GATE CS 1995 Q2.24 — Regular and Context-free Languages

To determine the nature of the languages $L \cup R$ and $R$, let us analyze them individually:

1. Analyzing $R$:

   The language $R = \{0^n1^n \mid n > 0\}$ requires matching the number of $0$s with the number of $1$s. Since a finite automaton has a finite number of states, it cannot keep track of an arbitrary count $n$. By the Pumping Lemma, $R$ is not regular (it is a deterministic context-free language).

2. Analyzing $L \cup R$:

   We are given $L = \Sigma^*$, which represents the set of all possible strings over the alphabet $\Sigma = \{0, 1\}$. By definition, any language over $\Sigma$ is a subset of $\Sigma^*$. Therefore, $R \subseteq L$.

   Taking the union of $L$ and $R$ gives:

   $$L \cup R = \Sigma^* \cup R = \Sigma^*$$

   Since $\Sigma^*$ is represented by the regular expression $(0+1)^*$, it is a regular language.

Thus, $L \cup R$ is regular and $R$ is not regular.