Replacing $ with \z is trival:

$regex = '^[z26\\\\\[\]\{\}\^\$]{26}\z';
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I've also worked out what the 'meta match' is for this expression (that is, $re2 below will only match $regex):

$re2 = '^\^\\[z26\\\\\\\\\\\\\[\\\\\]\\\\\{\\\\\}\\\\\^\\\\\$\\]\\{26\
+\}\\\\z\z';
[download]

So, the idea is your want, in the regex, to have $re2 attempt to match $regex, then have $regex left around that basically is a large character class with a {n} modifier (Thus, if we need to add any new characters like '(', ')', or '?', we work with $regex to add them, modify $re2 above, and we're set).

However, I can't figure out how to do this part. I'm still thinking about it, but this is a rather interesting problem.

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Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain