Chapter 2: Q24P (page 163)

Let $E = {a^{i} b^{j} | i \neq j a n d 2 i \neq j}$ . Show that $E$ is context-free language.

Short Answer

Expert verified

It is shown that $E$ is a context-free language.

Step by step solution

Define context-free language

The context-free language is generated by context-free grammar. These languages are accepted by Pushdown Automata. These are the supersets of regular languages.

Consider context-free languages $L_{1}$ described as $G_{1} = (V_{1}, S, R_{1}, S_{1}) .$

Consider context-free language $L_{2}$ described as $G_{2} = (V_{2}, S, R_{2}, S_{2}) .$

Explain the solution

Split it as unions of three languages.

$\{a^{i} b^{j} | i > j\} ̀ E \{a^{i} b^{j} | i < j a n d 2 i > j\} \cup \{a^{i} b^{j} | 2 i < j\} G^{1} = \{a^{i} b^{j} | i > j\} G^{2} = \{a^{i} b^{j} | i < j a n d 2 i > j\} G^{3} = \{a^{i} b^{j} | 2 i < j\}$

$G^{1} a n d G^{3} a r e q u i t e e a s y t o f i g u r e o u t . f o r G^{1} S \to a S b | a S^{1} S^{1} \to a S^{1} | ε$

for G³

$S \to a S b b | S^{1} b S^{1} \to S^{1} b | ε N o w t r i c k y p a r t i s G 2 .$

S \to a S b | a S^{1} b S^{1} \to a S^{1} b b | a S^{2} b b S^{2} \to ε

It is proved very clearly, that $S w i l l g e n e r a t e a^{n} a^{m} b^{2 m} b^{n} t h a t i s e q u i v a l e n t t o a^{n + m} b^{n + 2 m}$ ,

where $m, n m u s t b e g r e a t e r t h a n 0 .$

Hence, $E$ is a context-free language.

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Let $E = {a^{i} b^{j} | i \neq j a n d 2 i \neq j}$ . Show that $E$ is context-free language.

Short Answer

Step by step solution

Define context-free language

Explain the solution

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Let E={aibj|i≠jand2i≠j}. Show thatE is context-free language.

Short Answer

Step by step solution

Define context-free language

Explain the solution

One App. One Place for Learning.

Most popular questions from this chapter

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Let $E = {a^{i} b^{j} | i \neq j a n d 2 i \neq j}$ . Show that $E$ is context-free language.