The document discusses several topics related to regular languages and regular expressions: 1. It proves that the language L={0n2/n is an integer, n>1} is not regular by showing that the length of strings in L exceeds the number of states in any finite automaton recognizing L. 2. It defines regular expressions and describes their syntax and how they are used to represent regular languages. 3. It provides a regular expression (01)*+(10)*+0(10)*+1(01)* for the language of strings with alternating 0s and 1s. 4. It proves several example languages such as {02n|n>1} are not regular by using

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PART B
1. Show that L = {0n2 / n is an integer, n > 1} is not regular. (or) Show that the set
L = {0i2 |i > 1} not regular.(Nov/Dec2015) (Nov/Dec2014) (May/June2014)
SOLUTION:
Let i = n2 and L = { 0i / i > 1}.
The number of states is i > n.
Take string w = 0i
The length of the string |w| = i > n.
Assume xy = 0m, y = 0j , z = 0i-m
xykz = xy(y)k-1z
= 0m. 0j(k-1). 0i-m
= 0i. 0j(k-1)
Apply i = n2 xykz = 0n2 + j(k-1)
Apply k = 0 xykz = 0n2 – j L
Apply k = 0 xykz = 0n2 + j L
Thus the Language L is not a regular Language.
2. Discuss on regular Expression.(May/June 2016) (May/June2013)
REGULAR EXPRESSION:
The Language accepted by finite automata are easily described by simple
expression called regular expression. A regular expression is a string that describes
the whole set of strings according to certain syntax rules. These expressions are used
by many text editors and utilities to search bodies of text for certain patterns etc.
Let Σ be an alphabet. The regular expression over Σ and the sets they denote
are:
i. ε is a R.E, denotes empty set and Language L(ε) = { ε}.
ii. pie is a R.E denotes the set { } and Language L(pie) = { pie}.
iii. A variable represented in upper case like L is any language.
i. For each ‘a’ in Σ , ‘a’ is a R.E and denotes the set {a}.

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ii. If r and s are regular expression, then
i. If ‘r’ and ‘s’ are R.E denoting the languages R and S respectively
then (r+s),
ii. (rs) and (r*) are R.E that denote the sets RUS, RS and R*
respectively..
3. Write a regular expression for set of strings that consist of alternating 0’s and
1’s. (May/June 2016)
Consider the regular expression for the language consisting of a single string 01.
The star operator is used to get an expression of all strings of the form 0101…01
The basic rules for RE tells us that 0 and 1 are expressions denoting the languages
{0} and {1}, respectively. If we concatenate the two expressions, we get a regular
expression for the language {01};
RE = 01
Now, to get all strings consisting of zero or more occurrences of 01,
Regular expression (01)* and get L(01)*.
Since it only includes strings beginning with 0. To account for strings that starts with
1 and strings that end with 0 or with 1:
(01)* + (10)*+ 0 (10)*+ 1(01)*
4. Prove that the following languages are not regular (May/June2013)
(Nov/Dec2013)
1. {02n | n > 1}
SOLUTION:
Let L be a regular language. The number of states are 2n > n.
Let w = 0i, i > 1 where, i = 2n.
w = xyz
|w| = 2n > n.
Assume xy = 0m, y = 0j, z = 0i-m
xykz = xy(y)k-1z

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= 0m. 0j(k-1). 0i-m
= 0i + j(k-1)
Apply i = 2n xykz = 02n + j(k-1)
Apply k = 0 xykz = 02n - j L
Apply k = 1 xykz = 02n L
Apply k = 2 xykz = 02n + j L
Thus Language L = {02n | n > 1} is not a regular.
5. {ambnam+n | m > 1 and n > 1} (or) L = { 0m1n0m+n | m > 1 and n > 1}
SOLUTION:
Let L be a regular language. The number of states are 2m+2n > n.
.
w = ambnam+n
|w| = 2m+2n > n.
Assume xy = am, y = aj, z = an-m. bn1. am+n1
xykz = xy(y)k-1z
= am . aj(k-1) . an-m. bn1. am+n1
xykz = an + j(k-1) . bn1. am+n1
Apply k = 0 xykz = an - j . bn1. am+n1 L
Apply k = 1 xykz = an. bn1. am+n1 L
Apply k = 2 xykz = an + j . bn1. am+n1 L
Thus Language L = {ambnam+n |m>=1 and n>=1} is not a regular.

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6. Construction of - NFA from a regular expression
Basis: Automata for , and ‘a’ are (a),(b) and (c) respectively.
a) Accepting  b) Accepting  c) Accepting a
Induction: Automata for P+Q, PQ and P* are (d), (e) and (f) respectively.
d) P+Q
e) PQ
f) P*
Example: Construct -NFA for the regular expression (a|b)*|c
Solution: using Thompson's Construction. First we construct the union of a and b:
Next we construct the Kleene Star of the previous union:

5. Finally we create the union between this and the next symbol c:
7. Construction of regular expression from Finite Automata:
Arden’s theorem: Let P and Q be two regular expression over ∑. If ‘P’ does not contain 
then
the equation in R=Q+RP has unique solution (i.e only one solution) given by R=QP*
Method for finding regular expression of Finite automata in transition diagram
representation using Arden’s theorem:
The following assumptions are made regarding the finite automata.
i. The finite automaton does not have  - moves.
ii. It has only one initial state, say q0.
iii. It’s states are q0,q1.....qn
i) Qi is the regular expression representing the set of string accepted by the automata even
through qi is a final state.
ii)  ij denotes the regular expression representation the set of labels of edges from vi to vj when
there is no such edge  ij= .Consequently, we can get the following set of equation in Q1,
….Qn
Q1= Q1 11+ Q2 21+….+ Qn n1+
Q2= Q1  12+ Q2  22+….+ Qn  n2
……………………………………………..
.

6. ……………………………………………..
Qn= Q1 1n+ Q2 2n+…..+ Qn nn
By repeatedly applying substitutions and Arden’s theorem we can express Ri in terms of  ij’s for
getting the set of strings recognized by the automata, we have to take union of all Ri’sCorresponding
to final states.
Example1:
Derive a regular expression from the following given FA?
Sol:
q1= +q10................(1)
q2=q11+q21...............(2)
q3=q20+q30+q31 ..........(3)
(2)  q2=q11+q21
q2=q1
11* ...........(4)
(1)  q1= +q10 (Apply Arden’s theorem)
q1=0*
q1=0*
(4) q2=0*11*
q2=0*1*