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In DFA, is two arrows on each state necessary?

Or it depend on language alphabets? I mean if there is $\Sigma = \{a\}$ then there should be one arrows on each state.

OR if there are $\Sigma = \{a , b\}$ then there should be two arrows on each state.

OR if there are $\Sigma = \{a , b , c\}$ then there should be three arrows on each state.

and so on...

xskxzr
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Bilal Khan
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1 Answers1

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You are correct.

In an NFA, there might be more arrows (for epsilon transitions, or multiple transitions on the same symbol) or fewer arrows (for undefined transitions) than there are symbols in the alphabet. In a DFA, neither of these is allowed: in fact, a DFA is specifically defined as an NFA where neither of these things happens.

Therefore a DFA will always have exactly as many arrows from each state as there are symbols in the alphabet.

Draconis
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  • though in some cases the DFA is simplified by allowing less arrows as there are symbols, those missing transitions implicitly going to a non-accepting sink state. – ratchet freak Aug 03 '18 at 09:58