Question

Let \( G = (V, \Sigma, S, P) \) be a context-free grammar in Chomsky Normal Form with \( \Sigma = \{a, b, c\} \) and \( V \) containing 10 variable symbols including the start symbol \( S \). The string \( w = a^{30}b^{30}c^{30} \) is derivable from \( S \). The number of steps (application of rules) in the derivation \( S \to^* w \) is \_\_\_\_\_\_.

Answer 1

Step 1: Understand Chomsky Normal Form (CNF).
In CNF, every production is of the form \( A \to BC \) or \( A \to a \). To derive a string of length \( n \), we need \( n - 1 \) steps for concatenation and \( n \) steps to derive each terminal symbol. Step 2: Derive \( w = a^{30}b^{30}c^{30} \).
To generate 90 terminals (\( 30a + 30b + 30c \)), \( 90 \) steps are needed.
To combine these terminals, \( 90 - 1 = 89 \) steps are needed. Step 3: Total steps. \[ \text{Total steps} = 90 + 89 = 179. \] Final Answer: \[ \boxed{179} \]