0 votes 0 votes Made easy says answer is option C but i doubt it. Can somebody confirm ? User007 asked Feb 4, 2017 • edited Feb 4, 2017 by User007 User007 500 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply srestha commented Feb 4, 2017 reply Follow Share Why do u think it is incorrect? 0 votes 0 votes User007 commented Feb 4, 2017 reply Follow Share Made easy says answer is option C but i doubt it. 0 votes 0 votes saurabh rai commented Feb 4, 2017 reply Follow Share @srestha i think it same as language of balanced paranthesis. 0 votes 0 votes srestha commented Feb 4, 2017 reply Follow Share hm Actually here in pda we are taking number of a's first (as there is a loop in q0) ,so any number of a's first taking as input. Now, number of b's are same as number of a's. Because there b is taking with the transition of a. (b,a, $\epsilon$) always in one transition . That means similar number of a's and b's are there. Now, only taking $\epsilon$ q0 also can go to the final state So, $\left ( a^{n}b^{n} \right )^{*}$ is correct. But what (a,z0,QZ0) is doing here?Could u understand it @ saurabh 0 votes 0 votes saurabh rai commented Feb 4, 2017 reply Follow Share @srestha i think that should be (a,z0,aZ0) instead of (a,z0,QZ0) 0 votes 0 votes srestha commented Feb 4, 2017 reply Follow Share No, that means a is only taking part in transition without b. Because in that case, after taking a, it can accept $\epsilon$ and go to final state. Then "equal number of a's and b's"condition violated 0 votes 0 votes saurabh rai commented Feb 4, 2017 reply Follow Share @^ ll u xplain little bit more on Because in that case, after taking a, it can accept ϵϵ and go to final state. Then "equal number of a's and b's"condition violated i m nt getting 0 votes 0 votes srestha commented Feb 4, 2017 reply Follow Share if it will accept (a,z0,aZ0) ,then it will accept string anbna rt? 0 votes 0 votes saurabh rai commented Feb 4, 2017 reply Follow Share ^i think no bcoz top of stack is nt z0 0 votes 0 votes Please log in or register to add a comment.