GATE CS 2011 Q26 — Regular and Context-free Languages

Let us analyze the classification of each language given in the problem:

1. Language $L_1 = \{0^p 1^q \mid p, q \in N\}$:
   This language represents any sequence of $0$s followed by any sequence of $1$s. It can be represented by the regular expression $0^+1^+$ (or $0^*1^*$ depending on whether $N$ includes $0$). Since it can be defined by a regular expression, $L_1$ is a Regular Language.
2. Language $L_2 = \{0^p 1^q \mid p, q \in N \text{ and } p=q\}$:
   This is the language $\{0^n 1^n \mid n \in N\}$. It requires matching the number of $0$s with the number of $1$s, which requires a single stack memory. Thus, $L_2$ is a Context-Free Language (CFL), but not regular.
3. Language $L_3 = \{0^p 1^q 0^r \mid p, q, r \in N \text{ and } p=q=r\}$:
   This is the language $\{0^n 1^n 0^n \mid n \in N\}$. It requires comparing three quantities simultaneously ($p=q$ and $q=r$), which cannot be done using a single stack. Thus, $L_3$ is Context-Sensitive (CSL) but NOT Context-Free.

Now, let's evaluate each option:

• Option A: Push Down Automata (PDA) can be used to recognize $L_1$ and $L_2$.
  Since $L_1$ is regular and $L_2$ is context-free, both can be recognized by a PDA. (True)
• Option B: $L_1$ is a regular language.
  As shown above, $L_1$ is regular. (True)
• Option C: All the three languages are context free.
  Since $L_3$ is not context-free, this statement is NOT TRUE. (False)
• Option D: Turing machines can be used to recognize all the languages.
  All three languages are recursive (and thus recursively enumerable), meaning they can all be recognized by a Turing machine. (True)