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This question is about whether the definition of weighted Non-deterministic Finite state Automata (NFAs) excludes the possibility of infinitely many transitions.

The definition of Finite State Automata implies finitely many states, with a finite alphabet for possible transitions. However, in a weighted NFA, it is possible to have multiple arcs between the same two states with the same transition symbol but with different weights.

Now my question is: Could one define a NFA with two states and infinitely many transitions between them with the same symbol, e.g. one for each natural number?

J.-E. Pin
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languagen00b
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  • I think this is really a question of definition than anything else. I'll let someone who knows more about the automata then I do to provide an answer, but according the the Wikipedia article on weighted automatas, the set of transitions is supposed to be finite. – Keen Apr 09 '22 at 17:10
  • Does it really matter? If there are multiple arcs between two states $A$ and $B$, all with the same symbol and with weights $w_i$, you can always replace the multiple arcs with a single arc that has weight $\sum w_i$, without changing the behavior of the automaton. So there is nothing that you can model or analyze with an automaton with an infinite set of transitions that you can't already model or analyze with an equivalent automaton with only a finite set of transitions. – MJD Apr 09 '22 at 17:55

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The standard definition, as given in Wikipedia, in [1] or elsewhere, is that weighted non-deterministic finite automata only have finitely many transitions.

If you wanted to change the definition to allow infinitely many transitions, you would need infinite sums to define the behaviour of your automaton. This can be done in some semirings, like $({\cal P}{\Bbb N}), \cup, \cap)$, but not in every semiring.

Moreover, if you have two transitions, say $p \xrightarrow{a \mid x} q$ and $p \xrightarrow{a \mid y} q$, replacing them with the transition $p \xrightarrow{a \mid x+y} q$ gives you a weighted automaton with the same behaviour. Thus, without loss of generality, you may assume that for all $p$, $a$ and $q$, there is at most one transition of the form $p \xrightarrow{a \mid z} q$.

[1] M. Droste and D. Kuske, Weighted automata, Chap. 4 in Handbook of automata theory Vol. I. Theoretical foundations, 113-150, EMS Press, Berlin, (2021).

J.-E. Pin
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  • Thanks for your detailed answer! You are right that usually transitions are defined as a finite set too, and that it should not matter because transitions along the same arc can just be (semiring) summed together. I was just wondering whether without this restriction one could construct non-determanizable automata in a specific semiring that would otherwise always be determinalizable. The issue is that in the example I was thinking about, the infinite sum would not be defined in the semiring. However, it turns out adding multiplicative quasi-inverse to the semiring does the trick! – languagen00b Apr 09 '22 at 20:37