CMI2016-A-3

Question

For a regular expression $e$, let $L(e)$ be the language generated by $e$. If $e$ is an expression that has no Kleene star $\ast$ occurring in it, which of the following is true about $e$ in general?

• A.  $L(e)$ is empty
• B.  $L(e)$ is finite
• C. Complement of  $L(e)$ is empty
• D. Both  $L(e)$ and its complement are infinite

Comments

@Kabir5454

what if e itself is ε

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What are you implying?

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@Kabir5454

Given:

 let L(e) be the language generated by e. 

e can be 0 +1 or 0+00 or any other thing 

does e = ε possible.

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yes why not .

$L(e)=\left \{ \epsilon \right \}$.

The corresponding regular expression is = $\epsilon$ .

Here $L(e)$ is not empty but finite with a string of $0$ length .

Complement of $L(e)$=$(a+b)^{+}$  assuming $\sum =\left \{ a,b \right \}$ which is infinite not empty . so option C is also false .

Option D is also false as $L(e)$ is finite .

Answer 1

Suppose R.E $e  = 00 + 11$ 

Language generated by R.E is $L(e) = \left \{ {00,11}\right \}$

Clearly the given language is finite

So option B.

complement of L is $\Sigma ^*- \left \{ {00,11}\right \}$  not empty.

Comments

let me put in this way 

consider DFA with 2 states q0 and q1 where q0 is finial and q1 is final and let $\Sigma$={a}

	a
$\rightarrow$q0	q1
*q1	q1

now tell me is given DFA valid for this question 

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your dfa contain kleene  star 00*

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so what  you are trying to say that by any means we should not be able to generate 0⁺ directly or indirectly 

Answer 2