Question

QUESTION 1 Convert the GNFA below to the corresponding regular expression. Draw any intermediate diagrams and the final diagram. (pay attention to arrow directions) b a e start - ՎԻ qy n 9 h m 9: qw d

Answer 1

The GNFA given is :

a b e start qy n 9 h m qz qw k с d

The GNFA is transformed into its corresponding regular expression using the following steps of state elimination method :

Step 1 :

If there is an incoming edge to the initial state and an outgoing edge from the final state then create 2 new states S1 and S2 and assign them as the new initial and final states where S1 is connected to old initial state through an epsilon transition and S2 is connected to old final state through an epsilon transition.

The above property holds for the given GNFA and so the new GNFA is :

a e start S1 qx n qz qw S2 k с d

Step 2 : Eliminate state qx.

On eliminating state qx, the set of strings which can pass from state S1 to qy is represented by the regular expression a*e.

On eliminating state qx, the set of strings which can pass from state S1 to qz is represented by the regular expression a*h.

On eliminating state qx, the set of strings which can pass from state qz to qy is represented by the regular expression ga*e + n.

The GNFA is modified as :

a*e start S1 qy gate+n ath m E qz qw S2 k с

Step 3 : Eliminate state qy.

On eliminating state qy, the set of strings which can pass from state S1 to qz is represented by the regular expression a*eb*m + a*h.

On eliminating state qy, the set of strings which can pass from state S1 to qw is represented by the regular expression a*eb*i.

On eliminating state qy, the set of strings which can pass from state qz to itself is represented by the regular expression ((ga*e + n)b*m + c ).

The GNFA is modified as :

start S1 a*ebi ateb*m + ath qz qw S2 k ((ga*e + n)b*m + c) d

Step 4 : Eliminate state qz.

On eliminating state qz, the set of strings which can pass from state S1 to qw is represented by the regular expression (( a*eb*m + a*h )(ga*e + n)b*m + c )* l +  a*eb*i ).

On eliminating state qz, the set of strings which can pass from state qw to itself is represented by the regular expression ( k(ga*e + n)b*m + c )*l + d ).

The GNFA is modified as :

start S1 (a*eb*m + ath (ga*e + n)b*m + c)*1 + a*ebi ) qw S2 (k(ga*e + n)b*m + c)*+ d)

Step 5 : Eliminate state qw.

On eliminating state qw, the set of strings which can pass from state S1 to S2 is represented by the regular expression (( a*eb*m + a*h )(ga*e + n)b*m + c )* l +  a*eb*i ) ( k(ga*e + n)b*m + c )*l + d )*.

start S1 (( a*eb*m + ath )(ga*e + n)b*m + c)*] + a*ebi) (k(ga*e + n)b*m + c)*+ d )* S2

So, the required regular expression is (( a*eb*m + a*h )(ga*e + n)b*m + c )* l +  a*eb*i ) ( k(ga*e + n)b*m + c )*l + d )*.

Hence, the regular expression is (( a*eb*m + a*h )(ga*e + n)b*m + c )* l +  a*eb*i ) ( k(ga*e + n)b*m + c )*l + d )*.