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(a) L₁ = {w = {0, 1}* ||w| ≥ 6, and the first and fourth symbols of w are the same as each other} (b) L₂= {aab, aba, bb} (You can assume here that the input alphabet is {a, b}. Your finite automaton should accept the three strings in L2 and should not accept any other strings of 'a's and 'b's.) (c) L3 = {w€ {a,b,c}* ||w| ≥ 3, and the symbol in every odd position (1st, 3rd, 5th, etc.) is an 'a'}

(a) L₁ = {w = {0, 1}* ||w| ≥ 6, and the first and fourth symbols of w are the same as each other} (b) L₂= {aab, aba, bb} (You can assume here that the input alphabet is {a, b}. Your finite automaton should accept the three strings in L2 and should not accept any other strings of 'a's and 'b's.) (c) L3 = {w€ {a,b,c}* ||w| ≥ 3, and the symbol in every odd position (1st, 3rd, 5th, etc.) is an 'a'}

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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For each of the following languages, draw the state diagram of a deterministic finite automaton that accepts the language. (Reminder: Deterministic means that there is exactly one transition defined for each input symbol in each state.) (a) L₁ = {w = {0, 1}* ||w| ≥ 6, and the first and fourth symbols of w are the same as each other} (b) L₂ = {aab, aba, bb} (You can assume here that the input alphabet is {a, b}. Your finite automaton should accept the three strings in L2 and should not accept any other strings of 'a's and 'b's.) (c) L3 = {w € {a, b, c}* ||w| ≥ 3, and the symbol in every odd position (1st, 3rd, 5th, etc.) is an 'a'}
Transcribed Image Text:For each of the following languages, draw the state diagram of a deterministic finite automaton that accepts the language. (Reminder: Deterministic means that there is exactly one transition defined for each input symbol in each state.) (a) L₁ = {w = {0, 1}* ||w| ≥ 6, and the first and fourth symbols of w are the same as each other} (b) L₂ = {aab, aba, bb} (You can assume here that the input alphabet is {a, b}. Your finite automaton should accept the three strings in L2 and should not accept any other strings of 'a's and 'b's.) (c) L3 = {w € {a, b, c}* ||w| ≥ 3, and the symbol in every odd position (1st, 3rd, 5th, etc.) is an 'a'}
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