It doesn't have non-deterministic paths.
Every transition is one-to-one.
Accepted states are defined/guaranteed on specific inputs.
Not sure why its not an FA
Edit: 0.0.2
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2There are no transitions from the accepting state. – saulspatz Mar 30 '19 at 22:23
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@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left. – Gerry Myerson Mar 30 '19 at 23:09
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It might help if we could see 0.0.2 – Gerry Myerson Mar 30 '19 at 23:10
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@GerryMyerson I added 0.0.2, still don't understand why its not a FA. – A_for_ Abacus Mar 30 '19 at 23:28
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Nvm, figured it out.
The transition function $\delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $\delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $\delta(q_1, 0)=q_2$.
A_for_ Abacus
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1OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole". – Gerry Myerson Mar 31 '19 at 02:26
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2I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$". – Tanner Swett Mar 31 '19 at 03:59