Consider the following languages. $$L_1 = \{wxyx \mid w, x, y \in (0 + 1)^+\}$$ $$L_2 = \{xy \mid x, y \in (a + b)^*, |x| = |y|, x \neq y\}$$ Which one of the following is TRUE?

### Analysis of $L_1$: $L_1 = \{wxyx \mid w, x, y \in (0 + 1)^+\}$

Notice that if $x$ is chosen to be a single character $c \in \{0, 1\}$, the string has the form $w c y c$ where $w, y \in (0+1)^+$. Any string in $L_1$ has $x$ ending in either $0$ or $1$. Let $x = x^{\prime}c$ with $c \in \{0, 1\}$, then:

Since $w, y \in (0+1)^+$, $wx^{\prime} \in (0+1)^+$ and $yx^{\prime} \in (0+1)^+$. Thus, any string in $L_1$ matches the regular expression:

Conversely, any string matching this regular expression belongs to $L_1$ by setting $x = c$. Because $L_1$ can be expressed by a regular expression, $L_1$ is regular.

### Analysis of $L_2$: $L_2 = \{xy \mid x, y \in (a + b)^*, |x| = |y|, x \neq y\}$

$L_2$ consists of all even-length strings where the first half is not equal to the second half. $x \neq y$ implies there is some position $i$ ($1 \le i \le |x|$) such that the $i$-th character of $x$ differs from the $i$-th character of $y$.

This can be generated by a Context-Free Grammar (CFG):

• $S \to AB \mid BA$

• $A \to CAC \mid a$

• $B \to CBC \mid b$

• $C \to a \mid b$

Here, $A$ generates strings of odd length with 'a' in the middle, and $B$ generates strings of odd length with 'b' in the middle. The concatenation $AB$ or $BA$ generates strings of even length $2n$ where the $i$-th character from the start and the $(n+i)$-th character from the start differ. Since a CFG exists, $L_2$ is context-free (non-deterministic CFL, not regular).

Therefore, $L_1$ is regular and $L_2$ is context-free, making A the correct option.

*(Note: Option D was printed on the succeeding page and is omitted as per partial extraction rules.)*