2 votes 2 votes L=${(a^{n}b^{n})^*c^{^{m}} | n,m\geqslant 1}$. Is this a cfl ?? ans given : it is not cfl pm9999 asked Jan 21, 2018 edited Jan 21, 2018 by pm9999 pm9999 612 views answer comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments Anu007 commented Jan 21, 2018 reply Follow Share L = (anbn|n>=0} and L*What you are thing But is not same as L =(anbn)* 0 votes 0 votes Inspiron commented Jan 21, 2018 reply Follow Share No its CSL : let $*=2$ and $n=2$ then $aabbaabb$ cannot be parsed if yest draw a small dpda if you feel its dcfl. 0 votes 0 votes Saswat Senapati commented Jan 21, 2018 reply Follow Share @inspiron you are considering n=2 fixed for every iteration..... Then n value of (i +1)th iteration is completely independent of n value of i th iteration.. Now as you said taking n=2 , a2b2 a2b2 will be string i.e. aabbaabb Here parsing will be first all a's pushed then when we see b pop a's .....now again when you see a's doesn't matter what was a's previous n value , it starts pushing all a's then on seeing b it pops a's. Expression means it should generate all kind of language and we have to see all those are accepted by PDA or not, if accepted then lang will be CFL... So a2b2a2b2 is an INSTANCE of that language which will be accepted by PDA. 0 votes 0 votes Please log in or register to add a comment.