1 votes 1 votes L1 = {a^nb^nC^n | n>=0} L2 = {a^nb^mc^k | k=m+n , n,m>=0} L1.(L2)* = ?? Theory of Computation theory-of-computation identify-class-language + – Ravi_1511 asked Jan 23, 2017 edited Jun 16, 2022 by Arjun Ravi_1511 569 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes L1 is CSL because comparison ar like this {a=b, b=c,a=c} L2 is DCFL as comparison is deterministic L1.(L2)* CSL(DCFL)* DCFL is not closed under KLEEN's closure so push it up in chomsky heiarchy CSL.(CFL)* CSL.CFL CFL is closed under KLEEN's closure and also in concatenation and every CFL is also CSL SO >> CSL ANSWER. correct me .. focus _GATE answered Jan 23, 2017 focus _GATE comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Lokesh . commented Jan 23, 2017 reply Follow Share @Prajwal Bhat he changed the ques little bit, https://gateoverflow.in/109381/regular-and-cfl-testbook-test-series-2#c109385 1 votes 1 votes Prajwal Bhat commented Jan 23, 2017 reply Follow Share Thanks Lokesh, Got it why L4 was regular 0 votes 0 votes Ravi_1511 commented Jan 24, 2017 reply Follow Share Yes I asked what I dint get. 0 votes 0 votes Please log in or register to add a comment.