made easy test 1 2022

Question

how   option 2 is correct becz regular lang. dosnt accept comparision can anyone explain?

Answer 1

The question says that #a in x = #b in y.

I think the answer is Σ*

let's take the members of  Σ* and check

Σ* ={€,a,b,aa,ab,ba,bb,..........}

for €: 0 a's in x and 0 b's in y;

     a: x=€ and y=a; 0 a's in x and 0 b's in y;

    b: x=b and y=€; 0 a's in x and 0 b's in y;

    aa: x=€and y=aa; 0 a's in x and 0 b's in y;

    bb: x=bb and y=€; 0 a's in x and 0 b's in y;

    ab: x=a and y = b ; 1 a in x and 1 b in y;

    ba: x=b and y = a ; 0 a in x and 0 b in y;

and so on;

It is denoting the set of (a+b) *

Comments

nope it’s not (a+b)*   in (a+b)* we accept all language which have no of a equal to no of b an which have a>b  as wall as b>a,

my doubt is that in regular language we dont compare,