MCQ Analysis: Regularity of Languages L1 and L2

The question asks us to determine whether the languages $L_1$ and $L_2$ are regular. The languages are defined over the alphabet $\{a, b\}$ using the structure $\alpha\beta\alpha$. Let's analyze each language.

Understanding Language L1

Definition: $L_1 = \{\alpha\beta\alpha | \alpha \in \{a,b\}^+ \text{ AND } \beta\in \{a,b\}^+ \}$.

In this definition:

• $\alpha$ represents a non-empty string (at least one character) consisting of symbols 'a' and 'b'.
• $\beta$ represents a non-empty string (at least one character) consisting of symbols 'a' and 'b'.
• The structure $\alpha\beta\alpha$ implies that the string must start with a sequence $\alpha$, followed by a sequence $\beta$, and end with the *exact same* sequence $\alpha$.

Analysis of Regularity:

To determine if $L_1$ is regular, we can use the Pumping Lemma for regular languages. Let's consider a specific string in $L_1$. Choose a pumping length $p$. Let the string be $s = a^p b b a^p$.

This string $s$ is in $L_1$ because we can choose $\alpha = a^p$ (which is in $\{a,b\}^+$ since $p \ge 1$) and $\beta = bb$ (which is in $\{a,b\}^+$).

According to the Pumping Lemma, $s$ can be divided into $xyz$ such that $|y| \ge 1$, $|xy| \le p$, and for any $k \ge 0$, the string $xy^kz$ must also be in $L_1$.

Since $|xy| \le p$, the combined part $xy$ must occur within the first $a^p$ segment of $s$. This means $x$ and $y$ consist only of 'a's. Let $x = a^i$ and $y = a^j$, where $i \ge 0, j \ge 1$, and $i+j \le p$. The remaining part is $z = a^{p-(i+j)} b b a^p$.

Now, consider pumping $s$ with $k=2$. The resulting string is $xy^2z = a^i a^{2j} a^{p-(i+j)} b b a^p = a^{p+j} b b a^p$.

For this new string $a^{p+j} b b a^p$ to be in $L_1$, it must be of the form $\alpha'\beta'\alpha'$, where the prefix $\alpha'$ is identical to the suffix $\alpha'$. However, the prefix is $a^{p+j}$ and the suffix is $a^p$. Since $j \ge 1$, $p+j \neq p$. Therefore, the prefix and suffix are different.

This contradiction shows that $xy^2z$ is not in $L_1$. By the Pumping Lemma, $L_1$ cannot be a regular language.

Examples in L1: 'aba' ($\alpha=a, \beta=b$), 'aabaa' ($\alpha=aa, \beta=b$), 'ababa' ($\alpha=ab, \beta=a$), 'babab' ($\alpha=ba, \beta=b$).

Understanding Language L2

Definition: $L_2 = \{\alpha\beta\alpha | \alpha \in \{a\}^+ \text{ AND } \beta\in \{a,b\}^+ \}$.

In this definition:

• $\alpha$ represents a non-empty string consisting of only 'a's. So, $\alpha = a^n$ for some integer $n \ge 1$.
• $\beta$ represents a non-empty string consisting of symbols 'a' and 'b'.
• The structure $\alpha\beta\alpha$ implies the string must start with $a^n$, end with the *exact same* $a^n$, and have a non-empty string $\beta$ in between.

Analysis of Regularity:

A string $w$ in $L_2$ has the form $w = a^n \beta a^n$, where $n \ge 1$ and $\beta \in \{a,b\}^+$.

Let's consider the language $L' = a(a|b)^+a$. This language consists of all strings that start with 'a', end with 'a', and have a length of at least 3. This language is regular and can be described by the regular expression $a(a|b)^+a$.

Now, let's see if $L_2$ is equivalent to $L'$.

1. Is $L_2 \subseteq L'$?

If $w \in L_2$, then $w = a^n \beta a^n$ with $n \ge 1$ and $\beta \in \{a,b\}^+$. This string $w$ clearly starts with 'a' (since $n \ge 1$) and ends with 'a'. The length of $w$ is $n + |\beta| + n$. Since $n \ge 1$ and $|\beta| \ge 1$, the length is at least $1 + 1 + 1 = 3$. So, $w$ is in $a(a|b)^+a$. Thus, $L_2 \subseteq L'$.

2. Is $L' \subseteq L_2$?

If $w \in L'$, then $w = a x a$ where $x \in \{a,b\}^+$. We need to show that $w$ can be written as $a^n \beta a^n$ with $n \ge 1$ and $\beta \in \{a,b\}^+$. We can choose $n=1$. Then $w = a \beta a$. This requires $\beta = x$. Since $x \in \{a,b\}^+$, the condition on $\beta$ is satisfied. Therefore, any string $w$ in $L' = a(a|b)^+a$ is also in $L_2$. Thus, $L' \subseteq L_2$.

Since $L_2 \subseteq L'$ and $L' \subseteq L_2$, we have $L_2 = L'$. As $L' = a(a|b)^+a$ is a regular language, $L_2$ is also a regular language.

(Note: If we tried to apply the Pumping Lemma to $L_2$ with a string like $s = a^p b a^p$, the logic would show non-regularity. However, the equivalence $L_2 = a(a|b)^+a$ is derived from the definition and shows regularity.)

Examples in L2: 'aba' ($\alpha=a, \beta=b$), 'aabaa' ($\alpha=aa, \beta=b$), 'aaabaaa' ($\alpha=aaa, \beta=b$), 'aabaabaa' ($\alpha=aa, \beta=aba$).

Conclusion

Based on the analysis:

• $L_1$ requires the prefix and suffix ($\alpha$) to be identical and non-empty, using any symbols from $\{a,b\}$. This structure leads to non-regularity, as demonstrated by the Pumping Lemma.
• $L_2$ requires the prefix and suffix ($\alpha$) to be identical strings of only 'a's ($a^n, n \ge 1$) and the middle part ($\beta$) to be any non-empty string from $\{a,b\}$. This language simplifies to $a(a|b)^+a$, which is a regular language.

Therefore, $L_1$ is not a regular language, but $L_2$ is a regular language.