Talk:Fibonacci word: Difference between revisions
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Them That Be. (I stopped after 20 because the algorithm I use for locating |
Them That Be. (I stopped after 20 because the algorithm I use for locating |
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repeated substrings is pretty naive and slow - hmmm, maybe <i>that</i> should be a separate |
repeated substrings is pretty naive and slow - hmmm, maybe <i>that</i> should be a separate |
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RosettaCode task?* |
RosettaCode task?*). |
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In fact, thinking about it, the longest repeating substring has got to be <i>at least</i> |
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length(fword(n-2)) for all n > 2. Even expressing in hexidecimal isn't going to help much. |
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Revision as of 22:22, 13 July 2013
Uh, are you sure that any of the Fibonacci words actually have rep-strings?
- It would seem not. I have replaced rep string with repeated substring.--Nigel Galloway (talk) 12:09, 13 July 2013 (UTC)
I'm not sure that's much better. I've modified the Icon/Unicon solution to print the lengths of the longest repeated substring for the first 20 Fibonacci Words. It looks like the lengths of those substrings are going to get you in trouble (again) with Them That Be. (I stopped after 20 because the algorithm I use for locating repeated substrings is pretty naive and slow - hmmm, maybe that should be a separate RosettaCode task?*).
In fact, thinking about it, the longest repeating substring has got to be at least length(fword(n-2)) for all n > 2. Even expressing in hexidecimal isn't going to help much.