2 2 votes Minimum number of states in DFA over Ʃ = {0, 1} with each string contains odd number of 0’s or odd number of 1’s. Theory of Computation finite-automata + – himgta 3.6k views answer comment Share Follow Print See all 4 Comments 4 4 Comments reply Anand. commented Jul 30, 2018 i edited by Anand. Jul 30, 2018 reply Follow flag $3$ states 0 0 replyShare abhishekmehta4u commented Jul 30, 2018 reply Follow flag 010 is not accepted 1 1 replyShare Anand. commented Jul 30, 2018 reply Follow flag @abhishek :thanks for correcting. Can you check if it is correct now ? 0 0 replyShare abhishekmehta4u commented Jul 30, 2018 reply Follow flag 011, 100..... Is not accepted 2 2 replyShare Please log in or register to add a comment.
2 2 votes 4 state abhishekmehta4u answered Jul 30, 2018 abhishekmehta4u comment Share Follow See all 2 Comments 2 2 Comments reply Aditya Gangwar commented May 9, 2020 reply Follow flag How did you reach to this solution? 0 0 replyShare Yedilmaagemore commented Jun 9, 2025 reply Follow flag @Aditya Gangwar where there is some clue and its saying you to focus on couting.Always Consider mod machine.In this question they said that the string should have odd no of 0's or odd no of 1's so it means count should be used. 0 0 replyShare Please log in or register to add a comment.
0 0 votes 4 States.....can be construted using product automata Dfa that accepts odd no of zeros contain 2 states and dfa that accepts odd no of 1 will also contain 2 states so product automata will contain 2×2=4 states Priyanka Agarwal answered Jul 31, 2018 Priyanka Agarwal comment Share Follow 0 reply Please log in or register to add a comment.
