in reply to Re^3: Another regex to solve ...
in thread Another regex to solve ...

Hm ... yes, of course ... the explicit disjunction would work. I guess we both agree on the elegance of this solution (or lack thereof) ;-)

Thanks for posting it, nonetheless!

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Re^5: Another regex to solve ...
by AR (Friar) on Aug 18, 2011 at 16:58 UTC

    Two regexes work:

    $w =~ /([aeiou])p\b/ && $w !~ /${1}${1}p\b/
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