91Ó°ÊÓ

Theorem 1.49 states that “The class of regular languages is closed under the star operation."

Consider the data,

A language A is recognized by the automata$N_1=\left(Q_1,∑,n_{1,}{q}_1,F_1\right)$

Assume N is the non-deterministic finite automaton that recognizes the language $A_1*$.

We can start step 2 with an example to easily understand. So, let’s see the below example.

Example

Assume a language $A_1=\left\{\left(00\right)*1\right\}$ .

The finite state automata$N_1$, which recognizes the language$A_1$is as follows:

The following procedure is used to construct the finite state automata N, which recognizes the language $A_1*:$

a. States of N are the states of $N_1$.

States of $N_1$are $\left\{1,2,3\right\}$ .

So, States of N are $\left\{1,2,3\right\}$ .

b. The start state of localid="1663213421606" $N$is same as start state of localid="1663213414314" $N_1$.

Start state of localid="1663213426131" $N_1$is localid="1663213431322" $\left\{1\right\}$.

So, Start state of localid="1663213439797" $N$is localid="1663213435395" $\left\{1\right\}$

c. . localid="1663213443717" $F=\left\{q_1\right\}∪F_1$The accept states for are the accept states of including the start state. So, the accept state for are 1and 3.

d. Define the transition localid="1663213448158" $\mathrm{δ}foranyq∈Q_{1}andanya∈\underset{∈}{∑}$ by using the following transition:

localid="1663213453268" $$\begin{gathered} \mathrm{δ}(q,a)=\left\{{\mathrm{δ}}^1(q,a)   q∉F_1  and  a≠ ∈ \\ {\mathrm{δ}}^1(q,a) ∪\left\{q_1\right\}   q∈F_1  and  a= ∈ \\ \right\} \end{gathered}$$

localid="1663213457964" $$\begin{gathered} \mathrm{δ}(1,∈)={\mathrm{δ}}_1(1,∈)∪\left\{1\right\} \\                 =âˆ\dots ∪\left\{1\right\} \\                 =\left\{1\right\} \end{gathered}$$

localid="1663213462341" $$\begin{gathered} \mathrm{δ}(3,∈)={\mathrm{δ}}_1(3,∈)∪\left\{1\right\} \\                 =âˆ\dots ∪\left\{1\right\} \\                 =\left\{1\right\} \end{gathered}$$

localid="1663213468584" $$\begin{gathered} \mathrm{δ}(1,0)={\mathrm{δ}}_1(1,0) \\                 =2 \\ \mathrm{δ}(1,1)={\mathrm{δ}}_1(1,1) \\                 =3 \\ \mathrm{δ}(2,0)={\mathrm{δ}}_1(2,0)  \\                 =1 \\ \mathrm{δ}(2,1)={\mathrm{δ}}_1(2,1)  \\                 =âˆ\dots \\ \mathrm{δ}(3,0)={\mathrm{δ}}_1(3,0)  \\                 =âˆ\dots \\ \mathrm{δ}(3,1)={\mathrm{δ}}_1(3,1)  \\                 =âˆ\dots \end{gathered}$$

The state diagram for the Finite State Automata is as follows:

Its beginning state is added to the set of accept states in the aforementioned finite automata, as well as some more undesirable strings and the recognised language. The automaton does not get a new beginning state that is also an accept state. As a result, the new state is not added to the automata, resulting in automata that are distinct from either the original automata.