Question

Consider the following finite state Automation

The language accepted by this automation is given by the regular expression

• A. $b^{\ast }a{b}^{\ast }a{b}^{\ast }a{b}^{\ast }$
• B. $b^{\ast }a{b}^{\ast }a{b}^{\ast }$
• C. $(a+b{)}^{\ast }$
• D. $b^{\ast }a(a+b{)}^{\ast }$

Answer 1

The correct option is D $b^{\ast }a(a+b{)}^{\ast }$


$q_3$ is unreachanble from starting state and hence can be deleted to give the following diagram.






$q_0$ is the initial state. From $q_{0}to{q}_1$ the regular expression is $b^{\ast }a$. After that, every combination of a and b is accepted by $q_{1}or{q}_2.$ So $q_{1}and{q}_2.$ together can be collapsed into a permanent accept state and now the diagram becomes.







$\therefore r=b^{\ast }a(a+b{)}^{\ast }$