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Which of the following statements is correct?

  1. $A=\{a^nb^n\mid n= 0,1,2,3\dots \}$ is regular language
  2. Set $B$ of all strings of equal number of $a$'s and $b$'s defines a regular language
  3. $L(A^*B^*) \cap B$ gives the set $A$
  4. None of these.
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4 Answers

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Option A and B requires memory to calculate. So they are not regular.

Now coming to Option C they told that the language is intersection of two languages. The property of first language is a's is followed by b's, and the property of second language is No. of a's and No. of b's are equal. Their intersection means the resultant language should have both the properties, which is nothing but set A.

Hence option C is correct.
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Here if we look at the options

A :  Here given that {a^n b^n | n >= 0 }

Now this clearly required a stack or a memory to keep track of count hence non regular .

B: This is like the upper case and hence non regular .

C : This is any no of a followed by any no of b and hence requires nothing to keep track of and then further is intersected by B which gives  A and hence true .
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Option C) is correct,

both option A and option B requires comparison (therefore needs stack) ,so not regular.

(A* B*) intersection B gives A .
Answer:

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