I didn't realise anyone had had a go at this sort of thing
already, though it is inevitable really!
I'll anwser tilly's question with an example:-
My code gives (for the list 1..255)
[1-9]|(?:[1-9]|1\d|2[0-4])\d|25[0-5]
Whereas your code gives
((?:1(?:|0(?:|0|1|2|3|4|5|6|7|8|9)|1(?:|0|1|2|3|4|5|6|7|8|9)|2(?:|0|
+1|2|3|4|5|6|7|8|9)|3(?:|0|1|2|3|4|5|6|7|8|9)|4(?:|0|1|2|3|4|5|6|7|8|9
+)|5(?:|0|1|2|3|4|5|6|7|8|9)|6(?:|0|1|2|3|4|5|6|7|8|9)|7(?:|0|1|2|3|4|
+5|6|7|8|9)|8(?:|0|1|2|3|4|5|6|7|8|9)|9(?:|0|1|2|3|4|5|6|7|8|9))|2(?:|
+0(?:|0|1|2|3|4|5|6|7|8|9)|1(?:|0|1|2|3|4|5|6|7|8|9)|2(?:|0|1|2|3|4|5|
+6|7|8|9)|3(?:|0|1|2|3|4|5|6|7|8|9)|4(?:|0|1|2|3|4|5|6|7|8|9)|5(?:|0|1
+|2|3|4|5)|6|7|8|9)|3(?:|0|1|2|3|4|5|6|7|8|9)|4(?:|0|1|2|3|4|5|6|7|8|9
+)|5(?:|0|1|2|3|4|5|6|7|8|9)|6(?:|0|1|2|3|4|5|6|7|8|9)|7(?:|0|1|2|3|4|
+5|6|7|8|9)|8(?:|0|1|2|3|4|5|6|7|8|9)|9(?:|0|1|2|3|4|5|6|7|8|9)))
My aim was to get rid of as many alternations as possible
(which are slow) and turn them into character classes (which
are fast). I wanted also to factor the regexp as much as possible.
If you change my code replacing all \d's with \w or whatever
it should work fine for any list of words, but I designed and tested
it with numeric lists in mind.
My first attempt at this problem used a trie like data structure
but I abandonded it once I had the idea of using backtracking
regexps - the irony of using regexps to optimise regexps was
irresistable!
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