Here's a great idea which i'm about to patent:
Look the nick up in a dictionary. If not found; kick the person telling them to correct the spelling error.

Seriously though...
Have you tried using soundex? I think it would work much better than just relying on the random string not containing certain chars. I suggest using a soundex "dictionary", a list or array of soundexes based on the words from a real dictionary + common names. This dictionary could then be searched to see if the soundex of a nick == some existing word's soundex, which should weed out most random strings.

Then you could have a mechanism for re-entering the room that you tell about in the kick reason, so those not meant to be kicked can rejoin. (a /msg command or what ever)

This is of course just theory and fairly cpu/memory intensive stuff, but IMO worth a try

That's the beauty of my method (I think/hope), because even if they're not real words most nicks SOUND like real words. Except strange nicks like where a number becomes a couple of chars in a word etc...but these could be translated to chars with a 'string map' before the soundex is made (in most cases) and the rest would be allowed to re-enter through the /msg command

Statistics: Posted by awyeah — Wed Jul 11, 2007 8:34 pm

Statistics: Posted by Sir_Fz — Wed Jul 11, 2007 7:15 am

The Soundex code for a name consists of a letter followed by three numbers: the letter is the first letter of the name, and the numbers encode the remaining consonants.

Similar sounding consonants share the same number so, for example, the labial B, F, P and V are all encoded as 1. Vowels can affect the coding, but are never coded directly unless they appear at the start of the name.

The exact algorithm is as follows:

Retain the first letter of the string
Remove all occurrences of the following letters, unless it is the first letter: a, e, h, i, o, u, w, y
Assign numbers to the remaining letters (after the first) as follows:
b, f, p, v = 1
c, g, j, k, q, s, x, z = 2
d, t = 3
l = 4
m, n = 5
r = 6
If two or more letters with the same number were adjacent in the original name (before step 1), or adjacent except for any intervening h and w (American census only), then omit all but the first.

Return the first four characters, right-padding with zeroes if there are fewer than four.
Using this algorithm, both "Robert" and "Rupert" return the same string "R163" while "Rubin" yields "R150".

Statistics: Posted by awyeah — Mon Jul 09, 2007 10:46 pm