The only problem with that is that a dictionary is required for it to work because each "symbol" can have multiple translations. Taking info from the wikipedia[1]: a final "s" can be changed to "z" to get the l33t, but to reverse it you have to check first with the "z" because it might be an actual "z". Then if it is not a dictionary word perform the translation and check for a word ending in "s".
For example, given the l33t word "h4x0rz", an algorithm would have to perform something like the following translations, checking each one till it finds a dictionary entry if any:
(done by hand and I don't know much about l33t, so...)
h4x0rz h4x0rs h4xorz h4xors h4xerz h4xers h4ck0rz h4ck0rs h4ckorz h4ckors h4ckerz h4ckers h4cks0rz h4cks0rs h4cksorz h4cksors h4ckserz h4cksers hack0rz hack0rs hackorz hackors hackerz hackers => BINGO
(More permutations here, but we already found a dictionary word, so we stop.)
The basic algorithm for anyone who want to try it, and it's pretty commonly seen in parsing, so it's relatively straigtforward:
scan string till you reach the end of a "word" check dictionary for the "word" LOOP: back up apply conversion(s) check dictionary repeat until success or no more permutations END LOOP:
This would probably make a good QotW, or rather the original question would make a good quiz while the above would be one possible solution. So would implementing an efficient dictionary lookup without loading the entire dictionary in memory.
Randy.
1. <http://en.wikipedia.org/wiki/Leetspeak>
Wow, this is more difficult than I first thought. I think I'm just going to drop the whole idea as the channel is relatively low traffic and it was more for fun than usefulness. Thanks for all the suggestions, though.
-- Andrew Gaffney Network Administrator Skyline Aeronautics, LLC.
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