Question 46

Analysis of $L_1$:

• Consider the patterns $110$ and $011$.
• A $110$ pattern occurs when a sequence of at least two 1s transitions into a 0 (i.e., ...110...).
• A $011$ pattern occurs when a 0 transitions into at least two 1s (i.e., ...011...).
• In any binary string, to have another $110$, the string must transition back through $011$ to enter a block of 1s again.
• Consequently, occurrences of $110$ and $011$ strictly alternate along the string, meaning that for any prefix $w$:
  $$|\#_{110}(w) - \#_{011}(w)| \le 1.$$
• Because the difference between the two counts is strictly bounded by 1, a finite automaton with a small number of states can keep track of the difference.
• Therefore, $L_1$ is regular.

Analysis of $L_2$:

• Consider strings in the regular subset $0^* 1^*$, specifically strings of the form $w = 0^{n+2} 1^{m+2}$ for $n, m \ge 0$.
• In $0^{n+2}$, the substring $000$ occurs exactly $n$ times.
• In $1^{m+2}$, the substring $111$ occurs exactly $m$ times.
• There are no overlapping $000$ or $111$ cross-patterns between 0 and 1.
• For $w \in L_2$, we must have $n \ge m$. The language $L_2 \cap 0^* 1^* = \{0^{n+2} 1^{m+2} \mid n \ge m\}$ is a well-known non-regular context-free language.
• Since intersection with a regular language yields a non-regular language, $L_2$ CANNOT be regular.

Conclusion:
$L_1$ is regular but not $L_2$, matching option (A).