Cfg Solved Examples May 2026

Derivation for abba : [ S \Rightarrow aSbS \Rightarrow a\varepsilon bS \Rightarrow abS \Rightarrow abbSaS \Rightarrow abb\varepsilon a\varepsilon = abba ] Language : Valid arithmetic expressions with a, b, +, *, (, )

: [ 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 ) cfg solved examples

: [ S \Rightarrow SS \Rightarrow (S)S \Rightarrow ((S))S \Rightarrow (())S \Rightarrow (())(S) \Rightarrow (())() ] 4. Example 3 – ( a^n b^n ) (equal number of a’s and b’s) Language : ( a^n b^n \mid n \ge 0 ) Derivation for abba : [ S \Rightarrow aSbS

Better approach — known correct grammar: [ S \to aSb \mid aSbb \mid \varepsilon ] For m=3, n=2: S → aSbb → a(aSb)bb → aa(ε)bbbb? No — that’s 4 b’s. So maybe n=2, m=3 not possible? Actually it is: ( a^2 b^3 ) = a a b b b. Let’s test: No — that’s 4 b’s

Better: [ S \to aaS \mid abS \mid baS \mid bbS \mid \varepsilon ] But that forces pairs. Actually, simpler:

Check: ( S \Rightarrow aA \Rightarrow abS \Rightarrow ab\varepsilon = ab ) (length 2). Works. Language : All strings of ( and ) that are balanced.