Understanding the Grammar Rules

The given grammar G defines a language using the following production rules:

• S → abScB | e
• B → bB | b

Here, S is the start symbol, and B is another non-terminal. The rule S → e means that S can derive the empty string (epsilon).

Analyzing the Non-terminal B

Let's first determine the strings generated by the non-terminal B:

• B can directly produce 'b' (using B → b).
• B can produce 'bb' (using B → bB → bb).
• B can produce 'bbb' (using B → bB → b bB → bbb), and so on.

Therefore, B generates strings consisting of one or more 'b's. This language is denoted as b⁺.

Analyzing the Start Symbol S

Now, let's examine the strings generated by the start symbol S using its production rules:

• S → e: This rule generates the empty string.
• S → abScB: This is a recursive rule. It introduces the fixed sequence 'ab' at the beginning and the sequence 'cB' at the end. Since we know B generates b⁺, this rule effectively behaves like S → ab S c b⁺.

Let's trace some derivations to understand the structure:

1. Base Case: If we apply S → e immediately: S ⇒ e The empty string is generated.
2. One Recursive Step: Apply S → abScB once, followed by S → e: S ⇒ ab S c B Substituting S → e: S ⇒ ab (e) c B = abcB Since B generates b⁺, this produces strings like 'abcb', 'abcbb', etc. This structure matches $ (ab)^1 c b^{p_1} $ where $ p_1 \ge 1 $.
3. Two Recursive Steps: Apply S → abScB twice, then S → e: S ⇒ ab S c B Substitute S → abScB again: S ⇒ ab (ab S c B) c B = (ab)² S (cB)² Substitute S → e: S ⇒ (ab)² (e) (cB)² = (ab)² (cB)² Since B generates b⁺, this produces strings like '(ab)² c b^p₁ c b^p₂', where $ p_1 \ge 1 $ and $ p_2 \ge 1 $. For example, 'abab cb cb' or 'abab cbb cb'.

Generalizing, if we apply the recursive rule S → abScB exactly $ k $ times and then use S → e, the generated strings have the following structure:

where $ k \ge 0 $ represents the number of times the recursive rule was applied, and each $ p_i \ge 1 $ because B generates one or more 'b's.

Matching the Generated Language to the Options

The language generated by the grammar can be described as:

This means the language consists of zero or more repetitions of 'ab', followed by an equal number of 'c's, where each 'c' is followed by at least one 'b'. Let's compare this with the provided options:

• Option 1: \({ (ab)^n (cb)^m | n = m }\). This option implies a fixed structure where each 'c' is followed by exactly one 'b', which is too restrictive compared to our derived structure ($ p_i \ge 1 $).
• Option 2: \({ (a^n b)^n (cb)^m | n ¹ m }\). This option includes the term $ a^n b $, which is not produced by the grammar's $ ab $ sequence.
• Option 3: \({ (ab^n (cb)^m )^n | n ≥ 0, m > 0 }\).
• Option 4: \({ (ab)^n (cb)^m )^n | n ≥ 0, m ≥ 0 }\). The structure $ (...)^n $ suggests block repetition, which doesn't match the nesting derived from $ S \to abScB $.

Option 3 provides the correct description for the language generated by the grammar.

• The parameter $ n \ge 0 $ in Option 3 corresponds to the number of recursive applications $ k $ of the rule $ S \to abScB $. This matches the $ k \ge 0 $ found in our derived structure.
• The condition $ m > 0 $ ensures that there is at least one 'b' following 'c', which aligns with $ B $ generating $ b^+ $ ($ p_i \ge 1 $).
• While the notation $ ab^n $ might seem different from the literal 'ab' sequence in the grammar, it represents the $ n $ occurrences of the 'ab' prefix. The outer exponent $ ^n $ reflects the nesting depth achieved through the $ n $ recursive steps.

Conclusion

Considering the parameters $ n \ge 0 $ and $ m > 0 $, and the structural pattern derived from the grammar rules $ S \to abScB $ and $ B \to bB | b $, Option 3 accurately represents the language generated by the grammar G.