0 votes 0 votes Theory of Computation made-easy-test-series theory-of-computation regular-language + – vaishali jhalani asked Jan 26, 2017 • edited Mar 5, 2019 by adeebafatima1 vaishali jhalani 404 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes L = a^m(bc^k)^n Consider second part of the language i.e(bc^k)^n Number of c's are dpends on number of b's . Here one count is require . It is CFL Ex . b=1 c>=1 b=2 c>=2 b=3 c>=3 ...................................... Answer is D Dhananjay2017 answered Jan 26, 2017 • edited Jan 26, 2017 by Dhananjay2017 Dhananjay2017 comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Dhananjay2017 commented Jan 28, 2017 reply Follow Share L = a^mb^nc^(kn) We can not select any number of c . abbc is not in language because n =2 and if k =0 then c must 0 . If k=1 then number of c's must be 2 . 0 votes 0 votes Sushant Gokhale commented Jan 28, 2017 reply Follow Share No, its not like that. (bck)n = (bck)(bck)(bck).......n times 0 votes 0 votes Abbas2131 commented Jan 28, 2017 reply Follow Share am (bck)n m,n,k >0 and are independent. U meantioned If b=1 then c>=1 b=4 then c>=4 Well this is not true. Consider this (bc2)4 This will generate bccbccbccbcc notice that c is independent of b. A simple regular expression can even define this language. L= a+(bc+)+ Isnt it ? 0 votes 0 votes Please log in or register to add a comment.
