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--- lfa:2022:lab09-cfl [2022/12/04 20:46]
alexandra.udrescu01
+++ lfa:2022:lab09-cfl [2022/12/17 11:05] (current)
alexandra.udrescu01
@@ Line 6: / Line 6: @@
 **9.1.1.** $ L = \{\: w \in \{A,B\}^* \ | \:w \text{ is a palindrome}\} $.
 <hidden>
@@ Line 11: / Line 12: @@
 $ S \leftarrow ASA | BSB | A | B | \epsilon $
 </note>
+{{ :lfa:2022:lfa2022_lab9_1_1.png?400 |}}
 </hidden>
 **9.1.2.** $ L = \{ A^{m} B^{m+n} C^{n} \ | \: n, m \geq 0 \} $
 <hidden>
@@ Line 21: / Line 27: @@
 $ Y \leftarrow BYC | \epsilon $
 </note>
+{{ :lfa:2022:lfa2022_lab9_1_4.png?400 |}}
 </hidden>
 **9.1.3.** $ L = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \} $
 <hidden>
 <note warning>
 Try #1:\\
-$ S \leftarrow aBS | baS | \epsilon $ \\
+$ S \leftarrow abS | baS | \epsilon $ \\
-ambiguuous because of: “ababab” \\
+incomplete because of: “aabb” \\
 Try #2:\\
-$ S \leftarrow $ aSb | bSa | SS | $ \epsilon $ \\
+$ S \leftarrow aSb | bSa | SS | \epsilon $ \\
-ambiguuous because of: “abab”
+incomplete because of: “abab”
 </note>
+**Solution 1:**
+We can either generate one ''a'' followed by one ''b'', as well as one ''b'' followed by one ''a''. At the same time, other sequences of equal number of a and b can appear freely, hence one solution is:
+$math[S \leftarrow aSbS \mid bSaS \mid \epsilon]
+/*
+**Solution 2:**
 <note important>
-**Idea:** A = rule that promises that one letter 'a' should come without a pair; B = rule that promises that one letter 'b' should come without a pair \\
+**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 $
 </note>
-{{ :lfa:2022:lfa2022_lab9_1_3.png?300 |}} accept by empty stack
+{{ :lfa:2022:lfa2022_lab9_1_3.png?300 |}}
+We can easily check by induction (over the length of the derivation sequence) that $math[A \Rightarrow^* w] iff $math[\#_A(w) = 1 + \#_B(w)].
+The basis case (length 1) is straightforward as $math[A \Rightarrow a]. Now suppose $math[A \Rightarrow bAA \Rightarrow^* bw_1w_2]. By induction hypothesis: $math[\#_A(bw_1w_2) = \#_A(w_1) + \#_A(w_2) = \#_B(w_1) + 1 + \#_B(w_2) + 1 = \#_B(bw_1w_2)+1].
+*/
 </hidden>
 **9.1.4.** $ L = \{w \in \{a, b\}^* | \#_a(w) \neq \#_b(w) \} $
 <hidden>
+/*
+Solution 1:
 <note important>
 $ S \leftarrow A | B $ \\
-$ A \leftarrow G |GG $ \\
+$ A \leftarrow G |GA $ \\
-$ B \leftarrow H | HH $ \\
+$ B \leftarrow H | HB $ \\
 $ G \leftarrow Ea $ \\
 $ H \leftarrow Eb $ \\
 $ E \leftarrow aME | bNE | \epsilon $ \\
 $ M \leftarrow b | aMM $ \\
-$ N \leftarrow a | bNN $
+$ N \leftarrow a | bNN $ \\
+Non-terminal E generates words with the same number of 'a' and 'b', as it was presented in the previous exercise. \\
+Non-terminal A generates words that have more "a", while B generates words with fewer "a". \\ Then S generates words with either more, or fewer 'a' than 'b'.
 </note>
-{{ :lfa:2022:lfa2022_lab9_1_4.png?400 |}} accept by empty stack
+{{ :lfa:2022:lfa2022_lab9_1_4.png?400 |}}
+Solution 2:
+*/
+Start with a grammar that generates $math[L_= = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \}], for instance $math[S \leftarrow aSbS | bSaS | \epsilon]. Next, we generate two grammars, one for $math[L_> = \{w \in \{a, b\}^* | \#_a(w) > \#_b(w) \}], and another for $math[L_< = \{w \in \{a, b\}^* | \#_a(w) < \#_b(w) \}], and we can combine them into our response. We illustrate writing a rule for the former.
+The language $math[L_>] can be described by the following language operations: $math[L_> = L_=(\{a\}L_=)^+]. Note that $math[\epsilon] is member of $math[L_=]. If $math[S] is the start symbol for $math[L_=], then our grammar is: $math[S_> \leftarrow SaST, T \leftarrow aST | \epsilon].
 </hidden>
 **9.1.5.** $ L = \{a^ib^jc^k | i = j \lor j = k \} $ \\
 <hidden>
@@ Line 74: / Line 116: @@
 {{ :lfa:2022:lfa2022_lab9_1_5.png?600 |}}
 </hidden>
 ===== 9.2. Ambiguous grammars =====
@@ Line 84: / Line 128: @@
 $ A \leftarrow aA | B $ \\
 $ B \leftarrow bB | \epsilon $
 <hidden>
@@ Line 95: / Line 141: @@
 Repaired, unambiguous grammar:
-$ S \leftarrow A $ \\
+$ S \leftarrow A $  \\
 $ A \leftarrow aA | B $ \\
 $ B \leftarrow bB | \epsilon $
 </note>
 </hidden>
 **9.2.2.**
@@ Line 108: / Line 156: @@
 $ A \leftarrow 0A1 | 01 | B $\\
 $ B \leftarrow B1 | \epsilon $
@@ Line 117: / Line 166: @@
 Repaired: \\
-$ S \leftarrow AS | \epsilon $ \\
+$ S \leftarrow 1BS' | S' $ \\
-$ A \leftarrow 0A’1B $ \\
+$ S' \leftarrow ABS' | \epsilon $ \\
-$ A’ \leftarrow 0A’1 | \epsilon $ \\
+$ A \leftarrow 0A1 | 01 $ \\
 $ B \leftarrow B1 | \epsilon $
 </note></hidden>
 **9.2.3.**
@@ Line 130: / Line 181: @@
 $ A \leftarrow aA | \epsilon $\\
 $ B \leftarrow bB | \epsilon $
@@ Line 138: / Line 190: @@
   * S ⇒ ASB ⇒ SB ⇒ B ⇒ $ \epsilon $
+The grammar actually generates the language {a,b}*:
+$ S \Rightarrow ASB \Rightarrow aASB \Rightarrow aSB \Rightarrow aS $ \\
+$ S \Rightarrow BSA \Rightarrow bBSA \Rightarrow bSA \Rightarrow bS $
 Repaired: \\
-$ S \leftarrow aSb | bSa | b | a | \epsilon $
+$ S \leftarrow aS | bS | \epsilon $
 </note>
 </hidden>
 **9.2.4.** Write an ambiguous grammar for $ L(a^*) $.
 <hidden><note important>
@@ Line 153: / Line 212: @@
   * S => aaS => aa
 </note></hidden>