Differences

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--- lfa:2022:lab02-dfa [2022/10/06 09:18]
alexandra.udrescu01
+++ lfa:2022:lab02-dfa [2023/10/17 00:01] (current)
mihai.calitescu added exercises for DFA lab
@@ Line 1: / Line 1: @@
 ====== 2. Deterministic Finite Automata ======
-===== 2.1. Demo =====
+===== 2.1. DFA practice =====
-<code>
-TODO
-</code>
-===== 2.2. DFA practice =====
 <note>
   * Blue = final state
   * Orange = non-final state
-  * Sink state = a state that is not final; once reached, there is no transition to leave it, thus the DFA will reject; there is a transition that loops in this state for each possible character
+  * Sink state = a state that is not final; once reached, there is no transition that leaves it, thus the DFA will reject; there is a transition that loops in this state for each possible character
 </note>
 === Write DFAs for the following languages: ===
-**2.2.1.** $ L=\{w \in \{0,1\}^* \text{ | w contains an odd number of ones} \} $
+**2.1.1.** $ L=\{w \in \{0,1\}^* \text{ | w contains an odd number of ones} \} $
 <hidden>
@@ Line 25: / Line 25: @@
 </hidden>
-**2.2.2.** The language of binary words which contain **exactly** two ones
+**2.1.2.** The language of binary words which contain **exactly** two ones
 <hidden>
@@ Line 35: / Line 37: @@
 </hidden>
-**2.2.3.** The language of binary words which encode odd numbers (the last digit is least significative)
+**2.1.3.** The language of binary words which encode odd numbers (the last digit is least significative)
 <hidden>
@@ Line 42: / Line 46: @@
 </hidden>
-**2.2.4.** (hard) The language of words which encode numbers divisible by 3
+**2.1.4.** (hard) The language of words which encode numbers divisible by 3
 <hidden>
@@ Line 49: / Line 55: @@
   * State 1: 3k + 1
   * State 2: 3k + 2
-  * When we read a 0, the number we had so far is simply multiplied by 2:
+  * When we read a 0, the number we had so far is multiplied by 2:
       * 3k --(*2)--> 3k
       * 3k + 1 --(*2)--> 3k + 2
       * 3k + 2 --(*2)--> 3k + 1
-  * When we read a 1, the number we had so far is simply multiplied by 2 and we also add 1:
+  * When we read a 1, the number we had so far is multiplied by 2 and we also add 1:
       * 3k --(*2)--> 3k --(+1)--> 3k + 1
       * 3k + 1 --(*2)--> 3k + 2 --(+1)--> 3k
@@ Line 59: / Line 65: @@
 </hidden>
-**2.2.5.** The language of words that encode a calendar date (DD/MM/YYYY)
+**2.1.5.** The language of words that encode a calendar date (DD/MM/YYYY)
  (do not overthink about the actual number of days in a month)
 <hidden>
 {{:lfa:2022:lfa2022_lab2_ex5.png?600|}}
+  * Note: Checking for months with 30 or 31 days can be done with a slightly bigger DFA.
   * Note: Even checking for bisect years (29 Feb) can still be done with a DFA, but it's too big to do as a seminar example.
 </hidden>
-**2.2.6.** (HARD) $ L=\{w \in \{0,1,2,3\}^* \text{ | w follows the rule that every zero is immediately followed by a sequence of at least 2 consecutive threes and every one is immediately followed by a sequence of at most 2 consecutive twos} \} $
+**2.1.6.** (HARD) $ L=\{w \in \{0,1,2,3\}^* \text{ | w follows the rule that every zero is immediately followed by a sequence of at least 2 consecutive threes and every one is immediately followed by a sequence of at most 2 consecutive twos} \} $
 <hidden>
+{{:lfa:2022:lfa2022_lab2_ex6_v2.png?400|}}
 {{:lfa:2022:lfa2022_lab2_ex6.png?400|}}
-  * injuraturile redirectati-le catre AU
+  * 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>
 === Describe the language of the following DFAs and simulate them on the given inputs: ===
-**2.2.7.** 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|}}
 <hidden>
   * The language of binary words where there are no sequences in which the same character repeats 3 or more times.
@@ Line 84: / Line 108: @@
   * (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>