6. Dfa to Regex conversions

Consider the following DFAs:

DFA1

DFA2

Convert the given DFAs to a Regex (using the state-elimination strategy). Hint: is it easier to apply conversion on another DFA?

Janusz Brzozowski worked out a very elegant (and more computationally efficient) way to convert Dfa's to Regexes. It relies on an observation called Arden's Lemma:

Proposition (Arden's Lemma). Let $ X, A$ and $ B$ be languages, such that $ X = A\cdot X \cup B$ . Then $ X = A^*B$ .

• Note that we can apply Arden's lemma in various settings, for instance, let $ e_A, e_B$ be regexes such that $ X = L(e_A)\cdot X \cup L(e_B)$ . Then $ X = L(e_A^*e_B)$ .

For each state $ q$ , build an equation of the form: $ q = c_1 q_1 \cup c_2 q_2 \ldots c_n q_n$ , such that: $ \delta(q,c_i) = q_i$ . Here $ c_i\in\Sigma$ , thus $ q_i$ are the $ c_i$ -successors of $ q$ . Additionally, if $ q$ is a final state, add an $ \epsilon$ : $ q = c_1 q_1 \cup c_2 q_2 \ldots c_n q_n \cup \epsilon$ .

Example
Consider the Dfa from the left figure. The equations that we get are:
$ q_1 = A q_1 \cup B q_2$
$ q_2 = (A\cup B)q_2 \cup \epsilon$

Equations are expressions containing regexes together with state variables (like $ q_1$ ). Which are the unknowns. An equation of the form $ q = e\cdot q'$ signifies the fact that $ L(A_q) = L(e) L(A_{q'})$ , where $ L(A_q), L(A_{q'})$ are the languages accepted by our Dfa, starting from states $ q$ and $ q'$ , respectively.

We can choose any equation except that corresponding to the initial state, and eliminate it, by exploiting Arden's Lemma:

• the solution to any equation of the form $ q = e\cdot q \cup e'$ is $ q = e^*e'$ .

Example

Going back to the previous system of equations, we can find the solution to $ q_2$ which is: $ (A \cup B)^*$ . Next, we can replace the solution to $ q_2$ in $ q_1$ which yields:

• $ q_1 = A q_1 \cup B(A\cup B)^*$ .
• We apply Arden's Lemma one more time and yield: $ q_1 = A^*B(A \cup B)^*$ .

The initial set of equations is:
$ q_1 = A q_1 \cup B q_2$
$ q_2 = A q_2 \cup B q_1 \cup \epsilon$

• We reduce the system by removing the second equation:
• $ q_2 = A^*(Bq_1\cup \epsilon)$
• We replace $ q_2$ into the first equation:
• $ q_1 = A q_1 \cup BA^*(B q_1 \cup \epsilon) = (A \cup BA^*) q_1 \cup BA^*$
• and we apply Arden's lemma again: $ q_1 = (A \cup BA^*)^*BA^*$

6.2.1. Apply Brzozowsky's method to find a regex for the following DFA: