Understanding the Language \(L\)

The given language is \(L = \left\{ {{a^n}{b^{n - 3}}\ | n > 2} \right\}\) over the alphabet ∑ = {a, b}.

This means strings in \(L\) consist of some number of 'a's followed by some number of 'b's. The number of 'a's is denoted by \(n\), and the number of 'b's is \(n-3\). The condition is that \(n\) must be an integer strictly greater than 2, which means \(n \ge 3\).

Let's list the first few strings in the language \(L\) for increasing values of \(n\):

• For \(n=3\): \(a^3 b^{3-3} = a^3 b^0 = aaa\)
• For \(n=4\): \(a^4 b^{4-3} = a^4 b^1 = aaaab\)
• For \(n=5\): \(a^5 b^{5-3} = a^5 b^2 = aaaaabb\)
• For \(n=6\): \(a^6 b^{6-3} = a^6 b^3 = aaaaaabbb\)

The language \(L\) is the set {aaa, aaaab, aaaaabb, aaaaaabbb, ...}. Notice that the number of 'a's is always 3 more than the number of 'b's, and there are at least 3 'a's.

Analyzing the Given Grammars

We need to find which of the given grammars generates exactly this language \(L\). A grammar generates a language by starting with the start symbol (usually S) and applying production rules until only terminal symbols (a, b, λ) remain.

Grammar 1: \(S \to aA | a, A \to aAb | b\)

Let's explore strings generated by this grammar:

• Using \(S \to a\): Generates 'a'. This string has 1 'a' and 0 'b's. For \(n=1\), \(n-3 = -2\), which doesn't fit the language definition. 'a' is not in \(L\).
• Using \(S \to aA\):
  • \(S \to aA \to a(b) = ab\). (1 'a', 1 'b'. Not in L).
  • \(S \to aA \to a(aAb) = aaAb \to aa(b) = aab\). (2 'a's, 1 'b'. Not in L).
  • \(S \to aA \to a(aAb) \to aaAb \to aa(aAb) = aaaAb \to aaa(b) = aaab\). (3 'a's, 1 'b'. For \(n=3\), \(L\) requires 'aaa'. Not in L).
  • \(S \to aA \to a(aAb) \to aaaAb \to aaa(aAb) = aaaaAb \to aaaa(b) = aaaab\). (4 'a's, 1 'b'. This is \(a^4 b^1\), which is in \(L\) for \(n=4\)).
  • \(S \to aA \to \dots \to aaaaAb \to aaaa(aAb) = aaaaaAb \to aaaaa(b) = aaaaabb\). (5 'a's, 2 'b's. This is \(a^5 b^2\), which is in \(L\) for \(n=5\)).
  The grammar generates some strings in \(L\) (for \(n \ge 4\)), but also generates strings not in \(L\) ('a', 'ab', 'aab', 'aaab'). It also fails to generate 'aaa', which is in \(L\) (for \(n=3\)). Thus, Grammar 1 is incorrect.

Grammar 2: \(S \to aaA | \lambda, A \to aAb | \lambda\)

First, let's see what strings can be generated from the non-terminal A using the rule \(A \to aAb | \lambda\). We can apply \(A \to aAb\) zero or more times, followed by \(A \to \lambda\):

• \(A \to \lambda\) (0 applications of \(aAb\)). Generates \(\lambda\).
• \(A \to aAb \to a\lambda b = ab\) (1 application of \(aAb\)). Generates \(ab\).
• \(A \to aAb \to a(aAb)b = aaAbb \to aa\lambda bb = aabb\) (2 applications of \(aAb\)). Generates \(aabb\).
• In general, applying \(A \to aAb\) \(k\) times followed by \(A \to \lambda\) generates \(a^k b^k\), for any \(k \ge 0\). The language generated by A is \(\{a^k b^k \mid k \ge 0\} = \{\lambda, ab, aabb, aaabbb, ...\}\).

Now consider the rules for S:

• Using \(S \to \lambda\): Generates \(\lambda\). This string is not in \(L\).
• Using \(S \to aaA\): A generates \(a^k b^k\). So S generates strings of the form \(aa (a^k b^k) = a^{2+k} b^k\) for \(k \ge 0\). Let \(m = 2+k\). Since \(k \ge 0\), \(m \ge 2\). The generated strings are of the form \(a^m b^{m-2}\) for \(m \ge 2\). For \(m=2\): \(a^2 b^0 = aa\). (Not in L). For \(m=3\): \(a^3 b^1 = aaab\). (Not in L). For \(m=4\): \(a^4 b^2 = aaaabb\). (Not in L).

The language generated by Grammar 2 is \(\{ \lambda, aa, aaab, aaaabb, ...\}\). This does not match the language \(L\). Grammar 2 is incorrect.

Grammar 3: \(S \to aaaA | a, A \to aAb | \lambda\)

As shown for Grammar 2, \(A \to aAb | \lambda\) generates \(a^k b^k\) for \(k \ge 0\).

Now consider the rules for S:

• Using \(S \to a\): Generates 'a'. This string is not in \(L\).
• Using \(S \to aaaA\): A generates \(a^k b^k\). So S generates strings of the form \(aaa (a^k b^k) = a^{3+k} b^k\) for \(k \ge 0\). Let \(n = 3+k\). Since \(k \ge 0\), \(n \ge 3\). The generated strings are of the form \(a^n b^{n-3}\) for \(n \ge 3\). This precisely matches the definition of language \(L = \left\{ {{a^n}{b^{n - 3}}\ | n > 2} \right\}\).

The language generated by Grammar 3 includes all strings in \(L\) from the \(S \to aaaA\) rule, but it also includes the string 'a' from the \(S \to a\) rule. Since 'a' is not in \(L\), Grammar 3 is incorrect.

Grammar 4: \(S \to aaaA, A \to aAb | \lambda\)

As shown for Grammar 2 and 3, \(A \to aAb | \lambda\) generates strings of the form \(a^k b^k\) for \(k \ge 0\).

The only rule for S is \(S \to aaaA\).

S generates \(aaa (a^k b^k)\) where \(a^k b^k\) is any string generated by A. This means S generates strings of the form \(a^{3+k} b^k\) for any \(k \ge 0\).

Let's check if this matches \(L = \left\{ {{a^n}{b^{n - 3}}\ | n > 2} \right\}\).

Let \(n = 3+k\). Since \(k\) can be any integer \( \ge 0\), \(n\) can be any integer \( \ge 3\). This satisfies the condition \(n > 2\).

The number of 'a's is \(3+k\), which is \(n\).

The number of 'b's is \(k\). Since \(n = 3+k\), \(k = n-3\). The number of 'b's is \(n-3\).

So, the strings generated are of the form \(a^n b^{n-3}\) for \(n \ge 3\), which is exactly the language \(L\).

Let's trace derivations for the first few strings in \(L\):

• For \(n=3\) (k=0): \(S \to aaaA \to aaa\lambda = aaa\). Correct.
• For \(n=4\) (k=1): \(S \to aaaA \to aaa(aAb) \to aaa(a\lambda b) = aaaab\). Correct.
• For \(n=5\) (k=2): \(S \to aaaA \to aaa(aAb) \to aaa(a(aAb)b) = aaa(aaAbb) \to aaa(aa\lambda bb) = aaaaabb\). Correct.

Grammar 4 generates exactly the language \(L\).

Conclusion

Based on the analysis, Grammar 4, with rules \(S \to aaaA, A \to aAb | \lambda\), is the only grammar that generates exactly the language \(L = \left\{ {{a^n}{b^{n - 3}}\ | n > 2} \right\}\).

Revision Table: Comparing Grammars for Language \(L\)

Additional Information: Context-Free Grammars and Formal Languages

A Context-Free Grammar (CFG) is a system for describing a formal language. It consists of a set of terminal symbols (the alphabet), a set of non-terminal symbols, a set of production rules, and a start symbol. The production rules show how non-terminals can be replaced by strings of terminals and/or non-terminals.

Formal languages are sets of strings defined by precise mathematical or logical rules. In this case, the language \(L\) is defined by a pattern involving the count of symbols ('a' and 'b') and a condition on that count (\(n > 2\)).

The task of finding a grammar that generates a specific language is a fundamental problem in the study of formal languages and automata theory. Grammars provide a way to define the structure of valid strings in a language. Context-Free Grammars are powerful enough to describe many common language structures, including the syntax of most programming languages.

The recursive rule \(A \to aAb\) is key here. It's a common pattern in CFGs used to generate languages like \(\{a^n b^n \mid n \ge 0\}\) or variations. This rule adds one 'a' and one 'b' symmetrically around the non-terminal A, maintaining a certain balance or relationship between the counts of 'a's and 'b's as A is expanded. The initial 'aaa' from the S rule in Grammar 4 provides the necessary offset (3 'a's) to match the language \(L\).