Use the procedure described in Lemma 1.60to convert the following finite automata to regular expressions.

Consider Finite Automata:

Now, inside the methods following, transform such a finite automaton to something like a regular expression:

To turn the initial allow state a non-accepting state, add the start state (S) and new accept state (F) as follows:

Mostly in second stage, remove state(1); there is no need to construct a loop for state 1; instead, add a loop to state (2) directly, as well as create the expressions by transferring state (S) to state (2).

Every circuit is defined as a union step:

So, for the above finite automata, the notation is$a*b(a\cup ba*b)*$

Allow the 2nd finite automata is

Therefore, in the following stages, transform this finite automaton to a regular expression:

Add a new start state (S) and a new accepted state (A) (F). If you change the admit states to non-accepting states, the Finite Automata becomes:

Perform union on the edge from state 1 to 2 state

From the above step 2 , there are no unions or loops for the state 1 , So eliminate the state 1 as follows:

Perform unions on edges from state 3 to state 2 and from state 3 to the final state, Then the Automata becomes as below:

To minimise the automata, remove 2 then conduct a federation on 3, then transfer the state (S) statement to state(3), then apply a loop on state (3) with the state (S) expression (2

Eliminate state 3 and write the expression from state(S) to state (F), because there are no loops and unions.

Perform union on the edge from state S to state F.

So, the regular expression for the given finite automaton is in the diagram.