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lfa:2024:lab10 [2024/12/16 18:47]
cata_chiru
lfa:2024:lab10 [2024/12/16 19:34] (current)
cata_chiru
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 **9.1.3.** $ L = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \} $ **9.1.3.** $ L = \{w \in \{a, b\}^* | \#_a(w) = \#_b(w) \} $
 +
 +<hidden Solution 9.1.3>
 +
 +$ S \leftarrow aBS\ |\ bAS\ |\ \epsilon $ \\
 +$ A \leftarrow a\ |\ bAA $ \\
 +$ B \leftarrow b\ |\ aBB $ \\
 +
 +</​hidden>​
 +
  
 **9.1.4.** $ L = \{w \in \{a, b\}^* | \#_a(w) \neq \#_b(w) \} $ **9.1.4.** $ L = \{w \in \{a, b\}^* | \#_a(w) \neq \#_b(w) \} $
 +
  
 **9.1.5.** $ L = \{a^ib^jc^k | i = j \lor j = k \} $ \\ **9.1.5.** $ L = \{a^ib^jc^k | i = j \lor j = k \} $ \\
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 <hidden Solution 9.1.5> <hidden Solution 9.1.5>
  
-$ S \leftarrow C \| \A $ \\+$ S \leftarrow C\ |\ A $ \\
 $ C \leftarrow Cc\ |\ B $ \\ $ C \leftarrow Cc\ |\ B $ \\
 $ B \leftarrow aBb\ |\ \ \epsilon $ \\ $ B \leftarrow aBb\ |\ \ \epsilon $ \\
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 $ A \leftarrow aA\ |\ B $ \\ $ A \leftarrow aA\ |\ B $ \\
 $ B \leftarrow bB\ |\ \epsilon $ $ B \leftarrow bB\ |\ \epsilon $
 +
 +<hidden Solution 9.2.1>
 +
 +$ S \leftarrow A $ \\
 +$ A \leftarrow aA\ |\ B $ \\
 +$ B \leftarrow bB\ |\ \epsilon $
 +
 +</​hidden>​
  
 **9.2.2.** **9.2.2.**
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 $ A \leftarrow 0A1\ |\ 01\ |\ B $\\ $ A \leftarrow 0A1\ |\ 01\ |\ B $\\
 $ B \leftarrow B1\ |\ \epsilon $ $ B \leftarrow B1\ |\ \epsilon $
 +
 +<hidden Solution 9.2.2>
 +
 +$ S \leftarrow AS\ |\ A $ \\
 +$ A \leftarrow 0A1\ |\ B $\\
 +$ B \leftarrow B1\ |\ \epsilon $
 +
 +</​hidden>​
 +
  
 **9.2.3.** **9.2.3.**
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 $ A \leftarrow aA\ |\ \epsilon $\\ $ A \leftarrow aA\ |\ \epsilon $\\
 $ B \leftarrow bB\ |\ \epsilon $ $ B \leftarrow bB\ |\ \epsilon $
 +
 +<hidden Solution 9.2.3>
 +
 +$ S \leftarrow aS\ |\ bS\ |\ \epsilon $ \\
 +
 +</​hidden>​
 +
 +
  
 **9.2.4.** Write an ambiguous grammar for $ L(a^*) $. **9.2.4.** Write an ambiguous grammar for $ L(a^*) $.
  
 +<hidden Solution 9.2.4>
 +
 +$ S \leftarrow aS \ |\ A\ |\ \epsilon $\\
 +$ A \leftarrow aA\ |\ \epsilon $\\
 +
 +</​hidden>​
 ===== 9.3 Regular Grammars ===== ===== 9.3 Regular Grammars =====
  
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 <hidden Solution 9.3.1> <hidden Solution 9.3.1>
  
-Y produces 1^*(0 | \epsilon)+Y produces ​1^* (0 | \epsilon) ​
  
-X produces 0*(1Y\rightarrow 0^*1^+(0 | \epsilon)+X produces ​0^* 1Y \rightarrow 0^* 1^+ (0 | \epsilon) ​$
 </​hidden>​ </​hidden>​