Is my attempt is true or where am I wrong?
DFA : The set of strings over $\{a, b\} $ that have odd length or end with $aaa$.
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Did you try finding a smaller automaton? – k.stm Mar 24 '16 at 20:29
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@k.stm Yes, I tried but I couldn't find.Is this true? – zxccxz Mar 24 '16 at 20:30
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1Haven’t checked it. Actually, you probably do need eight states. Have you tried organizing them? Can you tell us what each of the states $1$, …, $8$ describe? (Oh, you have nine states – I think you can toss one.) – k.stm Mar 24 '16 at 20:33
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@k.stm there are 8 states.It is hard to describe what each of states describe. – zxccxz Mar 24 '16 at 20:41
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It looks correct to me, but the best thing to do is to run it in your computer – 3SAT Mar 24 '16 at 20:58
2 Answers
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Your automaton is correct but it is not minimal. The minimal automaton only has 5 states. Actually, you can obtain it by minimising your automaton. You will then identify states 2 and 4, states 3 and 5 and states 6 and 8.
J.-E. Pin
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Of states $1$ and $4$, only one is an accepting state,so identifying them will lead to an inequivalent automaton. You mean $2$ and $4$, I presume. – k.stm Mar 25 '16 at 09:26
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There are two states 1 in the picture. Let us call $0$ the initial state to avoid any confusion. – J.-E. Pin Mar 25 '16 at 09:36
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Your DFA is correct; it consists of eight states, as you have, one for every combination of "Is the string odd or even length?" and "Have I most recently seen 0,1,2, or 3+ consecutive A's"? You can arrange your DFA like this for easier readability:
And there's an equivalent DFA with fewer states that might be easier to check:
user326210
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