Question:

Consider a context-free grammar \(G\) with the following 3 rules: \[ S \to aS, \quad S \to aSbS, \quad S \to c \] Let \(w \in L(G)\). Let \(n_a(w), n_b(w), n_c(w)\) denote the number of times \(a, b, c\) occur in \(w\), respectively. Which of the following statements is/are TRUE?

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For context-free grammars, analyze the relationship between terminal symbols by considering the structure of derivations.
Updated On: Jan 23, 2025
  • \(n_a(w)>n_b(w)\)
  • \(n_a(w)>n_c(w) - 2\)
  • \(n_c(w) = n_b(w) + 1\)
  • \(n_c(w) = n_b(w) * 2\)
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The Correct Option is B

Solution and Explanation

The grammar \(G\) generates strings of the form: \[ w = a^p c b^q, \] where the derivation of \(c\) adds one \(b\) and one \(a\). Hence, the relationships between \(n_a(w), n_b(w), n_c(w)\) can be analyzed:
Option (A): Incorrect. While \(n_a(w)\) is often greater than \(n_b(w)\), there is no guarantee, as \(n_a(w)\) and \(n_b(w)\) depend on the specific derivation.
Option (B): Correct. Since the derivation adds at least two \(a\)'s for every \(c\), the inequality \(n_a(w)>n_c(w) - 2\) holds.
Option (C): Correct. Each derivation of \(S \to aSbS\) adds one \(b\) for every \(c\), making \(n_c(w) = n_b(w) + 1\).
Option (D): Incorrect. The number of \(c\)'s is not necessarily twice the number of \(b\)'s. Final Answer: \[ \boxed{\text{(B), (C)}} \]
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