Re: How to get a concatenation of the negations with rx (ex: [^a][^b])?

[<tomas@tuxteam.de>]

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On Mon, Nov 13, 2023 at 08:46:15AM +0000, Anders Munch wrote:
> tomas@tuxteam.de wrote:
> > OK -- this was roughly my train of thought: build the NFA, then invert that... OMG. Then I decided this is better left as an exercise to the reader.
> 
> At the DFA level it's easy.  So you can just convert the NFA to a DFA and work from there.  I did that exercise once upon a time, let me see if I can remember it.
> 
> First convert to DFA.

Which already involves a power set. Uh, oh ;-)

>                       For every node, add an outgoing edge to the acceptance state for every character that doesn't already have an outgoing node. Remove all edges to the acceptance state that were in the original DFA.
> 
> The main problem is not implementation.  It's that it's not obvious what to use them for in the variable-length searches that regexes are typically used for.  It's just confusing that the string "abz" is a match for the regular expression "not ab", and if you were looking for a two-character string that is not "ab", then a general negation operator isn't going to help you, at least not by itself.

Yes, I guess this is more or less what I hand-waved away with
my "modulo corner cases" (ain't natural language wonderful? ;)

For your example, one would have to append .* to all non-end
anchored (i.e. those not ending with $) regexps to better match
usual expectations. But who knows whether that's all.

Cheers
-- 
t

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