Made Easy Test Series:TOC-DFA

Question

How many number of $DFA$ states(minimal DFA) required which accepts the language $L=\left \{ a^{n}:n=\text{3 or n>= 2m for all m>= 1} \right \}$ ___________

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Answer will be $3$ or $6?$

Answer 1

Answer will be 3 assuming that $\Sigma = \{ a\}$ Or It will be $4$ if $\Sigma \supset \{ a \} $

The language $L = \Sigma^* - \{ \in,a\}$ 

If $\Sigma = \{ a\}$ then the Minimal DFA would look like the following :

If $\Sigma \supset \{ a \} $ then Minimal DFA will look like the above but with one Dead State extra. 

Comments

@ Satbir yeah thanks I did not notice n>=2m(greater than sign), but if n=2m where  m>=1 or n=3 then minimal dfa would have 6 states right?

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yes

$*$ represents the final state.

$\rightarrow A\overset{a}{\rightarrow}B\overset{a}{\rightarrow}C^*\overset{a}{\rightarrow}D^*\overset{a}{\rightarrow}E^*\overset{a}{\leftrightarrow}F$

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thanks bro