why dead state? I havenot understand this point

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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.

0

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

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

0

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.

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

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

+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.

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.

0

@ 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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