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Is it theoretically possible to have a nondeterministic finite state machine without any initial state or does it need at least one initial state?

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  • $\begingroup$ In the standard definition, it's allowed. However, it is pretty pointless (algorithmically), since you can easily verify it. This is in contrast, for example, to an NFA whose language is empty since there are no reachable accepting states. $\endgroup$
    – Shaull
    Commented May 28, 2013 at 18:30

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In my favourite definition a finite state automaton is specified as 5-tuple $(Q,\Sigma,\delta,q_0,F)$, where $Q$ is the finite set of states, $\Sigma$ is a (finite, nonempty) alphabet, $q_0\in Q$ the initial state, $F\subseteq Q$ the set of final states, and $\delta\subseteq Q\times\Sigma\times Q$ the set of transitions. In that definition we would have exactly one initial state.

It is however very well possible that, for reasons of symmetry, one specifies a set $I\subseteq Q$ of initial states. There you can have one, several, of no initial state.

Short answer: consult the book, lecture notes, or scientific paper you are using. They set the rules for the game. Not us, not wikipedia.

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    $\begingroup$ +1. There are useful, reasonable definitions of nondeterministic finite automata that may require one state, or allow a non-empty set of states, or even let you specify no initial state. It's a definition that has little to do with the model of computation NFAs embody. $\endgroup$
    – Patrick87
    Commented May 28, 2013 at 20:49

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