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AUTOMATA

ANDCOMPILER DESIGN

B.Tech III IT B

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UNIT - I

Formal Language and Regular Expressions:

Languages Definition

regular expressions

Regular setsidentity rules.

Finite Automata:

DFA

NFANFA with transitionsSignificance

acceptance of languages

NFA to DFA conversion

minimization of DFA

Finite Automata with outputMoore and Mealy machines

Constructing finite Automata for a given

regular expressions

Conversion of Finite Automata to Regular

expressions.

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What is Automata Theory?

Study of abstract computing devices, ormachines

Automaton = an abstract computing device

Note:A device need not even be a physicalhardware!

A fundamental question in computer science: Find out what different models of machines can do

and cannot do

The theory of computation

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Alan Turing (1912-1954)

Father of Modern ComputerScience

English mathematician

Studied abstract machines

called Tur ing mach ineseven before computersexisted

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The Central Concepts of

Automata Theory Alphabet

Strings (words)

Language

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Alphabet

An alphabet is a finite, non-empty set ofsymbols

We use the symbol (sigma) to denote an

alphabet Examples:

Binary: = {0,1}

All lower case letters: = {a,b,c,..z}

Alphanumeric: = {a-z, A-Z, 0-9}

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Strings

A string or word is a finite sequence ofsymbols chosen from

Ex: The string 01011 is formed over analphabet { 0,1 }

Empty st r ing is (or epsilon)

Length of a string w,denoted by |w|, is equalto the number of (non-) characters in the string

E.g., x = 010100 |x| = 6

x = 01 0 1 00 |x| = ?

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Language

A language, L, is simply any set of strings over a fixed alphabet.

Alphabet Languages

{0,1} {0,10,100,1000,100000}

{0,1,00,11,000,111,}{a,b,c} {abc, aabbcc,

aaabbbccc,}

{A, ,Z} {TEE,FORE,BALL,}

{FOR,WHILE,GOTO,}

{A,,Z,a,,z,0,9, { All legal C programs}

+,-,,,}

Special Languages: - EMPTY LANGUAGE- contains string only

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Regular Expressions

A declarative way to express the pattern of any string

over an alphabet

or

Regular expressions are an algebraic way to describelanguages.

If Eis a regular expression, then L(E)is the language it

defines.

A Language denoted by a regular expression is said tobe

regular set.

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Operation on these two languages are:1. unionof L and D written L D: is the set of

letters and digits

2. con catenat ion of L and D written LD :is the set of

strings consisting of letter followed by a digit.3. L4 is the set of all four letter string

4. Kleene clos ureof L written L*:is the set of all

strings of letters, including the empty string ()

(denotes zero or more concatenations of L)

5. pos i t ive c losureof D written D+: D+is the set of all

strings of one or more digits. (denotes one or more

concatenations of D)

Ex: Let L be the set consists of alphabetsLet D be the set consists of digitsL={ A,B,Z , a, b,...z}D = {0,1,..9}

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Regular Expressions

Let = { a , b}

1. The regular expression (a+b) denotes following set {a, b}

2. (a+b)(a+b) = { aa, ab ,ba ,bb}

3. a*= {, a, aa ,aaa , ..}

4. (a+b)* = {, a , b ,ab, ba ,aaa ,bbb, aba ,aab ,bbabb }

5. (ab)* = {,ab ,abab, ababab}

6. a*b*= {,a,b,aaaa, bbb, aaabbbb,abbb,aaaaab..}

7. a*|b* = {, aaaa , bbbbbbb}

8. a+= {a , aa, aaa , ..}

9. a(a+b)* denotes any no. of as and any no. of bs but the string

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Examples

1. 01* = {0, 01, 011, 0111, ..}

2. (01*)(01) = {001, 0101, 01101, 011101, ..}

3. (0+1)*

4. (0+1)*01(0+1)*

5. ((0+1)(0+1)+(0+1)(0+1)(0+1))*

6. ((0+1)(0+1))*+((0+1)(0+1)(0+1))*

7. (1+01+001)*(+0+00)

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Regular Expression

Examples

1. All Strings that start with tab or end with bat:

tab{A,,Z,a,...,z}*|{A,,Z,a,....,z}*bat

2. All Strings in Which Digits 1,2,3 exist in ascending

numerical order:

{A,,Z}*1 {A,,Z}*2 {A,,Z}*3 {A,,Z}*

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Find a regular expressions:

1.The set of bit strings with even length

(00 +01 +10 +11)*

2.Set of bit strings ending with a 0not containing11not the null string

(0 +10)*(0+10) or (0+10)+

3.The set of bit strings containing and oddnumber of 0s

1*01*(01*01*)*

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Following rules are used to

simplifying Regular Expression

1. +R= R

2. R =R3.R+R =R4.R*R*=R*

5.(R*)*=R*

6. + RR*=R*7. (P+Q)*= (P*Q*)* = (P*+Q*)*

8.RR*=R+

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Transition Diagrams

The language is recognized by using diagrammatic representationcalled Transition Diagram.

The Transition diagram is made up of set of statesand transitionsfrom one state to another.

The Transition Diagram(TD) has:

States: Represented by Circles

Transitions(actions): Represented by Arrowsbetween states Start State : Beginning of a pattern (Arrowhead)

Final State(s) : End of pattern (Concentric Circles)

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- transition

Each edge in TD is labeled with i/p character scanned by the machine

The -transition in TD is used to move from one state to next state

without reading any input character.

A B

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1) a+b

2) ab

3) a*

4) (a+b)*

5) (ab)*

q0q1a/b

q0q2q1

a b

q1

a

q1

a/b

q0q2q1

a b

Regular Expression Transition Diagram

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1) a+ =(aa*)

2) a*|b*

3) a*b*

4) a(a+b)*

q0q1a

q1

a

a

q1

q1

q0

b

a q1

a/b

q2

q0

a

b

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Finite AutomataThe generalized transition diagram is called Finite Automata

Formally a Finite Automata is defined as a five tuple set

M={Q,, , q0 F}

Where Q Finite set of states

input symbol (an alphabet)

Transition function specifies from which state on which i/p

symbol ,where the transition goes. It maps

(p , a) = q

where p, q are states

a is i/p symbolq0Initial state

F Set of Final states FQ

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The Finite Automata is classified into two ways

Finite Automata

Finite Automata

Without output

Finite Automata

With output

NFA DFA Mealy Moore

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Finite Automaton without output

(Language Recognizers)

Input

Accept

or

Reject

String

Finite

Automaton

NFA or DFA

Output

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Finite Automata : A recognizer that takes an input string &determines whether its a valid sentence of the

language

Non-Deterministic : Has more than one alternative action for the

same input symbolNon-Deterministic Finite Automata (NFAs)

easily represent regular expression, but are

somewhat less precise.

Deterministic : Has at most one action for a given input

symbol. Deterministic Finite Automata (DFAs)require more complexity to represent regular

expressions, but offer more precision.

Both types are used to recognize regular expressions.

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Non-Deterministic Finite

Automata

An NFA is a mathematical model that consists of :

N= {Q ,, , q0, F}

Q, a set of states

, the symbols of the input alphabet

() a transition function.

(state, symbol) set of states

: Q {} power(Q) (2Q

)A state, q0Q, the start state

F Q, a set of finalor accepting states.

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Representing NFAs

Transition Diagrams :

Transition Tables:

Number of states (circles),

arcs, final states,

More suitable

representation within a

computer

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NFA -Example

Given the regular expression : (a|b)*abb

start0 3b21 ba

a

b

Q = { 0, 1, 2, 3 }

Q0= 0

F = { 3 }

= { a, b }

EXAMPLE:Input: ababb

(0, a) = 0

(0, b) = 0

(0, a) = 1

(1, b) = 2

(2, b) = 3

ACCEPT !

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NFA - Example

32

bca

1

6

7

c

a

c

4 b

a (b*c)

a (b | c+)

Given the regular expression : (a (b*c)) | (a (b | c+))

5

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(a (b*c)) | (a (b | c+))

32

bca

1

6

5

7

c

a

c

4 b

0

Input : abbc

abcc

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Transition table:

A tabular representation can also be used to represent the finite automata iscalled transition table. In Transition table there is a row for each stateand

column for each input symbol

The entry for row i and symbol ain the table is a set of states that can be

reached by transition from state ion input a.

q1q0

0/1

0

1

states Input symbol

0 1

q0 {q0 ,q1} q0

q1 q1

Transition Table

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An NFA accepts an input string x if and only if there is some path in

the transition diagram from the initial state to final state.

q0

0/1

0

1

q1 q20

1

String : 001001

q0 q1 q20 0

q0 q1 q1 q2

q0 q0 q1 q2 q2

q0 q1 q1

q0 q0

0

0

0

0 0 0

0

0

0

1

1

1

1

1 accepted

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Thomsons construction of an NFA from

a Regular Expression

start i f

1. The NFA for

astart i f

2. NFA for a input symbol

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3. NFA for R1R2 where R1, R2 are two regular expressions

R1start i

R2f

3. NFA for R1|R2 where R1, R2 are two regular expressions

i f

R1

R2

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5. NFA for R*

Rstart i f

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R1

R2

i f

6. NFA for (R1|R2)*

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The NFA for (a|b)*abb using Thomsons construction

1

2 3

54

6

a

b

2a 3

4b

5

1. r1 = a

2. r2 = b

3. r3 = r1|r2 i.e a|b

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7 9 10a b b

8

5. r5 = abb

0 1

2 3

54

6 7

a

b

start

4. r4 = (r3)* i.e (a|b)*

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0 1

2 3

54

6 7 8 9

10

a

a

b

b

b

start

6. r6= r4r5 i.e : (a|b)* abb

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Regular Expression : (a|b)*abb

NFA without moves

NFA with moves

start0 3b21 ba

a

b

0 1

2 3

54

6 7 8 9

10

a

a

b

b

b

start

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Deterministic Finite

Automata

A DFAis defined as Five tuple set with the following properties

D= {Q ,, , q0, F}

Where transition function defines mapping from QQ

i.e in DFA

i) No state has an - transitionii) There exists only one transition from a state on same input symbol

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DFA simulated as

Q= { q0 , q1}

={0,1}

q0 ={q0}

F={ q1}

(q0,0)q1

(q0,1)q0

(q1,0)q1

(q1,1)q0

q0 q1

0

1

1 0

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Example -DFA

start

0 3b21 ba

b

ab

aa

Regular Expression : (a|b)*abb

DFA:

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Conversion of an NFA (without moves) into equivalent DFA

For every NFA there exists an equivalent DFA.Consider an NFA for the regular expression (a/b)*abb

start q0q3

bq2q1 ba

a

b

Q ={ q0 ,q1 ,q2 ,q3}

= {a,b}

q0 = {q0}

F= {q3}

a b

q0

q1q2

q3

{q0,q1}

q0

q2q3

Transition Table

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a b

q0

q1

q2

q3

{q0,q1}

{q0,q2}

{q0,q3}

{q0,q1}

{q0,q1}

{q0,q1}

{q0,q1}

q0

q2

q3

{q0,q2}

{q0,q3}

q0

The DFA transition table DFA Transition Diagram

q0 q0,q1

q2q1 q3

q0,q2 q0,q3

b

b

b

b

b b

a

a

a

a

This part is eliminated

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-closure(s) : set of NFA states reachable from NFA state son transitionsalone.

Move(T,a): set of NFA states to which there is a transition on input symbol afrom

some NFA state s in T

1

2 3

54

6

a

b

-closure(1): { 1 , 2 ,4}-closure(3) : {3, 6}-closure(3 ,5) : {3 ,5 , 6 }

move(2 ,a) : {3}Move( {1 ,2 ,3} , a) : { 3}

NFA (with moves) to DFA

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NFA (with moves) to DFA

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NFA (with moves) to DFA(subset construction)

First we calculate: -closure(0) (i.e., state 0)

-closure(0) = {0, 1, 2, 4, 7} (all states reachable from 0on -moves)

Let A={0, 1, 2, 4, 7} be a state of new DFA, D.

0 1

2 3

54

6 7 8 9

10

a

a

b

b

b

start

Start with NFA: R.E: (a | b)*abb

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b: -closure(move(A,b)) = -closure(move({0,1,2,4,7},b))

adds {5} ( since move(4,b)=5)

From this we have : -closure({5}) = {1,2,4,5,6,7}(since 56 1 4, 6 7, and 1 2 all by -moves)

Let C={1,2,4,5,6,7} be a new state. Define Dtran[A,b] = C.

2nd, we calculate : a : -closure(move(A,a)) andb : -closure(move(A,b))

a: -closure(move(A,a)) = -closure(move({0,1,2,4,7},a))}adds {3,8}( since move(2,a)=3 and move(7,a)=8)

From this we have : -closure({3,8}) = {1,2,3,4,6,7,8}(since 36 1 4, 6 7, and 1 2 all by -moves)

Let B={1,2,3,4,6,7,8} be a new state. Define Dtran[A,a] = B.

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3rd, we calculate for state B on {a,b}

a: -closure(move(B,a)) = -closure(move({1,2,3,4,6,7,8},a))}= {1,2,3,4,6,7,8} = BDefine Dtran[B,a] = B.

b: -closure(move(B,b)) = -closure(move({1,2,3,4,6,7,8},b))}= {1,2,4,5,6,7,9} = D

Define Dtran[B,b] = D.

4th, we calculate for state C on {a,b}

a: -closure(move(C,a)) = -closure(move({1,2,4,5,6,7},a))}= {1,2,3,4,6,7,8} = B

Define Dtran[C,a] = B.

b: -closure(move(C,b)) = -closure(move({1,2,4,5,6,7},b))}= {1,2,4,5,6,7} = C

Define Dtran[C,b] = C.

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5th, we calculate for state D on {a,b}

a: -closure(move(D,a)) = -closure(move({1,2,4,5,6,7,9},a))}= {1,2,3,4,6,7,8} = BDefine Dtran[D,a] = B.

b: -closure(move(D,b)) = -closure(move({1,2,4,5,6,7,9},b))}= {1,2,4,5,6,7,10} = E

Define Dtran[D,b] = E.

Finally, we calculate for state E on {a,b}

a: -closure(move(E,a)) = -closure(move({1,2,4,5,6,7,10},a))}= {1,2,3,4,6,7,8} = B

Define Dtran[E,a] = B.

b: -closure(move(E,b)) = -closure(move({1,2,4,5,6,7,10},b))}= {1,2,4,5,6,7} = C

Define Dtran[E,b] = C.

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DstatesInput Symbol

a b

A B C

B B D

C B C

E B C

D B E

A

C

B D Estart bb

b

b

b

aa

a

a

This gives the transition table Dtranfor the DFA of:

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Mi i i i b f t t i

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Minimizing number of states in

DFA

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1.All states in DFA are divided into two groups:Final states

Non final states

2. If for every input symbol a, two states sand tin G have

transitions on a to the same group, then sand tstay in the

same group.

Otherwise, divide Gand put sand tto different groups

3. Repeat the division, until no changes on grouping

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DFA Minimization

(a|b)*abb

A, C, E has

the same transitionsCan be merged

But E is a final

state and A and

C are not

Merge A and C A

A

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DstatesInput Symbol

a b

A B A

B B D

E B A

D B E

A B D Estart bb

bb

aa

a

a

DFA after minimizing the states

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Regular expression from

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Regular expression from

finite automata0

To calculate regular expression for this automata will apply following procedure

q0= + q01+q11 ..(1)q1=q00+q10 .(2)

apply ardens theorem

q1=q000*

q1= q00+ (3)

Use (3) in (1)q0= + q01+ q00+1apply ardens theorm

q0 =(1+0+1)* ..(4)

q1 = (1+0+1)*00*

q0 q10

1

1

Arden theorem :

If the equation in the form

R= Q+RP if P doesnt contain

where P ,Q are 2

regular expressions

ThenR=QP*

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Fi it t t ith t t

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Finite automata with output

Mealy Machine: In this output is associated with input symbol.

Moore Machine: In this output is associated with state.

Both the machines define a six tuple ( Q ,,,,,q0)

where

Q Set of states

Input symbol

output symbol

transition function Q X Q

output function (Q in Moore and Q X in Mealy)

q0 initial state

Both the machines are deterministic in nature.

No final states.

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Construction of Mealy and Moore machines to generate

the output , same as binary input.

Mealy Machine Moore Machine

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q0

0/0 , 1/1

q0/0 q1/11

0

0 1

0NS O/P

1NS O/P

q0 q0 0 q0 1

O/P 0 1

q0 0 q0 q1

q1 1 q0 q1

Design a Mealys Machine which will increment the value of a given binary

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Design a Mealy sMachine which will increment the value of a given binary

number by 1.

Design a MealysMachine to obtain the 2s complement of a given binary

number.

Design a Moore Machine to determine the mod 3 for each binary string

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Design a Moore Machine to determine the mod 3 for each binary string

treated as Binary Integer

In Moore machine if nbit length input is received , produces n+1 bit output

Assignment - 1

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1. Define the following

(a) Alphabet (b) String (c) Language

2. Explain in detail about closed properties and identity rules of Regular sets.

3. Write the regular expressions for the following languages

(a) All strings of lowercase letters that contain the five vowels in order.(b) All strings of lowercase letters in which the letters are in ascending lexicographi

order.

4. Construct NFA with -moves for the following regular expressions.

(i) (11+0)* (00+1)* (ii) 10 + (0+11)0*1 (iii) (a + b)* (aa+bb)(a + b)*

5. Construct Minimum state DFA for the following regular languages.

(i) (0+1)*1 (0+1)*(ii) Let L be the set of all binary strings whose last two symbols are same.

6.Obtain the regular expression for the following Finite Automata.

7. Construct Moore and Mealy machines that accepts all binary strings as input

and produces output y if the string ends with two consecutive symbols of same

type, n otherwise.

8 Prove (a*|b*)* and ((|a)b*)* are equal