Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
lfa:2022:lab06-dfa-to-regex [2022/11/14 00:46] mihai.udubasa fix typo |
lfa:2022:lab06-dfa-to-regex [2022/11/19 10:34] (current) alexandra.udrescu01 |
||
|---|---|---|---|
| Line 5: | Line 5: | ||
| 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). | ||
| Line 168: | Line 164: | ||
| </hidden> | </hidden> | ||
| - | */ | + | |
| ==== 6.2. Brzozowsky's algebraic method ==== | ==== 6.2. Brzozowsky's algebraic method ==== | ||
| Line 195: | Line 191: | ||
| === 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)^*]. | ||
| Line 221: | Line 217: | ||
| ^ {{ :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> | ||
| Line 271: | Line 267: | ||
| </hidden> | </hidden> | ||
| - | */ | + | |