages, Automata and Computability – p.20/74


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Type2 language

Definition If in a grammar, the production rules are of the form, A → α,

where A ∈ N and α ∈ (N ∪ T )∗, the grammar is called a type 2 grammar or

context-free grammar. The language generated is called a type 2 language or

context-free language.


T 2 l


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Type2 language

Definition If in a grammar, the production rules are of the form, A → α,

where A ∈ N and α ∈ (N ∪ T )∗, the grammar is called a type 2 grammar or

context-free grammar. The language generated is called a type 2 language or

context-free language.

Definition If the rules are of the form A → αB,A → β, A,B ∈ N, α, β ∈

T

∗

, the grammar is called a right linear grammar or type 3 grammar and the

language generated is called a type 3 language or regular set. We can even

put the restriction that the rules can be of the form A → aB,A → b, where

A,B ∈ N, a ∈ T, b ∈ T ∪ . This is possible because a rule A → a1 . . . akB

can be split into A → a1B1, B1 → a2B2, . . . , Bk−1 → akB by introducing

new nonterminals B1, . . . , Bk.


E l


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Example

Let G = (N,T,P,S ) where N = {S }, T = {a, b}

P consists of the following rules.

1. S → aS

2. S → bS 3. S →

This grammar generates all strings in T ∗. For example, the string abbaab is generated as follows:

S ⇒ aS (rule 1)

⇒ abS (rule 2)

⇒ abbS (rule 2)

⇒ abbaS (rule 1)

⇒ abbaaS (rule 1)

⇒ abbaabS (rule 2)

⇒ abbaab (rule 3)



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Derivation tree

We have considered the definition of a grammar andderivation. Each derivation can be represented by atree called a derivation tree (sometimes called parsetree). A derivation tree for the derivation consideredin previous example with grammarS → aSc, S → aAc, A → b is

a S c

S

Sa

a

c

c

b

A


E l


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Example

Consider the following CFG, G = (N , T , P , S ),N = {S,A,B}, T = {a, b}. P consists of the

following productions1. S → aB

2. B → b

3. B → bS

4. B → aBB

5. S → bA6. A → a

7. A → aS

8. A → bAAIntroduction to Formal Languages, Automata and Computability – p.24/74


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contd..

The derivation tree for aaabbb is as follows,

a B

S

Ba

a

B

b

B B

b

b


E l


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Example

Consider the grammarG = ({S }, {a, b}, {S → SaSbS , S → SbSaS ,

S → }, S ). The language generated by this grammaris the same as the language generated by the grammarG = (N , T , P , S ), N = {S,A,B}, T = {a, b}, P consists of the following productionsS → aB, B → b, B → bS, B → aBB, S → bA, A →a, A → aS, A → bAA, except that , the empty stringis also generated here.


E ample


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Example

Consider the grammarG = ({S }, {a, b}, {S → SaSbS , S → SbSaS ,

S → }, S ). The language generated by this grammaris the same as the language generated by the grammarG = (N , T , P , S ), N = {S,A,B}, T = {a, b}, P consists of the following productionsS → aB, B → b, B → bS, B → aBB, S → bA, A →a, A → aS, A → bAA, except that , the empty stringis also generated here.

0

1

2

−1

−2

X


td


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contd..

Consider a string w having equal number of a’s andb’s. We use induction.Basis|w| = 0, S ⇒ |w| = 2, it is either ab or ba, then

S ⇒ SaSbS

∗

⇒ abS ⇒ SbSaS ∗⇒ ba.


td


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contd..

Induction Assume that the result holds up to stringsof length k − 1. Prove that the result holds for stringsof length k. Draw a graph where the x axis representsthe length of the prefixes of the given string. y axisrepresents the number of a’s - number of b’s. For thestring aabbabba, the graph will look as given in theprevious figure. For a given string w with equalnumber of a’s and b’s there are 3 possibilities.

1. The string begins with ‘a’ and ends with ‘b’.

2. The string begins with ‘b’ and ends with ‘a’.

3. Other two cases (begins with a and ends with a,

begins with b and ends with b)Introduction to Formal Languages, Automata and Computability – p.28/74

td


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contd..

In the first case w = aw1b and w1 has equal numbersof a’s and b’s. So we have

S ⇒ SaSbS ⇒ aSbS ⇒ aSb

∗

⇒ aw1b as S

∗

⇒ w1 byinductive hypotheses. A similar argument holds forcase 2. In case 3, the graph mentioned above willcross the x axis.Consider w = w1w2 where w1, w2 have equal numberof a’s and b’s. Let us say w1 begins with a, and w1corresponds to the portion where the graph touches

the x axis for the first time. In the above examplew1 = aabb and w2 = abba.


td


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contd..

In this case we can have a derivation as follows:

S 1⇒ SaSbS 3

⇒ aSbS ∗

⇒ aw1bS (w1 = aw1b)

∗⇒ aw1bw2∗

⇒ w1w2.

S ∗⇒ w follows from inductive hypothesis.


CS


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CS

Theorem Every context-sensitive language is length increasing and

conversely.

That every context-sensitive language is length increasing can be seen

from definitions.

Every length increasing language is context-sensitive can be seen from

the following construction.

Let L be a length increasing language generated by G = (N , T , P , S ).

Without loss of generality, one can assume that the productions in P

are of the form X → a, X → X 1 . . . X m, X 1 . . . X m → Y 1 . . . Y n,

2 ≤ m ≤ n, X, X 1, . . . , X m, Y 1, . . . , Y n ∈ N , a ∈ T . Productions in

P which are already context-sensitive productions are not modified.

Hence consider, a production of the form

X 1 . . . X m → Y 1 . . . Y n, 2 ≤ m ≤ nIntroduction to Formal Languages, Automata and Computability – p.31/74

contd


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contd..

It is replaced by the following set of context-sensitiveproductions:

X 1 . . . X m → Z 1X 2 . . . X mZ 1X 2 . . . X m → Z 1Z 2X 3 . . . X m

..

.Z 1Z 2 . . . Z m−1X m → Z 1Z 2 . . . Z mY m+1 . . . Y nZ 1Z 2 . . . Z mY m+1 . . . Y n → Y 1Z 2 . . . Z mY m+1 . . . Y n

...

Y 1Y 2 . . . Y m−1Z mY m+1 . . . Y n → Y 1Y 2 . . . Y mY m+1 . . . Y n

where Z k, 1 ≤ k ≤ m are new nonterminals.Introduction to Formal Languages, Automata and Computability – p.32/74

contd


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contd..

Each production that is not context-sensitive is to bereplaced by a set of context-sensitive productions asmentioned above. Application of this set of rules hasthe same effect as applying X 1 . . . X m → Y 1 . . . Y n.Hence a new grammar G thus obtained iscontext-sensitive that is equivalent to G.Example Let L = {anbmcndm/n,m ≥ 1}.The type 1 grammar generating this CSL is given byG = (N , T , P , S ) with N = {S,A,B,X,Y },

T = {a,b,c,d}


contd


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contd..

and P =

S → aAB|aB

A → aAX |aX

B → bBd|bY d

Xb → bX XY → Y c

Y → c.


contd


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contd..

Sample Derivations

S ⇒ aB ⇒ abY d ⇒ abcd.

S ⇒ aB ⇒ abBd ⇒ abbY dd ⇒ ab2cd2.S ⇒ aAB ⇒ aaXB ⇒ aaXbY d

⇒ aabXY d

⇒ aabY cd

⇒ a2bc2d.


contd


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contd..

S ⇒ aAB ⇒ aaAXB

⇒ aaaXXB⇒ aaaXXbY d

⇒ aaaXbXY d

⇒ aaabXXY d⇒ aaabXY cd

⇒ aaabY ccd

⇒ aaabcccd.


Exercise


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Exercise

Consider the length increasing grammar withproductions P =S → aSBc, S → abc, cB → Bc, bB → bb. All rulesexcept the rule cB → Bc are context-sensitive. Thefollowing grammar is a context-sensitive grammarequivalent to the above grammar.

S → aSBC

S → abc

C → cCB → DB

DB → DC

DC → BC.Introduction to Formal Languages, Automata and Computability – p.37/74

Ambiguity


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Ambiguity

Definition Let G = (N , T , P , S ) be a CFG. A word win L(G) is said to be ambiguously derivable in G, if it

has two or more different derivation trees in G.Since the correspondence between derivation treesand leftmost derivations is a bijection, an equivalentdefinition in terms of leftmost derivations can be

given.


Ambiguity


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Ambiguity

Definition Let G = (N , T , P , S ) be a CFG. A word win L(G) is said to be ambiguously derivable in G, if ithas two or more different derivation trees in

G.

Since the correspondence between derivation treesand leftmost derivations is a bijection, an equivalentdefinition in terms of leftmost derivations can be

given.DefinitionLet G = (N , T , P , S ) be a CFG. A word win L(G) is said to be ambiguously derivable in G, if ithas two or more different leftmost derivations in G.


Ambiguity


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g y

Definition Let G = (N , T , P , S ) be a CFG. A word win L(G) is said to be ambiguously derivable in G, if ithas two or more different derivation trees in

G.

Since the correspondence between derivation treesand leftmost derivations is a bijection, an equivalentdefinition in terms of leftmost derivations can be

given.DefinitionLet G = (N , T , P , S ) be a CFG. A word win L(G) is said to be ambiguously derivable in G, if ithas two or more different leftmost derivations in G.

DefinitionA CFG is said to be ambiguous if there is a

word w in L(G) which is ambiguously derivable. Oth-erwise it is unambiguous. Introduction to Formal Languages, Automata and Computability – p.38/74

Example


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p

Consider the grammar G with rules S → SS, S → awhere S is the nonterminal and a is the terminalsymbol. L(G) = {an/n ≥ 1}.This grammar isambiguous as a3 has two different derivation trees asfollows


Example


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p

Consider the grammar G with rules S → SS, S → awhere S is the nonterminal and a is the terminalsymbol. L(G) = {an/n ≥ 1}.This grammar isambiguous as a3 has two different derivation trees asfollows

S

S

a

a

S

S

S

a

S

S

a

a

S

S

S

a


Ambiguity contd..


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g y

Definition A CFL L is said to be inherently ambiguous if all the grammars

generating it are ambiguous or in other words, there is no unambiguous

grammar generating it.

Example L = {an

bn

c p

/n,m,p ≥ 1, n = m or m = p}.This can be looked at as L = L1 ∪ L2

L1 = {anbnc p/n,p ≥ 1}

L2 = {anbmcm/n,m ≥ 1}.

L1 and L2 can be generated individually by unambiguous grammars, but any

grammar generating L1 ∪ L2 will be ambiguous. Since strings of the form

anbncn will have two different derivation trees, one corresponding to L1 and

another corresponding to L2. Hence L is inherently ambiguous.


Ambiguity contd..


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g y

Definition A CFL L is said to be inherently ambiguous if all the grammars

generating it are ambiguous or in other words, there is no unambiguous

grammar generating it.

Example L = {an

bn

c p

/n,m,p ≥ 1, n = m or m = p}.This can be looked at as L = L1 ∪ L2

L1 = {anbnc p/n,p ≥ 1}

L2 = {anbmcm/n,m ≥ 1}.

L1 and L2 can be generated individually by unambiguous grammars, but any

grammar generating L1 ∪ L2 will be ambiguous. Since strings of the form

anbncn will have two different derivation trees, one corresponding to L1 and

another corresponding to L2. Hence L is inherently ambiguous.


contd..


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co td..

Theorem It is undecidable to determine whether agiven CFG G is ambiguous or not.


contd..


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Theorem It is undecidable to determine whether agiven CFG G is ambiguous or not.Theorem It is undecidable to determine whether a

given CFL L is ambiguous or not.


contd..


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given CFL L is ambiguous or not.DefinitionA CFL L is bounded if there exists stringsw1, . . . , wk such that L ⊆ w

∗1w

∗2 . . . w

∗k.


contd..


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∗1w

∗2 . . . w

∗k.

TheoremThere exists an algorithm to find out whethera given bounded CFL is inherently ambiguous or not.


contd..


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∗1w

∗2 . . . w

∗k.

TheoremThere exists an algorithm to find out whethera given bounded CFL is inherently ambiguous or not.

DefinitionLet G = (N , T , P , S ) be a CFG then the de-

gree of ambiguity of G is the maximum number of

derivation trees a string w ∈ L(G) can have in G.


contd.


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We can also use the idea of power series and find outthe number of different derivation trees a string canhave. Consider the grammar with rules

S → SS, S → a write an equation S = SS + aInitial solution is S = a, S 1 = aUse this in the equation for S on the right-hand side

S 2 = S 1S 1 + a

= aa + a.

S 3 = S 2S 2 + a= (aa + a)(aa + a) + a

= a4 + 2a3 + a2 + a.


contd.


57/88

S 4 = S 3S 3 + a

= (a

4

+ 2a

3

+ a

2

+ a)

2

+ a= a8 + 4a7 + 6a6 + 6a5 + 5a4 + 2a3 + a2 + a.

We can proceed like this using S i = S i−1S i−1 + aIn S i, upto strings of length i, the coefficient of the

string will give the number of different derivation trees

it can have in G. For example, in S 4 coefficient of a4 is

5 and a4 has 5 different derivation trees in G. The co-

efficient of a3 is 2 - the number of different derivationtrees for a3 is 2 in G. Introduction to Formal Languages, Automata and Computability – p.43/74

Simplification


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Definition Let G = (N , T , P , S ) be a CFG. A variable X in N is said

to be useful if and only if there is at least a string α ∈ L(G) such that

S

∗

⇒ α1Xα2∗

⇒ α,

where α1, α2 ∈ (N ∪ T )∗ i.e., X is useful because it appears in at least

one derivation from S to a word α in L(G). Otherwise X is useless.

Consequently the production involving X is useful.

One can understand the ‘useful symbol’ concept in two steps.

Step 1 For a symbol X ∈ N to be useful it should occur in some

derivation starting from S , i.e., S ∗⇒ α1Xα2.


contd.


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Step 2 Also X has to derive a string α ∈ T ∗ i.e.,

X ∗⇒ α

These two conditions are necessary. But they are notsufficient. These two conditions may be satisfied stillα1 or α2 may contain a nonterminal from which aterminal string cannot be derived. So the usefulness of

a symbol has to be tested in two steps as above.

Lemma Let G = (N , T , P , S ) be a CFG such that

L(G) = φ. Then there exists an equivalent context-free grammar G = (N , T , P , S ) that does not con-

tain any useless symbol or productions.


contd.


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The context-free grammar G is obtained by thefollowing elimination proceduresI. First eliminate all those symbols X such that X does not derive any string over T ∗. LetG2 = (N 2, T , P 2, S ) be the grammar thus modified.As L(G) = φ, S will not be eliminated. The followingalgorithm identifies symbols not to be eliminated.Algorithm GENERATINGStep 1 Let GEN = T ;

Step 2 If A → α and every symbol of α belongs toGEN , then add A to GEN .

Remove from N all those symbols that are not in the

set GEN and all the rules using them.Introduction to Formal Languages, Automata and Computability – p.46/74

contd.


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Let the resultant grammar be G2 = (N 2, T , P 2, S ).

II. Now eliminate all symbols in the grammar G2 that are not occurring

in any derivation from ‘S ’. i.e., S ∗

⇒ α1Xα2.

Algorithm REACHABLE

Let REACH = {S }.

If A ∈ REACH and A → α ∈ P then every symbol of α is in

REACH .

The above algorithm terminates as any grammar has only finite set of

rules. It collects all those symbols which are reachable from S through

derivations from S .

Now in G2 remove all those symbols that are not in REACH and

also productions involving them. Hence one gets the modified gram-

mar G = (N , T , P , S ), a new CF G. Introduction to Formal Languages, Automata and Computability – p.47/74

contd.


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without useless symbols and productions using them.The equivalence of G with G can be easily seen sinceonly symbols and productions leading to derivation of

terminal strings from S are present in G. HenceL(G) = L(G).Theorem Given a CFG G = (N , T , P , S ). Procedure Iof the previous lemma is executed to getG2 = (N 2, T , P 2, S ) and procedure II of previouslemma is executed to get G = (N , T , P , S ). Then

G contains no useless symbols.

Suppose G contains a symbol X (say) which is use-

less. It is easily seen that N

⊆ N 2, T

⊆ T , P

⊆ P 2.Introduction to Formal Languages, Automata and Computability – p.48/74

contd.


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Since X is obtained after execution of II,

S ∗⇒ α1Xα2, α1, α2 ∈ (N

∪ T )∗. Every symbol of

N

is also in N 2

. Since G2

is obtained by execution of I, it is possible to get a terminal string from every

symbol of N 2 and hence from N : α1

∗⇒ w1 and

α2∗

⇒ w2, w1, w2 ∈ T ∗

. ThusS ∗⇒ α1Xα2

∗⇒ w1Xw2

∗⇒ w1ww2.

Clearly S is not useless as supposed. Hence G

contains only useful symbols.


Example


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Let G = (N , T , P , S ) be a CFG withN = {S,A,B,C }T = {a, b} andP = {S → Sa|A|C , A → a, B → bb, C → aC |B}First ‘GEN ’ set will be {S,A,B,a,b}. ThenG

2 = (N

2, T , P

2, S ) where N

2 = {S,A,B}

P = {S → Sa|A, A → a, B → bb}In G2, REACH set will be {S,A,a}. Hence

G

= (N

, T

, P

, S ) whereN = {S, A}T = {a}

P

= {S → Sa|A, A → a}L(G) = L(G) = {an|n ≥ 1}. Introduction to Formal Languages, Automata and Computability – p.50/74

-rule Elimination


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Definition Any production of the form A → is called

a -rule. If A ∗⇒ , then we call A a nullable symbol.

Theorem Let G = (N , T , P , S )

be a CFG such that

/∈ L(G). Then there exists a CFG without -rulesgenerating L(G).Before modifying the grammar one has to identify theset of null-able symbols of G. This is done by thefollowing procedure.Algorithm NULL

1. Let N U LL := φ,

2. If A → ∈ P , then A ∈ N U LL,

3. If A → B1 . . . Bt ∈ P , and each Bi is in N U LL,then A ∈ N U LL Introduction to Formal Languages, Automata and Computability – p.51/74

contd.


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Run algorithm N U LL for G and get the N U LL set.The modification of G to get G = (N , T , P , S ) withrespect to N U LL is given below.If A → A1 . . . At ∈ P , t ≥ 1 and if n (n < t) of theseAis are in N U LL. Then P will contain 2

n rules wherethe variables in N U LL are either present or absent in

all possible combinations. If n = t then removeA → from P . The grammar G = (N , T , P , S ) thusobtained is -free. To prove that a word w ∈ L(G) if

and only if w ∈ L(G). As G and G do not differ inN and T , one can equivalently show that,

A ∗

⇒G wIntroduction to Formal Languages, Automata and Computability – p.52/74

contd.


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if and only if

A ∗⇒G w

for A ∈ N . Clearly w = . Let A n⇒G w. n =

1, A ∗⇒G w then A → w is in P and hence in P

.

Then A ∗

⇒G w. Assume the result to be true for allderivations from A of length n − 1. Let A

n⇒G w

i.e., A ⇒G

Y 1, Y 2, . . . , Y kn−1

=⇒G w. Let w = w1 . . . wk

and let X 1, X 2, . . . , X m be those Y j’s in order such that

Y j∗

⇒ w j, w j = . Clearly k ≥ 1 as w = . Hence

A → X 1 . . . X m is a rule in G.Introduction to Formal Languages, Automata and Computability – p.53/74

contd.


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We can see that X 1 . . . X m∗⇒G w as some Y j derive

only . Since each Y j∗

⇒ w j, w j takes fewer than n

steps, by induction, Y j ∗⇒ w j, for w j = . HenceA ⇒G X 1 . . . X m

∗⇒ w.

Conversely if A ∗⇒

G w. Then to show that A

∗⇒

G w

also. Again the proof is by induction on the number of derivation steps.Basis If A ⇒

G

w, then A → w ∈ G. By the

construction of G, one can see that there exists a ruleA → α in G such that α and w differ only in nullable

symbols. Hence A ⇒G α ∗

⇒G w where for α ∗

⇒ w only-rules are used. Introduction to Formal Languages, Automata and Computability – p.54/74

contd.


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Induction

Assume A n⇒G w n > 1. Then let

A ⇒G

Y 1

. . . Y k

∗⇒

G w. For A ⇒

GY 1

. . . Y k

the

corresponding equivalent derivation by G will be

A ⇒

G

X 1 . . . X m∗

⇒G Y 1 . . . Y k as some of the X i are

nullable. Hence A ∗⇒G Y 1 . . . Y k

∗⇒G w1 . . . wk = w

by induction hypothesis. Hence A ∗⇒G w. Hence

L(G) = L(G

).


Example


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Consider G = (N , T , P , S ) whereN = {S,A,B,C,D}, T = {0, 1} andP = {S → AB0C , A → BC , B → 1|, C → D|,D → }The set N U LL = {A,B,C,D}.Then G will be withP = {S → AB0C |AB0|A0C |B0C |A0|B0|0C |0,

A → BC |B|C, B → 1, C → D}.


Procedure to Eliminate Unit Rules


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Definition Any rule of the form X → Y , X, Y ∈ N iscalled a unit rule. Note that A → a with a ∈ T is nota unit rule.

In simplification of context-free grammars anotherimportant step is to make the given context-freegrammar unit rule free i.e., to eliminate unit rules.

This is essential for the application in compilers. Asthe compiler spends time on each rule used in parsingby generating semantic routines, having unnecessary

unit rules will increase compiler time.Lemma Let G be a CFG, then there exists a CFG G

without unit rules such that L(G) = L(G).


contd.


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Let G = (N , T , P , S ) be a CFG. First find the sets of pairs of

nonterminals (A, B) in G such that A ∗⇒ B by unit rules. Such a pair

is called a unit-pair.

Algorithm UNIT-PAIR

1. (A, A) ∈ U N I T − PAIR, for every variable A ∈ N .

2. If (A, B) ∈ U N I T − PAIR and B → C ∈ P then(A, C ) ∈ U N I T − PAIR

Now construct G

= (N , T , P

, S ) as follows. Remove all unit produc-tions. For every unit-pair (X, Y ), if Y → α ∈ P is a nonunit rule, add

to P , X → α. Thus G has no unit rules.


contd.


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Let us consider leftmost derivations in G and G. If w ∈ L(G), then

S ∗⇒G w. Clearly there exists a derivation of w from S by G where

zero or more applications of unit rules are used. Hence S ∗⇒G w whose

length may be different from S ∗⇒G w.

If w ∈ L(G), then one can consider S ∗⇒G w by G. A sequence of

unit-productions applied in this derivation is to be written by a nonunit

production. Since this is a leftmost derivation, for such sequence there

exists one rule in G doing the same job. Hence one can see the simula-

tion of a derivation of G with G.


contd.


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Hence L(G) = L(G).Example Let G = (N , T , P , S ) be a CFG whereN = {X, Y }, T = {a, b} andP = {X → aX |Y |b, Y → bK |K |b, K → a}UNIT − PAIR ={(X, X ), (Y, Y ), (K, K ), (X, Y ), (Y, K ), (X, K )}Then G = (N , T , P , S ) whereP = {X → aX |bK |b|a, Y → bK |b|a, K → a}.

Remark Since removal of -rule can introduce unitproductions, to get a simplified CFG to generate

L(G) − {}, the following steps have to be used in

the order given.Introduction to Formal Languages, Automata and Computability – p.60/74

contd.


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1. Remove -rules

2. Remove unit-rules

3. Remove useless symbols.(i) Remove symbols not deriving terminal

strings.

(ii) Remove symbols not reachable from S .

Example It is essential that steps 3(i) and 3(ii) have tobe executed in that order. If step 3(ii) is executed firstand then step 3(i), we may not get the requiredreduced grammar. Consider the CFGG = (N , T , P , S ) where N = {S,A,B,C },

T = {a, b} and P = {S → ABC |a, A → a, C → b}= Introduction to Formal Languages, Automata and Computability – p.61/74

contd.


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Applying step 3(ii) first removes nothing. Then applystep 3(i) which removes B and S → ABC leavingS → a, A → a, C → b. Though A and C do not

contribute to the set L(G), they are not removed.On the other hand applying step 3(i) first, removes B,

S → ABC . Afterwards apply step 3(ii), removesA,C,A → a, C → b. Hence S → a is the only rule

left which is the required result.


Normal form


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The most popular normal forms are Weak ChomskyNormal Form, Chomsky Normal Form, StrongChomsky Normal Form, Greibach Normal Form.

Definition Let G = (N , T , P , S ) be a CFG. If eachrule in P is of the form A → ∆, A → a or A → ,where A ∈ N , ∆ ∈ N +, a ∈ T , then G is said to be in

Weak Chomsky Normal Form (WCNF).Example Let G = (N , T , P , S ) be a CFG whereN = {S,A,B}, T = {a, b} and

P = {S → ASB|AB, A → a, B → b}. G is inWCNF.


contd.


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Theorem For any CFG G = (N , T , P , S ) there exists aCFG G in WCNF such that L(G) = L(G).Let G = (N , T , P , S ) be a CFG. One can construct anequivalent CFG in WCNF as below. LetG = (N , T , P , S ) be an equivalent CFG whereN = N ∪ {Aa/a ∈ T }, none of Aa’s belong to N .P = {A → ∆|A → α ∈ P and every occurrence of asymbol from T present in α is replaced by Aa, giving∆} ∪ {Aa → a|a ∈ T }. Clearly ∆ ∈ N

+ and P getsthe required form. G is in WCNF. That G and Gequivalent can be seen easily.


Chomsky Normal Form


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Definition Let /∈ L(G) and G = (N , T , P , S ) be a CFG. G is said to

be in Chomsky Normal Form (CNF) if all its productions are of the

form A → BC or A → a, A,B,C ∈ N, a ∈ T .

Example The following CFGs are in CNF.

1. G1 = (N , T , P , S ) where N = {S,A,B,C }, T = {0, 1} and

P = {S → AB|AC |SS , C → SA, A → 0, B → 1}.

2. G2 = (N , T , P , S ) where N = {S,A,B,C }, T = {a, b} andP = {S → AS |SB, A → AB|a, B → b}.

No CFG in CNF can generate . If is to be added to L(G), then a new

start symbol S

is to be taken and S

→ should be added. For everyrule S → α, S → α should be added which make sure that the new

start symbol does not appear on the right-hand side of any production.


contd.


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Theorem Given CFG G, there exists an equivalent CFG G in CNF.

Let G = (N , T , P , S ) be a CFG without rules, unit-rules and useless

symbols and also /∈ L(G). Modify G to G such that

G = (N , T , P , S ) is in WCNF. Let A → ∆ ∈ P . If |∆| = 2, such

rules need not be modified. If |∆| ≥ 3, the modification is as below:

If A → ∆ = A1A2A3, the new set of equivalent rules will be:

A → A1B1

B1 → A2A3.

Similarly if A → A1A2 . . . An ∈ P , it is replaced by


contd.


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A → A1B1

B1 → A2B2

...

Bn−2 → An−1An.

Let P be the collection of modified rules andG = (N , T , P , S ) be the modified grammar whichis clearly in CNF. Also L(G) = L(G).


contd.


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Example Let G = (N , T , P , S ) be a CFG whereN = {S,A,B}, T = {a, b}P = {S → SAB|AB|SBC , A → AB|a,B → BAB|b, C → b}.Clearly G is not in CNF but in WCNF. Hence themodification of rules in P are as below:

For S → SAB, the equivalent rules areS → SB1, B1 → AB.For S → SBC , the equivalent rules are

S → SB2, B2 → BC .For B → BAB, the equivalent rules areB → BB3, B3 → AB.


contd.


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Hence G = (N , T , P , S ) will be withN = {S,A,B,C,B1, B2, B3}, T = {a, b}

P ={S → SB1|SB2|SA,B1 → AB, B2 → BC, B3 → AB,A → AB|a, B → BB3|b, C → b}.

Clearly G is in CNF.


Strong Chomsky Normal Form


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Definition A CFG G = (N , T , P , S ) is said to be inStrong Chomsky Normal Form (SCNF) when rules inP are only of the forms A → a, A → BC where

A,B,C ∈ N , a ∈ T subject to the followingconditions:(i) if A → BC ∈ P , then B = C .

(ii) if A → BC ∈ P , then for each ruleX → DE ∈ P , we have E = B and D = C .


contd.


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Theorem For every CFG G = (N , T , P , S ) thereexists an equivalent CFG in SCNF.

Let G = (N , T , P , S ) be a CFG in CNF. One can

construct an equivalent CFG.G = (N , T , P , S ) in SCNF as below.N = {S } ∪ {AL, AR|A ∈ N }

T = T


contd.


86/88

P ={AL → BLC R, AR → BLC R|A → BC ∈ P }

∪ {S

→ X LY R|S → XY ∈ P }∪ {S → a|S → a ∈ P }

∪ {AL → a, AR → a|A → a ∈ P, a ∈ T }.

Clearly L(G) = L(G) and G is in SCNF.


contd.

E l L ( ) b CFG h


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Example Let G = (N , T , P , S ) be a CFG whereN = {S,A,B}, T = {0, 1} andP = {S → AB|0, B → BA|1, A → AB|0}.Then G = (N , T , P , S ) in SCNF will be with

N = {S , S L, S R, AL, AR, BL, BR}

T = {0, 1}P = {S → ALBR|0, S L → ALBR|0,

S R → ALBR|0, AL → ALBR|0,AR → ALBR|0, BL → BLAR|1,

BR → BLAR|1}.


Greibach Normal Form

D fi i i L / L(G) d G (N T P S) b


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Definition Let /∈ L(G) and G = (N , T , P , S ) be aCFG. G is said to be in Greibach Normal Form(GNF), if each rule in P rewrites a variable into a

word in T N ∗ i.e., each rule will be of the formA → aα, a ∈ T , α ∈ N ∗.


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