Question

Which of the following string is not generated by the grammar :
$S \to SaSbS | \epsilon$ ?

• A. aabb
• B. abab
• C. aababb
• D. aaabb

Hint:

For grammars generating balanced strings, always check the count first. If $count(a) \neq count(b)$, the string is immediately invalid.

Answer 1

The Correct Option is D

Concept:
• The grammar \(S \rightarrow SaSbS \mid \epsilon\) generates a language where terminals 'a' and 'b' are balanced.
• Every application of the recursive rule \(SaSbS\) introduces exactly one 'a' and exactly one 'b'.
• This specific structure is the generator for the Dyck language, representing balanced strings of parentheses where 'a' is the opening and 'b' is the closing parenthesis.
• A necessary (but not sufficient) condition for any string \(w\) in this language is that the number of 'a's must equal the number of 'b's: \(n_a(w) = n_b(w)\).

Step 1: Analyze the fundamental counting property of the grammar
Each production rule \(S \rightarrow SaSbS\) adds exactly one 'a' and one 'b'.
The base case \(S \rightarrow \epsilon\) adds zero terminals.
Therefore, any string generated by this grammar must satisfy the property \(n_a(string) = n_b(string)\).

Step 2: Evaluate each option based on the counting property

• Option (A) \(aabb\): Contains 2 'a's and 2 'b's. \(2 = 2\). Matches property.
• Option (B) \(abab\): Contains 2 'a's and 2 'b's. \(2 = 2\). Matches property.
• Option (C) \(aababb\): Contains 3 'a's and 3 'b's. \(3 = 3\). Matches property.
• Option (D) \(aaabb\): Contains 3 'a's and 2 'b's. \(3 \neq 2\). Does not match.

Step 3: Verify derivations for the valid strings to ensure the grammar structure is met
For \(aabb\): \(S \Rightarrow SaSbS \Rightarrow aSbS \Rightarrow a(SaSbS)bS \Rightarrow aabS \Rightarrow aabb\).
For \(abab\): \(S \Rightarrow SaSbS \Rightarrow aSbS \Rightarrow abS \Rightarrow ab(SaSbS) \Rightarrow abab\).
Since \(aaabb\) fails the basic count equality, it cannot be generated regardless of the sequence of rules used.