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  1. POUsing finite automata as keys to a container
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    <p>I have a problem where I really need to be able to use finite automata as the keys to an associative container. Each key should actually represent an equivalence class of automata, so that when I search, I will find an equivalent automaton (if such a key exists), even if that automaton isn't structurally identical.</p> <p>An obvious last-resort approach is of course to use linear search with an equivalence test for each key checked. I'm hoping it's possible to do a lot better than this.</p> <p>I've been thinking in terms of trying to impose an arbitrary but consistent ordering, and deriving an ordered comparison algorithm. First principles involve the sets of strings that the automata represent. Evaluate the set of possible first tokens for each automaton, and apply an ordering based on those two sets. If necessary, continue to the sets of possible second tokens, third tokens etc. The obvious problem with doing this naively is that there's an infinite number of token-sets to check before you can prove equivalence.</p> <p>I've been considering a few vague ideas - minimising the input automata first and using some kind of closure algorithm, or converting back to a regular grammar, some ideas involving spanning trees. I've come to the conclusion that I need to abandon the set-of-tokens lexical ordering, but the most significant conclusion I've reached so far is that this isn't trivial, and I'm probably better off reading up on someone elses solution.</p> <p>I've downloaded a paper from CiteSeerX - <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.6.7310" rel="nofollow noreferrer">Total Ordering on Subgroups and Cosets</a> - but my abstract algebra isn't even good enough to know if this is relevant yet.</p> <p>It also occurred to me that there might be some way to derive a hash from an automaton, but I haven't given this much thought yet.</p> <p>Can anyone suggest a good paper to read? - or at least let me know if the one I've downloaded is a red herring or not?</p>
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