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I find lots of solution where you have an Automata and a input string , you can validate whether input string is accepted by automata or not.

Can we do the reverse ?

I am looking for solution which generates few possible inputs (not all) accepted by automata(NFA/DFA).

One of the Brute-force approach can be to randomly select inputs and validate with automata if passes then consider those as desired .

But really don't want that . What really am i looking for here, can we use automata itself for produce possible inputs which are accepted by automata.

Any approach also would be helpful.

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3 Answers 3

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Just do a BFS from the start state, and at every node keep track of what transitions you took to get there. Obviously this will only generate a finite number of words, as it doesn't follow cycles.

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  • $\begingroup$ This gives the shortest string(s) accepted. $\endgroup$
    – vonbrand
    Jun 20, 2014 at 14:37
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    $\begingroup$ If you mean that there is no string in the language that is shorter than a string in the ones we generate, that is not necessarily the case. There could be a cycle that gives many string shorter than one on a long path. $\endgroup$
    – gardenhead
    Jun 20, 2014 at 21:04
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A way to generate strings accepted by a given DFA is to start at the initial state, and follow a random transition out of the current state at each step, recording the symbol(s) that would give this transition. Output the result each time you reach a final state, backtrack when you hit a dead state. Collect a few strings each run, and start over to get another path through the DFA.

I'd minimize the DFA beforehand to make this more efficient.

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Another way to think of the process described in the other answers is as follows:

  1. Convert the DFA into a Regular Grammar (use the one that puts the nonterminal on the right, for convenience)
  2. Assign $L_0 = \{S\}$ where $S$ is the start symbol for your grammar.
  3. Iteratively compute $L_{i+1}$ from $L_i$. All sentences in $L_i$ will end with a single nonterminal; they are of the form $\alpha X$ where $\alpha$ consists only of terminals (maybe empty). Now, for each $\alpha X \in L_i$, consider all $\alpha \beta$ where your grammar has a production of the form $X \rightarrow \beta$. If $\alpha \beta$ consists only of terminals, it's a string in the language accepted by the DFA; otherwise, it belongs in $L_{i+1}$. If $L_{i}$ is ever empty, then your DFA accepts a finite language.

This might be a bit more natural, since grammars are typically thought of as generating languages anyway (whereas automata are usually thought of as recognizing or accepting strings).

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