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--- lfa:2022:lab06-dfa-to-regex [2022/11/13 20:36]
mihai.udubasa change 6.1. title
+++ lfa:2022:lab06-dfa-to-regex [2022/11/19 10:34] (current)
alexandra.udrescu01
@@ Line 5: / Line 5: @@
 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 On what DFA should the algorithm be applied?>
 It is recommended to apply the State elimination algorithm on the minimal DFA.
@@ Line 76: / Line 75: @@
   - 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_2.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_3.png?500 |}}
   - we eliminate the state 0,1
     - in:
@@ Line 90: / Line 91: @@
       * init --(0*)--> fin
       * init --(0*1)--> 2
-    * {{ :lfa:2022:lab6_1_dfa1_3.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_4.png?500 |}}
   - we eliminate the state 2
     - in:
@@ Line 106: / Line 107: @@
       * 3 --(10)--> 3
       * 3 --(11)--> 4
-    * {{ :lfa:2022:lab6_1_dfa1_4.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_5.png?500 |}}
   - we eliminate the state 4
     - in:
       * init --(0*11)--> 4
       * 3 --(11)--> 4
-      * 6,7 --(0)--> 4
+      * 6,7 --(1)--> 4
     - out:
-      * 4 --(1)--> 6,7
+      * 4 --(0)--> 6,7
     - we add:
-      * init --(0*111)--> 6,7
+      * init --(0*110)--> 6,7
-      * 3 --(111)--> 6,7
+      * 3 --(110)--> 6,7
-      * 6,7 --(01)--> 6,7
+      * 6,7 --(10)--> 6,7
-    * {{ :lfa:2022:lab6_1_dfa1_5.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_6.png?500 |}}
   - we eliminate the state 6,7
     - in:
-      * init --(0*111)--> 6,7
+      * init --(0*110)--> 6,7
-      * 3 --(111)--> 6,7
+      * 3 --(110)--> 6,7
-      * 5 --(0)--> 6,7
+      * 5 --(1)--> 6,7
     - out:
-      * 6,7 --(1)--> 5
+      * 6,7 --(0)--> 5
       * 6,7 --(ε)--> fin
     - loop:
-      * 6,7 --(01)--> 6,7
+      * 6,7 --(10)--> 6,7
     - we add:
-      * init --(0*111(01)*1)--> 5
+      * init --(0*110(10)*0)--> 5
-      * init --(0*111(01)*)--> fin
+      * init --(0*110(10)*)--> fin
-      * 3 --(111(01)*1)--> 5
+      * 3 --(110(10)*0)--> 5
-      * 3 --(111(01)*)--> fin
+      * 3 --(110(10)*)--> fin
-      * 5 --(0(01)*1)--> 5
+      * 5 --(1(10)*0)--> 5
-      * 5 --(0(01)*)--> fin
+      * 5 --(1(10)*)--> fin
-    * {{ :lfa:2022:lab6_1_dfa1_6.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_7.png?500 |}}
   - we eliminate the state 3
     - in:
@@ Line 142: / Line 143: @@
     - out:
       * 3 --(ε|1)--> fin
-      * 3 --(0|111(01)*1)--> 5
+      * 3 --(0|110(10)*0)--> 5
     - loop:
-      * 3 --(01)--> 3
+      * 3 --(10)--> 3
     - we add:
-      * init --(0*10(01)*(ε|1) )--> fin
+      * init --(0*10(10)*(ε|1) )--> fin
-      * init --(0*10(01)*(0|111(01)*1) )-->5
+      * init --(0*10(10)*(0|110(10)*0) )-->5
-    * {{ :lfa:2022:lab6_1_dfa1_7.png?500 |}}
+    * {{ :lfa:2022:lab6_1_dfa1_8.png?500 |}}
   - we eliminate the state 5
     - in:
-      * init --(0*111(01)*1|0*10(01)*(0|111(01)*1) )--> 5
+      * init --(0*110(10)*0|0*10(10)*(0|110(10)*0) )--> 5
     - out:
-      * 5 --(0(01)*)--> fin
+      * 5 --(1(10)*)--> fin
     - loop:
-      * 5 --(0(01)*1)--> 5
+      * 5 --(1(10)*0)--> 5
     - we add:
-      * init --( (0*111(01)*1|0*10(01)*(0|111(01)*1) ) (0(01)*1)*0(01)*)--> fin
+      * init --( (0*110(10)*0|0*10(10)*(0|110(10)*0) ) (1(10)*0)*1(10)*)--> fin
-    * {{ :lfa:2022:lab6_1_dfa1_8.png?500 |}}
   - the final regex is ''
-*|0*1|0*111(01)*|0*10(01)*(ε|1)|(0*111(01)*1|0*10(01)*(0|111(01)*1))(0(01)*1)*0(01)*''
+*|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 ====
@@ Line 176: / 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 189: / 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 214: / Line 216: @@
 .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>