Let \(L \subseteq \{0,1\}^*\) be an arbitrary regular language accepted by a minimal DFA with \(k\) states. Which one of the following languages must necessarily be accepted by a minimal DFA with \(k\) states?
Show Hint
- \(L - \{01\}\)
- \(L \cup \{01\}\)
- \(\{0,1\}^* - L\)
- \(L \cdot L\)
The Correct Option is C
Solution and Explanation
Step 1: Recall closure properties of regular languages.
Regular languages are closed under complement, union, difference, and concatenation. However, closure does not imply that the minimal DFA size remains the same.
Step 2: Analyze option (C): Complement of \(L\).
The language \(\{0,1\}^* - L\) is the complement of \(L\).
To obtain a DFA for the complement, we only need to swap accepting and non-accepting states of the minimal DFA for \(L\).
Step 3: Minimality of the complemented DFA.
Complementing a DFA does not change the number of states, and minimality is preserved because distinguishability of states remains unchanged.
Hence, the complement of \(L\) is necessarily accepted by a minimal DFA with exactly \(k\) states.
Step 4: Eliminate other options.
(A) Removing a single string may reduce or increase the number of states.
(B) Adding a string may require additional states.
(D) Concatenation generally increases the number of states.
Step 5: Conclusion.
Only the complement of \(L\) must be accepted by a minimal DFA with exactly \(k\) states.
Final Answer: (C)
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