2 votes 2 votes A = { 0k1u0k | k >=1 and u ∈ ∑* }. Is A is regular ? Theory of Computation regular-language theory-of-computation + – amrendra pal asked Aug 22, 2017 amrendra pal 381 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes A is CFG not Regular. because leftside 0^k cannot be merged with u. yogi_p answered Aug 23, 2017 yogi_p comment Share Follow See all 4 Comments See all 4 4 Comments reply amrendra pal commented Aug 23, 2017 reply Follow Share @yogi_p, A = { 0k1u0k | k >=1 and u ∈ ∑* } .Then A is not regular ? okk , confirm . but when B = { 0ku0k | k >=1 and u ∈ ∑* } . Then what about B regular or not ? 0 votes 0 votes yogi_p commented Aug 23, 2017 reply Follow Share B is regular 0 votes 0 votes amrendra pal commented Aug 23, 2017 reply Follow Share @yogi_p , can you explain why B is regular 0 votes 0 votes yogi_p commented Aug 23, 2017 reply Follow Share @amrendra pal In B u belongs to Σ* so it can take all the 0's of both side leaving just single 0 in start & at end . So its the language of all strings starting and ending with 0 0 votes 0 votes Please log in or register to add a comment.