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Prove that the following language is regular. L = {r | r € {0,1}*, r does not contain six consecutive 0's or six consecutive I's} QUESTION 2. Hint: You may take advantage of some closure properties.

Prove that the following language is regular. L = {r | r € {0,1}*, r does not contain six consecutive 0's or six consecutive I's} QUESTION 2. Hint: You may take advantage of some closure properties.

Database System Concepts
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
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I would like a detailed explanation on how to get to the right answer since I want to be able to understand how to solve it thank you The topic is Theory of Automata

 

--- ### Question 2 **Objective:** Prove that the following language is regular. \[ L = \{ x \mid x \in \{0, 1\}^*, x \text{ does not contain six consecutive 0's or six consecutive 1's} \} \] **Hint:** You may take advantage of some closure properties. --- **Explanation:** To prove that the given language \( L \) is regular, we need to show that there exists a finite automaton (deterministic or non-deterministic) that recognizes \( L \). The language \( L \) consists of strings over the alphabet \(\{0, 1\}\) that do not have six consecutive 0's or six consecutive 1's. **Steps to Prove:** 1. Construct a finite automaton that keeps track of the maximum sequence of consecutive 0's and 1's. 2. States can represent the count of consecutive 0's and 1's up to a maximum of five (since encountering six would lead to rejection). 3. Use both deterministic finite automaton (DFA) or non-deterministic finite automaton (NFA) based approach. By closing the construction of such finite automaton, you can prove and validate that the language is regular due to the finite nature of the states required to track the sequences. --- This explanation leverages the concept of closure properties of regular languages and finite automata theory to provide a structured approach to solve the problem.
Transcribed Image Text:--- ### Question 2 **Objective:** Prove that the following language is regular. \[ L = \{ x \mid x \in \{0, 1\}^*, x \text{ does not contain six consecutive 0's or six consecutive 1's} \} \] **Hint:** You may take advantage of some closure properties. --- **Explanation:** To prove that the given language \( L \) is regular, we need to show that there exists a finite automaton (deterministic or non-deterministic) that recognizes \( L \). The language \( L \) consists of strings over the alphabet \(\{0, 1\}\) that do not have six consecutive 0's or six consecutive 1's. **Steps to Prove:** 1. Construct a finite automaton that keeps track of the maximum sequence of consecutive 0's and 1's. 2. States can represent the count of consecutive 0's and 1's up to a maximum of five (since encountering six would lead to rejection). 3. Use both deterministic finite automaton (DFA) or non-deterministic finite automaton (NFA) based approach. By closing the construction of such finite automaton, you can prove and validate that the language is regular due to the finite nature of the states required to track the sequences. --- This explanation leverages the concept of closure properties of regular languages and finite automata theory to provide a structured approach to solve the problem.
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