Question:

A regular language \(L\) is accepted by a non-deterministic finite automaton (NFA) with \(n\) states. Which of the following statement(s) is/are FALSE?

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The conversion from NFA to DFA can cause an exponential increase in the number of states. However, some DFA constructions can still have fewer states depending on the language.
Updated On: Apr 4, 2025
  • \( L \) may have an accepting NFA with \(< n\) states.
  • \( L \) may have an accepting DFA with \(< n\) states.
  • There exists a DFA with \( \leq 2^n \) states that accepts \(L\).
  • Every DFA that accepts \(L\) has \(>2^n \) states.
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The Correct Option is B

Solution and Explanation

Option (A): This is true. An NFA can potentially accept a language using fewer states due to non-determinism, but for DFA equivalence, the number of states could be larger.

Option (B): This is false. A DFA may require more states than an NFA to accept the same language. For any language accepted by an NFA with \( n \) states, the equivalent DFA may have up to \( 2^n \) states, but it cannot have fewer states.

Option (C): This is true. There is always a DFA with \( 2^n \) states for any language accepted by an NFA with \( n \) states (since \( NFA \to DFA \) conversion can potentially lead to exponential state growth).

Option (D): This is false. A DFA does not always need more than \( 2^n \) states. It is possible for the DFA to have fewer states than \( 2^n \), depending on the language.

Thus, the correct answer is \( \boxed{D} \).
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