TAFL 1 (ECS-403)

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Unit- II

2.1 Regular Expression:

2.1.1 The Operators of Regular Expressions:

2.1.2 Building Regular Expressions

2.1.3 Precedence of Regular-Expression Operators

2.1.4 Algebraic laws for Regular expressions:

2.2 Conversions:

2.2.1 Regular Expression to F.A.

2.2.2 F.A. to Regular Expression

2.2.2.1 using formula

2.2.2.2 using Arden’s Theorem

2.3 Pumping Lemma

2.4 Closure Properties of Regular Languages.

2.5 Decision Properties of Regular Languages.

2.6 F.A with Output

2.6.1 Introduction to moore and mealy machine (Some Examples)

2.6.2 Conversion from moore to mealy machine

2.6.3 Conversion from mealy to moore machine

2.7 Applications and Limitations of F.A.

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2.1 Regular Expression:

Regular Expressions are another type of language-defining notation, Regular expressions

also may be thought of as a "programming language," in which we express some important

applications, such as text-search applications or compiler components. Regular expressions are

closely related to nondeterministic finite automata and can be thought of as a "user-friendly"

alternative to the NFA notation for describing software components.

Example:

01* + 10* denotes the language consisting of all strings that are either a single 0 followed by any

number of l's or a single 1 followed by any number of O's.

2.1.1 The Operators of Regular Expressions:

a) The union of two languages L and M, denoted L Ụ M, is the set of strings that are in

either L or M, or both. For example if L = {00l, 10,111} and M = {Ԑ, 001}, then L Ụ M =

{Ԑ, 10,001,111}.

b) The concatenation of languages L and M is the set of strings that can be formed by

taking any string in L and concatenating it with any string in M. For example, if L =

{001, 10,111} and M = {Ԑ, 001}, then L.M, or just LM, is {001, 10,111, 001001, 10001,

111001}.

c) The closure (or star, or Kleene closure) of a language L is denoted L* and represents

the set of those strings that can be formed by taking any number of strings from L,

possibly with repetitions (i.e., the same string may be selected more than once) and

concatenating all of them. For instance, if L = {0, 1}, then L* is all strings of O's and l's.

If L = {0, 11}, then L* consists of those strings of O's and 1's such that the l's come in

pairs, e.g., 011, 11110, and e, but not 01011 or 101. More formally, L* is the infinite

union Ui≥o Li, where L0 = {Ԑ}, L 1= L, and Li, for i > 1 is LL... L (the concatenation of i

copies of L). L2 = {00,011,110, 1111}.

L3 = {000, 0011,0110, 1100, 01111, 11011, 11110, 111111}

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2.1.2 Building Regular Expressions

Basis:

a) The constants Ԑ and ø are regular expressions, denoting the languages {Ԑ} and ø,

respectively. That is, L(Ԑ) = {Ԑ}, and L(ø) = ø.

b) If a is any symbol, than a is a regular expression. This expression denotes the language

{a}. That is, L(a) = {a}.

c) A variable, usually capitalized and italic such as L, is a variable: representing any

language.

Induction:

a) If E and F arc regular expressions, then E + F is a regular expression denoting the union

of L(E) and L(F). That is, L(E+ F) = L(E) U L(F).

b) If E and F are regular expressions, then EF is a regular expression denoting the

concatenation of L(E) and L(F). That is, L(EF) = L(E)L(F).

c) If E is a regular expression, then E* is a regular expression, denoting the closure of L(E).

That is, L(E*) = (L(E))*.

d) If E is a regular expression, then (E), a parenthesized E, is also a regular expression,

denoting the same language as E. Formally; L((E)) = L(E).

2.1.3 Precedence of Regular-Expression Operators

a) The star operator is of highest precedence.

b) Next in precedence comes the concatenation or "dot" operator. After grouping all stars

to their operands, we group concatenation operators to their operands.

c) Finally, all unions (+ operators) are grouped with their operands. Since union is also

associative, it again matters little in which order consecutive unions are grouped, but we

shall assume grouping from the left.

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2.1.4 Algebraic laws for Regular expressions:

a) Ø + R = R

b) ØR = RØ = Ø

c) ԐR = RԐ = R

d) Ԑ* = Ԑ and Ø* = Ԑ

e) R + R = R

f) R* R* = R*

g) RR* = R*R

h) (R*)* = R*

i) Ԑ + RR* = R* = Ԑ + R*R

j) (PQ)*P = P(QP)*

k) (P+Q)* = (P*Q*)* = (P* + Q*)*

l) (P+Q)R = PR+QR

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EXERCISE

1. Write regular expressions for the following languages:

a) The set of all strings over a,b of even length.

b) The set of all strings over a,b of length 4, starting with an a.

c) {a2n│ n ≥ 1}

d) The set of all strings over a,b having abab as a substring.

e) {0,1,2}

f) {12n+1 │n > 0}

g) {w Ԑ {a,b}* │ w has only one a}.

h) The set of all strings over {0,1} which has at most two zeros.

i) {a2, a5, a8,……}

j) {an │ n is divisible by 2 or 3 or n = 5}

k) The set of all string over a,b beginning and ending with a.

l) The set of all strings over a,b with three consecutive b’s.

m) The set of all strings over 0,1 beginning with 00.

n) The set of all strings over 0,1 ending with 00 and beginning with 1.

o) The set of all strings over a,b containing exactly 2a’s.

p) The set of all strings over a,b containing at least 2a’s.

q) The set of all strings over a,b containing at most 2a’s.

r) The set of all strings over a,b containing the substring aa.

2. Give English descriptions of the languages of the following regular expressions:

a) ab*+b*

b) a(a+b)*ab

c) a*b + b*a

d) (aa + b)*(bb + a)*

e) (a+b)*(aa+bb+ab+ba)*

f) (aa)*+(aaa)*

g) a+b(a+b)*

3. Show that

a) (1+00*1) + (1+00*1) (0+10*1)* (0+10*1) = 0*1(0+10*1)*

b) Ԑ+1*(011)*(1*(011)*)* = (1+011)*

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SOLUTIONS

1. Regular Expressions:

a) (ab+aa+bb+ba)*

b) a(a+b)(a+b)(a+b)

c) (aa)+

d) (a+b)*abab(a+b)*

e) 0+1+2

f) 1(11)+

g) b*ab*

h) 1*+1*01*+1*01*01*

i) aa(aaa)*

j) (aa)* + (aaa)* + aaaaa

k) a(a+b)*a

l) (0+1)*000(0+1)*

m) 00(0+1)*

n) 1(0+1)*00

o) b*ab*ab*

p) (a+b)*a(a+b)*a(a+b)* + b*

q) b*ab*ab*+b*ab*

r) (a+b)*aa(a+b)*

2. English Descriptions:

a) Starting with a and having no other a’s or having no a’s but only b’s.

b) The set of all strings starting with a and ending with ab.

c) The strings are either strings of a’s followed by one b or strings of b’s followed by

one a.

d) The set of all strings of aa or b followed by set of all string of bb or a.

e) The set of all strings of a or b followed by all set of even length of a and b.

f) {xԐ{a}* │ │x│ is divisible by 2 or 3.}

g) Accept only a or starting with b.

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3. Proof:

a) (1+00*1)+(1+00*1)(0+10*1)*(0+10*1)=0*1(0+10*1)*

L.H.S. = (1+00*1)+(1+00*1)(0+10*1)* (0+10*1)

= (1+00*1)(Ԑ + Ԑ (0+10*1)* (0+10*1)

= (1+00*1) (0+10*1)*

= 1(Ԑ+00*) (0+10*1)*

= 0*1(0+10*1)*

2.2 Conversions:

7

(1+00*1)+(1+00*1)(0+10*1)*(0+10*1)=0*1(0+10*1)*

(1+00*1)+(1+00*1)(0+10*1)* (0+10*1)

(0+10*1)* (0+10*1))

R.H.S.

b) Ԑ+1*(011)*(1*(011)*)* = (1+011)*

L.H.S. = Ԑ+1*(011)*(1*(011)*)*

= (1*(011)*)* (1*(011)*)*

= (1*(011)*)*

= (1+(011))*

= (1+011)* R.H.S.

(ECS-403)

RKGIT

+1*(011)*(1*(011)*)* = (1+011)*

Ԑ+1*(011)*(1*(011)*)*

= (1*(011)*)* (1*(011)*)*

= (1*(011)*)*

= (1+(011))*

= (1+011)* R.H.S.

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2.2.2 F.A. to Regular Expression

2.2.2.1 using formula

2.2.2.2 using Arden’s Theorem

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NOTE: Arden’s Theorem

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2.3 Pumping Lemma:

Regular Languages are the languages that can be represented by Regular Expressions.

But there are certain languages which cannot be represented using Regular Expressions. Such

Languages are recognized as

there is a theorem called pumping lemma

Theorem: Let L be a regular language. Then there exists a constant n (which depends on L) such

that for every string z in L such that

that:

• v ≠ Ԑ

• │uv│ ≤ n

• For all i ≥ 0, the string uv

Proof: Suppose L is regular. Then L = L(M

consider any string z of length n or more, say

input symbol. The transition function

If qm is in F i.e. q1,q2….qm is in L(

path from q0 to qm that goes throw q

δ(qo, a1, aj ak+1 ……am)

16

Regular Languages are the languages that can be represented by Regular Expressions.

But there are certain languages which cannot be represented using Regular Expressions. Such

Languages are recognized as non-regular languages. For recognizing non

pumping lemma is used.

Let L be a regular language. Then there exists a constant n (which depends on L) such

in L such that │z│≥ n, we can break z into three strings,

0, the string uviw is also in L.

ppose L is regular. Then L = L(M) for some DFA A. Suppose A has

gth n or more, say w = a1, a2....... am where m

The transition function δcould be written as δ(qo, q1, q2….. qi)

is in L(M) a1a2……aj ak+1 ak+2……am is also in L(M). Since there is

that goes throw qj but not around the loop labeled aj+1…..a

) = δ( δ (qo, q1, q2….. qj ), ak+1 ak+2……am)

= δ (qj, qk+1,……qm)

= δ (qk, qk+1……qm)

= qm

(ECS-403)

RKGIT

Regular Languages are the languages that can be represented by Regular Expressions.

But there are certain languages which cannot be represented using Regular Expressions. Such

. For recognizing non-regular languages

Let L be a regular language. Then there exists a constant n (which depends on L) such

into three strings, z = uvw, such

) for some DFA A. Suppose A has n states. Now,

where m ≥ n and each ai is an

) = qi,

is also in L(M). Since there is

…..ak. Thus:

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So any given string can be accepted FA, we should be able to find a substring near the beginning

of the string that may be pumped i.e. repeated as many times as we like and resulting string may

be accepted by F.A.

But reverse will not be true.

Example1: Show that L = {0n1n+1│ n > 0} is not regular.

For n=1 011

For n=2 00111

For n=3 0001111 and so on

Now consider z = 00111 such that u = 0, v = 01 and w = 11

uv2w = 0010111

uv3w = 001010111

we can see that uviw is not in language L i.e. L = {0n1n+1│ n > 0} is not regular

Exercise:

1. Show that L = {02n│ n > 0} is regular.

2. Show that L = {ap│ p is a prime} is not regular.

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2.4 Closure Properties of Regular Languages.

1. The union of two regular languages is regular.

2. The intersection of two regular languages is regular.

3. The complement of a regular language is regular.

4. The difference of two regular languages is regular.

5. The reversal of a regular language is regular.

6. The closure (star) of a regular language is regular.

7. The concatenation of regular languages is regular.

8. A homomorphism (substitution of strings for symbols) of a regular language is regular.

9. The inverse homomorphism of a regular language is regular.

2.5 Decision Properties of Regular Languages.

1. Membership Property: is a particular string w belongs to some language L?

2. Emptiness Property: is the given language empty?

3. Equivalence Property: Do the two descriptions of a language represent the same

language?

2.6 F.A with Output:

There are two types of FA with outputs and those are:

1. Moore Machine 2. Mealy Machine

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2.6.1 Introduction to moore and mealy machine

Moore Machine

Six Tuple (Q, ∑, ∆, δ, λ, q0)

Q: finite set of states

∑: finite set of input symbols

∆: an output alphabet

δ: an transition function Q X ∑ → Q

λ: output function Q → ∆

q0: initial state of machine

Example:

Current

State

Next State Output

λ 0 1

→q0 q1 q2 1

q1 q2 q1 0

q2 q2 q0 0

Mealy Machine

Six Tuple (Q, ∑, ∆, δ, λ, q0)

Q: finite set of states

∑: finite set of input symbols

∆: an output alphabet

δ: an transition function Q X ∑ → Q

λ: output function Q X ∑ → ∆

q0: initial state of machine

Example:

Current

State

Next State

a=0 a=1

State Output State Output

→q1 q2 A q3 A

q2 q2 B q3 A

q3 q2 A q3 B

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2.6.2 Conversion from moore to mealy machine

Current

State

Next State Output

λ 0 1

→q0 q0 q1 0

q1 q2 q0 1

q2 q1 q2 2

λ’(q,a) = λ(δ(q,a)

λ’(q0,0) = λ(δ(q0,0) = λ (q0) = 0

λ’(q0,1) = λ(δ(q0,1) = λ (q1) = 1

λ’(q1,0) = λ(δ(q1,0) = λ (q2) = 2

λ’(q1,1) = λ(δ(q1,1) = λ (q0) = 0

λ’(q2,0) = λ(δ(q2,0) = λ (q1) = 1

λ’(q2,1) = λ(δ(q2,1) = λ (q2) = 2

Current

State

Next State

a=0 a=1

State Output State output

→q0 q0 0 q1 1

q1 q2 2 q0 0

q2 q1 1 q2 2

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2.6.3 Conversion from mealy to moore machine

Current

State

Next State

a=0 a=1

State Output State output

→q1 q3 0 q2 0

q2 q1 1 q4 0

q3 q2 1 q1 1

q4 q4 1 q3 0

Current

State

Next State

a=0 a=1

State Output State output

→q1 q3 0 q20 0

q20 q1 1 q40 0

q21 q1 1 q40 0

q3 q21 1 q1 1

q40 q41 1 q3 0

q41 q41 1 q3 0

Current

State

Next State Output

λ 0 1

→q1 q3 q20 1

q20 q1 q40 0

q21 q1 q40 1

q3 q21 q1 0

q40 q41 q3 0

q41 q41 q3 1

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This moore machine accepts a zero-length sequence which is not accepted by the mealy machine

so we must add a new starting state.

Current

State

Next State Output

λ 0 1

→q0 q3 q20 0

q1 q3 q20 1

q20 q1 q40 0

q21 q1 q40 1

q3 q21 q1 0

q40 q41 q3 0

q41 q41 q3 1

2.7 Applications and Limitations of F.A.

Applications:

1. Text Editors: used for processing the text.

2. Lexical Analyzers: to scan the input programs

Limitations:

FA does not contain any memory so it cannot solve following types of problems:

1. Checking well-formedness of parenthesis

2. Checking the palindrome condition of given language.