regular grammar

Question

Consider the following finite automata F₁

Which of the following is not equivalent Regular expression?

1.   c(ac+b)*
2.   (cac)*+b*
3.   (cac)*+(cb)*
4.   None of the above

Answer 1

The question can be done by option elimination also . 

2) and 3) are accepting epsilon as minimal length strings . But according to the given automata , the minimal string is 'c' . 

The correct regular expression can be found as :

First we reach to state q1 which is done by reading 'c' ..Followed by this we consider the loops on this state which are constituting 'b' and 'ac' .. Hence the incoming loops can be written as : (b + ac)* ..So we have entire reg exp :   c(b + ac)*..

Option 1) is also the same.Hence it is correct regular expression for the given automata.So 2) and 3) are not equivalent regular expression..But there is no option like "2) and 3)"..

Hence D) should be the correct answer..

Comments

Yes 2 and 3 are not regular expression but there is no option which consist of choice 2 and 3 so most appropiate answer should be D) None of the above . 

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Doesn't None of the above, mean all the given regular expressions are equivalent to the given DFA?
D can't be the answer!

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@manu

yeaa exactly..that was my  doubt