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Here I found a example for a regular language.

L = { a^n | n>=2 } is regular. Clearly, we can draw a finite automaton with 3 states.

I was asking myself how this graph would look like. If I choose n=11, this means, that the language contains all words with a sequence of 11 a's. This can't be solved with a graph with 3 states, or am I wrong?

Johnny
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2 Answers2

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given language is L = { a^n | n>=2 }. here the minimum number of a's is 2, so the finite automata contains 3 states(2+1) =>(q0,a)->(q1,a)->(qf,a*). Similarly if the minimum number of a's required is 11(n>=11),so the finite automata contains 12 states (11+1).
Therefore L = { a^n | n>=11} cannot be solved using 3 states

Vyshnavi Koppisetty
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The given language L cannot be solved using 3 states for your condition i.e, 11a's. For every word to contain a sequence of 11 a's, there should be a minimum of 11 a's. So the graph should contain (11+1)=12 States. Therefore it can't be solved using 3 states.

Bharath K
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