[asterisk-dev] Better pattern matching

Steve Murphy murf at parsetree.com
Thu Aug 2 14:55:16 CDT 2007

On Thu, 2007-08-02 at 13:09 -0400, Kurt Lidl wrote:
> Vasil Kolev wrote:
> > В ср, 2007-08-01 в 16:08 -0600, Steve Murphy написа:
> > 
> >> We might be able to extend the current algorithm to handle a few more
> >> simple cases without resorting to such huge machinery. We shall see.
> >>
> > 
> > Going through this thread, I'm thinking - regular expressions are parsed
> > to finite state machines, and it shouldn't be hard (for some values of
> > hard, I have to brush on my discrete maths memories) to build one
> > automaton to handle the full dialplan for a context and to be able on
> > each digit to cut the possible places to go. There's some stuff that
> > can't be done this way for sure (as we all know, all kind of weird shit
> > is possible with regex), but at least the current syntax can get a bit
> > extended, and probably a lot of IVRs simplified.
> The data structure that you're grasping for is the 'trie':
> 	http://en.wikipedia.org/wiki/Trie
> At a previous job, one of my developers wrote a regex -> trie generator,
> so we could do high-speed lookups of prefix and suffix matches on
> usernames coming into a radius proxy.  Once we finally got it debugged,
> it was amazingly fast.  For that particular appplication we had several
> hundred prefix and suffix patterns, and generated one giant tree to
> handle them all.
> This was pretty slow to generate.  So we did the work in a helper
> program, and wrote the binary trie into a disk file.  The main app
> would stat() the file periodically, and if it was newer than
> the last one read in, just suck the entire thing into memory, and
> change the pointer to the root of the trie.  I guess we called
> free on the old one.  Seamless, fast and worked extremely well...
> Generating one trie per context would probably work really well.
> -Kurt


Yes, what I did in fast-ast is, according to definition, a trie.
And, yes, one per context. And, yes, it works really well.

No, it's not a binary tree. But that's not a requirement.


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