Lecture 3,4

Lecture # 3
Automata Theory and formal languages
(CSC-307)
Muhammad Shahzeb

Defining Languages Continued…
 Recursive definition of languages:
 The following three steps are used in recursive
definition
1. Some basic words are specified in the language.
2. Rules for constructing more words are defined
in the language.
3. No strings except those constructed in above,
are allowed to be in the language.

Example
Defining language of INTEGER
Step 1: 1 is in INTEGER.
Step 2: If x is in INTEGER then x+1 and x-1
are also in INTEGER.
Step 3: No strings except those constructed in
above, are allowed to be in INTEGER.

Example
Defining language of EVEN
Step 1: 2 is in EVEN.
Step 2: If x is in EVEN then x+2 and x-2 are
also in EVEN.
above, are allowed to be in EVEN.

Example
Defining the language factorial
Step 1: As 0!=1, so 1 is in factorial.
Step 2: n!=n*(n-1)! is in factorial.
above, are allowed to be in factorial.

Defining the language PALINDROME, defined over Σ
= {a,b}
Step 1: a and b are in PALINDROME
Step 2: if x is palindrome, then s(x)Rev(s) and
xx will also be palindrome, where s belongs to Σ*
above, are allowed to be in palindrome

Defining the language {an
bn
}, n=1,2,3,… , of strings
defined over Σ={a,b}
Step 1: ab is in {an
bn
}
Step 2: if x is in {an
bn
}, then axb is in {an
bn
}
Step 3: No strings except those
constructed in above, are allowed to be in {an
bn
}

Defining the language L, of strings ending in a ,
defined over Σ={a,b}
Step 1: a is in L
Step 2: if x is in L then s(x) is also in L, where s
belongs to Σ*
above, are allowed to be in L

Defining the language L, of strings beginning and
ending in same letters , defined over Σ={a, b}
Step 1: a and b are in L
Step 2: (a)s(a) and (b)s(b) are also in L, where
s belongs to Σ*

Defining the language L, of strings containing aa or
bb , defined over Σ={a, b}
Step 1: aa and bb are in L
Step 2: s(aa)s and s(bb)s are also in L, where s
belongs to Σ*

Defining the language L, of strings containing exactly
aa, defined over Σ={a, b}
Step 1: aa is in L
Step 2: s(aa)s is also in L, where s belongs to
b*

Q2) Prove that for any set of strings S
(S+
)*=(S*)*
Solution: In general Λ is not in S+ , while Λ does
belong to S*. Obviously Λ will now be in (S+
)*,
while (S*)* and S* generate the same set of
strings. Hence (S+
)*=(S*)*.

ii) (S+
)+
=S+
Solution: since S+ generates all possible strings
that can be obtained by concatenating the strings
of S, so (S+)+ generates all possible strings that
can be obtained by concatenating the strings of
S+ , will not generate any new string. Hence (S+)
+=S+

Is (S*)+
=(S+
)*
Solution: since Λ belongs to S* ,so Λ will belong
to (S*)+
as member of S* .Moreover Λ may not
belong to S+
, in general, while Λ will automatically
belong to (S+
)*. Hence
(S*)+
=(S+
)*

Lecture # 4
Automata Theory and formal languages
(CSC-307)
Date: 26 August 2015

Regular Expression
As discussed earlier that
a* generates Λ, a, aa, aaa, … and
a+
generates a, aa, aaa, aaaa, …, so the language
L1 = {Λ, a, aa, aaa, …} and
L2 = {a, aa, aaa, aaaa, …}
can simply be expressed by a* and a+
, respectively.
a* and a+
are called the regular expressions (RE) for
L1 and L2 respectively.
Note: a+
, aa* and a*a also generate L2.

Recursive definition of Regular Expression(RE)
 Step 1: Every letter of Σ including Λ is a regular expression
 Step 2: If r1 and r2 are regular expressions then
 (r1)
 r1r2
 r1 + r2 and
 r1*
are also regular expressions.
 Step 3: Nothing else is a regular expression

Method 3 (Regular Expressions)
Consider the language L={Λ, a, aa, aaa,…} of
strings, defined over Σ = {a}. We can write this
language as the Kleene star closure of alphabet Σ
or L=Σ*={a}* this language can also be expressed
by the regular expression a*.
Similarly the language L={a, aa, aaa,…}, defined
over Σ = {a}, can be expressed by the regular
expression a+
.

Example
Now consider another language L, consisting of all
possible strings, defined over Σ = {a, b}. This
language can also be expressed by the regular
expression (a + b)*

Example
Now consider another language L, consisting of all
possible strings, defined over Σ = {a, b}. This
language can also be expressed by the regular
expression (a + b)*
Now consider another language L, of strings
having exactly double a, having any number of b’s
defined over Σ = {a, b}, then it’s regular
expression may be
b*aab*

Example
Now consider another language L, of even length,
expression may be
((a+b)(a+b))*
Now consider another language L, of odd length,
expression may be
(a+b)((a+b)(a+b))* or
((a+b)(a+b))*(a+b)

Remark
It may be noted that a language may be
expressed by more than one regular expressions,
while given a regular expression there exist a
unique language generated by that regular
expression.

Example
Consider the language, defined over Σ={a, b} of
words having at least one a, may be expressed by
a regular expression
(a+b)*a(a+b)*
Consider the language, defined over Σ = {a, b} of
words having at least one a and one b, may be
expressed by a regular expression
(a+b)*a(a+b)*b(a+b)* + (a+b)*b(a+b)*a(a+b)*

Example
Consider the language, defined over Σ={a, b}, of
words starting with double a and ending in
double b then its regular expression may be
aa(a+b)*bb
words starting with a and ending in b OR
starting with b and ending in a, then its regular
expression may be
a(a+b)*b+b(a+b)*a

Practice – Regular Expression ?
words beginning with a
words beginning and ending in same letter

Practice – Regular Expression ?
words ending in b
words not ending in a

An important Example
The Language EVEN-EVEN :
Language of strings, defined over Σ={a, b} having
even number of a’s and even number of b’s.
i.e.
EVEN-EVEN = {Λ, aa, bb, aaaa,aabb,abab, abba,
baab, baba, bbaa, bbbb,…},
Its regular expression can be written as
( aa + bb + (ab+ba)(aa+bb)*(ab+ba) )*

Note
It is important to be clear about the difference of
the following regular expressions
r1=a*+b*
r2=(a+b)*
Here r1 does not generate any string of
concatenation of a and b, while r2 generates such
strings.

Equivalent Regular Expressions
Definition: Two regular expressions are said to
be equivalent if they generate the same language.
Example: Consider the following regular
expressions
r1= (a + b)* (aa + bb)
r2= (a + b)*aa + ( a + b)*bb then
both regular expressions define the language of
strings ending in aa or bb.

Note
 If r1 =(aa + bb) and r2 =( a + b) then
1. r1 + r2 =(aa + bb) + (a + b)
2. r1 r2 = (aa + bb) (a + b) = (aaa + aab + bba + bbb)
3. (r1)* =(aa + bb)*

Regular Languages
Definition:
The language generated by any regular
expression is called a regular language.
It is to be noted that if r1, r2 are regular
expressions, corresponding to the languages L1
and L2then the languages generated by r1 + r2 , r1r2
( or r2r1) and r1*( or r2*) are also regular
languages.

Note
 It is to be noted that if L1 and L2 are expressed by
r1 and r2, respectively then the language expressed
by
1. r1+ r2, is the language L1+ L2 or L1U L2
2. r1 r2, , is the language L1L2, of strings obtained by
prefixing every string of L1 with every string of L2
3. r1*, is the language L1*, of strings obtained by
concatenating the strings of L, including the null
string.

Example
If r1=(aa+bb) and r2=(a+b) then the language of
strings generated by r1+ r2, is also a regular
language, expressed by (aa+bb)+(a+b)
If r1=(aa+bb) and r2=(a+b) then the language of
strings generated by r1r2, is also a regular
language, expressed by (aa+bb)(a+b)
If r=(aa+bb) then the language of strings
generated by r*, is also a regular language,
expressed by (aa+bb)*

All finite languages are regular
Example:
Consider the language L, defined over Σ={a,b}, of
strings of length 2, starting with a, then
L={aa, ab}, may be expressed by the regular
expression aa+ab.
Hence L, by definition, is a regular language.

Note
It may be noted that if a language contains even
thousand words, its RE may be expressed, placing
‘ + ’ between all the words.
Here the special structure of RE is not important.
Consider the language L={aaa, aab, aba, abb, baa,
bab, bba, bbb}, that may be expressed by a RE
aaa+aab+aba+abb+baa+bab+bba+bbb, which is
equivalent to (a+b)(a+b)(a+b).

Lecture 3,4

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