1 votes 1 votes The number of states in 2’s complements Moore machine is ? a. 2 b. 3 c. 4 d. 1 garg div asked Nov 27, 2017 garg div 7.8k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Namit Dhupar commented Nov 27, 2017 reply Follow Share I guess 3 0 votes 0 votes garg div commented Nov 27, 2017 reply Follow Share Please explain how 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes I made this Moore State diagram for better understanding! Hope that helps! Namit Dhupar answered Nov 27, 2017 • selected Nov 27, 2017 by garg div Namit Dhupar comment Share Follow See all 9 Comments See all 9 9 Comments reply garg div commented Nov 27, 2017 reply Follow Share thanx 1 votes 1 votes Red_devil commented Nov 27, 2017 reply Follow Share @Namit Dhupar how will you deal with the extra number generated by moore machine which is the start symbol in A 0 votes 0 votes Namit Dhupar commented Nov 27, 2017 reply Follow Share If I understand your question right @Red_devil, the question here is asking for 2's complement of any binary number... Well, to do that, I learnt a little trick from somewhere, where given a binary number instead of finding a 1's complement and adding a 1 to LSB and then find the answer, what I do is- 1.) Take the given binary number and look at It's LSB 2.) while going from LSB to MSB all the 0's on the LSB as well as the last 1 from the LSB must be kept as it is and rest of the remaining digits should be complemented... Here, let me show an example say given binary is 11100 ----> So 2' compliment will be (00100) Highlighted digits are unchanged and rest complemented. or 1111 -----> 0001 (Even if there are no 0's the 1 at the LSB should remain as it is and rest 1's complemented) And no matter what and how long number you throw at it, this machine will accept it! 0 votes 0 votes Red_devil commented Nov 27, 2017 reply Follow Share @ Namit Dhupar thanks for replying but my question is that when you are start in state you are already generating 0 in your moore machine so your number will have an extra 0 in LSB , please correct me if i am wrong, 0 votes 0 votes Namit Dhupar commented Nov 27, 2017 reply Follow Share Oh, I forgot to mention, You have to feed the given binary number from LSB into the automata. 0 votes 0 votes Red_devil commented Nov 27, 2017 reply Follow Share @Namit Dhupar i am just asking what will your machine generate if no string is supplied to it.. i.e. epsilon. 0 votes 0 votes Namit Dhupar commented Nov 27, 2017 reply Follow Share Output of the initial state,0 0 votes 0 votes Red_devil commented Nov 27, 2017 reply Follow Share @ Namit Dhupar yes!! that is what i am asking.. so every thing you generate will have a 0 at LSB how will you overcome this problem please explain. 0 votes 0 votes Namit Dhupar commented Nov 27, 2017 i edited by Namit Dhupar Nov 27, 2017 reply Follow Share I gotta be honest buddy, what I did was I originally derived the question in it's Mealy form and then converted it to Moore. And the asker asked about the number of states which I immediately responded to, though you are right about Prefixes being always printed.... Can I have an (A,$\epsilon$) in the initial state with no transitions and make another state with say (X,0) with 0 as output, and then continue with rest of the automata, but this would make number of states as 4! and general rule says, for K states in Mealy, there are K+1 states in Moore... Other than that,What approach do you have in mind? 1 votes 1 votes Please log in or register to add a comment.