1 votes 1 votes If L1 = { a^n | n ≥ 0 } and L2 = { b^n | n ≥ 0 }, Consider then L1 . L2 wil be a) (ab)^n b) a^n b^n c) b^n a^n d)b^m a^n e) { a^m b^n | m ≥ 0, n ≥ 0 } why answer is d why not b ???? Theory of Computation theory-of-computation made-easy-test-series identify-class-language + – air1ankit asked Dec 9, 2017 edited Mar 7, 2019 by Aditi Singh air1ankit 793 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply abhishek tiwary commented Dec 9, 2017 i edited by abhishek tiwary Dec 9, 2017 reply Follow Share L1.L2=a*b* so d is not correct not b also 0 votes 0 votes joshi_nitish commented Dec 9, 2017 reply Follow Share D) is not correct. L1.L2 = a*b* = {ambn | m,n>=0} 1 votes 1 votes Please log in or register to add a comment.
Best answer 4 votes 4 votes 'n' of L1 is different from the 'n' of L2 (i.e these two 'n' are not equal), So due to this difference L1.L2 = {a^mb^n :m>=0,n>=0} will be the answer. Rajnish Kumar 1 answered Dec 9, 2017 selected Dec 27, 2017 by air1ankit Rajnish Kumar 1 comment Share Follow See 1 comment See all 1 1 comment reply air1ankit commented Dec 9, 2017 reply Follow Share got it thanks bro ..! 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes L1 and L2 are both regular languages, therefore concatenation of both i.e L1.L2 is also regular language. And this language is "Number of a's followed by number of b's". Gawade Sahadev answered Dec 28, 2017 Gawade Sahadev comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes option e is right abhishekmehta4u answered Mar 22, 2018 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes no option is correct:- D must be like this D. ambn and m,n >=0 which also represent a*b*... L1={null,a,aa,aaa,......}=a* L2={null,b,bb,bbb,......}=b* hs_yadav answered Dec 9, 2017 hs_yadav comment Share Follow See all 3 Comments See all 3 3 Comments reply air1ankit commented Dec 9, 2017 reply Follow Share now one option is correct , tell me one thing please if it should be a^n b^n then whats wrong here .... 0 votes 0 votes joshi_nitish commented Dec 9, 2017 reply Follow Share anbn is CFL, as count of 'a' = count of 'b' ,but ambn is regular, because count of 'a' and 'b' does not depend on each other. 1 votes 1 votes air1ankit commented Dec 9, 2017 reply Follow Share got it according to you we , lets say one example we use stack for a^n b^n (FOR CFL)?? correct?? 0 votes 0 votes Please log in or register to add a comment.