✎ lfa:2022:lab06-dfa-to-regex [books]

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====== 7. Dfa to Regex conversions ======

==== 7.1. TBD ====

Consider the following DFAs:

**DFA1**

{{ :lfa:screenshot_2021-11-04_at_15.33.10.png?400 |}}

**DFA2**

{{ :lfa:2022:lfa2022_lab5_ex2_4_cerinta.png?300 |}}

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

<hidden DFA1>

</hidden>

<hidden minDFA2>

{{ :lfa:2022:lfa2022_lab6_mindfa2_1.png?300 |}}

**Step 1:**

If there are multiple final states, make them non-final, create a new final state and add ε-transitions to it.

If a final state has any transitions going out of it, create a new final state and add ε-transition.

If the initial state has any transitions going into it, create a new initial state and add ε-transition.

{{ :lfa:2022:lfa2022_lab6_mindfa2_2.png?300 |}}

** Step 2:**

Pick a state that is not initial and not final. For each way to reach its direct successors from its direct predecessors, add a new transition. Then remove the state.

Repeat until there is only one final state and the initial state left.

Pick **state 1**:
  - in: 
    - 0 --(ε)--> 1
    - 2 --(0)--> 1
  - out:
    - 1 --(ε)--> 4
    - 1 --(1)--> 2
  - loop: 
      - 1 --(0)--> 1
  - Add new transitions:
    - 2 --(00*)--> 4
    - 2 --(00*1)--> 2
    - 0 --(0*)--> 4
    - 0 --(0*1)--> 2

{{ :lfa:2022:lfa2022_lab6_mindfa2_3.png?300 |}}

Pick **state 3**:
  - Because state 3 has no direct successor, it can simply be removed without adding new transitions.

Pick **state 2**:
  - in:
    - 0 --(0*1)--> 2
  - out:
    - 2 --(ε U 00*)--> 4
  - loop:
    - 2 --(00*1)--> 4
  - Add new transitions:
    - 0 --(0*1(00*1)*(ε U 00*))--> 4
    - This transition can be written more simply: 0 --((0*1)(0+1)*0*)--> 4

{{ :lfa:2022:lfa2022_lab6_mindfa2_4.png?300 |}}
</hidden>

==== 7.2. Brzozowsky's algebraic method ====

[[https://en.wikipedia.org/wiki/Janusz_Brzozowski_(computer_scientist)|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**:

=== Arden's Lemma ===

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

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

=== Dfa to regex conversion ===

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

^ ^ ^
| **Example** | {{ :lfa:2022:screenshot_2022-11-09_at_09.35.36.png?200 |}} |
| Consider the Dfa from the left figure. The equations that we get are: | ::: |
| $math[q_1 = A q_1 \cup B q_2] | ::: |
| $math[q_2 = (A\cup B)q_2 \cup \epsilon] | ::: |

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

=== Reducing the system of equations ===

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 $math[q = e\cdot q \cup e'] is $math[q = e^*e'].

**Example**

Goind back to the previous system of equations, we can find the solution to $math[q_2] which is: $math[(A \cup B)^*]. Next, we can replace the solution to $math[q_2] in $math[q_1] which yields:
  * $math[q_1 = A q_1 \cup B(A\cup B)^*].
  * We apply Arden's Lemma one more time and yield: $math[q_1 = A^*B(A \cup B)^*].

=== Another example ===

| The initial set of equations is: | {{ :lfa:2022:screenshot_2022-11-09_at_09.36.24.png?200 |}}|
| $math[q_1 = A q_1 \cup B q_2] | ::: |
| $math[q_2 = A q_2 \cup B q_1 \cup \epsilon] | ::: |

  * We reduce the system by removing the second equation: 
  * $math[q_2 = A^*(Bq_1\cup \epsilon)] 
  * We replace $math[q_2] into the first equation: 
  * $math[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: $math[q_1 = (A \cup BA^*)^*BA^*]

=== Exercise ===

 7.2.1. Apply Brzozowsky's method to find a regex for the following DFA:
^ {{ :lfa:2022:screenshot_2022-11-09_at_15.46.23.png?200 |}} ^