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--- lfa:2022:lab06-dfa-to-regex [2022/11/11 09:31]
alexandra.udrescu01
+++ lfa:2022:lab06-dfa-to-regex [2022/11/19 10:34] (current)
alexandra.udrescu01
@@ Line 1: / Line 1: @@
-====== 7. Dfa to Regex conversions ======
+====== 6. Dfa to Regex conversions ======
-==== 7.1. TBD ====
+==== 6.1. State elimination ====
 Consider the following DFAs:
-**DFA1**
+^ **DFA1** ^ **DFA2** ^
+|{{ :lfa:screenshot_2021-11-04_at_15.33.10.png?400 |}}| {{ :lfa:2022:lfa2022_lab5_ex2_4_cerinta.png?300 |}} |
-{{ :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 DFA2>
+<hidden On what DFA should the algorithm be applied?>
+It is recommended to apply the State elimination algorithm on the minimal DFA.
 </hidden>
-<hidden minDFA2>
+<hidden DFA2>
 {{ :lfa:2022:lfa2022_lab6_mindfa2_1.png?300 |}}
@@ Line 38: / Line 33: @@
 {{ :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.
+  - 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 |}}
+** Step 3:**
+Repeat step 2 and stop when there is one final and one initial state left.
 </hidden>
-==== 7.2. Brzozowsky's algebraic method ====
+<hidden DFA1>
+  - first, we apply the minimisation algorithm
+    * {{ :lfa:2022:lab6_1_dfa1_1.png?500 |}}
+  - we add the new final and initial states
+    * {{ :lfa:2022:lab6_1_dfa1_2.png?500 |}}
+  - we eliminate the state 8
+    - as it has no outgoing transitions, we can just remove it
+    * {{ :lfa:2022:lab6_1_dfa1_3.png?500 |}}
+  - we eliminate the state 0,1
+    - in:
+      * init --(ε)--> 0,1
+    - out:
+      * 0,1 --(ε)--> fin
+      * 0,1 --(1)--> 2
+    - loop:
+      * 0,1 --(0)--> 0,1
+    - we add:
+      * init --(0*)--> fin
+      * init --(0*1)--> 2
+    * {{ :lfa:2022:lab6_1_dfa1_4.png?500 |}}
+  - we eliminate the state 2
+    - in:
+      * init --(0*1)--> 2
+      * 3 --(1)--> 2
+    - out:
+      * 2 --(ε)--> fin
+      * 2 --(0)--> 3
+      * 2 --(1)--> 4
+    - we add:
+      * init --(0*1)--> fin
+      * init --(0*10)--> 3
+      * init --(0*11)--> 4
+      * 3 --(1)--> fin
+      * 3 --(10)--> 3
+      * 3 --(11)--> 4
+    * {{ :lfa:2022:lab6_1_dfa1_5.png?500 |}}
+  - we eliminate the state 4
+    - in:
+      * init --(0*11)--> 4
+      * 3 --(11)--> 4
+      * 6,7 --(1)--> 4
+    - out:
+      * 4 --(0)--> 6,7
+    - we add:
+      * init --(0*110)--> 6,7
+      * 3 --(110)--> 6,7
+      * 6,7 --(10)--> 6,7
+    * {{ :lfa:2022:lab6_1_dfa1_6.png?500 |}}
+  - we eliminate the state 6,7
+    - in:
+      * init --(0*110)--> 6,7
+      * 3 --(110)--> 6,7
+      * 5 --(1)--> 6,7
+    - out:
+      * 6,7 --(0)--> 5
+      * 6,7 --(ε)--> fin
+    - loop:
+      * 6,7 --(10)--> 6,7
+    - we add:
+      * init --(0*110(10)*0)--> 5
+      * init --(0*110(10)*)--> fin
+      * 3 --(110(10)*0)--> 5
+      * 3 --(110(10)*)--> fin
+      * 5 --(1(10)*0)--> 5
+      * 5 --(1(10)*)--> fin
+    * {{ :lfa:2022:lab6_1_dfa1_7.png?500 |}}
+  - we eliminate the state 3
+    - in:
+      * init --(0*10)--> 3
+    - out:
+      * 3 --(ε|1)--> fin
+      * 3 --(0|110(10)*0)--> 5
+    - loop:
+      * 3 --(10)--> 3
+    - we add:
+      * init --(0*10(10)*(ε|1) )--> fin
+      * init --(0*10(10)*(0|110(10)*0) )-->5
+    * {{ :lfa:2022:lab6_1_dfa1_8.png?500 |}}
+  - we eliminate the state 5
+    - in:
+      * init --(0*110(10)*0|0*10(10)*(0|110(10)*0) )--> 5
+    - out:
+      * 5 --(1(10)*)--> fin
+    - loop:
+      * 5 --(1(10)*0)--> 5
+    - we add:
+      * init --( (0*110(10)*0|0*10(10)*(0|110(10)*0) ) (1(10)*0)*1(10)*)--> fin
+  - the final regex is ''
+*|0*1|0*110(10)*|0*10(10)*(ε|1|110(10)*)|(0*110(10)*0|0*10(10)*(0|110(10)*0))(1(10)*0)*1(10)*''
+</hidden>
+==== 6.2. 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**:
@@ Line 52: / Line 178: @@
 === 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].
+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, add an $math[\epsilon]: $math[q = c_1 q_1 \cup c_2 q_2 \ldots c_n q_n \cup \epsilon].
 ^ ^ ^
@@ Line 65: / Line 191: @@
 === 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**:
+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:
+Going 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)^*].
@@ Line 88: / Line 214: @@
 === Exercise ===
 .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 |}} ^
+<hidden Solution>
+q0 = ε | Aq1 | Bq2
+q1 = ε | Aq1 | Bq2
+q2 = Bq2 | Aq0
+=> q2 = B*Aq0
+\\
+Replace q2:
+q0 = ε | Aq1 | BB*Aq0
+q1 = ε | Aq1 | BB*Aq0
+=> q1 = A*(ε | BB*Aq0)
+\\
+Replace q1:
+q0 = ε | AA*(ε | BB*Aq0) | BB*Aq0
+=> q0 = ε | A+ | (A+B+A | B+A)q0
+=> q0 = A* | (A+B+A | B+A)*q0
+=> q0 = (A+B+A | B+A)*A*
+\\
+From q1 = A*(ε | BB*Aq0)
+=> q1 = A*(ε | BB*A(A+B+A | B+A)*A*)
+=> q1 = A* | A*BB*A(A+B+A | B+A)*A*
+\\
+From q2 = B*Aq0
+=> q2 = B*A(A+B+A | B+A)*A*
+</hidden>