b part unit 1

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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}.
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..

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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
= 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.

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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.
6. 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*

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q1=0*
(4) 2=0*11*
q2=0*1*
Describe the language denoted by following regular expression
1. r.e. = (b* (aaa)* b*)*
Solution:
The language can be predicted from the regular expression by finding the meaning of it. We
will first split the regular expression as:
r.e. = (any combination of b's) (aaa)* (any combination of b's)
L = {The language consists of the string in which a's appear triples, there is no restriction on
the number of b's}
Write the regular expression for the language over ∑ = {0} having even length of the string.
Solution:
The regular expression has to be built for the language:
1. L = {ε, 00, 0000, 000000, ......}
The regular expression for the above language is:
1. R = (00)*

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