6 votes 6 votes The minimum number of states of the non-deterministic finite automaton which accepts the language $\{ a b a b^n \mid n \geq 0 \} \cup \{ a b a^n \mid n \geq 0 \}$ is 3 4 5 6 Theory of Computation ugcnetcse-dec2012-paper3 theory-of-computation finite-automata + – go_editor asked Jul 13, 2016 • recategorized Oct 10, 2018 by Pooja Khatri go_editor 9.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 10 votes 10 votes L={ababn∣n≥0}∪{aban∣n≥0} L={ab, aba , abaa ,abab ,ababb,..........} The minimum number of states of the non-deterministic finite automaton which accepts the language =5 Hence,Option(c)5 is the correct choice. LeenSharma answered Jul 13, 2016 • edited Jul 13, 2016 by LeenSharma LeenSharma comment Share Follow See all 14 Comments See all 14 14 Comments reply Show 11 previous comments LeenSharma commented Jul 13, 2016 reply Follow Share Praveen Saini Sir please check this 1 votes 1 votes Praveen Saini commented Jul 13, 2016 reply Follow Share yes now it is correct update in answer 0 votes 0 votes LeenSharma commented Jul 13, 2016 reply Follow Share Thank You sir 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes answer : 4 Correct me if I am wrong sh!va answered Jul 13, 2016 sh!va comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments One commented Jul 28, 2016 i reshown by gatecse May 14, 2021 reply Follow Share ok I got it in case of NFA it is fine 0 votes 0 votes sh!va commented Mar 10, 2017 i reshown by gatecse May 14, 2021 reply Follow Share Yes, This NFA wrongly accepts the string abaaaaabbb.. which does not belong to L Answer is wrong 1 votes 1 votes Priyanshu2602y commented Dec 10, 2023 reply Follow Share @sh!va thanks your point makes a difference here now i got it 0 votes 0 votes Please log in or register to add a comment.