0 votes 0 votes himgta asked Jul 14, 2018 himgta 254 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Shubham Shukla 6 commented Jul 14, 2018 reply Follow Share all are regular..! regular exprsn for i.) (aa)*(bb)*[even+even=even]+(aa)*a(bb)*b[odd+odd=even] ii.) (aa)*a(bb)*[odd+even=odd]+(aa)* b(bb)*[even+odd=odd] iii.) a*b*c* iv.) (aa)* 0 votes 0 votes abhishekmehta4u commented Jul 14, 2018 reply Follow Share In L4 it should be a*. u=a*,v=a* and conncatanation of both is also a*.???? 0 votes 0 votes Shubham Shukla 6 commented Jul 14, 2018 reply Follow Share @abhishek how can you accept 1,3,5 a's if you take u as aa than v is reverse of it so it will be aa too is that not so.? 1 votes 1 votes abhishekmehta4u commented Jul 14, 2018 reply Follow Share Yes i missed v is reverse of L5 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes All are regular language. iii) regular exp= a*b*c* iv) regular exp=(aa)* abhishekmehta4u answered Jul 14, 2018 edited Jul 14, 2018 by abhishekmehta4u abhishekmehta4u comment Share Follow See all 3 Comments See all 3 3 Comments reply himgta commented Jul 14, 2018 reply Follow Share @abhishekmehta4u @Shubham Shukla 6... my doubt is in L1 and L2 we have to perform unbounded counting against number of a's and b's then how it became regular though I understood the regular expressions given by you! 0 votes 0 votes Shubham Shukla 6 commented Jul 14, 2018 reply Follow Share here why to count regular exprsn will take care of it only languages falling in this exprsn would be accepted..regualr exrsn can generate infinte languages so why you thinking about unbounded here we dont need to do any comaprsion or checking...even and odd checking possible by regukar exprsn so thats why abve 2 are regular 0 votes 0 votes himgta commented Jul 14, 2018 reply Follow Share I am still confused...@Deepak Poonia(dee) please explain! 0 votes 0 votes Please log in or register to add a comment.