3 votes 3 votes Which of the following grammars violate the Chomsky Normal Form? $S \rightarrow AB$, $A \rightarrow a , B \rightarrow \epsilon \mid b$ $S \rightarrow AcB, A \rightarrow a, B \rightarrow b$ $S \rightarrow AB, \ A \rightarrow a, \ B \rightarrow b$ All of the above (i) only (i) and (ii) only (ii) and (iii) only Theory of Computation tbb-toc-1 + – Bikram asked Nov 26, 2016 • edited Aug 20, 2019 by Counsellor Bikram 365 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 5 votes 5 votes CNF: V->VV V->t where V is single variable and t is single terminal 1) CNF doesn't contain epsilon 2) S--> AcB not in the form of V->VV 3) CNF Vijay Thakur answered Jan 29, 2017 • selected Jan 29, 2017 by Bikram Vijay Thakur comment Share Follow See 1 comment See all 1 1 comment reply jaswanth431 commented Aug 22, 2021 reply Follow Share CNF do contain epsilon if epsilon is there in language. We will include epsilon in only start production to make sure we get epsilon in language. https://cs.stackexchange.com/questions/24012/chomsky-normal-form-epsilon-rule 0 votes 0 votes Please log in or register to add a comment.