Regular Expressions and Canonical Equivalence
Stephen E Slevinski Jr
slevin at signpuddle.net
Thu May 14 12:25:06 CDT 2015
On 5/14/15 5:58 AM, Philippe Verdy wrote:
> Yes it is problematic: (ab)* is not the same as (a|b)* as this
> requires matching pairs of letters "ab" in that order in the first
> expression, but random strings of "a" and "b" i nthe second one (so
> the second matches *more* input samples.
> Even if you consider canonical equivalences (where the relative order
> of "ab" does not matter for example because they have distinct
> non-zero canonical) this does not mean that "a" alone will match in
> the first expression "(ab*)", even though it MUST match in "(a|b)*".
> So the solution is just elegant to simplify the first level of
> analysis of "(ab)*" by using "(a|b)*" instead. But then you need to
> perform a second pass on the match to make sure it is containing only
> complete sequences "ab" in that order (or any other order if they are
> all combining with a non-zero combining class) and no unpaired "a" or "b".
If you always want to find "a" and "b" in a pair without regard to the
order, how about the regex:
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