The Greibach normal form grammar for the language L = {aⁿ bⁿ⁺¹ | n ≥ 0} is
2013
The Greibach normal form grammar for the language L = {aⁿ bⁿ⁺¹ | n ≥ 0} is
Answer: C. S → aSB / b B → b — Greibach Normal Form requires every production to start with a terminal followed by variables. For L = {a^n b^{n+1}}, the grammar must generate n 'a's and n+1…
- A.
S → aSB
B → bB / λ - B.
S → aSB
B → bB / b - C.
S → aSB / b
B → b - D.
S → aSb / b
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Correct answer: C
Greibach Normal Form requires every production to start with a terminal followed by variables.
For L = {a^n b^{n+1}}, the grammar must generate n 'a's and n+1 'b's correctly.
Option 2 satisfies GNF by ensuring terminals start productions and avoiding terminals in variable suffixes.
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