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**Part V. Regular Expressions**

We found in earlier sections that a language can be described by using an NFA or DFA. But a more concise way of describing a lnguage is by using regular expressions. Thus one way of describing a language is via the notion of regular expressions.

For example,

ab,

a,

ab^{∗},

(ab)^{+},

abcd^{∗},

x/y,

a/bcd^{∗},

letter(letter/digit)*,

a(b/c)d*

are different examples of regular expressions. We will learn the meaning of all these symbols later. A regular expression generates a set of strings.

For example, the regular expression, a^{∗ }generates the strings, ε, a, aa, aaa, aaaa.....

The regular expression, a(b/a) generates the strings ab, ba.

**7 Definition of a Regular Expression**

Let ∑ denotes the character set, ie. a, b, c....z.

The regular expression is defined by the following rules:

Rule 1: Every character of ∑ is a regular expression.

For example, a is a regular expression, b is a regular expression, c is a regular expression and so on.

Rule 2: Null string, ε is a regular expression.

Rule 3: If R1 and R2 are regular expressions, then R1 R2 is also a regular expression.

For example, let R1 denotes abc and

let R2 denotes (pq)*,

then abc(pq)* is also a regular expression.

Rule 4: If R1 and R2 are regular expressions, then R1 / R2 is also a regular expression.

For example, let R1 denotes (ab)^{+ }and

let R2 denotes cd,

then (ab)^{+}/(cd) is also a regular expression.

Rule 5: If R is a regular expression, then (R) is also a regular expression.

For example, let R denotes adc,

then (adc) is also a regular expression.

Rule 6: If R is a regular expression, then R*is also a regular expression.

For example, let R denotes ab,

then (ab)*is also a regular expression.

Rule 7: If R is a regular expression, then R^{+} is also a regular expression.

For example, let R denotes pqr,

then (pqr)^{+}is also a regular expression.

Rule 8: Any combination of the preceding rules is also a regular expression.

For example, let R1 denotes (ab)*,

let R2 denotes d^{+,}

let R3 denotes pq,

then ((ab)*/(d^{+}pq)*)* is also a regular expression.

**Strings Generated from a Regular Expression**

A regular expression generates a set of strings.

1. A regular expression, a* means zero or more repretitions of a.

So the strings generated from this regular expression are,

ε,

a,

aa,

aaa,

aaaa, and so on.

2. A regular expression, p^{+ }means one or more repretitions of a.

So the strings generated from this regular expression are,

p,

pp,

ppp,

pppp, and so on.

3. A regular expression, a/b means the string can be either a or b.

So the strings generated from this regular expression are,

a,

b.

4. A regular expression, (a/b)c means a string that starts with either a or b and will be followed by c.

So the strings generated from this regular expression are,

ac,

bc

5. A regular expression, (a/b)c(d/e) means that a string will begin with either a or b, followed by c, and will end with either d or e.

So the strings generated from this regular expression are,

acd,

ace,

bcd,

bce.

Following shows some examples for the strings generated from regular expressions:

Regular Expression Strings Generated

**Example 1:**

Write regular expression for the language,

L={aa, aaaa, aaaaaa, aaaaaaaa, ......}

Here aa repeats a number of times starting from 1. So the regular expression is,

(aa)^{+}

**Example 2:**

Write regular expression for the language,

L={0, 1, 2}

The regular expression is,

0/1/2

**Example 3:**

Write regular expression for the language, L over {0,1}, such that every string in L begins with 00 and ends with 11.

The regular expression is,

00(0/1)*11.

**Example 4:**

Write regular expression for the language, L over {0,1}, such that every string in L contains alternating 0s and 1s.

The regular expression is,

(0(10)*)/(0(10)*1)/(1(01)*)/(1(01)*0)

**Example 5:**

Write regular expression for the language, L over {0,1}, such that every string in L ends with 11.

The regular expression is,

(0/1)*11

**Example 6:**

Write regular expression for the language, L over {a,b,c}, such that every string in L contains a substring ccc.

The regular expression is,

(a/b/c)*ccc(a/b/c)***Example 7:**

Write regular expression for the language, L = {a^{n}b^{m}|n ≥ 4, m ≤ 3}

This means that language contains the set of strings such that every string starts with at least 4 a’s and end with at most 3 b’s.

The regular expression is,

a^{4}a*(ε/b/b^{2}/b^{3})

**Example 8:**

Write regular expression for the language, L = {a^{n}b^{m}|(n+m) is even }

n+m can be even in two cases.

Case 1: Both n and m are even,

The regular expression is,

(aa)^{∗}(bb)*

Case 2: Both n and m are odd,

The regular expression is,

(aa)^{∗}a(bb)*b

Then the regular expression for L is,

((aa)*(bb)*)/((aa)*a(bb)*b)

**8 Transforming Regular Expressions to Finite Automata**

For every regular expression, a corresponding finite automata exist.

Here we will learn how to construct a finite automation corresponding to a regular expression.

1. The NFA corresponding to regular expression, ε is

2. The NFA corresponding to regular expression, a is

3. The NFA corresponding to regular expression, a | b is

4. The NFA corresponding to regular expression, ab is

5. The NFA corresponding to regular expression, a* is

**Example 1:**

Construct a finite automation for the regular expression, ab*c.

The finite automation corresponding to b* is

Then the finite automation corresponding to ab*c is

**Example 2:**

Construct a finite automation for the regular expression, (ab)*.

The finite automation corresponding to ab is

The finite automation corresponding to (ab)* is,

**Example 3:**

Find the NFA corresponding to regular expression (a|b)*a.

The NFA for a|b is

The NFA corresponding to (a|b)* is

The NFA corresponding to (a|b)*a is**9 DFA to Regular Expression Conversion**

Note: In this section, we use + symbol instead of / operator. For example, a/b is written as a+b. We can find out the regular expression for every deterministic finite automata (DFA).

Arden’s Theorem

This theorem is used to find the regular expression for the given DFA.

Let P and Q are two regular expressions,

then the equation, R = Q + RP has the solution,

R = QP*.

Proof:

Q + RP = Q + (QP*)P,

since R = QP*.

= Q(ε + P*P)

= Q[ε/(P*P)]

= QP*

= R

Thus if we are given, R = Q + RP , we may write it as,

R = QP*

The process is demonstrated using the following examples:

**Example 1:**

Find the regular expression corresponding to the following DFA,

Here we write equations for every state.

We write,

q_{1} = q_{2}b+ q_{3}a + ε

The term q_{2}b because there is an arrow from q_{2} to q_{1}on input symbol b.

The term q_{3}a because there is an arrow from q_{3} to q_{1}on input symbol a.

The term ε because q_{1 }is the start state.

q_{2} = q_{1}a

The term q_{1}a because there is an arrow from q_{1} to q_{2}on input symbol a.

q_{3} = q_{1}b

The term q1b because there is an arrow from q_{1} to q_{3}on input symbol b.

q_{4} = q_{2}a + q_{3}b + q_{4}a + q_{4}b

The term q_{2}a because there is an arrow from q_{2} to q_{4}on input symbol a.

The term q_{3}b because there is an arrow from q_{3} to q_{4}on input symbol b.

The term q_{4}b because there is an arrow from q4 to q_{4}on input symbol b.

The final state is q_{1}.

Putting q_{2} and q_{3} in the first equation (corresponding to the final state), we get,

q_{1} = q_{1}ab + q_{1}ba + ε

q_{1} = q_{1}(ab + ba) + ε

q_{1} = ε + q_{1}(ab + ba)

From Arden’s theorem,

q_{1 }= ε(ab + ba)*

q_{1 }= (ab + ba)*

So the regular expression is, ((ab)/(ba))*

**Example 2;**

Find the regular expression corresponding to the following DFA,

Let us write the equations

q_{1} = q_{1}0 + q_{3}0 + ε

q_{2} = q_{1}1 + q_{2}1 + q_{3}1

q_{3} = q_{2}0

q_{2} = q_{1}1 + q_{2}1 + q_{3}1

q_{2} = q_{1}1 + q_{2}1 + (q_{2}0)1

q_{2} = q_{1}1 + q_{2}(1 + 01)

q_{2} = q_{1}1(1 + 01)*(From Arden’s theorem)

Consider the equation corresponding to final state,

q_{1} = q_{1}0 + q_{3}0 + ε

q_{1} = q_{1}0 + (q_{2}0)0 + ε

q_{1} = q_{1}0 + (q_{1}1(1 + 01)*)0)0 + ε

q_{1} = q_{1}(0 + 1(1 + 01)*)00) + ε

q_{1} = ε + q_{1}(0 + 1(1 + 01)*)00)

q_{1} = ε(0 + 1(1 + 01)*)00)*

q_{1} = (0 + 1(1 + 01)*)00)*

Since q_{1} is a final state, the regular expression is,

(0/(1(1 + 01)*)00))*

**Example 3:**

Find the regular expression corresponding to the following DFA,

The equations are,

S = Sc + ε

A = Ac + F a + Sa

F = Ab

S = Sc + ε

S = ε + Sc

S = εc*

S = c*

A = Ac + F a + Sa

A = Ac + F a + c*a

A = Ac + Aba + c*a

A = A(c + ba) + c*a

A = c*a + A(c + ba)

A = c*a(c + ba)*

F = Ab

F = (c*a(c + ba)*)b

There are two final states in the DFA, they are A and F.

The regular expression corrresponding to A is c*a(c + ba)*

The regular expression corrresponding to F is (c*a(c + ba)*)b

Then the regular expression for the DFA is

c*a(c + ba)∗ + (c*a(c + ba)*)b

That is, [c*a(c/(ba))*]/[(c*a(c/(ba))*)b]

**Exercises:**

1. Find the regular expression corresponding to the following DFA,

2. Find the regular expression corresponding to the following DFA,

3. Find the regular expression corresponding to the following DFA,

4. Find the regular expression corresponding to the following DFA,

5. Find the regular expression corresponding to the following DFA,

6. Find the regular expression corresponding to the following DFA,

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