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

Answer will be $3$ or $6?$

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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.

by Boss (27.9k points)
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0
why dead state? I havenot understand this point
+1
I think he means if the number of input symbol is >1 then we will need a dead state.
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I have one doubt, why the above language is not

L={a2,a3,a4,a6,a8.......} So that the states in minimal dfa is 6
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@codingo1234

why 6 states??

is this language accepts $a^{5}?$

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coz it is not n=2m ...it is n > =2m and m>=1

so any string greater than or equal to size 2 is accepted.

Also $n=3$ is of no use here.
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@Satbir

So, no dead state required. right??

+1
depends on whether only $a$ is there as an input string or $\{a,b,....\}$ are there in the input set.

dead state will be required to show transition on inputs other than $a$.

like when $b$ is given as input then it should lead to dead state.
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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?

+1
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

+1 vote