0 votes 0 votes what is the number of states in dfa of all language over {a,b} where n(a)mod3>=n(b)mod2 how to think in this kind of question ?? Theory of Computation testbook-test-series theory-of-computation finite-automata + – Aboveallplayer asked Dec 20, 2016 • edited Mar 11, 2019 by akash.dinkar12 Aboveallplayer 408 views answer comment Share Follow See 1 comment See all 1 1 comment reply Gate Mission 1 commented Dec 20, 2016 reply Follow Share For dfa with n(a)mod3 ..we have three states each representing mod values as 0,1,2. Similarly for dfa with n(b)mod2 ...we have two states each representing mod values as 0,1. So product automaton will have 6 states like {00,01,10,11,20,21} --> where first num in state represent n(a)mod3 and second num represent n(b)mod2 ..so acc to required condition n(a)mod3>=n(b)mod2...so final states in product automaton will be {00,10,11,20,21}. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes answer is 3*2= 6 if n(a) mod x >= n(b) mod y then no. of state is x*y RAJESHWAR YADAV answered Dec 20, 2016 RAJESHWAR YADAV comment Share Follow See 1 comment See all 1 1 comment reply Aboveallplayer commented Dec 20, 2016 reply Follow Share how does that work?any source or lil explanation would have helped me 0 votes 0 votes Please log in or register to add a comment.