retagged by
1,322 views
0 votes
0 votes
Problem 1 : what is the Language associated with this regular expression ?  a*b* write it down.

Problem 2: Does either a subset or Super-set of a regular language is always a regular ?

Problem 3 : What is the difference between a^n b^n and a*b*

Explain with a example .
retagged by

3 Answers

Best answer
6 votes
6 votes

problem1: language :a*b* means any no. of a's followed by any no. of  b's 

problem2:  subset or Super-set of a regular language is  not always a regular

for example 1: a*b* is a regular language but its subset a^nb^n is not regular.

for example 2:a^n  n>1 is regular whereas its superset a^n +a^nb^n is not regular

problem-3 a*b* means any no. of a followed by any no. b

    whereas a^nb^n means equal no. of a's  followed by equal no. of b's

selected by
3 votes
3 votes

(1) L={null ,a ,b,aab,aaab,................}

 L=Any number of a followed by any numbers of b

(2) Let us consider regular language L=0*1*

  For example, 0*1* is regular, but its subset {On1n : n >= 0} is not regular, but its subset {01, 0011, 000111} is regular again.

  For super set  Not necessarily regular (take $\Phi$  is regular )

  So we can say  subset /superset of regular language need not be regular.

  Regular languages are not closed under the subset/superset relation.

edited by
2 votes
2 votes

1.L={A^mB^n   |m,n>=0}

2.NO.....FROM UR 1ST QUESTION a*b* (regular) a subset will be {a^nb^n  |n=>0 } not regular

3.         a^nb^n=non regular......and  a*b*=regular

Related questions

1 votes
1 votes
2 answers
1
shekhar chauhan asked Jun 8, 2016
1,754 views
If r1 and r2 are 2 Regular Expression Such thatr1 = (a+b)* r2 = (a*+b*+a*b*+b*a*)What are the different case's in which r1 = r2 ?Please Explain with an example
2 votes
2 votes
2 answers
2
Kapil asked Jul 8, 2016
1,342 views
The equality of two regular expression is computed in? Give reasons also..Constant Timepolynomial timelogarithmic Polynomial timeExponential time
0 votes
0 votes
1 answer
3