: [ S \to aSb \mid \varepsilon ]
So the sequence of rules: aSbb then aSb then ε. Good. So grammar works. Language : ( w \in a,b^* \mid w = w^R ) cfg solved examples
Check: ( S \Rightarrow aA \Rightarrow abS \Rightarrow ab\varepsilon = ab ) (length 2). Works. Language : All strings of ( and ) that are balanced. : [ S \to aSb \mid \varepsilon ]
[ S \to aA \mid bA \mid \varepsilon ] [ A \to aS \mid bS ] Language : ( w \in a,b^* \mid w
: [ S \Rightarrow aSb \Rightarrow aaSbb \Rightarrow aaaSbbb \Rightarrow aaabbb ] 5. Example 4 – ( a^n b^m ) with ( n \le m \le 2n ) Language : ( a^n b^m \mid n \ge 0, m \ge n, m \le 2n )
: [ E \to E + T \mid T ] [ T \to T \times F \mid F ] [ F \to (E) \mid a \mid b ]