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--- lfa:2023:lab02 [2023/10/16 20:27]
mihai.udubasa copied old lab
+++ lfa:2023:lab02 [2024/10/06 21:30] (current)
stefan.sterea old revision restored (2023/10/31 11:44)
@@ Line 1: / Line 1: @@
-====== 2. Deterministic Finite Automata ======
+====== 2. Deterministic Finite Automata and Regular Expressions ======
 ===== 2.1. DFA practice =====
@@ Line 83: / Line 83: @@
 {{:lfa:2022:lfa2022_lab2_ex6_v2.png?400|}}
-{{:lfa:2022:lfa2022_lab2_ex6.png?400|}}
   * injuraturile redirectati-le catre AU :)) (si MP)
   * All missing transitions lead to a sink state (where there is a transition that loops in that state for any character)
+</hidden>
+**2.1.7.** The set of all binary strings having the substring 00101
+<hidden>
+{{:lfa:2022:ex_8_dfa.png?400|}}
+  * Every state corresponds to a certain current configuration (eg. state **B** translates to a sequence of **0**; **C** to a seq of **00**; **D** -> **001** and so on)
+  * Analyze how each incoming character changes the current available sequence. For example, if we are in state **E** and we read character **1** we reach a final state, but if we read **0** we go back to state **C** since the available seq will be **00**
 </hidden>
@@ Line 91: / Line 100: @@
 === Describe the language of the following DFAs and simulate them on the given inputs: ===
-**2.1.7.** w<sub>1</sub> = 10011110, w<sub>2</sub> = 00110010
+**2.1.8.** w<sub>1</sub> = 10011110, w<sub>2</sub> = 00110010
 {{:lfa:2022:lfa2022_lab2_ex8.png?400|}}
@@ Line 100: / Line 109: @@
   * (0, **10011110**) |--<sub>A</sub> (11, 0011110) |--<sub>A</sub> (01, 011110) |--<sub>A</sub> (02, 11110) |--<sub>A</sub> (11, 1110) |--<sub>A</sub> (12, 110) [There is no transition, so we go to a sink state which is not figured on the drawing.] |--<sub>A</sub> (sink, 10) |--<sub>A</sub> (sink, 0) |--<sub>A</sub> (sink, $math[\epsilon])  The sink state is not a final state, so this word is **rejected**.
   * (0, **00110010**) |--<sub>A</sub> (01, 0110010) |--<sub>A</sub> (02, 110010) |--<sub>A</sub> (11, 10010) |--<sub>A</sub> (12, 0010) |--<sub>A</sub> (01, 010) |--<sub>A</sub> (02, 10) |--<sub>A</sub> (11, 0) |--<sub>A</sub> (01, $math[\epsilon]) The word is **accepted** by the DFA because we reached a final state (blue).
+</hidden>
+===== 2.2. Regex practice =====
+For each of the exercises from DFA practice write a regex describing the same language.
+<hidden 2.2.1.>
+$math[0^*10^*(10^*10^*)^*]
+</hidden>
+<hidden 2.2.2.>
+$math[0^*10^*10^*]
+</hidden>
+<hidden 2.2.3.>
+$math[(0 \cup 1)^*1]
+</hidden>
+<hidden 2.2.4.>
+creating the regex directly is not easy, but the pattern can be derived by studying the dfa, and trying to 'replace' states with more complex transitions. For a step-by-step description of the method, wait until the lecture/lab about the dfa to regex transformation algorithm
+$math[((1(01^*0)^*1)\cup0)^*]
+</hidden>
+<hidden 2.2.5.>
+$(0[1-9])\cup((1\cup2)[0-9])\cup(3(0\cup1))/((0[1-9])\cup(1(0\cup1\cup2))/[0-9][0-9][0-9][0-9])$
+</hidden>
+<hidden 2.2.6.>
+This language is a good example of a language which is significantly easier to describe using a dfa rather than a regex. The regex is very hard to build intuitively and is very long. For a way to algorithmically build this regex from the dfa built during the first part, wait until the lab/lecture about the dfa->regex transformation.
+</hidden>
+<hidden 2.2.7>
+$math[(0 \cup 1)^*00101(0 \cup 1)^*]
 </hidden>