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Official PYQ · 2013

GATE CS 2013 Q41 — Undecidability

Explain with AI

Which of the following is/are undecidable?

  1. GG is a CFG. Is L(G)=ϕL(G) = \phi?
  2. GG is a CFG. Is L(G)=Σ∗L(G) = \Sigma^*?
  3. MM is a Turing machine. Is L(M)L(M) regular?
  4. AA is a DFA and NN is an NFA. Is L(A)=L(N)L(A) = L(N)?
Choose one option
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Answer: D

To determine which of the given problems are undecidable, let us analyze each statement individually:

  1. GG is a CFG. Is L(G)=ϕL(G) = \phi? (Emptiness Problem)
    This is decidable. We can determine if a Context-Free Grammar (CFG) is empty by checking if the start symbol can derive any terminal string. This is done using a bottom-up marking algorithm to find all "generating" variables. If the start symbol is generating, the language is non-empty; otherwise, it is empty.

  2. GG is a CFG. Is L(G)=Σ∗L(G) = \Sigma^*? (Universality Problem)
    This is undecidable. The problem of checking whether a CFG generates all possible strings over its alphabet is undecidable, which can be shown by reducing the Turing machine halting problem (via valid computation histories) to it.

  3. MM is a Turing machine. Is L(M)L(M) regular? (Regularity Problem)
    This is undecidable. According to Rice's Theorem, any non-trivial semantic property of recursively enumerable languages (languages recognized by Turing machines) is undecidable. Since "being regular" is a semantic property of the language, and it is non-trivial (some TMs accept regular languages, while others do not), this problem is undecidable.

  4. AA is a DFA and NN is an NFA. Is L(A)=L(N)L(A) = L(N)? (Equivalence Problem)
    This is decidable. Regular languages are closed under complementation, intersection, and union. We can construct a DFA for the symmetric difference (L(A)∩L(N)‾)∪(L(A)‾∩L(N))(L(A) \cap \overline{L(N)}) \cup (\overline{L(A)} \cap L(N)) and test if this resulting language is empty, which is a decidable property.

Thus, only statements 2 and 3 are undecidable.

Correct Option: D

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