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1 Deterministic Finite Automata Consider a DFA M = (Q, E, §, s, f) with States Q = {s, q1, q2, ƒ}, where s is the start and ƒ is the final state; Alphabet £= {0, 1}; and transition function 8. 1. %3D Construct a state transition table for 8 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings 0, 11, 110, 1001, 1100, .. would be accepted strings, but 1, 10, 100, 101, ... would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.)

1 Deterministic Finite Automata Consider a DFA M = (Q, E, §, s, f) with States Q = {s, q1, q2, ƒ}, where s is the start and ƒ is the final state; Alphabet £= {0, 1}; and transition function 8. 1. %3D Construct a state transition table for 8 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings 0, 11, 110, 1001, 1100, .. would be accepted strings, but 1, 10, 100, 101, ... would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.)

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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1 Deterministic Finite Automata Consider a DFA M = (Q, E, 8, s, f) with States Q = {s, q1, 42, f}, where s is the start and ƒ is the final state; Alphabet £ = {0, 1}; and transition function d. 1. Construct a state transition table for 8 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings 0, 11, 110, 1001, 1100, . .. would be accepted strings, but 1, 10, 100, 101, . would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.)
Transcribed Image Text:1 Deterministic Finite Automata Consider a DFA M = (Q, E, 8, s, f) with States Q = {s, q1, 42, f}, where s is the start and ƒ is the final state; Alphabet £ = {0, 1}; and transition function d. 1. Construct a state transition table for 8 (or you can draw a state transition diagram) that recognizes regular expressions that are binary strings and multiples of 3, for example, the strings 0, 11, 110, 1001, 1100, . .. would be accepted strings, but 1, 10, 100, 101, . would not be accepted. (Hint: Think, if n = 3k is a multiple of 3, then the next multiple of 3 is 3k + 3. this could be accomplished by a transition from the current state to a next state by scanning 3 ones.)
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