1 votes 1 votes Let L = {anblak: n = l or l ≠ k}. The language is A) regular B) DCFL but not regular. C) NDCFL but not DCFL. D) context sensitive but not CFL. The Option is C) right ...??? Please verify ... Theory of Computation theory-of-computation finite-automata + – Vicky rix asked Apr 4, 2017 • retagged Jun 4, 2017 by Arjun Vicky rix 433 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $L1=\left \{a^{n}b^{n}a^{*} \right \}$ $L2=\left \{ a^{*}b^{m}a^{n}|m\neq n\right \}$ both L1 and L2 individually are DCFL but $L1\bigcup L2$ are not we can't discriminate working of both on seeing a, 2018 answered Apr 4, 2017 2018 comment Share Follow See all 2 Comments See all 2 2 Comments reply Vicky rix commented Apr 4, 2017 reply Follow Share so it is a NDCFL right ? 0 votes 0 votes 2018 commented Apr 4, 2017 reply Follow Share @vignesh yes.. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes n=l or l!=k means it can be n=l, l!=k, n=l and l!=k..in last case 2 comparisons are required so it cannot be cfl..it should be csl..plz correct me if I am wrong. Purvi Agrawal answered Apr 4, 2017 Purvi Agrawal comment Share Follow See all 0 reply Please log in or register to add a comment.