Re: $100.00 Programming Contest
- Posted by euman at bellsouth.net
Mar 02, 2002
----- Original Message -----
From: "David Cuny" <dcuny at LANSET.COM>
To: "EUforum" <EUforum at topica.com>
Euman wrote:
>> I would just guess the faster algorythm being something that much like
>> a syntax highlighter in an editor would change every letter in a select
>> number of words until all those words match dictionaries then proceed
>> to the next small chunk of words replacing the already captured tokens
>> until these words matched then the next set of words until EOF or EOS
>This sort of thing of brute force generate and test approach is trivial in
>Prolog.
>The problem is when it encounters a word it doesn't have in the dictionary.
>That was why I asked about it earlier. It has to backtrack and run through
>every possible combination before it figures out that it can't solve the
>word. At that point, you can flag the word as unknown and continue. But
>(depending on how efficient your pruning is) you've ended up wasting a lot of
>time backtracking.
>-- David Cuny
Davis the TOKEN sequence only needs to reach a set limit of 27 once
this length has been reached youve figured out the algorythm that made the
cipher wich would be more sufficient on large text and no less
sufficient on small text.
Once a word is found in the dictionary each letter found will then reside
as a token which can be applied to the next set of words for consideration.
I cant wait to see how everyone is going to do this...
Euman
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