Self doubt

Question

Is dead state necessary in minimal DFA?

Comments

If the language is (a + b)*. Is there any need of dead state in minimum dfa?

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not for this language but what if incase of $(10+1)^{*}$

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i got your point Hemant but for some languages jflap minimizes the DFA and doesn't include trap states so are trap/dead states necessary?

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Yes in this case (10 + 1)*, we require dead state. Whenever we see 2 consecutive zeros in the input string we reached to the dead state.

In DFA, we need to write transition at each state for every input symbol. If we don't do this, then it is not a DFA. Having a dead state is not necessary but writing transition at each state for every input symbol is necessary in DFA.

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It totally removes q₃ 

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The minimized finite automata is not DFA, it is NFA. See you haven't defined in minimized, where you go when you see zero on q0 state that's why it is not DFA.

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there is not a single example where they have trap states in peter linz that's why i'm confused. I totally agree with you but jflap created confusion and i'm not able to find example from reliable sources

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jflap is showing wrong bro. See rule is simple for DFA, You have to define transition at every state for every alphabet.

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ok thanks

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No, Dead State is not a mandatory state in DFA.

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Chandan source?

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Try for a DFA that accepts all sets of Strings whose length is a multiple of 2 or 3 or 4.

You will get some non-final states but you will not get any dead state.

Answer 1

Dead state possibe becouse of we need all input transtion in each and every state.it is conditional of dfa that we have to transtion all input. So thats why dead state necessary in dfa if it is in dfa