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--- lfa:2022:lab11-cfl_3 [2022/12/15 14:11]
alexandra.udrescu01
+++ lfa:2022:lab11-cfl_3 [2022/12/18 23:26] (current)
mihai.udubasa add some notes to 11.4.2
@@ Line 3: / Line 3: @@
 ===== 11.1. Chomsky Normal Form =====
-**11.1.1** Write the following grammars in CNF:
+Write the following grammars in CNF:
-  - \\ $math[S \leftarrow 0SA \mid ASB \\ A \leftarrow 0BA \mid 1S \mid 0A \\ B \leftarrow B1 \mid 0B \mid 1 \mid 0 ]
-  - \\ $ S \leftarrow ABC \\ A \leftarrow aAb \mid \epsilon \\ B \leftarrow bBc \mid bc \\ C \leftarrow cC \mid c $
+**11.1.1** \\ $ S \leftarrow ABC \\ A \leftarrow aAb \mid \epsilon \\ B \leftarrow bBc \mid bc \\ C \leftarrow cC \mid c $
+<hidden>
+<note tip>**Step 1:** Eliminate isolated terminals \\
+$ X \leftarrow \alpha y \beta $ becomes $ X \leftarrow \alpha Y \beta, Y \leftarrow y $ \\ \\
+$ S \leftarrow ABC \\ A \leftarrow L_aAL_b | \epsilon \\ B \leftarrow L_bBL_c | L_bL_c \\ C \leftarrow L_cC | L_c \\ L_a \leftarrow a \\ L_b \leftarrow b \\ L_c \leftarrow c $
+</note>
+<note tip>**Step 2:** Eliminate long rules \\ A rule of 3 or more non-terminals becomes a set of several rules with only 2 non-terminals. \\ \\
+$ S \leftarrow AS_1 \\ S_1 \leftarrow BC \\ A \leftarrow L_aA_1 | \epsilon \\ A_1 \leftarrow AL_b \\ B \leftarrow L_bB_1 | L_bL_c \\ B_1 \leftarrow BL_c \\ C \leftarrow L_cC | L_c \\ L_a \leftarrow a \\ L_b \leftarrow b \\ L_c \leftarrow c $</note>
+<note tip>**Step 3:** Eliminate $ \epsilon $ rules
+There is one $ \epsilon $ rule and that is for non-terminal A: $ A \leftarrow \epsilon $. \\ Identify all rules $ X \leftarrow \alpha A \beta $ and add $ X \leftarrow \alpha A \beta | \alpha \beta $. Then remove the $ \epsilon $ rules. \\ \\
+$ S \leftarrow S_1 | AS_1 \\ S_1 \leftarrow BC \\ A \leftarrow L_aA_1 \\ A_1 \leftarrow AL_b | L_b \\ B \leftarrow L_bB_1 | L_bL_c \\ B_1 \leftarrow BL_c \\ C \leftarrow L_cC | L_c \\ L_a \leftarrow a \\ L_b \leftarrow b \\ L_c \leftarrow c $</note>
+<note tip>**Step 4:** Eliminate unit rules \\ In order to eliminate each rule $ Y \leftarrow X $, we must: \\ a. Identify all rules $ X \leftarrow \alpha_1 , ..., \alpha_n $, if none exists, simply remove $ Y \leftarrow X $ \\ b. otherwise replace (expand) $ Y \leftarrow X $ by the rule $ Y \leftarrow \alpha_1 | ... | \alpha_n $ \\ c. Repeat the process for all other rules \\ \\
+$ S \leftarrow BC | AS_1 \\ S_1 \leftarrow BC \\ A \leftarrow L_aA_1 \\ A_1 \leftarrow AL_b | b \\ B \leftarrow L_bB_1 | L_bL_c \\ B_1 \leftarrow BL_c \\ C \leftarrow L_cC | c \\ L_a \leftarrow a \\ L_b \leftarrow b \\ L_c \leftarrow c $</note>
+<note tip>**Step 5:** Eeliminate unused, cyclic non-terminals and their rules</note>
+</hidden>
+**11.1.2** \\ $math[S \leftarrow 0SA \mid ASB \\ A \leftarrow 0BA \mid 1S \mid 0A \\ B \leftarrow B1 \mid 0B \mid 1 \mid 0 ]
+<hidden>
+<note tip>**Step 1:** Eliminate terminals\\ $ S \leftarrow ZSA | ASB \\ A \leftarrow ZBA | US | ZA \\ B \leftarrow BU | ZB | U | Z \\U \leftarrow 1 \\ Z  \leftarrow 0 \\ $</note>
+<note tip>**Step 2:** Eliminate long rules\\ $ S \leftarrow ZS_1 | AS_2 \\ S_1 \leftarrow SA \\ S_2 \leftarrow SB \\ A \leftarrow ZA_1 | US | ZA \\ A_1 \leftarrow BA \\ B \leftarrow BU | ZB | U | Z \\ U \leftarrow 1 \\ Z  \leftarrow 0 \\ $</note>
+<note tip>**Step 3:** Eliminate $ \epsilon $ rules</note>
+<note tip>**Step 4:** Eliminate unit rules \\ $ S \leftarrow ZS_1 | AS_2 \\ S_1 \leftarrow SA \\ S_2 \leftarrow SB \\ A \leftarrow ZA_1 | US | ZA \\ A_1 \leftarrow BA \\ B \leftarrow BU | ZB | 1 | 0 \\ U \leftarrow 1 \\ Z \leftarrow 0 \\ $</note>
+<note tip>**Step 5:** Eliminate unused, cyclic non-terminals and their rules</note>
+</hidden>
 ===== 11.2. Regular Grammars =====
-**11.2.1** Give an example of a regular grammar that generates $ L(0^*1^*) $.
+**11.2.1** Give an example of a regular grammar that generates $ L(1^*0^*) $.
+<hidden><note tip>$ S \leftarrow I \\ I \leftarrow 1I | O \\ O \leftarrow 0O | \epsilon $</note></hidden>
 ===== 11.3. DFA to Regular Grammar =====
 **11.3.1** For each of the following DFAs, algorithmically create a regular grammar that generates the same language.
-  - {{:lfa:2022:lfa2022_lab2_ex4.png?200|}}
+  * {{:lfa:2022:lfa2022_lab2_ex4.png?200|}}
-  - {{:lfa:2022:lfa2022_lab6_mindfa2_1.png?200|}}
+<hidden>
+<note important>For each $ \delta(q, c) = q' $ build $ S_q \leftarrow cS_{q'} $ \\ For each final state q build $ S_q \leftarrow \epsilon $</note>
+<note tip> $ S_0 \leftarrow 0S_0 | 1S_1 | \epsilon \\  S_1 \leftarrow 0S_2 | 1S_0 \\  S_2 \leftarrow 1S_2 | 0S_1 $ </note>
+</hidden>
+  * {{:lfa:2022:lfa2022_lab6_mindfa2_1.png?200|}}
+<hidden><note tip> $ S_1 \leftarrow 1S_2 | 0S_1 | \epsilon \\ S_2 \leftarrow 0S_1 | 1S_3 | \epsilon \\ S_3 \leftarrow 0S_3 | 1S_3$ </note></hidden>
 ===== 11.4. Short Exercises =====
@@ Line 22: / Line 52: @@
 <hidden>
 Create a graph $math[G=(V,E)] where the nodes are **terminals and non-terminals**. For each rule of the form $math[X \leftarrow aY] create a directed edge $math[(X,Y)]. For each rule $math[X \leftarrow a], also generate an edge $math[(X,a)].
- - First, we need to check first if the grammar generates a language different from $math[\emptyset]. So we check if there is a path
+  - First, we need to check first if the grammar generates a language different from $math[\emptyset]. So we check if there is a path from the start symbol to some terminal.
-   from the start symbol to some terminal.
+  - Second, we need to check if there are //loops//. It suffices to check if the graph is a tree. MU: the condition is not sufficient as a grammar could generate a finite language while also having looping rules (if the grammar comes from the dfa transformation, these would be from the sink states of the dfa; see state 3 of the second dfa from 11.3.1)
- - Second, we need to check if there are //loops//. It suffices to check if the graph is a tree.
 </hidden>
 **11.4.3.** Prove that any DFA can be converted to a regular grammar.
-<hidden> See lecture</hidden>
+<hidden> See lecture </hidden>
 **11.4.4.** Is there a decidable algorithm to remove ambiguity from regular grammars?
 <hidden>
@@ Line 34: / Line 63: @@
 We can determinize the NFA. In the resulting DFA (in any DFA for that matter), it is not possible to derive the same word using two different transition sequences (each state and character uniquely determines the next-state). Hence, if we convert this DFA back to a regular grammar, we will get an unambiguous one.
 </hidden>
-**11.4.5.** Show the following grammar is ambiguous: $math[S \leftarrow aSbS | bSaS | epsilon]. Write a non-ambiguous equivalent.
+**11.4.5.** Show the following grammar is ambiguous: $ S \leftarrow aSbS | bSaS | \epsilon $. Write a non-ambiguous equivalent.
 <hidden>
-The grammar is indeed ambiguous: $math[S \Rightarrow aSbS \Rightarrow abS \Rightarrow abaSbS \Rightarrow ^* abab]. And also $math[S \Rightarrow aSbS \Rightarrow abSaSbS \Rightarrow^* abba]
+The grammar is indeed ambiguous: $math[S \Rightarrow aSbS \Rightarrow abS \Rightarrow abaSbS \Rightarrow ^* abab]. And also $math[S \Rightarrow aSbS \Rightarrow abSaSbS \Rightarrow^* abab]
 This grammar generates $math[\{ w \in \{0,1\}^* \mid \#_A(w) = \#_B(w) \}]. See the solution from the previous lab.
 See also [[https://cs.stackexchange.com/questions/64569/unambiguous-grammar-that-produce-equal-number-of-a-and-b | this]] for more details.
+\\
+**Idea:** A = rule that promises that exactly one letter 'a' is extra; B = rule that promises that exactly one letter 'b' is extra \\
+$ S \leftarrow aBS | bAS | \epsilon $ \\
+$ A \leftarrow a | bAA $ \\
+$ B \leftarrow b | aBB $
 </hidden>