2 votes 2 votes The context free grammar given by $S \rightarrow XYX$ $X \rightarrow aX \mid bX \mid \lambda$ $Y \rightarrow bbb$ generates the language which is defined by regular expression: $(a+b)^*bbb$ $abbb(a+b)^*$ $(a+b)^*(bbb)(a+b)^*$ $(a+b)(bbb)(a+b)^*$ Theory of Computation theory-of-computation regular-expression finite-automata expression ugcnetcse-dec2015-paper3 + – shekhar chauhan asked Jun 5, 2016 • recategorized Nov 8, 2021 by soujanyareddy13 shekhar chauhan 3.2k views answer comment Share Follow See 1 comment See all 1 1 comment reply ManojK commented Jun 5, 2016 i edited by ManojK Jun 5, 2016 reply Follow Share Its simple All string containing bbb as sub string . (a+b)*(bbb)(a+b)* 4 votes 4 votes Please log in or register to add a comment.
3 votes 3 votes looking at X-> aX | bX | eps we see that X is generating (a+b)* now simply replacing X by (a+b)* and Y by bbb in S-> XYX we get (a+b)*bbb(a+b)* option c Sanket_ answered Jun 22, 2016 Sanket_ comment Share Follow See all 0 reply Please log in or register to add a comment.