Questions are not graded i need b and c the topic is automata theory i included example 4.3
Questions are not graded i need b and c the topic is automata theory i included example 4.3
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Questions are not graded
i need b and c
the topic is automata theory
i included example 4.3
![### Example 4.3: Proving that the Language \( L_{pr} \) is Not Regular
Let us show that the language \( L_{pr} \) consisting of all strings of 1's whose length is a prime is not a regular language. Suppose it were. Then there would be a constant \( n \) satisfying the conditions of the pumping lemma. Consider some prime \( p \geq n + 2 \); there must be such a \( p \), since there are an infinity of primes. Let \( w = 1^p \).
By the pumping lemma, we can break \( w = xyz \) such that \( y \neq \epsilon \) and \( |xy| \leq n \). Let \( |y| = m \). Then \( |xz| = p - m \). Now consider the string \( xy^{p-m}z \), which must be in \( L_{pr} \) by the pumping lemma, if \( L_{pr} \) really is regular. However,
\[ |xy^{p-m}z| = |xz| + (p - m)|y| = p - m + (p - m)m = (m + 1)(p - m) \]
It looks like \( |xy^{p-m}z| \) is not a prime, since it has two factors \( m + 1 \) and \( p - m \). However, we must check that neither of these factors are 1, since then \( (m + 1)(p - m) \) might be a prime after all. But \( m + 1 > 1 \), since \( y \neq \epsilon \) tells us \( m \geq 1 \). Also, \( p - m > 1 \), since \( p \geq n + 2 \) was chosen, and \( m \leq n \) since
\[ m = |y| \leq |xy| \leq n \]
Thus, \( p - m \geq 2 \).
Again, we have started by assuming the language in question was regular, and we derived a contradiction by showing that some string not in the language was required by the pumping lemma to be in the language. Thus, we conclude that \( L_{pr} \) is not a regular language.
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Transcribed Image Text:### Example 4.3: Proving that the Language \( L_{pr} \) is Not Regular
Let us show that the language \( L_{pr} \) consisting of all strings of 1's whose length is a prime is not a regular language. Suppose it were. Then there would be a constant \( n \) satisfying the conditions of the pumping lemma. Consider some prime \( p \geq n + 2 \); there must be such a \( p \), since there are an infinity of primes. Let \( w = 1^p \).
By the pumping lemma, we can break \( w = xyz \) such that \( y \neq \epsilon \) and \( |xy| \leq n \). Let \( |y| = m \). Then \( |xz| = p - m \). Now consider the string \( xy^{p-m}z \), which must be in \( L_{pr} \) by the pumping lemma, if \( L_{pr} \) really is regular. However,
\[ |xy^{p-m}z| = |xz| + (p - m)|y| = p - m + (p - m)m = (m + 1)(p - m) \]
It looks like \( |xy^{p-m}z| \) is not a prime, since it has two factors \( m + 1 \) and \( p - m \). However, we must check that neither of these factors are 1, since then \( (m + 1)(p - m) \) might be a prime after all. But \( m + 1 > 1 \), since \( y \neq \epsilon \) tells us \( m \geq 1 \). Also, \( p - m > 1 \), since \( p \geq n + 2 \) was chosen, and \( m \leq n \) since
\[ m = |y| \leq |xy| \leq n \]
Thus, \( p - m \geq 2 \).
Again, we have started by assuming the language in question was regular, and we derived a contradiction by showing that some string not in the language was required by the pumping lemma to be in the language. Thus, we conclude that \( L_{pr} \) is not a regular language.
\[\Box\]
Transcribed Image Text:**Exercise 7.2.1**: Use the CFL (Context-Free Language) pumping lemma to show each of these languages not to be context-free:
* **a)** \(\{ a^i b^j c^k \mid i < j < k\}\).
* **b)** \(\{ a^n b^n c^i \mid i \leq n\}\).
* **c)** \(\{0^p \mid p \text{ is a prime}\}\). *Hint*: Adapt the same ideas used in Example 4.3, which showed this language not to be regular.
* **d)** \(\{0^i 1^j \mid j = i^2\}\).
* **e)** \(\{a^n b^n c^i \mid n \leq i \leq 2n\}\).
* **f)** \(\{ ww^R w \mid w \text{ is a string of 0's and 1's} \}\). That is, the set of strings consisting of some string \(w\) followed by the same string in reverse, and then the string \(w\) again, such as 001100001.
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