tag:blogger.com,1999:blog-1689592451067041352.post2579011692024397025..comments2023-05-08T07:41:01.071-07:00Comments on DiEbLog: A story of two weaselsDiEbhttp://www.blogger.com/profile/02099109109735165335noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-1689592451067041352.post-63101585922038344892009-08-25T01:26:15.254-07:002009-08-25T01:26:15.254-07:00If your fitness function (a kind of an oracle) onl...If your fitness function (a kind of an <i>oracle</i>) only states whether your string is correct or not, you need up to N^L trials. If the oracle gives you the number of correct letters (as in Dawkins's example), you can construct a search which needs at most N*L steps. And if it returns the place of the correct letters (as for Dembski), you have a variation of the <a href="http://en.wikipedia.org/wiki/Hangman_%28game%29" rel="nofollow">the hangman game</a>, so you need less than N trials.<br /><br />It isn't about finding a better algorithm, it's just about the properties of the ones which where presented.DiEbhttps://www.blogger.com/profile/02099109109735165335noreply@blogger.comtag:blogger.com,1999:blog-1689592451067041352.post-45205595590047128332009-08-24T07:06:52.330-07:002009-08-24T07:06:52.330-07:00How about:
Start with a string of 'n' let...How about:<br /><br />Start with a string of 'n' letters. <br /><br />A):Vary by random choice of one of: <br /><br />1) Cutting in two points at once (one cut could be either end of the string). And rearranging the pieces:<br /><br />ABCDEFGHI -> ABC|DEF|GHI -> DEFABCGHI<br /><br />2) Adding a random letter at a random point:<br /><br />ABCDEFGHI -> ABCjDEFGHI<br /><br />3) Changing one letter<br /><br />ABCDEFGHI -> ABCDjFGHI<br /><br />4) Removing a letter from a random point:<br /><br />ABCDEFGHI -> ABCDEFHI<br /><br />B): Examine the result against a dictionary to see if any subsection is now a valid word: if it is then lock it.<br /><br />Return to A).Terry (Regina Theresa)https://www.blogger.com/profile/02418010085438552289noreply@blogger.com