3 votes

Choose the correct statement -

- $A=\{a^nb^n \mid n= 1, 2, 3, \ldots\}$ is a regular language
- The set $B$, consisting of all strings made up of only $a's$ and $b's$ having equal number of $a's$ and $bs$ defines a regular language
- $L(A^*B)\cap B$ gives the set $A$
- None of the above

7 votes

Best answer

6 votes

(a) A={a^{n}b^{n}| n=1,2..} is DCFL .So,(a) is False

(b)The set B,consisting of all strings made up of only a's and b's having equal number of a's and b's is a DCFL. So,(b) is False.

(c) L(A*B) $\cap$ B

A= {a^{n}b^{n}| n=1,2..}

B=consisting of all strings made up of only a's and b's having equal number of a's and b's

L(A*B)= L( { B + AB + AAB + ...} )

Now, L(A*B) $\cap$ B = B

So,(c) is False.

Ans:(d) None of the above

1

@VS Here A,B defined as in the option then why you have taken random definition?

**Option C is incorrect.**

Here A=a`nb`n B=equal no. of a's and b's

On Expanding L(A*B)=L({B+AB+AAB+ ...})

Now on Taking Intersection with B results leads to SET B.

0

In my opinion Cfl intersection Cfl is not closed hence the resultant language belongs to Csl so answer is may be A or May be B but not a particular set . If i I am wrong rectify me

1

L(A*B) ∩ B = B

This would have been correct logic and thinking Since He hasn't mentioned what is A and B (Though from the Questions, It seems His intension was that We assume A and B as the language Set described in the Option 1 and 2 respectively)..But If You say that L(A*B) ∩ B = B, then You have self-assumed that A and B are some Random Regular expressions or Symbols in some alphabet...But even then, It is wrong. Because we can't intersect a Language with a Regular expression (Though People do because of "Without loss of generality")..Thus the expression must have been either

(A*B) ∩ B = B with Intersection operation defined on REs

or

L(A*B) ∩ L(B) = L(B)

So, It is clearly some Printing mistake or Not properly framed question. And It is now "The Interpretation Game"..

People Interpret what they interpret.