[ S \to aA \mid bA \mid \varepsilon ] [ A \to aS \mid bS ]
: [ E \to E + T \mid T ] [ T \to T \times F \mid F ] [ F \to (E) \mid a \mid b ] cfg solved examples
So to get m=3,n=2: S ⇒ aSbb (add a, b,b) Now S ⇒ aSb (add a, b) Total: a(aSb)bb ⇒ a(aεb)bb = a a b b b = 2 a, 3 b. Works. [ S \to aA \mid bA \mid \varepsilon
Check: ( S \Rightarrow aA \Rightarrow abS \Rightarrow ab\varepsilon = ab ) (length 2). Works. Language : All strings of ( and ) that are balanced. Let's fix: Derivation for a + b *
: [ S \to aS \mid bS \mid \varepsilon ] Wait — that gives any length. Let's fix:
Derivation for a + b * a : [ E \Rightarrow E+T \Rightarrow T+T \Rightarrow F+T \Rightarrow a+T \Rightarrow a+T\times F \Rightarrow a+F\times F \Rightarrow a+b\times a ] | Language pattern | CFG trick | |----------------|------------| | ( a^n b^n ) | ( S \to aSb \mid \varepsilon ) | | Matching parentheses | ( S \to SS \mid (S) \mid \varepsilon ) | | ( a^n b^m, n\le m ) | ( S \to aSb \mid bS \mid \varepsilon ) | | Palindromes | ( S \to aSa \mid bSb \mid a \mid b \mid \varepsilon ) | | ( a^i b^j c^i+j ) | Separate S for a’s + c’s, T for b’s + c’s | | Equal #a and #b (any order) | ( S \to aSbS \mid bSaS \mid \varepsilon ) | | Expression grammar | Left-recursive for left-assoc operators |