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lfa:2022:lab06-dfa-to-regex [2022/11/14 00:31]
mihai.udubasa unhide solutions temp 		lfa:2022:lab06-dfa-to-regex [2022/11/19 10:34] (current)
alexandra.udrescu01
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 Consider the following DFAs: 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). ​ Convert the given DFAs to a Regex (using the state-elimination strategy). ​
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 </​hidden>​ </​hidden>​
  
-*/+
  
 ==== 6.2. Brzozowsky'​s algebraic method ==== ==== 6.2. Brzozowsky'​s algebraic method ====
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 === Dfa to regex conversion === === 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].
  
 ^ ^ ^ ^ ^ ^
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 === Reducing the system of equations === === 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'​].   * the solution to any equation of the form $math[q = e\cdot q \cup e'] is $math[q = e^*e'​].
  
 **Example** **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)^*].   * $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)^*].   * We apply Arden'​s Lemma one more time and yield: $math[q_1 = A^*B(A \cup B)^*].
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 ^ {{ :​lfa:​2022:​screenshot_2022-11-09_at_15.46.23.png?​200 |}} ^ ^ {{ :​lfa:​2022:​screenshot_2022-11-09_at_15.46.23.png?​200 |}} ^
  
-/*+
  
 <hidden Solution>​ <hidden Solution>​
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 </​hidden>​ </​hidden>​
  
-*/+