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Consider the following Finite State Automaton. The language accepted by the automaton is given by the regular expression.

  1. $ab^*b^*$
  2. $a^*b^*$
  3. $b^*b$
  4. $b^*ab^*$
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3 Answers

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bab is accepted by given FA but can not be generated by option A, B, and C

can be generated through D

Correct answer is D
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A. Doesn't accept ba.

B. Doesn't accept ba. Also it produces aa not accepted by FA.

C. Doesn't accept a.

D. Correct.

So D is correct.
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According to the given state transition diagram, the regular expression is
Step-1: Starting state has loop with symbol ‘b’ . It means, we can get n number of times ‘b’.
Step-2: The symbol ‘a’ between the starting state and end state. It means we can get only one
‘a’ after b*.
Step-3: The final state also having self loop with symbol ‘b’.

Answer-D

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Here in the question, we are given with the Finite State Automation which accept the strings. In the question, it is aksing that which of the string (available options) is accepted by this automation.

String abb and bab is accepted by the Finite State Automation.

Option (A) and (D) are logically correct.
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