Talk:Fibonacci word: Difference between revisions
No edit summary |
mNo edit summary |
||
| Line 10: | Line 10: | ||
In fact, thinking about it, the longest repeating substring has got to be <i>at least</i> |
In fact, thinking about it, the longest repeating substring has got to be <i>at least</i> |
||
length(fword(n-2)) for all n > 2. |
length(fword(n-2)) for all n > 2. [I'll postulate, with only empirical evidence, that |
||
the longest repeated substring of fword(n) <i>is</i> fword(n-2).] |
|||
Even expressing in hexidecimal isn't going to help much. |
|||
Revision as of 23:34, 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. [I'll postulate, with only empirical evidence, that the longest repeated substring of fword(n) is fword(n-2).] Even expressing in hexidecimal isn't going to help much.