theory of computation unit-ii - rgcetpdy.ac.in year/theory of computation/uni… · equivalence of...

Theory Of Computation UNIT-II

J. Veerendeswari/IT/RGCET

Regular Expressions and Context Free Grammars: Regular expression formalism- equivalence with finite

automata-regular sets and closure properties- pumping lemma for regular languages- decision algorithms for

regular sets- applications. Context-Free Grammars – derivation trees, , ambiguous and unambiguous grammars-

equivalence of regular grammar and finite automata- Chomsky Normal Forms and Greibach Normal Forms

pumping lemma for Context free languages – applications.

REGULAR SETS

Regular languages can also be defined, from the empty set and from some finite number of singleton sets, by the

operations of union, composition, and Kleene closure. Specifically, consider any alphabet. Then a regular set

over is defined in the following way.

The empty set Ø, the set {} containing only the empty string, and the set {a} for each symbol a in Σ, are

regular sets.

If L1 and L2 are regular sets, then so are the union L1 L2, the composition L1L2, and the Kleene closure

L1*.

No other set is regular.

The Pumping Lemma for Regular Sets

Applications:

o The pumping lemma is a powerful tool for providing and proving certain language is regular or

not.

o It is also useful in the development of algorithms to answer certain questions concerning finite

automata, such as whether the language accepted by a given FA is finite or infinite.

Statement:

Let “L” be a regular set. Then there is a constant n such that if z is any word in L, and |z|≥n. We may

write z=uvw in such a way that |uv|≤n, |v|≥1 and for all i≥0, then uviw is in L.

Proof:

If a language is accepted by a DFA M= (Q, Σ, δ, q0, F) then it is regular with some particular number of

states, say n. Consider a input of n or more symbols a1a2…..am, m≥n and for i=1,2,…..,m.

Let δ (q0, a1a2…..ai) = qi

It is not possible for each of the n+1 states q0,q1,……qn to be distinct, since there are only different states. This

there are two integers j and k, 0≤j<k≤n, such that qj=qk. The path labeled a1a2….am in the transition diagram of

M. Since j<k, the string aj+1....ak is of length 1.

Aj+1…….ak

a1…….ajak+1………am

If qm is in F, that is a1a2……..am is in L(n), then a1a2……..ajak+1ak+2……am is also in L(n).

δ(q0,a1……..ajak+1……am)= δ(δ(q0,a1……..aj) ak+1……am)

=δ(qj, ak+1……am) (since δ(q0,a1……..aj)= qj)

= δ(qk, ak+1……am) (since qj= qk)

=qm

which is the final state of the automata to be accepted.

q0 qj=qk qm qm



Hence proved.

PROPERTIES OF REGULAR SETS

Property 1:

The regular sets are closed under union, concatenation, and Kleene closure.

Proof:

Let Σ be a finite set of symbols and let L, L1 and L2 be sets of strings from Σ*. The concatenation of L1

and L2, denoted L1L2, is the set {xy| x is in L1and y is in L2}. That is, the strings in L1L2 are formed by choosing a

string L1 and following it by a string in L2, in all possible combinations. Define L0= {ε} and L

i=LL

i-1 for i≥1. The

Kleene closure of L denoted L*, is the set

L*=∑i=0L

i

and the positive closure of L, denoted L+, is the set

L+=∑i=1L

i

That is, L* denotes words constructed by concatenating any number of words from L. L

+ is the same, but the case

of zero words, whose concatenation is defined to be ε, is excluded. Note that L+ contains ε if and only if L does.

Property 2:

The class of regular sets is closed under complementation. That is, if L is a regular set and L€Σ*, and then

Σ*-L is a regular set.

Proof:

Let L be L (M) for DFA M= (Q, Σ1, δ, q0, F) and L€Σ*. First, we may assume Σ1=Σ, for if there are

symbols in Σ1 not in Σ, we may delete all transitions of M on symbols not in Σ. The fact that L€Σ* assures us that

we shall not thereby change the language of M. If there are symbols in Σ not in Σ1, then none of these symbols

appear in words of L. We may therefore introduce a “dead state” d into M with δ (d, a) = d for all a in Σ and δ (q,

a) = d for all q in Q and a in Σ-Σ1.

Now, to accept Σ*-L, complement the final states of M. That is, let M

‟= (Q, Σ1, δ, q0, Q-F). Then M

‟

accepts a word w if and only if δ (q0, w) is in Q-F, that is, w is in Σ*-L. Note that it is essential to the proof that M

is deterministic and without ε moves.

Property 3:

The regular sets are closed under intersection.

Proof:

L1∩L2= (L1‟U L2

‟)‟, where the overbar denotes complementation with respect to an alphabet including the

alphabets of L1 and L2.Closure under intersection then follows from closure under union and complementation.

It is worth noting that a direct construction of a DFA for the intersection of two regular sets exists. The

construction involves taking the Cartesian product of states, and we sketch the construction as follows:

Let M1= (Q1, Σ, δ1, q1, F1) and M2= (Q2, Σ, δ2, q2, F2) be two deterministic finite automata. Let

M= (Q1* Q2, Σ, δ, [q1, q2], F1* F2)

where for all p1 in Q1, p2 in Q2, and a in Σ,

δ ([p1, p2], a) = [δ1 (p1, a), δ2 (p2, a)]

Property 4:

The class of regular sets is closed under substitution.



Proof:

Let R €Σ* be a regular set and for each a in Σ let Ra€∆

* be a regular set. Let f: Σ->∆

* be the substitutions

denoting R and Ra. Replace each occurrence of the symbol a in the regular expression R by the regular

expression for Ra. To prove that the resulting regular expression denotes f®, observe that the substitution of a

union, product or closure is the union, product or closure of the substitution. [Thus for example, f (L1 L2) = f

(L1)Uf (L2).] A simple induction on the number of operators in the regular expression completes the proof.

A type of substitution that is of special interest is the homomorphism. A homomorphism h is a

substitution such that h(a) contains a single string for each a. We generally take h(a) to be the string itself, rather

than the set containing that string. It is useful to define the inversehomomorphic image of a language L to be

h-1

(L)={λ| h(x) is in L}

We also use, for string w;

h-1

(w)={λ| h(w) = w}

Property 5:

The class of regular sets is closed under homomorphism and inverse homomorphism.

Proof:

Closure under homomorphism follows immediately from closure under substitution, since every

homomorphism is a substitution, in which h(a) has one member.

To show closure under inverse homomorphism, let M= (Q, Σ, δ, q0, F) be a DFA accepting L, and let h be

a homomorphism from ∆ to Σ*. We construct a DFA M

‟ that accepts

h-1

(L) by reading symbol a in ∆ and simulating M on h(a). Formally, let M‟= (Q, Σ, δ

‟, q0, F) and define δ

‟(q, a)

for q in Q and a in ∆ to be δ(q, h(a)). Note that h(a) may be long string, or ε, but δ is defined on all strings by

extension. It is easy to show by induction on |x| that δ‟(q0,x)=δ(q,h(x)). Therefore M

‟ accepts x if and only if M

accepts h(x). That is,

L(M‟)= h

-1(L(M)).

Property 6:

The class of regular sets is closed under quotient with arbitrary sets.

Proof:

Let M= (Q, Σ, δ, q0, F) be a finite automaton accepting some regular set R, and let L be an arbitrary

language. The quotient R/L is accepted b a finite automaton M‟= (Q, Σ, δ, q0, F

‟), which behaves like M except

that the final states of M‟ are all states q of M such that δ (q0, xy) is in F. Thus M

‟ accepts R/L.

CONTEXT FREE GRAMMARS

A context-free grammar is a 4-tuple (V, P, T, S), where

1. V is a finite set called the variables,

2. T is a finite set, disjoint from V, called the terminals,

3. P is a finite set of Productionrules, with each rule being a

variable and a string of variables and terminals, and

4. S is the start symbolε V



Rules to be followed while writing a CFG.

1. A single non terminal should be at LHS

2. Production should be always in the form of LHS RHS where RHS may be combination

ofnon terminal and terminal symbol.

3. The null derivation can be specified as NT ε

4. One of the NT should be start symbol.

Any language that can be generated bysome context-free grammar is called a context-free language (CFL).

Derivation

Derivation from S means generation of string ω from S

Types of derivation

1. Left most derivation

The derivation in which the left most non terminal is always replaced at each step (choose

leftmost non terminal in a sentential form )

Example : Given the grammar ( set of productions)

E E + E

E E * E

E id

Obtain the left most derivation for the string id*id

E => E * E => id * E => id * id

2. Right most derivation

The derivation in which the right most non terminal is always replaced at each step (choose

leftmost non terminal in a sentential form)


E E + E

E E * E

E id

Obtain the left most derivation for the string id*id

E => E * E => E * id => id * id

PARSE TREE OR DERIVATION TREE

The parse tree is the diagrammatic representation of a derivation, which can be defined in the following

way:

A vertex with a label which is a non terminal symbol is a parse tree.

If A → y1y2 … yn is a rule in R, then the treeis a parse tree




E E + E

E E * E

E ( E)

E id

Obtain the left most derivation for the string id*id+id

Left Most Derivation Tree

Right Most Derivation Tree

Example: Consider the following production

S->aB/bA

A->aS/bAA/a

B->bS/aBB/b

for string aaabbabbba find LMD and RMD

Left Most Derivation:

S->aB

->aaBB

->aaaBBB

->aaabBB

->aaabbB

->aaabbaBB

->aaabbabB



->aaabbabbS

->aaabbabbbA

->aaabbabbba

Right Most Derivation:

S->aB

->aaBB

->aaBbS

->aaBbbA

->aaBbba

->aaaBBbba

->aaaBbbba

->aaabSbbba

->aaabbAbbba

->aaabbabbba

SIMPLIFICATION OF CFG

All the grammars are not always optimized. That means grammar may consists of some extra symbol(non

terminal). Having extra symbols unnecessary increases the length of the grammar .simplification of grammar

means reduction of grammar by removing useless symbol.

The properties of the reduced grammar:

1. Each variable (ie non terminal) and each terminal of G appears in the derivation of some word in L.

2. There should not be any production as XY where X and Y arenon terminal

3. If 𝜀is not in the language L then there need not be the production X

Reduced grammar

Removal of useless symbol elimination of production removal of unit production

Removal of useless symbol

Any symbol is useful when it appears on the right hand side in the production rule and generates some

terminal string. If no such derivation exists then it is supposed to be an useless symbol.

A symbol p is useful if there exists some derivation in the following form

S=>aPB

apB=>W

Then p is said to be useful symbol.Where and may be some terminal or non terminal symbol and will help us

to derive certain string w in combination with p.

Example:1

G={V,T,P,S} where V={S,T,X}, T={0,1}

S0T|1T|X|0|1rule 1



T00 - rule 2

Now, in the above CFG the non terminal are S, T, X

To derive some string we have to start from start symbol S

S

0T S0T

000 T00

Thus we can reach to certain string after following these rules

But if SX then there is no further rule as a definition to X. That means there is no point in the rule

SX. hence we can declare that X is a useless symbol. And we can remove this so after removal of this

useless symbol CFG becomes

G=(V,T,P,S) where V={S,T}

T={0,1} and P={S0T|1T|0|1

T00}

S is start symbol

Example:2

Consider the CFG G= {V,T,P,S} where V={S,A,B} T={0,1}

P={SA11B|11A

SB|11

A0

BBB}

solution

Now in the given CFG if we try to derive any string A gives some terminal symbol as 0 but B does not

give any terminal string . By following the rules with B we simply get sample number of B and no

significant string.

Hence we can declare B as useless symbol and can remove the rules associated with it. Hence after

removal of useless symbol we get

S11A|11

A0

Elimination of 𝜀 production

If there is production we can remove it without changing the meaning of the grammar. Thus 𝜀production are not

necessary in a grammar.

Example:1

S0S|1S|𝜀

Then we remove production. But we have to take a care of meaning of CFG. Ie meaning of CFG should

not be get changed if we place S𝜀 in other rules we get S0 when S0S and S𝜀



As well as S1 when S1S and S𝜀

Hence we can rewrite the rules as

S0S|1S|0|1

Thus production is removed.

Example:2

SXYX

X0X|𝜀

Y1Y|𝜀

Now while removing production we are deleting the rules X𝜀 and Y𝜀 to preserve the meaning of CFG we

are actually placing 𝜀 at right hand side wherever X and Y have appeared.

Let us take

SXYX

If first X at right hand side is𝜀

Then SYX

Similarly if last X in RHS =𝜀

Then SXY

If Y= 𝜀 then

SXX

If Y and X are then

SX

SY when both X are replaced by 𝜀

SXY|YX|XX|X|Y

Now let us consider

X0X

If we place 𝜀 at right hand side for X then

X0

X0X|0

Y1Y|1

We can rewrite the CFG with removed 𝜀 production as

SXY|YX|XX|X|Y

X0X|0

Y1Y|1

Removing unit production:

The unit productions are the productions in which one non terminal gives another non terminal.

Eg X Y YZ

Then X,Y and Z are unit productions. To optimize the grammar we need to remove the unit production.

If A->B production we should add a rule A->x1x2x3x4…xn.



Example 1:

S0A|1B\C

A0S|00

B1|A

C01

SOLUTION:

Clearly SC is a unit production. But while removing SC we have to consider what C gives. So, we

can add a rule to S.

S0A\1B|01

Similarly BAis also a unit production so we can modify it as

B1|0S|00

Thus finally we can write CFG without unit production as

S0A|1B|01

A0S|00

B1|0S|00

C01

Example: 2

SA|0C1

AB|01|10

C𝜀 |CD

SOLUTION:

SABis a unit production

C is a null production.

CCD B and D are useless symbol.

Reducing a grammar we have to avoid all the above conditions

Let SAi.e. AB is a useless symbol because B is not defined further more

S01|10

i.e. S01|10|0C1

but C𝜀

hence ultimately S01|10

A but we can remove this production since B is a useless symbol.

Hence A01|10

But the start symbol S01\10

There is no A in the derivation of A so by considering A also as a useless symbol we get final CFG as

S01|10



CHOMSKY NORMAL FORM (CNF)

A Context-Free Grammar „G‟ is in Chomsky Normal Form if every production is of the form

Non Terminal->Terminal

Non Terminal-> Non Terminal*Non Terminal

The given CFG should be converted in the above format then we can say that grammar is in CNF. Before

converting the grammar into CNF it should be in reduced form. That means remove all useless symbol, ε

production and unit production from it. Then this reduced grammar is converted into CNF.

For Example:

Consider “G” whose productions are S->AB, A->a, B->b. Then G is in Chomsky Normal Form.

Reduction to Chomsky Normal Form

The steps involved in reduction of CFG to CNF are

Step 1: Elimination of null productions and unit productions.

Let the resultant grammar be G= (VN, Σ, P, S)

Step 2: Elimination of terminals on RHS.

We define G= (VN‟, Σ, P, S‟) where P and VN

‟ are constructed as follows:

(i)All the productions in P of the form A->a or A->BC are included in P1. All the variables in VN are

included in VN‟.

(ii) Consider A->X1X2…….Xn with some terminal on RHS of Xi is a terminal say ai, add a new variable

Cai to VN‟ and Ca->αi to P1. In production A->X1X2……Xn, every terminal on RHS is replaced by the

corresponding new variables on the RHS are retained. The resulting production is added to P1. Thus we get G1=

(VN‟, Σ, P1, S).

Step 3: Restricting the number of variables on RHS.

For any production in P1, the RHS consists of either a single terminal or two or more variables. We

define G= (VN‟, Σ, P2, S) as follows:

(i)All productions in P1 are added to P2 if they are in the required form. All the variables in VN‟‟.

(ii) Consider A->A1c1, C1->A2C2, cm-2->Am-1Am and new variables are c1, c2, ….., cm-2.

Example 1:

Find the grammar in CNF equivalent to SaAbB, AaA/a, BbB/b

Step 1:As there are no unit productions or null productions, we need not carry out step 1. We proceed to

step 2.



Step 2: Let G1= (VN‟, {a, b}, P, S) where P1 and VN

‟ are constructed.

Add productions A->a, B->b to P1.

Hence S->aAbB ~ (1)

replaced as

a) S->CaACbB

Add new productions Ca->a and Cb->b

b) A->aA becomes

A->CaA

c) B->bB becomes

B->CbB

Step 3:

a) S->CaACbB is converted as

S->CaD1 where

D1->ACbB

Again converted as D1->AD2

where D2->CbB

b) A->CaA

They are already in CNF

c) B->CbB

They are already in CNF

Hence the resultant CNF are

Example 2:

Find the CNF equivalent to the grammar S~S/[S∩S]/p/qwhere ~, [, ∩,], p, q are terminals.

Step 1: Consider the Grammar “G”.

S->~S/[S∩S]/p/q

Since there are no null productions and unit productions we can go for reduction.

S->CaD1

D1->AD2

D2->CbB

A->CaA

B->CbB

A->a

B->b



Step 2: Consider the production (1),

S->~S

Add new production C1->~

The production S->~S becomes S->C1S

Step 3: Consider the production (2),

S-> [S∩S]

Add new productions C2-> [

C3->∩

C4->]

Now S->C2SC3SC4

S-> C2D1

where D1-> SC3SC4

and D2-> C3SC4

So D1-> SD2

Again D2-> C3SC4

Let D3-> SC4

And hence D2-> C3D3

The resultant CNF productions are

Hence derived.

Note: NT-Terminal T-Terminal

1. Convert the following grammar to Chomsky Normal Form(CNF)

G:

SaAD

AaB/bAB

Bb

Dd

S->C1S

C1->~

S->C2D1

D1->SD2

D2->C3D3

D3->SC4

S->p

S->q



Solution:

Consider SaAD

The production is of the form NTT*NT*NT

Replace a by Ca. We get, SCaAD

Let D1AD. Then the production becomes SCa D1 which is in CNF.

Now consider AaB

This is of the form NTT*NT

Since Caa, A->aB becomes ACaB which is in CNF.

Let us now consider AbAB.

This production is of the form NTT*NT*NT

Replace b by Cb and AB by D2. We get, ACb D2.

The other two productions Bb and Dd are already in CNF.

Hence the resultant CNF productions are

2. Convert the following grammar to Chomsky Normal Form (CNF)

G:

SabSb/a/aAb

AbS/aAAb

Solution:

Consider SabSb

This production is of the form NTT*NT*T*NT

Replacing a by Ca and b by Cb we get, SCaCbSCb

This is of the form NTNT*NT*NT*NT.

So we replace CbSCb by D1. Hence the production becomes, SCa D1 which is of the CNF.

(Replace SCbby D to convert D1 into CNF. So D->SCb)

Since Sa is already in CNF we now consider SaAB

The above production is of the form NT->T*NT*T. Replace a by Ca and b by Cb.

So now, SCaACb.Replace ACbby D2 to convert the production to CNF.

The production now becomes SCa D2.

S->CaD1

D1->AD

A->CaB

A->CbD2.

D2.->AB

B->b

D->d



Let us now consider the production AbS. Replacing b by Cb the grammar gets converted into CNF,

ACbS.

The last production to be converted to CNF is AaAAb.

Replacing Ca and b by Cb we get ACaAACb. Let us now consider D3AD2 where

D2ACb.So the production becomes ACa D3 which is in CNF.



G:

SASA/bA

AB/S

BC

Solution:

Consider SASA.

The above production is of the form NTNT*NT*NT

Let D1SA. Replacing SA the production becomes SAD1 which is in CNF.

Now consider SbA. Replace b by Cb in order to convert the production to CNF.

S->bA becomes SCbA.

Consider the productions AB, AS and BC. They can also be written as AεB, AεS, BεC. If we

assume Caε then the productionsbecome ACaB, ACaS, BCaC which are in CNF.


D->SCb

D1->CbD

S->CaD1

D2->ACb

S->CaD2

A->CaD3

D3->AD2

D2->ACb

S->a

A->CbS.

D2.->AB

B->b

D->d




G:

S1A/0B

A1AA/0S/0

B0BB/1S/1

Solution:

Consider S1A. The above production is of the form NTT*NT. In order to convert it to CNF, we

replace 1 by Ca. Upon replacing we get, SCaA which is in CNF.

Now consider S0B. The above production is of the form NTT*NT. In order to convert it to CNF, we

replace 0 by Cb. Upon replacing we get, SCbB which is in CNF.

The next production to be considered is A1AA which is of the form NTT*NT*NT. Let us assume

that D1AA. Also we know Ca1. The production becomes A->CaD1 which is in CNF.

The next production A0S is of the form NTT*NT. We know Cb0. Upon replacing we get, ACbS

which is in CNF. The other production A0 is already in CNF.

Now take into account the production B->0BB which is of the form NTT*NT*NT. Let us assume that

D2BB. Also we know Cb0. The production becomes BCbD2 which is in CNF.

The next production B1S is of the form NTT*NT. We know Ca1. Upon replacing we get, BCaS

which is in CNF. The other production B1 is already in CNF.

Hence the resultant CNF production

SAD1

D1SA

ACaB

SCbA

ACaS

BCaC

S->CaA

S->CbB

D1->AA

A->CaD1

A->CbS

A->0

B->CaS

B->0



GREIBACH NORMAL FORM (GNF)

A Context-Free Grammar „G‟ is in Greibach Normal Form if every production is of the form

Aaα where α € VN* and a € Σ

(Non TerminalTerminal*any number of Non Terminals)

Aa (Non TerminalTerminal)

For Example:

SaAB

AbC

Bb

Cc

are in GNF.

Greibach Normal Form

Algorithm:

begin

for k:=1 to m do

begin

for j=1 to k-1 do

for each production of the form Ak->Ajα do

begin

for all productions Aj->β do

add production Ak->βα;

remove production Ak->Ajα

end;

for each production of the form Ak->Akα do

begin

add production Bk->α and Bk->α Bk;

remove production Ak->Akα

end;

for each production Ak->β, where β does

not begin with Ak do

add production Ak->β Bk

end

end



Reduction to Greibach Normal Form

Lemma (1):

Let G= (V,P,T, S) be a CFG. Let ABγ be an A-production in P. Let the B-production be

BB1|B2|…..|Bn

Lemma (2):

Let G= (V, Σ, P, S) be a CFG. Let A->Bγ be an A-production be

A->Aα1|….. Aαγ| B1|…..|Bn

then z be a new variable when P1 is defined as:

(i)The set of A-productions in P1 are

A->β1|β2|….|βn

A-> β1z| β2z|…….| βnz

(ii)The set of z-productions in P1 are

z->α1|α2|…|αn

z->α1z|α2z|…|αnz

Example:

Construct Equivalent GNF for the CFG S->AA/a,A->SS/b

Step 1: The given grammar is in CNF. S and A are renamed as A1 and A2.

Hence the productions become A1-> A2 A2/a, A2-> A1 A1/b

There is no need for null production or unit production elimination because the production is

already in CNF.

Step 2:The A1 productions are in the required form. They are A1-> A2 A2/a.

The production A2->b is also in the required form whereas we have to apply lemma 1 to convert

the production A2-> A1A1to the required form. Applying lemma 1 we get,

A2-> A2A2 A1 A2-> aA1

Step 3: We apply lemma 2 to A2 productions as we have A2-> A2A2 A1

Let z2 be a new variable. The resulting productions are

A2-> aA1 A2->b

A2-> aA1z2 A2->bz2

z2->A2 A1 z2 -> A2 A1 z2

Step 4: (i) The A2 productions are A2-> aA1/ b/ aA1z2

(ii)Among the A1 productions we retain A1->a and eliminate A1-> A2A2using lemma 1.

The resulting productions are A1-> aA1 A2/bA2 , A1-> aA1z A2/bz2A2.

The set of all A1-productions is A1->a/ aA1 A2/bA2/ aA1z A2/bz2A2



Step 5: The z2 productions to be modified are z2 -> A2A1 ,z2 -> A2 A1 z2.

Applying lemma 1 we get z2 -> aA1A1/bA1 /aA1z2 A1/ bz2A1

z2-> aA1A1z2/bA1 z2 /aA1z2 A1 z2/ bz2A1 z2

Hence the equivalent grammar is

Note: NT-Terminal T-Terminal

1. Let us convert to Greibach normal form the grammar

G= ({A1, A2, A3}, {a, b}, P, A1), where P consists of the following:

A1->A2A3

A2->A3A1/b

A3->A1A2/a

Solution:

Step 1: Since the right-hand side of the productions for A1 and A2 start with terminals or highest-

numbered variables, we begin with the production A3->A2A3 is the only production with A1 on the left.

The resultant set of productions is: A1->A2A3

A2->A3A1/b

A3->A2A3A2/a

Since the right side of the production A3->A2A3A2 begins with a lower-numbered variable, we substitute

for the first occurrence of A2 both A3A1 and b. Thus A3->A2A3A2

is replaced by A3->A3A1A3A2 and A3->bA3A2. The new set is

A1->A2A3

A2->A3A1/b

A3-> A3A1A3A2 / bA3A2/a

We now apply lemma 2 to the productions

A3-> A3A1A3A2 / bA3A2/a

Let z2 be a new variable. The resulting productions are

A1->A2A3

A2->A3A1/b

A3->bA3A2z2/az2/ bA3A2/a

z2-> A1A3A2 / A1A3A2z2

Step 2: Now all the productions with A3 on the left have right-hand sides that start with terminals. These

are used to replace A3 in the productions A2->A3A1 and then the productions with A2 on the left are used

to replace A2 in the production A1->A2A3. The result is the following.

A3->bA3A2z2 A3->bA3A2

A1->a/ aA1 A2/bA2/ aA1z A2/bz2A2

A2-> aA1/ b/ aA1z2/ bz2

z2 -> aA1A1/bA1 /aA1z2 A1/ bz2A1 z2

z2-> aA1A1z2/bA1 z2 /aA1z2 A1 z2/ bz2A1 z2



A3->az2 A3->a

A2->bA3A2z2A1 A2->bA3A2A1

A2->az2A1 A2->aA1

A2->b

A1->bA3A2z2A1A3 A1->bA3A2A1A3

A1->b z2A1A3 A1->b A1A3

A1->bA3

z2-> A1A3A2 z2-> A1A3A2z2

Step 3:The two z2productionsare converted to proper form, resulting in 10 more productions. That is, the

productions

z2-> A1A3A2 z2-> A1A3A2z2

are altered by substituting the right side of each of the five productions with A1 on the left for the first

occurrences of A1. Thus z2-> A1A3A2 becomes

z2->bA3A2z2A1A3 A3A2 z2->az2A1A3A3A2

z2-> bA3A3A2 z2-> bA3A2z2A1A3 A3A2

z2-> aA1A3A3A2

The other production for z2 is replaced similarly. The final set of productions is

A3->bA3A2z2 A3->bA3A2

A3->az2 A3->a

A2->bA3A2z2A1 A2->bA3A2A1

A2->az2A1 A2->aA1

A2->b

A1->bA3A2z2A1A3 A1->bA3A2A1A3

A1->b z2A1A3 A1->b A1A3

A1->bA3

z2-> bA3A2z2A1A3 A3A2 z2-> bA3A2z2A1A3 A3A2

z2-> az2A1A3A3A2z2 z2-> az2A1A3A3A2

z2-> bA3A3A2z2 z2-> bA3A3A2

z2-> bA3A2A1A3 A3A2z2 z2-> bA3A2A1A3 A3A2

z2-> aA1A3A3A2z2 z2-> aA1A3A3A2



AMBIGUOUS AND UNAMBIGUOUS GRAMMARS

Ambiguous Grammar: If the production contains more than one derivation or derivation tree then it is termed

as an ambiguous grammar.

For example consider the production, S->S+S/S*S/a/bWe have to derive the string a+a*b

The derivation is as follows:

S->S+S

->S+S*S

->a+a*b

Hence derived.

The same string can also be derived in the following manner

S->S*S

->S+S*S

->a+a*b

Hence from the above two derivations it is obvious that the string can be derived from more than one derivation.

Therefore it is known as ambiguous grammar.

Unambiguous Grammar: If the production contains exactly one derivation or one derivation tree then it is

termed as an unambiguous grammar.

Consider an example:

S->a/abSb/aAb

A->bS/aAAb, the word to be derived is W=abab

Derivation:

S->abSb

->abab

Hence derived.

Now consider

S->aAb

->abSb

->ababSbb

->ababbbb

S->abSb

->ababSbb

->ababbbb

A->aAAb

->abSbSb

->ababab

A->bS

->ba



From the above derivations it is obvious that the string abab can have only one derivation. Hence the grammar

can be termed as an unambiguous grammar.

THE MYHILL-NERODE THEROEM :ALGORITHM

PROCEDURE FOR MINIMIZATION OF AUTOMATA:

1. Write down all the states of DFA.

2. Divide all the states by accepting states and non- accepting states (Q10

AND Q2O =

Q- Q10) Which

Contains Final State S= Q10

U Q2O.

3. By using transition again divide the state and find the transition in number of same group.

4. Repeat the same procedure until there is no possibility to divide.

5. Finally take one representative from each group in transition table substitute representative for other

states in same group for transition.

EXAMPLE: Construct Minimal DFA

o

Begin

for p in F and q in Q-F do mark (p,q);

for each pair of distinct states (p,q) in F*F

or (Q-F)*(Q-F) do

if for some input a,(δ(p,a), δ(q,a)) is

marked then

begin

Mark(p,q)

recursively mark all unmarked pairs on the

list for (p,q) and on the lists of the other

pairs that are marked at this step.

end

else

for all input symbols a do

Put(p,q) on the list for ,(δ(p,a), δ(q,a))

unless δ(p,a)= δ(q,a)

end

a b c d e f g h o o

o o

o

o

o 1

1

1

1

1 1

1 1



Let DFA have a grammer, M= (Q, ∑, δ, q0, F)

Where, Q={a,b,c,d,e,f,g,h}

∑={0,1}

Initial state= a

Final state= d

Class A= contains final state A={d}

Class B= contains the states other than the final state B={a,b,c,e,f,g,h}

Transition table for all classes.

For class A

For class B

B a b c e f g h

0 B B A A B B B

1 B B B B B B A

For class C={c,e}

Class D={h}

B a b f g

0 B B C B

1 B C C B

For Class B

B a g

0 E E

1 B B

Class E={b,f}

Class B={a,g}

For Class C

A

d

O A

1 B



C c e

0 A A

1 E E

For Class D

D h

0 B

1 A

For Class E

E b f

0 B B

1 C C

Reduced Transition diagram

O 1

1 0

0 0 1

1 0 1

A B

C

D E

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