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--- lfa:2022:lab09-cfl [2022/12/04 20:13]
alexandra.udrescu01
+++ lfa:2022:lab09-cfl [2022/12/17 11:05] (current)
alexandra.udrescu01
@@ Line 10: / Line 10: @@
 <hidden>
 <note important>
+$ S \leftarrow ASA | BSB | A | B | \epsilon $
-S $ \leftarrow $ A S A | B S B | A | B | $ \varepsilon $
 </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>
-<note important>
-S $ \leftarrow $ X Y
-X $ \leftarrow $ A X B | $ \varepsilon $
-Y $ \leftarrow $ B Y C | $ \varepsilon $
+<hidden>
+<note important>
+$ S \leftarrow XY $ \\
+$ X \leftarrow AXB | \epsilon $ \\
+$ 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>
-S $ \leftarrow $ a B S | b a S | $ \varepsilon $
+Try #1:\\
+$ S \leftarrow abS | baS | \epsilon $ \\
+incomplete because of: “aabb” \\
+Try #2:\\
+$ S \leftarrow aSb | bSa | SS | \epsilon $ \\
+incomplete because of: “abab”
+</note>
-ambiguuous because of: “ababab”
+**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:**
-S $ \leftarrow $ a S b | b S a | S S | $ \varepsilon $
+<note important>
+**Idea:** A = rule that promises that exactly one letter 'a' is extra; B = rule that promises that exactly one letter 'b' is extra \\
-ambiguuous because of: “abab”
+$ S \leftarrow aBS | bAS | \epsilon $ \\
+$ A \leftarrow a | bAA $ \\
+$ B \leftarrow b | aBB $
 </note>
+{{ :lfa:2022:lfa2022_lab9_1_3.png?300 |}}
-<note important>
+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>
-**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
-S $ \leftarrow $ a B S | b A S | $ \varepsilon $
-A $ \leftarrow $ a | b A A
-B $ \leftarrow $ b | a B B
-</note></hidden>
 **9.1.4.** $ L = \{w \in \{a, b\}^* | \#_a(w) \neq \#_b(w) \} $
 <hidden>
+/*
+Solution 1:
 <note important>
-S $ \leftarrow $ A | B
+$ S \leftarrow A | B $ \\
+$ A \leftarrow G |GA $ \\
+$ B \leftarrow H | HB $ \\
+$ G \leftarrow Ea $ \\
+$ H \leftarrow Eb $ \\
+$ E \leftarrow aME | bNE | \epsilon $ \\
+$ M \leftarrow b | aMM $ \\
+$ 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 |}}
-A $ \leftarrow $ G |G G
+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.
-B $ \leftarrow $ H | H H
+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].
-G $ \leftarrow $ E a
+</hidden>
-H $ \leftarrow $ E b
-E $ \leftarrow $ a M E | b N E | $ \varepsilon $
-M $ \leftarrow $ b | a M M
+**9.1.5.** $ L = \{a^ib^jc^k | i = j \lor j = k \} $ \\
-N $ \leftarrow $ a | b N N
-</note>
-</hidden>
-**9.1.5.** $ L = \{a^ib^jc^k | i = j \lor j = k \} $ \\
 <hidden>
@@ Line 87: / Line 108: @@
 $ S \leftarrow X | Y $ \\
 $ X \leftarrow ZC $ \\
-$ Z \leftarrow aZb | \varepsilon $ \\
+$ Z \leftarrow aZb | \epsilon $ \\
-$ C \leftarrow cC | \varepsilon $ \\
+$ C \leftarrow cC | \epsilon $ \\
 $ Y \leftarrow AT $ \\
-$ A \leftarrow aA | \varepsilon $  \\
+$ A \leftarrow aA | \epsilon $  \\
-$ T \leftarrow bTc | \varepsilon $
+$ T \leftarrow bTc | \epsilon $  \\
 </note>
+{{ :lfa:2022:lfa2022_lab9_1_5.png?600 |}}
 </hidden>
 ===== 9.2. Ambiguous grammars =====
@@ Line 104: / Line 128: @@
 $ A \leftarrow aA | B $ \\
 $ B \leftarrow bB | \epsilon $
 <hidden>
@@ Line 115: / 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 128: / Line 156: @@
 $ A \leftarrow 0A1 | 01 | B $\\
 $ B \leftarrow B1 | \epsilon $
@@ Line 137: / 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 150: / Line 181: @@
 $ A \leftarrow aA | \epsilon $\\
 $ B \leftarrow bB | \epsilon $
@@ Line 158: / 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^*) $.
+<hidden><note important>
+$ S \leftarrow aS | aaS | \epsilon $\\
+Ambiguuous: yes\\
+  * S => aS => aaS => aa
+  * S => aaS => aa
+</note></hidden>