Analyzing Formal Languages and Context-Free Languages

The question asks us to determine which of the given languages, L₁ and L₂, is a Context-Free Language (CFL).

A Context-Free Language is a language that can be generated by a Context-Free Grammar (CFG) or recognized by a pushdown automaton. We will analyze each language separately.

Understanding Language L₁

Language L₁ is defined as: $L_1 = \{a^{n+m} b^n a^m \mid n, m \ge 0\}$.

Let's look at some examples of strings in L₁ by choosing different values for n and m:

• If $n=0, m=0$: $a^{0+0} b^0 a^0 = a^0 b^0 a^0 = \epsilon$ (empty string).
• If $n=1, m=0$: $a^{1+0} b^1 a^0 = a^1 b^1 a^0 = ab$.
• If $n=0, m=1$: $a^{0+1} b^0 a^1 = a^1 b^0 a^1 = aa$.
• If $n=1, m=1$: $a^{1+1} b^1 a^1 = a^2 b^1 a^1 = aaba$.
• If $n=2, m=1$: $a^{2+1} b^2 a^1 = a^3 b^2 a^1 = aaaba$.
• If $n=1, m=2$: $a^{1+2} b^1 a^2 = a^3 b^1 a^2 = aabaa$.
• If $n=2, m=2$: $a^{2+2} b^2 a^2 = a^4 b^2 a^2 = aaaabaa$.

The structure of strings in L₁ is a block of 'a's (length n+m), followed by a block of 'b's (length n), followed by a block of 'a's (length m). Notice that the number of 'b's (n) corresponds to the first 'n' 'a's in the initial block, and the number of 'm' 'a's in the final block corresponds to the last 'm' 'a's in the initial block. The initial block of 'a's can be thought of as $a^n a^m$. The string structure is $a^n a^m b^n a^m$.

We can construct a Context-Free Grammar (CFG) for L₁. Consider the grammar with starting symbol S and non-terminals A:

S → a S a | A

A → a A b | ε

Let's trace a derivation:

• Start with S.
• Using S → a S a repeatedly $m$ times, we get $S \implies a S a \implies aa S aa \implies \dots \implies a^m S a^m$.
• Now, replace S with A using S → A: $a^m S a^m \implies a^m A a^m$.
• Using A → a A b repeatedly $n$ times, we get $A \implies a A b \implies aa A bb \implies \dots \implies a^n A b^n$.
• Now, replace A with ε using A → ε: $a^n A b^n \implies a^n \epsilon b^n = a^n b^n$.
• Substitute $a^n b^n$ back into the string derivation for S: $a^m A a^m \implies a^m (a^n b^n) a^m = a^{m+n} b^n a^m$.

This derivation produces strings of the form $a^{m+n} b^n a^m$, which is exactly the definition of L₁. Since we found a CFG for L₁, L₁ is a Context-Free Language.

Understanding Language L₂

Language L₂ is defined as: $L_2 = \{a^{n+m} b^{n+m} a^{n+m} \mid n, m \ge 0\}$.

Let $k = n+m$. Since $n \ge 0$ and $m \ge 0$, $k$ can be any non-negative integer ($k \ge 0$). So, L₂ can be rewritten as: $L_2 = \{a^k b^k a^k \mid k \ge 0\}$.

Examples of strings in L₂:

• If $k=0$ (e.g., $n=0, m=0$): $a^0 b^0 a^0 = \epsilon$.
• If $k=1$ (e.g., $n=1, m=0$ or $n=0, m=1$): $a^1 b^1 a^1 = aba$.
• If $k=2$ (e.g., $n=2, m=0$ or $n=1, m=1$ or $n=0, m=2$): $a^2 b^2 a^2 = aabbaa$.
• If $k=3$: $a^3 b^3 a^3 = aaabbbaaa$.

The structure of strings in L₂ is a block of 'a's, followed by a block of 'b's, followed by a block of 'a's, where all three blocks have the exact same length, k.

Languages requiring the equality of counts across three or more separated parts are generally not Context-Free Languages. We can prove that L₂ is not a CFL using the Pumping Lemma for Context-Free Languages.

Proving L₂ is Not Context-Free Using the Pumping Lemma

The Pumping Lemma for CFLs states that if L is a CFL, then there exists a pumping length $p \ge 1$ such that for any string $s \in L$ with $|s| \ge p$, $s$ can be divided into five parts $s = uvwxy$ such that:

1. $|vwx| \le p$
2. $|vx| \ge 1$
3. For all $i \ge 0$, the string $uv^i w x^i y$ is in L.

Let's assume L₂ is a CFL. Let $p$ be the pumping length given by the lemma.

Consider the string $s = a^p b^p a^p$. This string is in L₂ (with $k=p$), and its length is $|s| = 3p$, which is $\ge p$. According to the lemma, $s$ can be divided into $uvwxy$ satisfying the conditions.

Condition 1: $|vwx| \le p$. Since the string $s = a^p b^p a^p$ consists of three blocks of length p separated by boundaries, a substring of length at most $p$ can lie entirely within one block (a^p, b^p, or a^p) or span across the boundary of at most two adjacent blocks (a's and b's, or b's and a's). It cannot span across all three blocks (i.e., contain 'a's from the first block and 'a's from the second block) because the distance between the first 'a' block and the second 'a' block is at least p (the length of the 'b' block), and $|vwx| \le p$.

Condition 2: $|vx| \ge 1$. This means that at least one of v or x is non-empty.

Now consider Condition 3: $uv^i w x^i y$ must be in L₂ for all $i \ge 0$. Let's test with $i=2$ (pumping up) and $i=0$ (pumping down).

Let's analyze the possible positions of the segment $vwx$ based on $|vwx| \le p$ in $s = a^p b^p a^p$:

• Case A: $vwx$ is entirely within the first block of 'a's.

  $s = \underbrace{a \dots a}_{u} \underbrace{a \dots a}_{v} \underbrace{a \dots a}_{w} \underbrace{a \dots a}_{x} \underbrace{a \dots a b^p a^p}_{y}$

  Here, $v$ and $x$ only contain 'a's from the first block. Since $|vx| \ge 1$, at least one 'a' is in $v$ or $x$. When we pump with $i=2$, we get $uv^2 w x^2 y$. This increases the number of 'a's in the first block, resulting in a string like $a^{p+k} b^p a^p$ where $k > 0$. The counts $(p+k, p, p)$ are not equal. This string is not in L₂.

• Case B: $vwx$ is entirely within the block of 'b's.

  $s = \underbrace{a^p b \dots b}_{u} \underbrace{b \dots b}_{v} \underbrace{b \dots b}_{w} \underbrace{b \dots b}_{x} \underbrace{b \dots b a^p}_{y}$

  Here, $v$ and $x$ only contain 'b's. Since $|vx| \ge 1$, at least one 'b' is in $v$ or $x$. Pumping with $i=2$ results in $a^p b^{p+k} a^p$ where $k > 0$. The counts $(p, p+k, p)$ are not equal. This string is not in L₂.

• Case C: $vwx$ is entirely within the second block of 'a's.

  $s = \underbrace{a^p b^p a \dots a}_{u} \underbrace{a \dots a}_{v} \underbrace{a \dots a}_{w} \underbrace{a \dots a}_{x} \underbrace{a \dots a}_{y}$

  Here, $v$ and $x$ only contain 'a's from the second block. Since $|vx| \ge 1$, at least one 'a' is in $v$ or $x$. Pumping with $i=2$ results in $a^p b^p a^{p+k}$ where $k > 0$. The counts $(p, p, p+k)$ are not equal. This string is not in L₂.

• Case D: $vwx$ spans across the boundary between the first 'a' block and the 'b' block.

  Here, $vwx$ contains some final 'a's from the first block and some initial 'b's from the 'b' block. So $v$ and $x$ (or $vwx$) must contain characters from both 'a' and 'b' blocks. For example, $v$ could be a string of 'a's and $x$ a string of 'b's (with $|vx| \ge 1$). When we pump $uv^2 w x^2 y$, we increase the number of 'a's and 'b's involved in $v$ and $x$. This will result in a string $a^{p+k} b^{p+l} a^p$ where $k, l \ge 0$ and $k+l \ge 1$. The counts $(p+k, p+l, p)$ will not all be equal (unless $k=l=0$, which is not possible as $|vx| \ge 1$). Thus, the pumped string is not in L₂.

• Case E: $vwx$ spans across the boundary between the 'b' block and the second 'a' block.

  Similar to Case D, pumping $uv^2 w x^2 y$ will result in a string $a^p b^{p+k} a^{p+l}$ where $k, l \ge 0$ and $k+l \ge 1$. The counts $(p, p+k, p+l)$ will not all be equal. Thus, the pumped string is not in L₂.

In all possible cases where $|vx| \ge 1$, pumping the string $s=a^p b^p a^p$ with $i=2$ produces a string that does not have an equal number of 'a's in the first block, 'b's in the middle block, and 'a's in the last block. Thus, the pumped string is not in L₂.

Since we found a string $s \in L_2$ with $|s| \ge p$ that cannot be pumped according to the lemma, our assumption that L₂ is a CFL must be false. Therefore, L₂ is not a Context-Free Language.

Conclusion

Based on our analysis:

• L₁ = $\{a^{n+m} b^n a^m \mid n, m \ge 0\}$ is a Context-Free Language because we found a CFG for it.
• L₂ = $\{a^{n+m} b^{n+m} a^{n+m} \mid n, m \ge 0\} = \{a^k b^k a^k \mid k \ge 0\}$ is not a Context-Free Language, as proven by the Pumping Lemma for CFLs.

Therefore, only L₁ is a Context-Free Language.

Comparing with Options

Let's evaluate the given options:

• Option 1: Only L₁ is Context Free Language. This matches our conclusion.
• Option 2: Both L₁ and L₂ are not Context Free Language. This is incorrect as L₁ is a CFL.
• Option 3: Only L₂ is Context Free Language. This is incorrect as L₂ is not a CFL.
• Option 4: Both L₁ and L₂ are Context Free Language. This is incorrect as L₂ is not a CFL.

The correct option is the one stating that only L₁ is a Context Free Language.

Revision Table: Language Properties

Additional Information: Chomsky Hierarchy and Language Classes

Formal languages are often classified into a hierarchy called the Chomsky hierarchy, based on the type of grammar required to generate them (or the type of automaton required to recognize them). The main levels are:

• Type 0: Recursively Enumerable Languages. Generated by unrestricted grammars. Recognized by Turing machines. Example: The Halting Problem.
• Type 1: Context-Sensitive Languages (CSL). Generated by context-sensitive grammars. Recognized by linear-bounded automata. Example: $\{a^n b^n c^n \mid n \ge 1\}$.
• Type 2: Context-Free Languages (CFL). Generated by context-free grammars. Recognized by pushdown automata. Example: $\{a^n b^n \mid n \ge 0\}$, balanced parentheses.
• Type 3: Regular Languages (RL). Generated by regular grammars (left-linear or right-linear). Recognized by finite automata (DFA or NFA). Example: $\{a^n \mid n \ge 0\}$, languages described by regular expressions.

These language classes form a strict hierarchy: Regular Languages ⊂ Context-Free Languages ⊂ Context-Sensitive Languages ⊂ Recursively Enumerable Languages. This means every regular language is context-free, every context-free language is context-sensitive, and so on.

The Pumping Lemma for CFLs is a useful tool to prove that a language is not context-free. There is also a Pumping Lemma for Regular Languages (used to show a language is not regular). For CSLs, there is a more complex pumping lemma, and for Type 0 languages, there is no pumping lemma.

The language $L_2 = \{a^k b^k a^k \mid k \ge 0\}$ is a classic example of a language that is Context-Sensitive but not Context-Free. It fits the pattern of CSLs that require correlating counts of three or more symbols across the string.