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# ch8ans - Because L1 µ L2 = Ga ± b ± c ±,n ³ 0´ is not...

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Formal Language Assignment Answer Chapter 8.1 2. 7. (d) We use pumping lemma to prove L is not context-free. There exist a string W belong to L. Let W g a G b G c G±² . b |vxy| ³ m , |vy| ´ 1 Case1: If vy g a µ or vy g b µ , then we can pick i ´ 2 . Case2: If vy g c µ and 1 ³ k ¶ · , then we can pick i g 0 . Case3: If vy g a µ b ¸ and 1 ³ k ¹ j ¶ · , then we can pick i ´ 2 . Case4: If vy g b µ c ¸ ,1 ³ k,j and k ¹ j ¶ · , then we can pick i g 0 . Note: uv º xy º z g a G b G»µ c G±²»¸ , 1 ³ j

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Chapter 8.2 11. Let L1 g Ga ± b ± c ² ,n ³ 0,m ³ 0´ , L2 g Ga ± b ² c ² ,n ³ 0,m ³ 0´ We know L1 and L2 are both deterministic context-free.
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Unformatted text preview: Because L1 µ L2 = Ga ± b ± c ± ,n ³ 0´ is not deterministic context-free. The family of context-free languages is not closed under intersection By textbook page 198 and 199 L3 g Ga ± b ± ,n ³ 0´ , L4 g Ga ± b ¶± ,n ³ 0´ are both deterministic context-free. Because L3 · L4 = Ga ± b ² ,m g n or m g 2n,n ³ 0´ is not deterministic context-free. The family of deterministic context-free languages is not closed under union. 17....
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ch8ans - Because L1 µ L2 = Ga ± b ± c ±,n ³ 0´ is not...

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