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Let L be a language over the alphabet Σ = {a, b}. Prove that

Let L be a language over the alphabet Σ = {a, b}. Prove that

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
Problem 1PE
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A The pumping length of the following deterministic finite automata is 4 because there are four states. 0 0,1 Q3 * 0 0 1 1 Q2 Answer the following: 1. State a string of length 4 or greater that is accepted by the deterministic finite automaton 2. Decompose the string into the x, y, and z substrings so that the conclusions of the pumping lemma holds, and, 3. Highlight a loop in the state diagram that corresponds to the substring y. B Let L be a language over the alphabet Σ = {a, b}. Prove that L = {akb2k for any integer k ≥ 0} is not regular. Proof: Assume that L is regular, so that the pumping lemma applies. Therefore L has a pumping length p. Let s = Since s belongs to This string is a member of L because L and has length longer than p, there exist strings x, y, and z such that s = xyz, where * xz is a member of L, *xy'z is a member of L for all positive integers i, * y >0, and |xy| ≤p. Now consider the following argument: (You finish the rest of the proof.)
Transcribed Image Text:A The pumping length of the following deterministic finite automata is 4 because there are four states. 0 0,1 Q3 * 0 0 1 1 Q2 Answer the following: 1. State a string of length 4 or greater that is accepted by the deterministic finite automaton 2. Decompose the string into the x, y, and z substrings so that the conclusions of the pumping lemma holds, and, 3. Highlight a loop in the state diagram that corresponds to the substring y. B Let L be a language over the alphabet Σ = {a, b}. Prove that L = {akb2k for any integer k ≥ 0} is not regular. Proof: Assume that L is regular, so that the pumping lemma applies. Therefore L has a pumping length p. Let s = Since s belongs to This string is a member of L because L and has length longer than p, there exist strings x, y, and z such that s = xyz, where * xz is a member of L, *xy'z is a member of L for all positive integers i, * y >0, and |xy| ≤p. Now consider the following argument: (You finish the rest of the proof.)
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