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S→ aB/abA→aAB/aB→ABb/bIs the given grammar G a context-free grammar? Justify your answer with anexplanation. Additionally, using the string "aaaabbbb," determine if the grammar G isambiguous or not by analysing its production rules and derivation.

Question

S→ aB/abA→aAB/aB→ABb/bIs the given grammar G a context-free grammar? Justify your answer with anexplanation. Additionally, using the string "aaaabbbb," determine if the grammar G isambiguous or not by analysing its production rules and derivation.

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Solution

To determine if the given grammar G is a context-free grammar, we need to analyze its production rules.

The production rules in the given grammar G are as follows:

  1. S → aB
  2. A → abA
  3. A → aB
  4. B → ABb
  5. B → b

A context-free grammar consists of production rules where the left-hand side contains only a single non-terminal symbol, and the right-hand side can contain a combination of terminal and non-terminal symbols.

In the given grammar G, all the production rules satisfy this condition. The left-hand side of each rule contains only a single non-terminal symbol (S, A, or B), and the right-hand side contains a combination of terminal symbols (a or b) and non-terminal symbols (A or B). Therefore, the given grammar G is a context-free grammar.

Now, let's analyze the string "aaaabbbb" using the production rules and derivation of grammar G to determine if it is ambiguous or not.

Starting with the start symbol S, we can derive the string "aaaabbbb" as follows: S → aB (using rule 1) → aaBb (using rule 1) → aABbb (using rule 4) → aaABbbb (using rule 1) → aaaABbbbb (using rule 1) → aaaabbbb (using rule 5)

From this derivation, we can see that there is only one possible derivation for the string "aaaabbbb" using the given production rules. There are no alternative derivations or multiple ways to derive the same string. Therefore, the grammar G is not ambiguous.

In conclusion, the given grammar G is a context-free grammar, and it is not ambiguous when deriving the string "aaaabbbb."

This problem has been solved

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