2 votes 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 asked Jul 30, 2018 himgta 2.9k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Anand. commented Jul 30, 2018 i edited by Anand. Jul 30, 2018 reply Follow Share $3$ states 0 votes 0 votes abhishekmehta4u commented Jul 30, 2018 reply Follow Share 010 is not accepted 1 votes 1 votes Anand. commented Jul 30, 2018 reply Follow Share @abhishek :thanks for correcting. Can you check if it is correct now ? 0 votes 0 votes abhishekmehta4u commented Jul 30, 2018 reply Follow Share 011, 100..... Is not accepted 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes 4 state abhishekmehta4u answered Jul 30, 2018 abhishekmehta4u comment Share Follow See 1 comment See all 1 1 comment reply Aditya Gangwar commented May 9, 2020 reply Follow Share How did you reach to this solution? 0 votes 0 votes Please log in or register to add a comment.
0 votes 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 See all 0 reply Please log in or register to add a comment.